Signals and Systems Lecture Notes

ECE 370 SIGNALS AND SYSTEMS - FALL 2012 Text: These lecture notes and board notes. Optional Texts: H.P. Hsu, Signals & Systems. Shaum's Outline: McGraw{Hill, 1995. This is essentially a good source of worked examples, all in the standard EE notation that we use in ECE 370. Reasonably priced! C.L. Phillips, J.M. Parr, and E.A. Riskin, Signals, Systems, & Transforms. 4th Edition, Prentice-Hall, 2008. Only get this if you are inclined to read additional books! Prerequisites: ECE 225 Electric Circuits, MA 238 Di®erential Equations. Course Goals: To provide electrical engineers with physical understanding and competence in mathematical signal analysis and systems theory. Instructor: Robert W. Scharstein, Houser 209 phone 205-348-1761, e-mail [email protected] O±ce hours: to be announced and by appointment. Lecture: 12:00 noon { 12:50 pm, Mon, Wed, Fri; SERC 1013. Problem Session: Wednesday 7:00-8:00 pm, Houser 301, optional. Bring your questions! Daily Homework: Homework problems and textbook reading will be assigned for each class. FFF !!!THESE ARE YOUR PRIMARY RESPONSIBILITY!!! FFF Tests will consist of homework problems, modi¯ed homework problems, new problems, and concepts and derivations from the text and lecture. Details and discussion of the MATLAB homework problems will appear on tests. Tentative Grades: All quizzes and tests are closed-book. Eleven Friday quizzes : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 1/3 Test 1, Friday September 21 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 1/6 Test 2, Friday November 2 : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 1/6 Final Exam, Monday December 10 11:30 am { 2:00 pm : : : : : : : : : : : : : : : : : : : : : : : : : : 1/3 Instructor may determine ¯nal course grades using any combination of the above. However, the ¯nal exam will not be weighted less than 1/3. Friday quiz dates: 1: Aug 24, 2: Aug 31, 3: Sept 7, 4: Sept 14, 5: Sept 28, 6: Oct 12 7: Oct 19, 8: Oct 26, 9: Nov 9, 10: Nov 16 11: Nov 30

\Our goal is not to develop all the applications, but to prepare for them - and that preparation can only come by understanding the theory." Gilbert Strang Linear Algebra and Its Applications, Academic Press, 1976 page x

Typeset by AMS-TEX 0

ECE 370 SIGNALS AND SYSTEMS - ROUGH LECTURE NOTES Robert W. Scharstein 23 July 2012 CONTENTS BACKGROUND AND TIME-DOMAIN SYSTEMS Complex Numbers : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 1. A Couple of Singularity Functions : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 7. Symmetry : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 10. A Note on Mathematical Functions and Units/Physical Dimensions : : : : : : : : : : : : : : : : : 13. Linear System : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 15. Time-Invariant System : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 16. Input/Output Relationship for a LTI System : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :17. Convolution : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 18. Another Convolution Example: The Convolution of Two Gaussian Pulses : : : : : : : : : : : 21. Yet Another Convolution Example : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 22. Causal System : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 28. Bounded Input/Bounded Output Stability : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 28. Singular Functions and Integrals : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 29. Homework Problems : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :35. Relationship Between Impulse Response and Step Response : : : : : : : : : : : : : : : : : : : : : : : : : 37. Time-Domain RC Circuit Analysis : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 38. Step Response of an RC Circuit : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 41. Step Response of Another RC Circuit : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 43. Homework Problems : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :45. Step and Impulse Response of Series RLC Circuit : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 46. Homework : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :53. FOURIER INTEGRAL Evaluation of an Important Integral via the Laplace Transform : : : : : : : : : : : : : : : : : : : : : : 54. Spectral Form of the Dirac-Delta Function : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 57. Fourier Transform : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 58. Fourier Transform Exercises : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 59. Tables of Integral Transforms : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 60. Fourier Transform of the Gaussian : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 63. Fourier Transform of the Heaviside Unit Step Function : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 65. Duality Property of the Fourier Transform : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 68. Half-Power Bandwidth or Pulse width : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 70. Homework : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :73. Uncertainty Principle for the Fourier Integral : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 74.

i

*Riemann-Lebesgue Lemma (two serious treatments) : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 76. Riemann-Lebesgue Lemma (one reasonable treatment) : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 78. Convergence of the Fourier Integral Representation : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :79. System Transfer Function : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :81. Superposition Interpretation of the System Transfer Function : : : : : : : : : : : : : : : : : : : : : : : 83. Transfer Function for Linear Electric Circuits : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 84. Parseval's Theorem for the Fourier Integral : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 87. Other Conventions for the Fourier Transform Pair : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :88. Cascade of Two LTI Systems : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 90. Linear Dispersionless Filter : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 90. Rectangular Pulse Response of Ideal Low Pass Filter : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 91. Ideal Band-Pass Filter : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 95. Complex Phasors for Time-Harmonic Circuit Analysis : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :96. Homework Problem on Uncertainty Principle : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 98. *Approximate Analysis of Dispersion : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 99. Fourier Integral Solution of Potential Problem : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 103. FOURIER SERIES Fourier Transform of the comb : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :106. Fourier Transform of a Periodic Signal : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 107. Periodic Signals and Fourier Series : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :110. Orthogonality : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 110. Kronecker delta : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 111. Orthogonality of the Fourier basis : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 111. Derivation of Fourier Coe±cients by Minimizing the Mean Square Error : : : : : : : : : : : : 112. Fourier Coe±cients by a Direct Inner Product : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 114. Trigonometric Form of the Fourier Series : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 114. Two Standard Notations for the Trig Fourier Series : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 116. Riemann-Lebesgue Lemma: Fourier Coe±cients : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 117. Gibbs' Phenomenon: Example Square Wave : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 118. Pointwise Convergence of the Fourier Series : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 119. On the Convergence and Summation of Some Series : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :126. Kummer Transform : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 129. Homework Problem: Convergence of Fourier Series : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 131. Fourier Series of Full-Wave Recti¯ed Sine Wave : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 132. Parseval's Theorem for the Fourier Series : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 134. Power Supply Performance : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 135. Periodic Excitation of a Simple Parallel GLC Circuit : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 139. Homework Problems : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 140. Simple Low-Pass and Hi-Pass Filter Response to a Periodic Input : : : : : : : : : : : : : : : : : : 141. The Vibrating String : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 147. Homework Problems : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 150. Poisson Sum Formula : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 151.

ii

TRANSITION TO DISCRETE-TIME SYSTEMS Sampling and Reconstruction : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 153. Discrete-Time Signals, Systems, and Transforms : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 157. Exercises : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 159. Discrete-Time Systems : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 159. Linear Time-Invariant Discrete-Time Systems : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 160. Input/Output Relationship : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 160. Example of a LTI Discrete-Time System : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 161. The Z-Transform : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 163. Fibonacci Sequence - Di®erence Equation and Z-Transform : : : : : : : : : : : : : : : : : : : : : : : : 165. Convolution Theorem of the Z-Transform : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 166. Z-Transform of a Delayed Sequence : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 166. System Transfer Function : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 166. Discrete-Time Fourier Transform : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 167. Discrete Fourier Transform : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 167. Exercises : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 169. DFT Approximation of the Continuous-Time Fourier Transform : : : : : : : : : : : : : : : : : : : :170. Numerical Example of Using the FFT to Evaluate the Fourier Integral : : : : : : : : : : : : : 171. Example: DFT Approximation of a Fourier Sine Integral : : : : : : : : : : : : : : : : : : : : : : : : : : : 174. Discrete-Time Approximation for the Step Response of Series RLC Circuit : : : : : : : : : 177.

* Sections marked by the asterisk are A+ level, so put these last in your prioritized list of study topics.

\... mathematics is learned by doing it, not by watching other people do it ... " M. Reed and B. Simon, Functional Analysis, Academic Press, 1980, page ix.

iii

Complex Numbers p j = + ¡1 y ... .. .. .. .. ... ... ... ... . .. . . .. ... ... .... .. .................. . ... . ..... . .. ..... ... ... ... ... . ... ... . .. . . . .. . ... ... ..... ... .. .. ... ... .... ... ... .. ... ... ... . . . . ......................................................................................................................................................................................... ... ... .. ... ... .. .. ... .. ... .. ... ... .. .. ... .

²

z

r

Á

x

complex z = x + jy plane

If z 2 C and x; y; r; Á 2 R, then we write the complex number z in rectangular and polar form as z = x + jy = rejÁ : Refzg = x

and

Imfzg = y

The modulus or magnitude of z is jzj = r and the argument or angle of z is argfzg = Á: The complex conjugate of z = x + jy = rejÁ is z ¤ = x ¡ jy = re¡jÁ : z + z ¤ = 2x

and

z ¡ z ¤ = 2jy

zz ¤ = jzj2 = r2 Euler's identity:

ejÁ = cos Á + j sin Á Taylor series for the exponential function, written as a function of the complex variable z is 1 X zn z e = : n! n=0 1

Let z = jy and separate the real and imaginary parts via the even and odd powers in 1 1 1 X X X (jy)n (jy)2m (jy)2m+1 = + ejy = n! (2m)! (2m + 1)! n=0 m=0 m=0 Note that j 2m = (¡1)m and j 2m+1 = j(¡1)m so that 1 1 X X (¡1)m y 2m (¡1)m y 2m+1 jy +j = cos y + j sin y e = (2m)! (2m + 1)! m=0 m=0 There was nothing special about the real number y, and we also write ejÁ = cos Á + j sin Á: The sum of ejÁ and its conjugate gives ejÁ + e¡jÁ = 2 cos Á and the di®erence gives ejÁ ¡ e¡jÁ = 2j sin Á

which are the important representations

ejÁ + e¡jÁ 2 jÁ e ¡ e¡jÁ sin Á = 2j Many (all?) of the trig identities are easily seen in complex form: Consider ³ ´³ ´ ³ ´³ ´ jÁ 2 jÁ ¡jÁ 1 = je j = e e = cos Á + j sin Á cos Á ¡ j sin Á = cos2 Á + sin2 Á: cos Á =

Now observe

£ ¤ z = x + jy = rejÁ = r cos Á + j sin Á

Equating real and imaginary parts separately gives x = r cos Á 2

¤

and

2

y = r sin Á:

jzj = zz = r = (x + jy)(x ¡ jy) = x2 + y 2 p r = + x2 + y 2 ³y ´ Á = tan¡1 x Note that the two-argument arctangent function (that maintains full information about quadrant) is required. For example, observe that μ ¶ p j3¼=4 1 3¼ ¡1 z1 = ¡1 + j = 2e where tan = ¡1 4 is certainly di®erent from μ ¶ p ¡j¼=4 ¡1 ¼ ¡1 z2 = 1 ¡ j = 2e where tan =¡ 1 4 Your calculator has the two-argument arctangent function as part of its rectangular-topolar conversion. In MATLAB, it's called atan2(y,x) just like Fortran and probably C§ . 2

Consider z1 = x1 + jy1 and z2 = x2 + jy2 . Then our four basic operations are: addition

z1 + z2 = x1 + jy1 + x2 + jy2 = (x1 + x2 ) + j(y1 + y2 )

multiplication by a real constant c:

cz1 = c(x1 + jy1 ) = cx1 + jcy1

multiplication z1 z2 = (x1 + jy1 )(x2 + jy2 ) = (x1 x2 ¡ y1 y2 ) + j(x1 y2 + x2 y1 ) and division

z1 z2 = (r1 ejÁ1 )(r2 ejÁ2 ) = r1 r2 ej(Á1 +Á2 )

(x1 x2 + y1 y2 ) + j(x2 y1 ¡ x1 y2 ) z1 x1 + jy1 (x1 + jy1 )(x2 ¡ jy2 ) = = = z2 x2 + jy2 (x2 + jy2 )(x2 ¡ jy2 ) x22 + y22 and

z1 r1 ejÁ1 r1 = = ej(Á1 ¡Á2 ) jÁ 2 z2 r2 e r2

Generally, since the real number line is a subset of the complex plane, all of our ordinary functions f (x) of a real variable x have extensions (or continuations) to a function f (z) of a complex variable z. One notable di®erence between the real number line and the more general complex plane is the lack of ordering on z: That is, there is no inequality on the complex plane. Inequality is only de¯ned on the reals, i.e. x1 > x2 ;

y1 · y2 ;

7 < 13;

but something like z1 < z2 is meaningless. Triangle inequalities. jz1 + z2 j · jz1 j + jz2 j ¯ ¯ jz1 ¡ z2 j ¸ ¯jz1 j ¡ jz2 j¯

3

etc

Proof of the ¯rst triangle inequality: jzj2 = x2 + y 2 jzj =

p x2 + y 2 ¸ jxj

jzj ¸ jRe(z)j jzj ¸ Re(z)

jz1 z2¤ j ¸ Re(z1 z2¤ )

(1)

jz1 + z2 j2 = (z1 + z2 )(z1¤ + z2¤ ) = z1 z1¤ + z2 z2¤ + z1 z2¤ + z1¤ z2 = jz1 j2 + jz2 j2 + 2Re(z1 z2¤ )

(jz1 j + jz2 j)2 = jz1 j2 + jz2 j2 + 2jz1 jjz2 j = jz1 j2 + jz2 j2 + 2jz1 z2¤ j compare (2) and (3) in view of (1) jz1 + z2 j2 · (jz1 j + jz2 j)2

........... .......... ..... ........... ......... .......... . . . . . . . . . . . . . 1 2..................... ..... ..... . .......... ..... .... ........... . . . . . . . . . . . . . .. ..... .......... ..... .......... 2 ..... ........... .......... ..... . . . . . . . . . . . . . ..... .. ......................................................................................................................

z +z

jz1 + z2 j · jz1 j + jz2 j

z

z1 Extensions ¯N ¯ N ¯X ¯ X ¯ ¯ f (z) · jfn (z)j ¯ ¯ n ¯ ¯ n=0

n=0

¯ b ¯ ¯Z ¯ Zb ¯ ¯ ¯ f (t) dt¯ · jf (t)j dt ¯ ¯ ¯ ¯ a

a

(Sometimes called the monotonicity property of the integral.) F Prove the second triangle inequality. Hint: let z1 = ³1 ¡ ³2 and z2 = ³2 in the ¯rst triangle inequality.

4

(2)

(3)

Multivalued functions. The nth roots of unity 1 = ej2k¼ 11=n = ej2k¼=n

(k = 0; §1; §2; : : : )

(k = 0; 1; 2; : : : ; n ¡ 1) distinct roots ... .. .. .. ......................... . . . . . . . . . . ....... ... . . . . .... . ..... .... .... .. ... ... .. ... ... . . . . . ... . . . . ... ... .... .. .. ... ................................................................................................................... ... . .. . . . ... ... .... ... . . . . ... .. .... .... ... ..... .. ...... ..... ......... ... .............. ...................... .. .. .. ... .

Examples. 1=2

1

11=3

=

½

ej0 = 1 (k = 0) ej¼ = ¡1 (k = 1)

8 j0 e =1 > > p > > < j2¼=3 ¡1 + j 3 = = e 2 p > > > > ¡1 ¡ j 3 : ej4¼=3 = 2

²

²

.. .. .. .. . ..................... . . . . . . . . . . . ....... ... . . . . . ..... .... .. . . .... . .. . .. ... ... ... ... . . . . ... . . . . . ... . . . . ... .... .... .................................................................................................................. ... ... .. . ... . .. . ... .. ... . ... ... . . ... .. ... ..... .. ..... ...... ... ...... ........ . . . . . . . . .......................... .. .. .. ... .

(k = 0)

²

(k = 1)

²

(k = 2)

²

logarithm ¤ £ ¤ £ ln(z) = ln rejÁ = ln rej(Á+2k¼) = ln(r) + j(Á + 2k¼)

(k = 0; §1; §2; : : : )

The value for k = 0 is called the principal branch of the logarithm.

5

powers: Let z = x + jy and w = u + jv be two complex numbers, that is z; w 2 C and x; y; u; v 2 R. £ ¤ £ ¤ z w = exp ln(z w ) = exp w ln z

and we know how to interpret/evaluate £ ¤ ez = ex+jy = ex cos y + j sin y

Example. ln(j) = j(¼=2 + 2k¼) j j = ej¢j(¼=2+2k¼) = e¡(¼=2+2k¼) principal value is j j = exp(¡¼=2) = 0:2079 ¢ ¢ ¢ μ Problem: Show that Hint: use the ¯nite geometric series

n¡1 P

a

k

¶ on page 164

k=0

n¡1 X

ej2¼k=n = 0

k=0

Fundamental theorem of algebra. The polynomial Pn (z) = an z n + an¡1 z n¡1 + ¢ ¢ ¢ + a2 z 2 + a1 z + a0 with an 6 = 0 is said to be of degree n. It has exactly n roots (zeros) counting multiplicity. Proposition: If the coe±cients ak (k = 0; 1; 2; : : : ; n) of the polynomial Pn (z) are all real, then any complex roots must occur in complex conjugate pairs. That is, if Pn (z) = 0 then Pn (z ¤ ) = 0. proof: an z n + an¡1 z n¡1 + ¢ ¢ ¢ + a2 z 2 + a1 z + a0 = 0 take the complex conjugate an (z ¤ )n + an¡1 (z ¤ )n¡1 + ¢ ¢ ¢ + a2 (z ¤ )2 + a1 z ¤ + a0 = 0 6

A Couple of Singularity Functions (or \Distributions" or \Generalized Functions") u(t)

±(t)

1

................................................................. ... .. ... .. ............................................................................................

...... ......... ...........................................................................

t

0 Heaviside unit-step function ½ 0; t < 0 u(t) = 1; t > 0

(1) ...... N ......... ... 0

u(t ¡ T ) =

=)

½

t

0; t < T : 1; t > T

Dirac delta-function (de¯nition) ±(t) = 0

if

t6 =0

AND

Z1

±(t) dt = 1

¡1

From this strange de¯nition, the sampling or sifting property follows: Zb a

f (t)±(t ¡ t0 ) dt =

½

f (t0 ); a < t0 < b 0; o.w.

if f (t) is continuous at t = t0 . This is e®ectively an operational de¯nition of the Dirac-delta. It is under the operation of integration that the Dirac delta-function enjoys meaning, and thus it is sometimes said that \every Dirac-delta function is begging to be integrated." Relationship: Zt

±(¿ ) d¿ = u(t)

¡1

()

±(t) =

du(t) dt

Dirac-Delta Sequences: There are lots (1!) of ordinary functions that satisfy, in some limit, the two-statement de¯nition of the Dirac-delta function. Here are several: μ ¶ t 1 lim ¦ = ±(t) ²!0 ² ² ²=¼ lim 2 = ±(t) ²!0 t + ²2 r ® ¡®t2 lim e = ±(t) ®!1 ¼ sin −t 1 = lim lim −!1 ¼t −!1 2¼

μ ¶ ½ 1 jtj < T =2 t where ¦ , u(t + T =2) ¡ u(t ¡ T =2) = T 0 jtj > T =2 Cauchy or Lorenz distribution Gaussian distribution Z−

e§j!t d! = ±(t)

¡−

7

Dirichlet kernel

Study the time behavior of the above \Dirac-delta sequences" to see how they approach the ideal delta function. Also examine the Heaviside unit-step sequence lim 1 [1 ®!1 2

+ tanh ®t] = u(t)

and its time-derivative. Note: we don't ever try to evaluate ±(0). You might want to, and some books might say that \±(0) = 1 ", but I would prefer we avoid this point. We don't ever have to attach a value to the Dirac delta function when its argument is zero. But, in the words of Lennon & McCartney, it's \nothin' to get hung about" from Strawberry Fields Forever. Another note: (along the same line) If a Dirac-delta function appears in the integrand of some integral, the integral is evaluated from the sampling property. Warning: the integration limits cannot \split" a Dirac-delta function \down the middle," for example Z1 ±(t) dt 0

should be either

Z1

±(t) dt = 1

or

0¡

Z1

±(t) dt = 0:

0+

You either capture all of the delta-function or you don't get any of it. It cannot be divided. Or you will have to agree on some scheme to cut it up while minimizing bloodshed. Proof of the SAMPLING PROPERTY: No harm is done in taking t0 = 0, so that we want to show ½ Zb f (0); a < 0 < b f (t)±(t) dt = 0; o.w. a

for f (t) continuous in a neighborhood of t = 0. For de¯niteness, use the limit of the rectangular pulse to represent μ ¶ t 1 ±(t) = lim ¦ ²!0 ² ² and represent the continuous f (t) by its Taylor series about t = 0: f (t) = f (0) + tf 0 (0) +

t2 00 t3 f (0) + f 000 (0) + O(t4 ): 2! 3!

Then the integral of interest is, for a < 0 < b, Zb a

1 f (t)±(t) dt = lim ²!0 ²

¸ Z²=2 · t2 00 0 f (0) + tf (0) + f (0) + : : : dt: 2! ¡²=2

8

Note Z²=2 ¡²=2

and so

Zb a

dt = ²;

Z²=2

t dt = 0;

¡²=2

Z²=2

²3 t dt = 12 2

¡²=2

¸ · 1 ²3 00 5 f (t)±(t) dt = lim ²f (0) + f (0) + O(² ) = f (0): ²!0 ² 24

If, for a < b, either 0 < a or b < 0, then the integral is zero since ±(t) = 0 for t 6 = 0.

¥

F Provide an alternate proof of the sampling property by appealing to the mean value theorem of the integral.

\Heavy formalism is not required to get across fundamental ideas. If it can't be said simply, it's not worth being said at all." Bernard H. Lavenda, Statistical Physics, 1991, p. viii.

\The pursuit of excellence is gratifying and healthy. The pursuit of perfection is frustrating, neurotic, and a terrible waste of time." Edwin Bliss

9

Symmetry1

symmetric or even fe (¡x) = fe (x)

asymmetric or odd fo (¡x) = ¡fo (x)

cos(x)

sin(x)

... .. . ......... ...... . . . . . ........ . ... . . . . .. ... .. .... ... .. .. .... .... ... ... .. .. . . . . . . . .. .. . . ... .. .. ... .... .... . ... . ... . . . . ... ... . ... .. ... ... ... ... .. . . . . . . ... .. . . . . . . . . . . . .. .. .. . . . . .. . . . . . .................................................................................................................................................................................................................................... .. .. . . . . . . . . . .. .. .. . . . . . . . . . . ... ... ... .. .... .. ... ... ... ... . . . . .... . .. ... ... . . .. . . . . ... . .. .. . . .. . . . . .. ... ... . . . . . . ... . . . . .. .. . . . . . . .. .. . . ... .. ... ... ... .. . ... . . ........ ........ ....... ...... .... ... ..

... ... .. . .... ......... .. .......... . ... ... . ... .. ... .. .. .... . ... . ... . ... ... .. .. . . . . . . . . .. .. .. ... .. ... .... ... .. ... ... . . . . . . . ... ... ... ... .. .. ... .... ... ... ... ... . . . . . . .. .. .. .. . .. .. ....... .. .. .. . .. . .. . . . . . . ..................................................................................................................................................................................................................................... .. .. . . . . . . . . . . . .. .. .. .. . . . . . ... ... . . . . . ... ... .. ... .. .... ... ... ... . . . . . . . . . ... ... ... . .. . . . . . . . . .. . .. .. . . . . . . . . . .. .. . ... ... ... .. ... .... ... ... . . .. .... . . . ... .. .. ... .. ... .. ....... ....... ...... .... .. ...

... .. ... ... ... ... . ... . . 2 ... ... ... ... ... ... . ... . ... .. ... ... .. ... .. ... ... .. ... . ... . ... ... ... .. ... ... .. ... ... ... ... ... . .... . .. .... .... .. .... .... ... ..... .... .. .... .... . . . . .................................................................................................................................................................................................................................. ..... .... .... ..... ..... ..... .. ..... ...... ... ..... . ....... . . . . . ... ........ ............................................ .... ... .. .. ... ..

x

x

x ¡ 25

1 4 8 (35x

1 \Symmetry

x

.... ... ........ ... ....... .. ....... . . . . . ... . . ........ .. ....... .. ....... ... ........ . . . . .. . . .... ... ....... ... ........ .. ........ .. ............. . . ..................................................................................................................................................................................................................................... ... . ....... .... . . . . . . . ........ .... ....... .. ....... .. ........ . . . . . ... . ... . . . . . . .. .. . . . . . ... . . ... . . . . . .. . ... . . . .. . . . .. . . . . . . ... . .. . . . . . .. . . ....

1 3 2 (5x . ¡

¡ 30x2 + 3)

... .. ... ... .. ... ... ... .. ... . . . .. ... .... .. ... .. .. ... .. ... ... . . ............... .. ... ........ ... .............. ... ... ...... ..... .... ... ..... ..... ... . . . . . . . . . .. ... . ... ................................................................................................................................................................................................................................................. ..... ... .. ... .... ..... ..... ... . . . . . . . . . ..... .... .. .. ..... .... ..... ..... .... ....... .. ........................ .................. .. .. ... ... .. .. ... ..

x

x

3x)

.. ... .. ... ... .. . ... . .. ... .. ... ... ... . . . . . . . . . . . . . . . . . . . . . . . . ........ . ....... ....... .... ... ..... ...... ... .... ...... .... ... ..... ..... .. . . . . . ... . .. . ............................................................................................................................................................................................................................................ . . .. .... .......... ... .... . . . . . . . . . ...... . .... ....... ..... ... ........ ..... .. ... ........................... ... ... . . ... .... ... ... .. .. . .. . .. ... . .. ...

is immunity to a possible change." J. Rosen, Symmetry Rules, Springer, 2008, p. 4 10

x

x

If a function f (x) is continuous to all orders (the function f (x) and all of its nth derivatives f (1) (x); f (2) (x); : : : are continuous) then f (x) can be represented by a Taylor series about the origin 1 X f (n) (0) n f (x) = x : n! n=0 The Taylor series of an even function will have only even powers of x 1 X f (2n) (0) 2n fe (x) = x (2n)! n=0 and therefore all of its odd derivatives (at the origin) vanish ¯ d2n+1 f (x) ¯¯ (2n+1) fe (0) = = 0: dx2n+1 ¯ x=0

In particular, note that

fe0 (0)

= 0 (if fe (x) is continuous and di®erentiable at the origin).

Similarly, the Taylor series of an odd function will have only odd powers of x 1 X f (2n+1) (0) 2n+1 x fo (x) = (2n + 1)! n=0 and therefore all of its even derivatives (at the origin) vanish ¯ d2n f (x) ¯¯ (2n) fo (0) = = 0: dx2n ¯ x=0

In particular, note that fo (0) = 0 (if fo (x) is continuous at the origin).

Even-Order Derivatives Preserve Symmetry and Odd-Order Derivatives Reverse Symmetry f (t + ¢t=2) ¡ f (t ¡ ¢t=2) ¢t!0 ¢t f (¡t + ¢t=2) ¡ f (¡t ¡ ¢t=2) f 0 (¡t) = lim ¢t!0 ¢t Even function fe (t) = fe (¡t) f 0 (t) = lim

fe (t + ¢t=2) ¡ fe (t ¡ ¢t=2) ¢t!0 ¢t

fe0 (t) = lim

fe (¡t + ¢t=2) ¡ fe (¡t ¡ ¢t=2) fe (t ¡ ¢t=2) ¡ fe (t + ¢t=2) = lim ¢t!0 ¢t!0 ¢t ¢t fe (t + ¢t=2) ¡ fe (t ¡ ¢t=2) d = ¡ lim = ¡fe0 (t) =) fe (t) is odd. ¢t!0 ¢t dt

fe0 (¡t) = lim

Similarly, d fo (t) is even where fo (¡t) = ¡fo (t) is an odd function. dt This is also seen by appealing to the Taylor series for even and odd functions. 11

Consider the integral of f (t) between limits that are symmetrically disposed about the origin, that is look at ZT f (t) dt ¡T

where T is some ¯xed, positive constant. fo (t) .

.... ... .................................................... ............. .. .......... .... ... ........ . . . . . . . .. . . .. ........ ... ....... .. ... ....... .. ... ....... . . . . .. . ... ... . . . ... . .. . . ... .. ............ .. ....... . . . . . . . ......................................................................................................................................................................................................................................................................................................................... . . ... ... ... . . . . . ... ....... ... ... ....... ... ... ...... .. ...... . ... . . . . .. . ... . . ... . . . ... . ... . . . ... . . . .. ... . . . . ... . . .. . ...................... .. . .. . . ............... . . . ........................................ ... .

¡T

T

ZT

t

fo (t) dt = 0

¡T

fe (t) .

.... ........... .......................... .... ....................................... . . . . . . . . . . . . . ........... . ......... ........... .... ........ ......... .. ........ ........ ... ....... ........ . . . . . . . ...... .. . . . . . . ......... . . ... ... . . . . . ... ........ . . . .. ... . . . . ...... . . .. . .. . . ...... . . . . . . . . .. ... . . . . . . . . .................................................................................................................................................................................................................................................................................................................................. ..... . .. . . . . ..... . . .. . . . ..... . . . ..... ... .... ..... ..... . . . . . ..... .. . . . . . ..... . ..... . .... .. .

¡T

T

t

ZT

fe (t) dt = 2

¡T

ZT

fe (t) dt

0

Any function f (t) (neither symmetric nor antisymmetric) can always be decomposed into a linear combination of a symmetric plus an antisymmetric part since f (t) = Example.

f (t) + f (¡t) f (t) ¡ f (¡t) + = fe (t) + fo (t): 2 2

..... .. .. .. ... ... .... .. .. .. ... ... .. .. .. .. ... .. ... .. ... ... .. . ...........................................................................................................................

0

T

.. .... ... ... .. .. .. .. ... .. .. .. ... . ... ... . ................................................................................................................................................................................ ... ..

t = +

¡T

T

... .. .. .. .... ... ... .. .. .. ... . ................................................................................................................................................................................. .... ... .. .. .. ... ... .. .. ... ..

¡T

T

t

t

F Homework. Decompose the Heaviside unit step function u(t) into the sum of a symmetric plus antisymmetric signal. 12

A Note on Mathematical Functions and Units/Physical Dimensions Consider almost any \ordinary" or \regular" function2 of an independent variable x, such as f (x) = cos(x) or sin2 (x) or 3x2 ¡ 7x or ex :

The argument or independent variable x is always dimensionless. To see this, examine the Taylor series expansion, such as f (x) = cos(x) = 1 ¡

x2 x4 + ¡ :::: 2! 4!

If x has units such as meters (m), then the sum would be nonsense since the ¯rst term is dimensionless, the second term has units (m)2 , the third term has units (m)4 , etc. Also note that the function f (x) itself is dimensionless, too. This may seem strange and somewhat contrary to your previous experience. For example, you have dealt with a time-domain voltage signal such as v(t) = 12 cos(t + 45± ) or 12 cos(t + ¼=4): If the time variable t is measured in (s), then there really is a transparent (or hidden) frequency ! = 1 (s¡1 ) multiplying the t so that v(t) = V0 cos(!t + Á): Now the argument !t + Á of the cosine is clearly dimensionless. The constant V0 = 12 (V) carries the dimensions of the voltage signal v(t). Also note that we often say ! has units of (rad/s), but a radian is not dimensioned: Its use in (rad/s) is simply a reminder that we are using the angular frequency ! and not the common experimentalist's frequency f = !=2¼ (cycle/s=Hz). The operators di®erentiation and integration impart units: Consider the timedomain current signal i(t) = I0 cos(!t): In our standard mksC (SI) units, the current amplitude I0 is in (A), the frequency ! is in (s¡1 ), and the time t is in (s). Di®erentiation with respect to the dimensioned variable t imparts the units of 1=t or (s¡1 ) since di(t) = ¡!I0 sin(!t): dt We can also see that the di®erential operator has to have units of inverse time by appealing to the de¯nition of the derivative df (t) f (t + ¢t) ¡ f (t) = lim : ¢t!0 dt ¢t The (s¡1 ) comes from the denominator ¢t that is in (s). 2 Not

±(x) or u(x) or jxj, to name a few exceptions. 13

Integration with respect to the dimensioned variable t imparts the units of t or (s) since Zt t0

Zt

i(¿ ) d¿ =

I0 cos(!¿ ) d¿ =

t0

¤ I0 £ sin(!t) ¡ sin(!t0 ) : !

Note the good practice of using a dummy time variable for the integration, to avoid confusion with the ¯nal time that is the upper integration limit. We can also see that the integral operator has to have units of time by appealing to the de¯nition of the integral in terms of a limit on the Riemann sum Zb

f (t) dt = lim

N !1

a

N X

f (tn )¢tn :

n=1

The prime notation for derivatives. When dealing with a generic function such as f (t) or g(x), we commonly denote derivatives using the prime df (t) = f 0 (t) dt

and

dg(x) = g 0 (x): dx

The prime denotes di®erentiation to the entire argument of the function, as in f 0 (») = Note that

df (») d»

or

f 0 (~) =

df (~) : d~

d d» df (») d» f (») = = f 0 (») : dt d» dt dt

14

Linear System ................................................. ... ... .. ... .. ... ... . . ........................... ........................... .... .... .. .. ... .. . . ................................................

x(t)

y(t)

T

Assume the zero-state response y(t) of a system to the input x(t) is described by an operator T , such that y(t) = T fx(t)g: Another name for T is system transformation rule. Examples might be T1 fx(t)g = Ax(t);

T2 fx(t)g = x(t ¡ t0 );

μ ¶ d2 d T4 fx(t)g = c2 2 + c1 + c0 x(t); dt dt T6 fx(t)g = x2 (t);

T3 fx(t)g =

T5 fx(t)g =

Zt

d x(t) dt

x(¿ ) d¿

¡1

T7 fx(t)g = Ax(t) + B:

Linear System: If the input/output relationship or system operator or system transformation rule satis¯es the superposition principle T fc1 x1 (t) + c2 x2 (t)g = c1 T fx1 (t)g + c2 T fx2 (t)g; then the system is said to be linear. F HW 1

Test the examples T1 ; T2 ; : : : ; T7 above for linearity.

15

Time-Invariant System:

.................................................. .. ... .. ... ..... . .. . ............................ ............................ .... .... .. .. . ... ....................................................

x(t)

y(t)

T

................................................. ... ... ... ... ... .. . . .............................. ............................. ... ... .. .. ... ... .. . ...............................................

x(t ¡ t0 )

y(t ¡ t0 )

T

Let the response of a system to the input signal x(t) be the output signal y(t). If the response to the delayed input x(t ¡ t0 ) is the delayed response y(t ¡ t0 ), the characteristics of the system did not change in time, and the system is said to be time-invariant. Note: In our linear RLC circuit analysis, the equation-of-motion is typically a linear integrodi®erential equation, with constant coe±cients, such as R

L

. . . ...... ..... ..... ..... ............................................................................................................. .... .... .... ................................ . . . ... .. . . .. . ... .................................................. .. ... .......... ................ .. . . . . . . ... . . ....... ... .... ...... .. ....... ... . . . . . . . . .... .. ....... . ... . . . . . . . . . . . ..... ... .. . . . . .. . . ... .. ... . . . . . . . . ......... .... . .. . ... ..... .. ... . . . . . . . . ... . ........... . . ... . . ........ .......... ... . .... ..... ... ... ..... ..... ... . . ...... . ... . .. . ........ . ... . . . . ... ........... . . . . ... . ..... . . . . ........................ ......................... .. ... ........ ... . .......................................................................................................................................................................

i(t)

vg (t)

+ ¡

C

Zt

di(t) 1 L + Ri(t) + dt C

vg (t)

............................................ ... ... ... .. ............................... .............................. .... ... .. . .............................................

i(t)

i(¿ ) d¿ = vg (t)

¡1

which is easily converted to an ordinary di®erential equation by di®erentiating w.r.t. t so that di(t) 1 d2 i(t) +R + i(t) = vg0 (t): L 2 dt dt C Again, note the constant coe±cients L, R, and 1=C. If the equation-of-motion of a system is a di®erential equation with non-constant (i.e. time varying) coe±cients, then the system is a time-variant system.

F HW 2

Give an example of a linear, time-varying system.

16

Input/Output Relationship for a LTI System (zero-state or driven response) x(t)

................................................. ... .... ... .. ... ... .. .......................... ........................ .... ... .. .. .. ... . . .................................................

T

y(t) = T fx(t)g

This fundamental concept is easy to see and derive, starting with a little \trick" by way of the notation. First, invoke the sampling property of the Dirac-delta function to write Z1

x(t) =

¡1

x(¿ )±(t ¡ ¿ ) d¿:

Now exploit the linearity of the operator T and note that T f¢g operates only on functions of time t. In particular, T treats x(¿ ) as a constant, and acts only on the time-domain function ±(t ¡ ¿ ) 8 1 9 Z1 0

f (¿ ) = e¡a¿ u(¿ )

... ... .. ........ ... ..... .. ...... ..... .. ...... ... ...... .. ...... .. ....... .. ........ ......... ... .......... .. ............. .. .................. ... .............................. ... . .....................................................................................................................................................................................................................................................................................................................................................................................................................

¿

0

) g(t ¡ ¿ ) = e¡b(t¡¿ u(t ¡ ¿ ) .

.. .. .. .. ..... ... ..... .. ... .... . ... .. . ... ... ... .. ... .... .. .... ... . . ... . ... ... ... ..... ... .. ..... . . . ... .. . . .. . . . . . ... . ... . ... . . . . . . . . .. . . . . . . ... .................................................................................................................................................................................................................................................................................................................................................................................................................................

¿

t

0

f (¿ )g(t ¡ ¿ )

.... .. .. .. ... .. ... ... .. ..... .. ....... .. ... ....... .... ... ... ........ . . . . . . . .. . . ... ..... .. ............. ... ..................................... . . ..............................................................................................................................................................................................................................................................................................................................................................................

t

0

s(t) =

Z1 ¡1

=

Z0

f (¿ )g(t ¡ ¿ ) d¿ 0 d¿ +

¡1

= e¡bt

Zt

¡a¿ ¡b(t¡¿ )

e

e

d¿ +

t

0

Zt

Z1

e(b¡a)¿ d¿ =

0

19

e¡at ¡ e¡bt b¡a

0 d¿

¿

Now combine both expressions (¯rst one for t < 0 and second one for t > 0) into a single equation valid for all t e¡at ¡ e¡bt u(t): s(t) = b¡a s(t)

.... .. .. ....... .. ....... . ................ ... ....... ..... ... ... ... ...... ...... .. .. ..... ...... .. .. ... ....... ... .... . ....... .. . . ........ ....... ... ... .... ........ ... .. .......... .. ..... ............ ................ ... .... ......................... . ..... . . ...................................................................................................................................................................................................................................

0

tm

t

Let's look for the maximum of s(t) analytically: The derivative be¡bt ¡ ae¡at e¡at ¡ e¡bt u(t) + ±(t) b¡a b¡a be¡bt ¡ ae¡at = u(t) b¡a

s0 (t) =

since

¯ e¡at ¡ e¡bt ¯¯ = 0: b ¡ a ¯t=0

Recall that Ã(t)±(t) = Ã(0)±(t). The derivative vanishes at time t = tm , so write s0 (tm ) = 0 so that ae¡atm = be¡btm b e(b¡a)tm = a tm =

ln(b=a) b¡a

20

Another Convolution Example: The Convolution of Two Gaussian Pulses Evaluate z(t) = f (t) ~ g(t) with f (t) = exp(¡at2 ) and g(t) = exp(¡bt2 ). Both f (t) and g(t) are \on" for all time, so we don't have to worry about sketching them to see when they turn on or o®.

z(t) =

Z1 ¡1

f (¿ )g(t ¡ ¿ ) d¿ = 2

= e¡bt

Z1

e¡[(a+b)¿

2

Z1

2

2

e¡a¿ e¡b(t¡¿ ) d¿

¡1

¡2bt¿ ]

d¿

¡1

complete the square μ ¶2 p bt (bt)2 (a + b)¿ ¡ 2bt¿ = a + b¿ ¡ p ¡ a+b a+b 2

z(t) = exp

·μ

" μ ¶ ¸ Z1 ¶2 # p bt b2 2 ¡b t exp ¡ a + b¿ ¡ p d¿ a+b a+b ¡1

let x=

p bt a + b¿ ¡ p a+b p dx = a + b d¿

¿ = §1

¡!

x = §1

recall or use (\and it is a trick!"see page 63) Z1

2

e¡x dx =

p ¼

¡1

·μ

¶ ¸ Z1 2 b2 dx z(t) = exp ¡ b t2 e¡x p a+b a+b ¡1 r · μ ¶ ¸ b2 ¼ = exp ¡ b ¡ t2 a+b a+b r · ¸ ¼ ab 2 = exp ¡ t a+b a+b The convolution of two Gaussians is a third Gaussian. 21

Yet Another Convolution Example Evaluate s(t) = f (t) ~ g(t) where f (t) and g(t) are the given rectangular pulses: f (t) = u(t) ¡ u(t ¡ a) = g(t) = u(t) ¡ u(t ¡ b) =

½ ½

1; 0 < t < a 0; t < 0 or

t>a

1; 0 < t < b 0; t < 0 or

t>b

Here the pulse width a of f (t) has been arbitrarily selected to be greater than the pulse width b of g(t). .. .. .. .. .................................................................................................................................. ... ... .. .... .. ... ... ... .. ... .. ... ... ... .. .. ... .. . ........................................................................................................................................................................................................................................................................................................................................................................................................................... .. .

f (t)

1

a

0

... ... .. ... ...................................................................................... ... .. ... .. ... ... .. ... .. .... ... .... .. .. ... ... .. . ........................................................................................................................................................................................................................................................................................................................................................................................................................... ... .

g(t)

t

1

0

b

Z1

s(t) = f (t) ~ g(t) =

¡1

t

f (¿ )g(t ¡ ¿ ) d¿

Since the arguments of our individual signals are ¿ (and a shifted, rotated version of ¿ ) in the convolution integral, let's redraw our functions f and g as functions of the dummy variable ¿ . . .. .. .. .. ................................................................................................................................... ... ... .. .... .. ... ... ... .. ... .. ... ... ... .. .. ... . ............................................................................................................................................................................................................................................................................................................................................................................................................................ .. .

f (¿ )

1

a

0

.... ... .. .................................................................................... .. ... ... .. ... .. ... ... .. ... .. .... ... .... .. .. ... ... .. . .......................................................................................................................................................................................................................................................................................................................................................................................................................... ...

g(¿ )

¿

1

0

b

22

¿

Let's leave f (¿ ) alone (as in the version of the convolution integral above), and shift and rotate g(¿ ) (as it appears in the version of the convolution integral above). ... ... .. ... ...................................................................................... .. .. ... .. ... ... .. .. ... ... .. ... ... ... .. ... .. ... ... .. . ... ... ... ... .. . . . .. . . ........................................................................................................................................................................................................................................................................................................................................................................................................................... ... .

g(¿ ¡ t)

t

0

t+b

..... .. .. .. ... ....................................................................................... ... ... .. ... .. ... .. .. ... ... .. ... .. . ... ... .... ... ... .. ... .. ... ... .. . . . . . .......................................................................................................................................................................................................................................................................................................................................................................................................................... ....

¿

g(t ¡ ¿ )

0 t¡b

t

¿

Now let's vary the parameter t (it survives the ¿ integration), compute (graph!) the product of f (¿ ) and g(t ¡ ¿ ), and evaluate the in¯nite range integral from ¿ = ¡1 to ¿ = +1. That is, we must calculate the total area under the curve of f (¿ )g(t ¡ ¿ ). case: t < 0

..... .. .. ................................................................................................................................ .. ... ... .. ... ... ... .. .. ... ... .... .. .... .. ... ... .. .. . . .......................................................................................................................................................................................................................................................................................................................................................................................................................... ....

f (¿ )

0

a

b

... .. .. .. ... ................................................................................... ... .... ... ... .. ... .. .. ... .. ... ... .. ... ... .. ... ... ... .. ... ... .. .. ... .. ... .. .. .......................................................................................................................................................................................................................................................................................................................................................................................................................... .. .

¿

g(t ¡ ¿ )

t¡ b

t

0

... .. .. .. ... .. ... ... .. .. ... .. ... ... ......................................................................................................................................................................................................................................................................................................................................................................................................................... ....

¿

f (¿ )g(t ¡ ¿ )

0

s(t) = f (t) ~ g(t) =

Z1 ¡1

f (¿ )g(t ¡ ¿ ) d¿ =

23

Z1 ¡1

0 d¿ = 0

¿

case: 0 < t < b

..... .. .. ................................................................................................................................ .. ... ... .. ... ... ... .. .. ... ... .... .. .... .. ... ... .. .. . . .......................................................................................................................................................................................................................................................................................................................................................................................................................... ....

f (¿ )

0

a

b

... .. .. .. ..................................................................................... .... .. ... .. .. .... ... ... ... .. .. ... .. .. ... ... ... ... .. .. .. .. ... ... ... .... .. .. .......................................................................................................................................................................................................................................................................................................................................................................................................................... .. ..

¿

g(t ¡ ¿ )

t¡ b

0

t

.... .. .. .. ........................................................... .. ... ... .... .. ... .. ... ... ... .. ... .. ... ... .. ... . ............................................................................................................................................................................................................................................................................................................................................................................................................................... ....

¿

f (¿ )g(t ¡ ¿ )

0

s(t) = f (t) ~ g(t) =

Z1 ¡1

t

a

b

f (¿ )g(t ¡ ¿ ) d¿ =

Zt

¿

d¿ = t

0

case: b < t < a

... ... .. ... .................................................................................................................................. ... .. ... .. ... ... .. ... .. .... ... .... .. .. ... ... ... . ............................................................................................................................................................................................................................................................................................................................................................................................................................. ... .

f (¿ )

0

a

b

..... .. .. .. ... ....................................................................................... ... ... .. .. ... ... .. .. ... ... .. ... .. ... ... ... .. ... ... .. ... .. ... ... .. . . . . . ........................................................................................................................................................................................................................................................................................................................................................................................................................... ....

¿

g(t ¡ ¿ )

0 t¡ b

t

... .. .. .. ... ................................................................................... ... .... ... ... .. ... .. .. ... .. ... ... .. . ... ... .... ... ... .. ... .. ... .. ... . . ... ... ................................................................................................................................................................................................................................................................................................................................................................................................................................ .. .

¿

f (¿ )g(t ¡ ¿ )

0 t¡ b

s(t) = f (t) ~ g(t) =

Z1 ¡1

b

f (¿ )g(t ¡ ¿ ) d¿ =

t a

Zt t¡b

24

d¿ = b

¿

case: a < t < a + b

... .. .. .. ................................................................................................................................... ... ... .. .... .. ... ... ... .. ... .. ... ... ... .. .. ... . . .............................................................................................................................................................................................................................................................................................................................................................................................................................. ... .

f (¿ )

0

a

b

a+b

.... ... .. ... .. ........................................................................................ .. ... .. .. ... ... .. ... ... ... .. ... ... .. ... ... .. ... .. . ... ... ... ... .. . . . . . . . . ............................................................................................................................................................................................................................................................................................................................................................................................................................ ...

¿

g(t ¡ ¿ )

t¡b

0

a

t

a+b

.. ... .. ... .. ...................................... .. .... ... .. ... ... .. ... ... .. ... ... ... .. ... ... .. ... .. . ... ... ... ... .. . . .... ... ................................................................................................................................................................................................................................................................................................................................................................................................................................. .. .

¿

f (¿ )g(t ¡ ¿ )

t¡b

0

s(t) = f (t) ~ g(t) =

Z1 ¡1

a

f (¿ )g(t ¡ ¿ ) d¿ =

t Za

a+b

¿

d¿ = a + b ¡ t

t¡b

case: t > a + b

... .. .. .. .................................................................................................................................. .. ... ... .... .. ... .. ... ... ... .. ... .. ... ... .. ... .. . . . . ............................................................................................................................................................................................................................................................................................................................................................................................................................ .. .

f (¿ )

0

a

b

a+b

.... .. .. .. ... ..................................................................................... ... .. ... .. ... ... .. .. ... ... .. ... .. ... ... ... .. ... ... .. ... .. ... ... .. . .. .. . . . . . .............................................................................................................................................................................................................................................................................................................................................................................................................................. ... .

¿

g(t ¡ ¿ )

a

0

t¡b a + b

t

.... .. .. .. ... .. ... ... .. .. ... .. ... .. ........................................................................................................................................................................................................................................................................................................................................................................................................................... ...

¿

f (¿ )g(t ¡ ¿ )

0

s(t) = f (t) ~ g(t) =

Z1 ¡1

f (¿ )g(t ¡ ¿ ) d¿ =

25

Z1 ¡1

0 d¿ = 0

¿

All together now. 8 0; > > > > > < t; s(t) = f (t) ~ g(t) = b; > > > a + b ¡ t; > > : 0;

t 0);

¡1

sin(!0 ¿ )±(¿ ) d¿;

¡1

Z1

Z1 ¡1

Z1

Z¼

cos(!0 ¿ )±(¿ ) d¿;

¡1

cos x dx; x

Z1 ¡1

0

Z2¼ · p ¡ ¼

u(t)e¡¾t e¡j!t dt

cos x dx; x

sin(x2 ) ±(x ¡ 2¼) dx ln(x=¼)

0

Z1

dx ; x2

1

Z1 0

dx ; x2

Z1

dx p ; x

0

¸ exp[j sin2 (!0 t)] ±(t + ¼) ¼ 2 sin(3t=2) ±(t ¡ ¼) ¡ 2 dt t3 tan(7t) + ln(sin t) t + 2t ¡ 4 sin¡1 (t=¼) 35

Z1 0

dx p x

3. Prove that convolution is commutative: f (t) ~ g(t) = g(t) ~ f (t): 4. Evaluate u(t) ~ u(t). 5. Show that ±(t) = ±(¡t) is an even function. 6. If f (t) is continuous at t = t0 , show that f (t)±(t ¡ t0 ) = f (t0 )±(t ¡ t0 ).

36

Relationship Between Impulse Response and Step Response ................................................. ... ... .. ... .. ... ... . . ........................... ............................. .... .... .. .. ... .. . . ................................................

d ±(t) = u(t) dt

±(t)

Zt

u(t)

u(t) =

h(t)

h(t) =

................................................. ... .... ... .. ... .. . . ............................ ............................ ... .. .. ... ... .. .. . .................................................

±(¿ ) d¿

¡1

s(t) s(t) =

Zt

d s(t) dt

h(¿ ) d¿

¡1

To see this, consider the general linear system input/output relationship. ................................................. .. ... .. ... ... ... .. . . ........................... ........................... .... .... .. .. ... ... . ... ..............................................

x(t)

y(t)

h(t)

Z1

y(t) = x(t) ~ h(t) =

¡1

d y(t) = dt

Z1 ¡1

x(¿ )h(t ¡ ¿ ) d¿ =

d x(¿ ) h(t ¡ ¿ ) d¿ = dt

Z1 ¡1

Z1 ¡1

h(¿ )

h(¿ )x(t ¡ ¿ ) d¿

d x(t ¡ ¿ ) d¿ dt

= x(t) ~ h (t) = x (t) ~ h(t) 0

0

The output of these two systems is identical, since the order of di®erentiation and passing through the linear system having a given impulse response h(t) does not matter. More concisely, the operators h(t)~ and d=dt commute.

x e(t)

x e(t)

..... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ..... .. .. .. .. .. ............................................... ............................................... .. ... ... .... .... .. ... .. .. .. .. .. ... ... .. .. . .. . . .............................................. ....................................... ........................................ . . . .. ... . . ... . .. ... .... .. ... .... ... .. .. .. .. .. .. . . . . .. .............................................. .............................................. . .. .. .. .... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .....

d dt

h(t)

..... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .... . .. .. .. ... .. .............................................. .............................................. . . . . .. .. ... . ... . . . . . . ... .. .... ... .... ... . . . . . . . . . . . . . . . . . . . . . . . . . ................................. ....................................... ..................................... . . . . . . .. ... .... ... .... ... . . . . . . .. .... .... .. ... ... .. ............................................... ............................................... .. .. .. .. ..... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .....

d dt

h(t)

37

ye(t)

ye(t)

Time-Domain RC Circuit Analysis R

... . ... .......... .. ...... ...... ...... ........ ........................................ .......... ...................................... ..... ..... ..... ..... ..... ............................................. ....... ..... ..... ..... .. ... . .. ... ... ... .. ... ... .................. . . . . ... .. ............... . . . . . . . . . . . . . . . . . . ... .... .. . ... . . ... .............................. .. ...... ... ................... ... ... ... ... ... .. .. .. .................................................................................................................................................................

+

t=0

V0 cos(!t)u(t) = vg (t)

+ ¡

C

v(t) ¡

The capacitor is initially charged, such that v(t) = Vc0 prior to the closing of the switch at time t = 0. Find an expression for the signal v(t) that is valid for all time t ¸ 0. The constants R, C, !, V0 , and Vc0 are given and thus known. The initial condition on the voltage v(t) that applies to our circuit for time t ¸ 0 is v(0+ ) and v(0+ ) = v(0¡ ) = Vc0 since the voltage across a capacitor must be a continuous function of time. That is, if the capacitor current is bounded, because the voltage/current terminal relationship of a capacitor is d iC (t) = C vC (t) (1) dt where iC (t) and vC (t) are related by the passive sign convention. Inverting this, that is, writing the voltage as a function of the current, we have 1 vC (t) = C

Zt ¡1

1 iC (¿ ) d¿ = C

Z0

1 iC (¿ ) d¿ + C

¡1

Zt

1 iC (¿ ) d¿ = vC (0) + C

0

Zt

iC (¿ ) d¿: (2)

0

The Kirchho® current law applied to the (single unknown) node of potential v(t), for time t ¸ 0 is dv(t) v(t) ¡ vg (t) C + =0 (3) dt R which is rearranged as dv(t) 1 V0 + v(t) = cos(!t)u(t): (4) dt RC RC Here we are already working in the time zone t ¸ 0, so with the stipulation that t ¸ 0, the Heaviside unit step is probably unnecessary so we can write dv(t) 1 V0 + v(t) = cos(!t): dt RC RC

(5)

The homogeneous solution is the solution to 1 dvh (t) + vh (t) = 0 dt RC 38

(6)

and it is vh (t) = Ke¡t=RC

(7)

where the constant K cannot be determined until we construct the complete solution. The particular solution is some speci¯c or particular function vp (t) that yields the right-hand side (forcing term) when operated upon by the di®erential operator μ ¶ V0 d 1 + vp (t) = cos(!t): (8) dt RC RC There are no unknown constants (unknown by the di®erential operator) in the particular solution.3 Since the forcing function is cos(!t) and since cos(!t) is not a homogeneous solution, then we see that the particular solution must be a linear combination of cos(!t) and sin(!t), as in (9) vp (t) = A cos(!t) + B sin(!t): The constants A and B come directly from the di®erential equation (8) , which readily surrenders them (to us): All we have to do is insert the form (9) into the di®erential equation (8). To do this, ¯rst write the required derivative dvp (t) = !B cos(!t) ¡ !A sin(!t): dt Insertion of (9) and (10) into (8) yields μ ¶ μ ¶ A B V0 + !B cos(!t) + ¡ !A sin(!t) = cos(!t): RC RC RC

(10)

(11)

The only way this can be true for all t ¸ 0 is if the coe±cients of the cos(!t) and the sin(!t) terms are separately equal on both sides of the equation, so that A V0 + !B = RC RC

(12)

B ¡ !A = 0: RC The solution to this two-by-two system of simultaneous, linear equations is A=

V0 1 + (!RC)2

and

B=

!RCV0 ; 1 + (!RC)2

(13)

(14)

and so our particular solution (9) is vp (t) =

£ ¤ V0 cos(!t) + !RC sin(!t) : 1 + (!RC)2

3 In

(15)

other words, the particular solution comes only from the di®erential equation, that is, the particular solution is completely independent and ignorant of any initial conditions. 39

Again, note that vp (t) comes completely from the di®erential equation alone. We now can assemble the complete solution v(t) = vh (t) + vp (t) or v(t) = Ke¡t=RC +

£ ¤ V0 cos(!t) + !RC sin(!t) : 1 + (!RC)2

(16)

Application of the one initial condition v(0) = v(0+ ) = Vc0 (note that it is the plus side of zero that applies to our di®erential equation, because the di®erential equation is valid for time t ¸ 0) to (16) gives Vc0 = K +

V0 1 + (!RC)2

K = Vc0 ¡

or

V0 : 1 + (!RC)2

(17)

Let's write out the complete solution in two forms. Firstly, observe that for t ¸ 0 · ¸ V0 v(t) = Vc0 ¡ e¡t=RC 1 + (!RC)2 | {z } homogeneous solution

+

£ ¤ V0 cos(!t) + !RC sin(!t) 1 + (!RC)2 | {z }

(18)

particular solution

is exactly in the form in which we constructed the solution, that is, the sum of the homogeneous plus particular solutions. If we instead group terms proportional to Vc0 and V0 , we have for t ¸ 0 v(t) =

¡t=RC

Vc0 e | {z

}

zero-input response

+

h i V0 ¡t=RC cos(!t) + !RC sin(!t) ¡ e : 1 + (!RC)2 | {z }

(19)

zero-state response

If there is no input signal (or generator voltage vg (t) or forcing term), then V0 = 0 and the output signal v(t) is caused solely by the initial energy (or non-zero initial state of the system) stored in the capacitor, and is proportional to the non-zero initial condition voltage Vc0 . On the other hand, if the initial state of the system is zero, so that Vc0 = 0, then the response due to the input vg (t) that is proportional to V0 is called the zero-state response.

40

Step Response of an RC Circuit

R

..... ..... ..... ... .. ... .. ... .. ............................................................................... ......... ......... ..................................................... .. .. .. .. ... .. ... .. ... . ... ................. .... ...... . . ............................... ... ..... . . ... .. ... .............................. ... ..... ... .................... ... ... ... .. ... .. .. ... ..................................................................................................................................................................

+ ¡

V0 u(t) = vg (t)

C

+ v(t) ¡

The step response is, by de¯nition, a zero-state response so the capacitor C is initially uncharged prior to the turning on of the input step function at time t = 0. Therefore, by the continuity of capacitor voltage we have the initial condition v(0+ ) = v(0¡ ) = 0:

(20)

Ultimately, the step response s(t) = y(t) should be the response due to a non dimensioned input signal u(t) = x(t), but in doing circuit analysis I ¯nd it better to explicitly express all physical signals such as voltages and currents in appropriate mksC (SI) units. At the end we can normalize by V0 , for example. In systems engineering (\black-box") notation, we have the general form of a zero-state response y(t) to some arbitrary input x(t) like this:

x(t)

............................................................................. ... ... ... .. .. ... .. ........................................ ...................................... ... ... ... ... .. ... . ...........................................................................

y(t)

............................................................................. ... .... ... .. ... .. . . ................................... . .................................... ... .. .... .. ... . .... .. ...........................................................................

s(t)

For the step response, we have: u(t)

Our ensuing circuit analysis regards the input as x(t) = vg (t) = V0 u(t) and our output signal is y(t) = v(t). The Kirchho® current law applied to the node of potential v(t) in the circuit is for time t ¸ 0 C

dv(t) v(t) ¡ vg (t) + =0 dt R

(21)

which is rearranged as 1 V0 dv(t) + v(t) = u(t): dt RC RC 41

(22)

Here we are already working in the time zone t ¸ 0, so with the stipulation that t ¸ 0, the Heaviside unit step is probably unnecessary so we can write 1 V0 dv(t) + v(t) = : dt RC RC The homogeneous solution is (still) of the form vh (t) = Ke¡t=RC

(23)

(24)

where K is some (new) unknown constant. But now the particular solution, due to a constant forcing term, is itself a constant, call it vp (t) = D:

(25)

This constant D is, again, completely determined (that is known by) the di®erential equation (24). Insertion of (25) into (23) gives the rather trivial V0 D = or D = V0 : RC RC The complete solution is now v(t) = vh (t) + vp (t) or v(t) = Ke¡t=RC + V0 ; whereupon the initial condition (20) gives K = ¡V0 so that £ ¤ v(t) = V0 1 ¡ e¡t=RC (t ¸ 0): A better (systems engineering style!) way to write this is £ ¤ v(t) = V0 1 ¡ e¡t=RC u(t):

(26)

(27)

(28)

(29)

The unit step not only eliminates the need for the disclaimer t ¸ 0, but its inclusion is invaluable in emphasizing the causal nature of the response, and it is critically important if (when!) we di®erentiate the response. Now we can normalize by V0 , as advertised, and obtain the true, non dimensional step response £ ¤ s(t) = 1 ¡ e¡t=RC u(t): (30)

The impulse response is the derivative of the step response £ ¤ d 1 ¡t=RC h(t) = s(t) = e u(t) + 1 ¡ e¡t=RC ±(t) dt RC 1 ¡t=RC e u(t): (31) = RC Note that the units of h(t) are [(−)(F)]¡1 = (s)¡1 . In this particular instance, the chainrule di®erentiation that also di®erentiated the u(t) to get a ±(t) might appear super°uous, since the contribution £ ¤ £ ¤¯ ¡t=RC ¡t=RC ¯ 1¡e ±(t) = 1 ¡ e ±(t) = 0±(t) = 0 (32) ¯ t=0

is zero. But that will not happen in general.

42

Step Response of Another RC Circuit + vC (t) ¡ ... ... .. .. .. .. . .................................................................... . ... .. ....................................................................... .. .. .. .. .. ... .... ... ... .. .. .. .. ..... ...... ... ... ....... . . . . . . . . . . . . . . . ....... ..... ......... . . . . . ..... . ... ... .. ........ ............ ..... .. ........ ... ... . .. .... ....... ....................... ............. ... . ... .... ... .. ... ... ... .. ... ... ... .......................................................................................................................................................

+

C

V0 u(t) = vg (t)

+ ¡

R

v(t) ¡

In this circuit, we take our output signal to be the voltage v(t) across the resistor, but the capacitor voltage vC (t) must also be speci¯cally addressed because of the importance of continuity of capacitor voltage for determining initial conditions. The zero-state response demands that there is no initial energy storage in the capacitor so vC (0¡ ) = 0 prior to the unit step input turning on at time t = 0. By the continuity of capacitor voltage vC (0+ ) = vC (0¡ ) = 0:

(33)

The application of the Kirchho® current law to the node of potential v(t) gives ¤ d£ v(t) +C v(t) ¡ vg (t) = 0 R dt

(34)

or

dv(t) 1 dvg (t) + v(t) = : (35) dt RC dt The forcing term is now the derivative of the voltage generator, which puts us in the unfamiliar realm of solving a di®erential equation with a wild Dirac-delta forcing term 1 dv(t) + v(t) = V0 ±(t): dt RC

(36)

If we stay away from t = 0, then the delta function is zero and we have (note the time restriction) 1 dv(t) + v(t) = 0 (t > 0): (37) dt RC It looks as though we have lost our excitation: The di®erential equation is now homogeneous. Information about the excitation, embodied by the source strength V0 , is included in the initial condition. Observe in the circuit that vg (t) = vC (t) + v(t) 43

(38)

and at time t = 0+ this is vg (0+ ) = vC (0+ ) + v(0+ ):

(39)

The initial condition (33) on the capacitor voltage together with vg (0+ ) = V0 reveals that the non-zero initial condition on our output signal v(t) is v(0+ ) = V0 . No particular solution is required for the complete solution to our homogeneous di®erential equation (37), which is readily seen to be v(t) = V0 e¡t=RC

(t > 0):

(40)

Inclusion of the Heaviside unit step function gives v(t) = V0 e¡t=RC u(t)

(41)

so that the time zone restriction t > 0 can be removed. The non dimensional step response is therefore (42) s(t) = e¡t=RC u(t) and the corresponding impulse response is h(t) =

since

1 ¡t=RC d s(t) = e¡t=RC ±(t) ¡ e u(t) dt RC 1 ¡t=RC e = ±(t) ¡ u(t) RC ¯ ¯ ¡t=RC ¯ e ¯ ¯

= 1:

(43)

(44)

t=0

Observe that the proper use of chain-rule di®erentiation and explicitly including the u(t) in the step response s(t) is necessary to get the true impulse response h(t), which most de¯nitely contains a ±(t) term.

44

Relationship Between the Two RC Circuits

+

. . ..... ..... ...... ................................................ ..... .... ..... .... ..... .............................................................................. ..... ...... .... .... . .... .. ... . ........................

²

R

±(t) ¡

C

+

+

.... .. ................................................................ ... ..

..... .. ............................................................................................ .... ... .. .. ..... ..... ...... . ............. ....... .. ....... ............. ........ . ....... ............. ... .... . ......................................................................................................................................................................

C

ha (t)

±(t)

¡

¡

........................ .. .. ... .. .. .. ......................................................................................................................................................................

²

²

+

hb (t)

R

²

¡

In the lab, simply connect R and C in series and hang the series leg across the output terminals of a (ideal?) voltage source. Then we see that the output of circuit (a) is the voltage across C and the output of circuit (b) is the voltage across R. Kirchho®'s voltage law clearly shows the simple relationship that is displayed by comparing (30) and (42) and by comparing (31) and (43). ............................................................................ .. ... ... ... ........ ... ...... . . . . .... . . ....... ... ........ .. ... ........ ............. ..... ....... ... .. ... ....... .............. .... . . . . . . . . . . .. ........ ... ... ...... ... .. ... .. ... ..... .. . .. ... . . .. .... . . . .. . .................... ... ... .. .... . .. ............................... ... ... ... .............................. . .... ... ... .. ... .. .. .. .. ... ............................................................................

R

vg (t)

+ ¡

C

+

vb (t)

vg (t) = va (t) + vb (t)

¡

±(t) = ha (t) + hb (t)

+ va (t)

u(t) = sa (t) + sb (t)

¡

Homework Problems 1. Observe that the circuit on page 41 is a special case of the circuit on page 38, if Vc0 = 0 and ! = 0. Therefore, show for yourself that the response (29) is the corresponding special case of the response (19). 2. Use convolution to ¯nd the zero-state (or driven) response of the RC circuit on page 41 to a rectangular input pulse vg (t) = V0 [u(t)¡u(t¡T )]. Graph this response using MATLAB for several numerical values of the important (non dimensional) parameter T =RC. 3. Go backwards, and obtain the two step responses (30) and (42) by integrating the corresponding impulse responses (31) and (43), since d h(t) = s(t) dt

() 45

s(t) =

Zt ¡1

h(¿ ) d¿:

Step and Impulse Response of Series RLC Circuit

.... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .... . .. .. .. .. ... .. . .. .. .. .. .. C .. .. ... ... . .... .... . . . . . . . . . . . . . . . .. . . . . . .. . .. .. .. .. .. . ... . .... .......... ........... .......... ............................................. .... ....................................................................................................................... ................................................................................. . . . . . .. . . .. . . . . . . ... . .. . .. .......................................... .. . .. .... ... .. ................. ... ........... . . . . . . . . .. . .. . . ........ .. .. .............. ....... .. ... .. . . . . .. . . ...... ...... . . . .. ... . . . ..... ..... . . . . . . .. .... . . ....... . ... . . . . ... . . . . .................. . . . . . . . . . . . . . . ... .... . . ... .. ............. ... ... ... ... .. . .. . . . . . . . . ... ....... ... .. .. ... ..... . .. .. . . . . . . . . . . . .. . . . ... ... . . . . . . . . . . . . ........ ... ... .. . . . . . . . . . .... ... . . . . . . .. . . . . ....... .......... . . . ... . . .. ............ ............. .... .. ... . . ... . ...... . . .. .. . . . ...... . .. . .. . . . . . . . . ... ........... . . .. ... . . . . . .. ......... ... . . .. . . . . .. . .......... . . . . .. . ... ....... ............... .. .... . .............................................. .. .... . .. . . . . .. . .... . . . . . . . . . . ............................................................................................................................................................................................................................................................................................................ ... .. . .. .. ... .. .. .. . .. .. .. .. .. .. .. ... .. . ..... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .....

+ v (t) ¡

L

²

i(t)

vg (t)

+

C

+ ¡

R

²

vR (t)

¡

Fig. 1. Circuit layout.

vg (t) = x(t)

...................................................................................................................................................................... ... ... ... ... ... .. .. ... ... . .... .. .. .. ... ... . . . .............................................................. . . .......................................................... .. ... ... ..... .. .. .. ... ... .. .. ... .. ... . .... . ....................................................................................................................................................................

y(t) = Ri(t)

Fig. 2. System block diagram.

The system step response is a \zero-state" response: No energy is initially stored in the system prior to the onset of the exciting Heaviside step function that turns on at t = 0. For de¯niteness and to keep track of the physical units, let vg (t) = V0 u(t) 46

(1)

with the (real) constant V0 being the amplitude of the exciting step function. Since the instantaneous energy stored in the inductor and capacitor at time t are wL (t) = 12 Li2 (t)

2 wC (t) = 12 CvC (t);

and

(2)

respectively, zero initial (at time t = 0¡ ) energy storage requires i(0¡ ) = 0

vC (0¡ ) = 0:

and

(3)

If the inductor voltage di(t) dt is to be bounded, then its current must be a continuous function of time, and hence vL (t) = L

i(0+ ) = i(0¡ ):

(4)

Similarly, if the capacitor current dvC (t) dt is to be bounded, then its voltage must be a continuous function as in iC (t) = C

vC (0+ ) = vC (0¡ ):

(5)

The equation-of-motion for the series circuit of Fig. 1 is Kirchho®'s voltage law Zt

di(t) 1 + Ri(t) + L dt C

i(¿ ) d¿ = vg (t);

(6)

¡1

an integrodi®erential equation in the unknown series current i(t). With the forcing function vg (t) known, the unique solution of equation (6), whose highest derivative term is of degree one, requires the speci¯cation of one initial condition. The physics of the circuit dictates i(0+ ) = 0;

(7)

from (3) and (4). If we di®erentiate (6) once to eliminate the integral operator, the resulting second-order di®erential equation di(t) d2 i(t) 1 +R (8) + i(t) = vg0 (t) 2 dt dt C now requires two initial conditions for a complete (unique) solution. This second initial condition was already contained in (6), before we di®erentiated. Evaluation of (6) at time t = 0+ is Li0 (0+ ) + Ri(0+ ) + vC (0+ ) = vg (0+ ): (9) L

The capacitor voltage, from (3) and (5), starts o® at zero vC (0+ ) = 0;

(10)

so that together with (1) and (7), statement (9) gives V0 : L Note that the dimensions are consistent in that (A/s) = (V/H). X i0 (0+ ) =

47

(11)

The proper formulation of the above initial value problem, especially stating the correct number (two) and type of initial conditions (on the current and its derivative), required our detailed understanding of the electrophysics embodied in the circuit. In one view the ensuing construction of the solution to the di®erential equation may seem like a pure exercise in mathematics, but that is not the view we will take. Instead, we want to let each step in our solution be guided by the physics that the mathematics represents, so that our mathematics will be e±cient and keep us in touch with the important physics that is our quest. If we do this right, then our symbols and formulas and solution will be in a form that not only allows, but even begs for physical interpretation. Let's restate our initial value problem: L

1 di(t) d2 i(t) + i(t) = vg0 (t) +R 2 dt dt C i(0+ ) = 0 i0 (0+ ) =

(12) (13)

V0 : L

(14)

Since our voltage generator is vg (t) = V0 u(t), the forcing function on the right-hand side of (12) is a Dirac-delta function, i.e. L

d2 i(t) di(t) 1 + i(t) = V0 ±(t): +R 2 dt dt C

(15)

That looks troublesome, I believe, but if we stay away from t = 0, then the delta function is zero and our equation is homogeneous L

di(t) 1 d2 i(t) + R + i(t) = 0 dt2 dt C

(t > 0):

(16)

It looks as though the homogeneous di®erential equation (16) has lost the source, but note that initial condition (14) has information about the original source in the circuit. If there is no initial energy storage to drive the \dead" circuit, the excitation of a nonzero response has to come from an explicit input. And our input (1) is proportional to the constant V0 that explicitly appears in the nonzero (\nonhomogeneous") initial condition (14). Homogeneous constant-coe±cient linear ordinary di®erential equations (LODE's) are particularly easy to solve, since the solutions are exponentials, say of the form i(t) = exp(rt) where two such linearly independent forms (i.e. two di®erent r values), are required in our case of a second-order equation. Before proceeding, it is expedient to condense the notation and in fact, combine the element values R, L, and C by in a manner that anticipates the solution. Of course, no one could truly do this in advance if they had not already solved the problem (or read someone else's solution). But that's not the point: We choose the best path because it draws attention to what is important and because it is e±cient and 48

because it works. Firstly, normalize (16) so that the coe±cient of the highest derivative is unity 1 d2 i(t) R di(t) + i(t) = 0 + dt2 L dt LC and then introduce constants 1 LC

(17)

d2 i(t) di(t) + 2® + !02 i(t) = 0: 2 dt dt

(18)

®= to write

R 2L

and

!02 =

Insertion of i(t) = exp(rt) into (18) gives £ 2 ¤ r + 2®r + !02 ert = 0; and since the exponential cannot equal zero, we arrive at the characteristic polynomial for the two roots r2 + 2®r + !02 = 0: (19) The quadratic formula provides the two roots immediately r1;2

q = ¡® § ®2 ¡ !02 :

(20)

Two distinct possibilities exist: The case of an over-damped circuit having ® > !0 and an under-damped circuit when ® < !0 . The case when ® = !0 is called critically-damped, and can be obtained from either of the other two cases in the limit as ® ! !0 . Or, it can be constructed directly. Under-damped circuit. If ® < !0 , introduce yet another parameter (constant) q ¯ = + !02 ¡ ®2

(21)

so that the two roots (20) are a complex-conjugate pair r1;2 = ¡® § j¯:

(22)

The solution to the homogeneous di®erential equation (18) is a linear combination £ ¤ i(t) = K1 er1 t + K2 er2 t = K1 e¡(®¡j¯)t + K2 e¡(®+j¯)t = e¡®t K1 e¡j¯t + K2 e+j¯t = e¡®t [A cos(¯t) + B sin(¯t)] :

(23)

It is critically important to recognize that a linear combination of exp(§jx) is equivalent to a (di®erent) linear combination of cos x and sin x. Although either form is ¯ne, the 49

trig form is better here because it facilitates the application of initial conditions. Initial condition (13) gives i(0) = A = 0 (24) immediately! The derivative of the remaining form

is

i(t) = Be¡®t sin(¯t)

(25)

£ ¤ i0 (t) = Be¡®t ¯ cos(¯t) ¡ ® sin(¯t)

(26)

i0 (0) = ¯B:

(27)

which has initial value Initial condition (14) is now V0 L and the complete solution to the initial value problem is

(28)

¯B =

i(t) =

V0 ¡®t e sin(¯t) ¯L

(t > 0)

or, with attention drawn to its \turned-on" nature, i(t) =

V0 ¡®t e sin(¯t)u(t): ¯L

(29)

Note that the dimensions of ® and ¯ are both (s¡1 ) and that the dimensions of the product ¯L are (H/s= −), so that the current is in (A). The signal that is regarded as the output in the linear system diagram of Fig. 2 is the voltage vR (t) = Ri(t), so that when the input vg (t) is regarded as x(t) = V0 u(t); (30) the output is y(t) =

RV0 ¡®t e sin(¯t)u(t): ¯L

(31)

If we normalize both signals by the amplitude V0 , then the nondimensional input/output signals are R ¡®t x(t) = u(t) and y(t) = s(t) = e sin(¯t)u(t): (32) ¯L The response to the Heaviside unit step function is called the step response, and is denoted here as s(t). The derivative of the convolution z(t) = f (t) ~ g(t) =

Z1 ¡1

f (¿ )g(t ¡ ¿ ) d¿ = 50

Z1 ¡1

g(¿ )f (t ¡ ¿ ) d¿

(33)

is

z 0 (t) = f 0 (t) ~ g(t) = f (t) ~ g 0 (t):

(34)

Therefore, the system impulse response is the derivative of its step response h(t) = s0 (t):

(35)

The impulse response of the RLC circuit of Fig. 1 is therefore ¸ · R ¡®t ® h(t) = e cos(¯t) ¡ sin(¯t) : L ¯

(36)

Note that the units of h(t) are (s¡1 ), as are the units of ±(t).

1.00 0.75 0.50 ¯L s(t) R

0.25 0.00 ¡0:25 ¡0:50 ¡0:75

..... ... ... .. .... ... .. . ... ... .. .. .. ... .. .. .. ... .. ... .. .. .. .. .. ... ..... .. .. .. .. .. .. ........ .. .. .. .... ...... .. .. .... .. .. .. . ... ... ... .. .. .... . ... .. ... .. .. .. . .. ... ...... . . .. . ... . .. ... ... . . .......... . ... . .. .. . ... ..... ......... . .. . . .. .. ... ... ... . ... . . ... .. .. ... . . . . . . .. . ... ... .. ... . . .. . ... ... ....... .. .. ... . . . .... ..... ... ... . . ..... . . .... .. . . . ................................................................................................................................................................................................................................................................................................................................................................................................................................................................ .... .. . ... ... . ... .. .. .... . ... ... ... ..... ... .... .. .... ... ... ... ... .. .. ... ... ... ... ... . . . . . . . . . .... . .. .. . .. .... ..... .. .. .. ...................... ... .. ... .. .. .. ... . .. .. . . . .. ... .. .. ... .. ... .. ... ..... ....... .. ........ ... . . .. . .. .. .. ... ... ... .. . ... ... .. .. .. ... ..... ..........

0

2

4

6

8

10

12

14

16

¯t Fig 2. Normalized step response of underdamped RLC circuit. Sold curve: ®=¯ = 0:1; Dashed curve: ®=¯ = 0:5

51

18

20

Over-damped circuit. If ® > !0 , then both roots of the characteristic quadratic polynomial (19) are real, so with r1 = ¡®1

q with ®1 = ® ¡ ®2 ¡ !02

and r2 = ¡®2

with ®2 = ® +

q

®2 ¡ !02 ;

(37)

(38)

we see that 0 < ®1 < ®2 and the solution to the homogeneous di®erential equation (16) can be written as the linear combination i(t) = k1 e¡®1 t + k2 e¡®2 t

(t > 0):

(39)

Initial condition (13) is i(0+ ) = 0 so that k2 = ¡k1 and therefore £ ¤ i(t) = k1 e¡®1 t ¡ e¡®2 t :

(40)

£ ¤ i0 (t) = k1 ®2 e¡®2 t ¡ ®1 e¡®1 t

(41)

The required derivative is

and at time t = 0+ this must be £ ¤ V0 i0 (0+ ) = k1 ®2 ¡ ®1 = L and therefore

(42)

V0 : (®2 ¡ ®1 )L

(43)

£ ¡®1 t ¤ V0 e ¡ e¡®2 t u(t) (®2 ¡ ®1 )L

(44)

k1 = The solution for the current is now i(t) =

and the nondimensional step response is s(t) =

£ ¡®1 t ¤ R e ¡ e¡®2 t u(t): (®2 ¡ ®1 )L

52

(45)

Homework. Given that a (scaled) rectangular pulse 1 x(t) = ¦ T

μ ¶ t T

behaves as the Dirac-delta function in the limit as T ! 0, how small does T have to be so that the response y(t) of a system to the pulse x(t) is a good approximation to the true system impulse response h(t) ? Consider the underdamped RLC circuit with parameters ® and ¯. (Note that specifying ® and ¯ is equivalent to, and in fact more illuminating than, specifying the circuit element values R, L, and C.) Use MATLAB to compare graphs of h(t) with y(t) for several values of the pulse-width T . The important parameter(s) of the problem should be nondimensional.

Homework Exercises. Find the impulse response of these four circuits.

+

. . . ...... ...... ...... ................................................. .... .... .... .... .... ............................................................................... ...... ...... ...... .... .... .. ... ........................

²

R

±(t) ¡ +

C

±(t) ¡

¡

¡

+

+

..... ..... ..... .................................................. ..... ..... ..... ..... ..... .............................................................................. ...... ...... ...... .. .. .... .... . ......... .. . ........ .. . ......... .. . ....... .. ... .......................................................................................................................................................................

L

²

²

C

±(t)

²

R

+

... .. ........................................................................................ ... ... .. ... .. . .. ....... . ........ ............ ........ .. ....... ............. ....... .. . . ..... ............. ... ... . .......................................................................................................................................................................

ha (t)

........................ .. .. .. .. ... . .......................................................................................................................................................................

²

+

.. ... ................................................................ .... .

hb (t)

R

²

........ .......... ......... .......... ................................................. .... ..... .... .............................................................................. .. .. ... ....... ....... . . . . . . . ...... ........ .. ....... ............. ....... . ....... ............. ... ... . .......................................................................................................................................................................

²

L

hc (t)

±(t)

¡

¡

What is the relationship between ha (t) and hb (t) ?

+

¡ +

hd (t)

R

²

¡

Between ha (t) and hd (t) ?

Homework. Find the step and impulse responses of the series RLC circuit that is critically damped: when ® = !0 and the characteristic quadratic polynomaial has a single, repeated root.

53

Evaluation of an Important Integral via the Laplace Transform

1.0 0.8 0.6 sin(x) x

0.4 0.2 0.0 ¡0:2 ¡0:4

..... ... ... .. ... .. .... . ... .. ... .. ... ... ... ... .. .. .. .. .. .. .. .. .. .. ... .. ... ... ... .. ... .... .. ... ... ... .. ... .. .. ... .. .... ... .. ... .. .. ... ... ... ... .. .. .. .. . . . . . ....... .. . .... ........ . . ..... ........ . .. . . . . . . . ... . . . . .. . . . ... . ... . . ....................... .... ........ . . . ........ . . ... . . . . . . . . .... . . . . ........ ........................................................................................................................................................................................................................................................................................................................................................................................................................... . . . . . ...... ...... ....... ....... . . .... .... ... . ... . . . . . ............ . . . . . . ..... ...... . . . . . . . . . . ...... ..... .. ... . ........... ........ ... ... ... .. .. .. .. ... ... ..... ... .... . ... ... ............ .......

¡6¼

¡4¼

¡2¼

0

2¼

4¼

6¼

x

· ¸ 1 2n+1 sin(x) 1X 1 x3 x5 n x sinc(x) , = (¡1) = x¡ + ¡ ::: x x n=0 (2n + 1)! x 3! 5! · ¸ x2 x4 = 1¡ + ¡ ::: 3! 5! sin(x) =1 x!0 x

sinc(0) = lim

sinc(¡x) = sinc(x) We want to show that

Z1

is even

sin(x) dx = ¼: x

(1)

(2) (3)

(4)

¡1

Since sinc(x) is even, observe that Z1

sin(x) dx = 2 x

¡1

Z1 0

54

sin(x) dx: x

(5)

Our approach is to ¯nd the Laplace transform of the function Si(t) ,

Zt

sin(x) dx; x

(6)

0

called the \sine integral." Our desired integral (half of it!) is the ¯nal value of Si(t), that is Si(1). Perhaps it is helpful to write Si(t) using ¿ as the dummy variable of integration Si(t) ,

Zt

sin(¿ ) d¿: ¿

(7)

0

We will use the following Laplace transform properties. If L[f (t)] = F (s), then ·

¸ Z1 f (t) L = F (¸) d¸ t

(8)

2 t 3 Z F (s) L 4 f (¿ ) d¿ 5 = s

(9)

s

0

lim f (t) = lim sF (s)

t!1

¯nal value theorem

s!0

(10)

The Laplace transform of g(t) = sin(t)u(t)

(11)

1 : +1

(12)

is G(s) =

s2

The Laplace transform of h(t) =

g(t) = sinc(t)u(t) t

is then H(s) =

Z1

d¸ : +1

¸2

(13)

(14)

s

This integral can be found in a table (say in your calculus book, or you can use a symbolic mathematical tool such as MAPLE, or you can use a web-based integrator4 ). Although not important for our purposes, let's do it ourselves just for fun. Consider y = tan(x) so that dy d sin(x) = = 1 + tan2 (x) dx dx cos(x) 4 If

you know one, tell the class. If you don't know one, ¯nd one! 55

Z Z1 s

dy 1 + y2

dy = dx 1 + y2 Z = dx = x = tan¡1 (y)

dy ¼ = tan¡1 (1) ¡ tan¡1 (s) = ¡ tan¡1 (s) = tan¡1 (1=s) 2 1+y 2

.... ........ .. ....... .... ....... . . . . . . ... . ...... ... ....... .. ....... . . . . . . .. . ... . . . . . .. . ... . . . .. . . . .. . . . . . . ... . ... . . . . .. . . ... . . . . . . ... .. . . . . . ... . . ... . . . . . .. . ... . . . .. . . . .. . . . . . . ... . ... . . . . .. . . ... . . . . . . . . . . . . . . . . ... . . . . . . . .. . . ..........................................................................................................................................................................

¼=2 ¡ μ

tan¡1 (s) + tan¡1 (1=s) =

s

¼ 2

μ

1

Therefore the Laplace transform of h(t) = sinc(t)u(t) is H(s) = tan¡1 (1=s):

(15)

The Laplace transform of f (t) = Si(t) =

Zt

sinc(¿ ) d¿

(16)

0

is now F (s) =

tan¡1 (1=s) s

(17)

and the ¯nal value theorem gives Si(1) = f (1) = lim sF (s) = lim tan¡1 (1=s) = s!0

s!0

56

¼ : ¥ 2

(18)

Proposition (Spectral form of the Dirac-delta). This Fourier integral satis¯es the de¯nition of the Dirac-delta (distribution or generalized function): 1 2¼

Z1

ej!t d! = ±(t):

(1)

¡1

Given the integral

Z1

sin x dx = ¼; x

(2)

¡1

Dirichlet's discontinuous integral falls out Z1 ¡1

sin ®x dx = ¼ sgn(®) = §¼ x

(® ? 0):

(3)

Integrate the left hand side of (1) from a to b: Zb a

1 2¼

Z1

j!t

e

¡1

1 = 2¼

1 d! dt = 2¼

Z1 · ¡1

Z1 Zb ¡1 a

sin b! sin a! ¡ ! !

j!t

e

¸

1 dt d! = 2¼

Z1

ejb! ¡ eja! d! j!

¡1

1 d! = [¼ sgn(b) ¡ ¼ sgn(a)] = 2¼

This is the sampling property of ±(t) ¤:

57

½

1; a < 0 < b 0; o.w.

(4)

Fourier Transform

F (!) = F[f (t)] =

Z1

1 f (t) = F ¡1 [F (!)] = 2¼

F

f (t)e¡j!t dt

()

¡1

Z1

F (!)ej!t d!

¡1

¯ 1 ¯ ¯Z ¯ Z1 Z1 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯f (t)e¡j!t ¯ dt = ¯f (t)¯ dt f (t)e¡j!t dt¯¯ · ¯ ¯ ¯ ¡1

¡1

¡1

Therefore a su±cient (not necessary) condition for the existence of F (!) is the absolute integrability of f (t) Z1 ¯ ¯ ¯f (t)¯ dt < 1: ¡1

Note that the zero frequency value of the Fourier transform (or spectrum)

F (0) =

Z1

f (t) dt

¡1

is the DC or \average value" of the signal f (t). We can easily demonstrate the legitimacy of the Fourier transform pair by taking the inverse transform of the forward transform 2 3 Z1 Z1 n £ o ¤ 1 4 F ¡1 F f (t) = f (¿ )e¡j!¿ d¿ 5 ej!t d! 2¼ ¡1 ¡1 {z } | F (!)

=

Z1

¡1

=

Z1 ¡1

f (¿ )

Z1

1 ej!(t¡¿ ) d! d¿ 2¼ ¡1 | {z } ±(¿ ¡t)

f (¿ )±(¿ ¡ t) d¿ = f (t):

58

Fourier Transform Exercises Find an expression for the Fourier transform of the following time-domain signals, and sketch BOTH f (t) and F (!): 0. f (t) = e¡at u(t) causal, decaying exponential (a > 0) 1. f (t) = A¦(t=T ) rectangular pulse 2. f (t) = B¤(t=T ) triangular pulse 3. f (t) = C constant 4. f (t) = ±(t ¡ t0 ) delayed Dirac impulse 2

5. f (t) = Ae¡®t Gaussian signal \and it is a trick" 6. f (t) = e¡®jtj ½ ¾ cos 7. f (t) = (!0 t) sin 8. f (t) = u(t) Heaviside unit-step function (why is this one so hard?) hint: write u(t) = lim e¡at u(t)

and note lim

a!0 a2

a!0

9. f (t) = sgn(t)

a = ¼±(!) + !2

signum (or signature or sign) function

10. f (t) = m(t) cos(!0 t) AM modulated signal i.e. !0 =2¼ = 1250 (kHz) and m(t) = \Your Cheatin' Heart" 11. f (t) = sinc(®t) 12. f (t) = ¦(t=T ) ~ sgn(t) 13. f (t) = t¦(t=T ) 14. f (t) = D combT (t) = D

1 P

n=¡1

15. f (t) =

1 P

k=¡1

16. f (t) = A

±(t ¡ nT ) Woodward's comb function

g(t ¡ kT ) periodic replication where F[g(t)] = G(!)

1 P

k=¡1

¦[(t ¡ kT )=¿ ] rectangular pulse train

59

Fourier Series f (t) = a0 +

1 X

cn ejn!0 t

(!0 = 2¼=T )

n=¡1

n=1

1 a0 = T

1 X

[an cos n!0 t + bn sin n!0 t] = tZ 0 +T

½

f (t) dt

an bn

¾

2 = T

t0

tZ 0 +T

f (t)

½

cos sin

¾

(n!0 t) dt

t0

1 cn = T

tZ 0 +T

f (t)e¡jn!0 t dt

t0

Fourier Transform F (!) = F[f (t)] =

Z1

¡j!t

f (t)e

¡1

Properties 1. superposition 2. time delay

7. multiplication

f (®t) F (t)

10. integration

Rt

¡1

c1 F1 (!) + c2 F2 (!)

F (!)G(!) 1 F (!) ~ G(!) 2¼

f (t)g(t)

¡jtf (t)

F (!)ej!t d!

F (! ¡ !0 )

f (t) ~ g(t)

9. frequency di®erentiation

Z1

2¼f (¡!)

f (t)ej!0 t

df (t) dt

1 [F (!)] = 2¼

e¡j!t0 F (!) 1 ³!´ F j®j ®

f (t ¡ t0 )

8. time di®erentiation

¡1

Transform

c1 f1 (t) + c2 f2 (t)

4. duality

6. convolution

dt () f (t) = F

Signal

3. time scale

5. frequency translation

F

j!F (!) d F (!) d!

f (¿ ) d¿

¼F (0)±(!) +

¡1

60

1 F (!) j!

Z1 ¡1

1 jf (t)j dt = 2¼

Z1

2

F

u(t) () F

¦(t=¿ ) () ¿

1 + ¼±(!) j! sin(!¿ =2) = ¿ sinc(!¿ =2) !¿ =2

F

sinc(¯t) () F

combT (t) ()

¡1

jF (!)j2 d!

¼ ¦(!=2¯) ¯

2¼ comb2¼=T (!) T

Z-Transform Zfx[n]g = X(z) = Properties 1. superposition 2. time delay

1 X

x[n]z ¡n

n=¡1

Signal

Transform

c1 x1 [n] + c2 x2 [n]

z ¡m X(z)

x[n ¡ m]

3. modulation

x[n]³ n

4. convolution

x[n] ~ y[n]

5. frequency di®erentiation

nx[n]

6. accumulation

n P

c1 X1 (z) + c2 X2 (z)

X(z=³) X(z)Y (z) d X(z) dz 1 X(z) 1 ¡ z ¡1 ¡z

x[k]

k=¡1

61

Lff (t)g = F (s) =

Z1

f (t)e¡st dt

is the transform pair

L

f (t) , F (s)

0

property

t-domain

linearity

®f (t) + ¯g(t)

s-domain ®F (s) + ¯G(s) e¡st0 F (s)

f (t ¡ t0 )u(t ¡ t0 )

time delay time scaling

1 ® F (s=®)

f (®t)

exponential modulation

e¡®t f (t)

time di®erentiation

df (t)=dt

time integration time integration s-domain di®erentiation

sF (s) ¡ f (0)

s2 F (s) ¡ sf (0) ¡ f 0 (0) F (s)=s F (s)=s + ¡dF (s)=ds R1 F (¸) d¸ s

tf (t)

division by t

f (t)=t

t-domain convolution

f (t) ~ g(t)

F (s)G(s)

Initial value theorem

lim f (t) = lim sF (s)

t!0

s!1

lim f (t) = lim sF (s)

Final value theorem

t!1

s!0

Short Table of Laplace Transforms (s = ¾ + j!) ±(t) u(t) e¡at u(t) te¡at u(t) tn u(t) cos(®t)u(t) sin(®t)u(t)

(® > 0)

F (s + ®)

d2 f (t)=dt2 Rt f (¿ ) d¿ 0 Rt f (¿ ) d¿ ¡1

reprise

(t0 ¸ 0)

1 1 s 1 s+a 1 (s + a)2 n! n+1 s s 2 s + ®2 ® 2 s + ®2

62

(8s) (¾ > 0) (¾ > ¡Re[a]) (¾ > ¡Re[a]) (¾ > 0) (¾ > 0) (¾ > 0)

1 s

R0

¡1

f (t) dt

Fourier Transform of the Gaussian r ¼ then F (!) = exp[¡! 2 =4®]. ®

If f (t) = exp[¡®t2 ]

This one is a bit involved to show, especially without complex variable theory. But it's so pervasive in signal analysis and communications that we use it almost daily, so here is a classical derivation that uses nothing more than integration-by-parts and a simple ¯rst order di®erential equation. A preliminary integral

Z1

I=

2

e¡x dx =

p ¼

¡1

is required. The trick5 to derive this is to consider Z1

2

I =

¡x2

e

dx

¡1

Z1

¡y 2

e

Z1 Z1

dy =

¡1

e¡(x

2

+y 2 )

dx dy:

¡1 ¡1

Conversion to polar coordinates via x = r cos Á

and

y = r sin Á

gives 2

I =

Z2¼Z1 0

¡r2

e

μ ¶ 1 =¼ r dr dÁ = (2¼) 2

and so

I=

0

Our desired Fourier integral is F (!) =

Z1

¡®t2 ¡j!t

e

e

dt = 2

Z

1

2

e¡®t cos !t dt

0

¡1

since f (¡t) = f (t) is even. Di®erentiation w.r.t. ! gives dF (!) = ¡2 d!

Z1

2

te¡®t sin !t dt:

0

Integration by parts

5 \And

Z

u dv = uv ¡

it is a trick!" 63

Z

v du

p ¼:

with u = sin !t

du = ! cos !t dt 2

e¡®t v= ®

¡®t2

dv = ¡2te gives

¯1 Z ¯ ! 1 ¡®t2 1 ¡®t2 dF (!) ¯ sin !t¯ ¡ e cos !t dt = e d! ® ® 0 t=0

or

dF (!) ! = ¡ F (!) d! 2® dF (!) d! = ¡ ! F (!) 2® d ! ln F (!) = ¡ d! 2® 2 ! +C ln F (!) = ¡ 4® F (!) = eC e¡! Z1

C

e = F (0) =

2

=4®

¡®t2

e

dt =

r

¼ ®

¡1

F (!) =

r

¼ ¡!2 =4® e ®

solid curves: ® = 1, dashed curves: ® = 2 f (t) 1.0 0.8 0.6 0.4 0.2 0.0

F (!) 2.0

........... .... .... .... ...... .... ..... . ..... .. .. .. ... .. . .. ... ... ... . . .. ... .. .. . .. ... ... .. .. .... .. .. .. ... ... .. .. ... . . .. . .. ... . .. . .. ... ... .. .. .... .. .. .. ... .. .. .. ... . .. ... .. ... . . . .. ... .. .. . .. .... . .. . .. ... . .. . . . .. ... . . . .. .. .. .. .. ... . . .. . .. .... . .. . .. ... . . . . . . .. ... . . .. . .. ... .. ... .. ... . . . .. .. . . . . . .. .... . . . .. . . .. . . . . . . ... ........ .. . .. ..... ......... ... ..... . . . . . . . . . . . . ... ... ........................................... . . . . . . .................. .. ... ...

¡3

¡2

¡1

0

1

2

1.5 1.0 0.5 0.0

.... ... ... .. .... .. .. . .. .. ... ... .. ... .. .. . .. ... .. .. .. ...... .... ..... .. .. .. ... .. .. ... .. .. ... . ..... .. .. ..... ..... .... ..... .. . ... ... . ... ..... .... .... .... .. . . . . .. .. . . ... .. . .. ... .. .. ... .. .. .. ... .. .. .. .. ... . . .. . ... .. . ... . ... .. . .. . ... ... . ... . .. .... .. ... . ... ... . .. . . . .... .... ... . . . . ......... ... ... .. . ......... . . . . . . ............................................ . . . . . . . . . . . . ................. .. ..............

¡8 ¡6 ¡4 ¡2 0 !

3

t 64

2

4

6

8

Fourier Transform of the Heaviside Unit Step Function

F (!) = F[f (t)] =

Z1

¡j!t

f (t)e

F

()

dt

¡1

f (t) = F

¡1

1 [F (!)] = 2¼

Z1

F (!)ej!t d!

¡1

Recall that a su±cient (but not necessary) condition for the Fourier transform F (!) of a signal f (t) to exist is that the signal be absolutely integrable, that is Z1 ¡1

jf (t)j dt < 1:

The Heaviside unit step function is not absolutely integrable Z1 ¡1

ju(t)j dt =

Z1

dt ¥ 1

0

and therefore we might anticipate that its Fourier transform will be both di±cult to evaluate and exhibit singular behavior. Recall or note that the periodic signal cos(!0 t) is not absolutely integrable, and its spectrum consists of Dirac-deltas 2 1 3 Z1 j!0 t Z Z1 ¡j!0 t e +e 1 e¡j!t dt = 4 F[cos(!0 t)] = e¡j(!¡!0 )t dt + e¡j(!+!0 )t dt5 2 2 ¡1 ¡1 ¡1 ¤ £ = ¼ ±(! ¡ !0 ) + ±(! + !0 ) : Two preliminary formulae or ideas: First:

lim

®=¼ = ±(!) + !2

®!0 ®2

(1)

This is the Cauchy or Lorentz Dirac-delta sequence (that we already studied back on page 7). Clearly ®=¼ lim 2 = 0 if ! 6 = 0. ®!0 ® + ! 2 The integral ® ¼

Z1 ¡1

d! 1 = 2 2 ® +! ¼

Z1 ¡1

¯1 ¯ dx 1 ¼=2 ¡ (¡¼=2) ¡1 ¯ = tan (x) = =1 ¯ 2 1+x ¼ ¼ x=¡1

gives unity independently of ®, so the truth of (1) is veri¯ed. 65

Rewrite it as

¡j! 1 = 2 +! j!

lim

Second:

(2)

®!0 ®2

1 1 = 2 ®!0 j![1 + (®=!) ] j! lim

which is clearly true if ! 6 = 0. The only other possibility is that our limit also has a Dirac-delta at the origin of strength or weight K 1 1 = + K±(!): ®!0 j![1 + (®=!)2 ] j! lim

Integrate both sides of this from ! = ¡² to ! = +² 1 j

Z²

d! ![1 + (®=!)2 ]

¡²

=

Z²

1 j

d! !

+

¡²

K

Z²

±(!) d!

¡²

and observe that the integral on the left side and the ¯rst integral on the right side vanish because of the odd symmetry of their integrands. Therefore the only possibility is K = 0 and the truth of (2) is established. One of our ¯rst examples (or exercises) of Fourier integral evaluations was the causal, decaying exponential ¤ £ 1 (® > 0): F e¡®t u(t) = ® + j! The condition that ® > 0 is necessary to ensure convergence of the integral, by making the time-domain function decay su±ciently fast as t ! 1. Although the restriction ® > 0 enabled us to evaluate the Fourier transform of the decaying exponential, let's try to get our desired Fourier transform of the plain Heaviside unit-step by taking the limit of the above result as ® ! 0. That is £ ¤ F[u(t)] = lim F e¡®t u(t) = lim

1 : ®!0 ® + j!

®!0

Rewrite

1 ® ¡ j! ® ¡j! = = 2 + 2 2 ® + j! (® + j!)(® ¡ j!) ® +! ® + !2

so that F[u(t)] = lim

®!0 ®2

® ¡j! 1 + lim 2 = ¼±(!) + 2 2 ®!0 ® + ! +! j!

from (1) and (2) above.

66

(F)

Since 2u(t) = 1 + sgn(t) we now have the Fourier transform of the signum function as · ¸ 2 1 F[sgn(t)] = 2F[u(t)] ¡ F[1] = 2 ¼±(!) + ¡ 2¼±(!) = : j! j! With F[u(t)] = ¼±(!) +

1 j!

then by the time-di®erentiation property of the Fourier transform ·

¸ df (t) F = j!F[f (t)] dt where ±(t) =

du(t) ; dt

we have the correct transform · ¸ 1 F[±(t)] = j! ¼±(!) + = 1: j! Using the convolution property of the Fourier transform F[f (t) ~ g(t)] = F (!)G(!) and the integral identity Zt

f (¿ ) d¿ = f (t) ~ u(t);

¡1

we have the Fourier transform of the integral as 2

F4

Zt

¡1

3

· ¸ 1 F (!) f (¿ ) d¿ 5 = F (!) ¼±(!) + = ¼F (0)±(!) + : j! j!

67

Duality Property of the Fourier Transform Recall our notation for the Fourier transform F (!) = F[f (t)] =

Z1

¡j!t

f (t)e

¡1

F

dt () f (t) = F

¡1

1 [F (!)] = 2¼

Z1

F (!)ej!t d!:

¡1

For our duality property, we want the Fourier transform of some signal F (t): F[F (t)] =

Z1

F (t)e¡j!t dt:

(1)

¡1

The fact that our time domain signal is a \big" F of t means that we interpret our big F as related to some \little" f via a direct Fourier transform

F (x) =

Z1

f (»)e¡jx» d»

(2)

¡1

and therefore the particular little f that must have spawned our big F must be Z1

1 f (») = 2¼

F (x)ejx» dx:

(3)

¡1

According to (2), we can also write

F (t) =

Z1

f (»)e¡jt» d»

(4)

¡1

and insertion of this into (1) yields

F[F (t)] =

8 Z1 < Z1 ¡1

:

f (»)e¡jt» d»

¡1

9 = ;

e¡j!t dt:

(5)

Reverse the orders of the » and t integrations

F[F (t)] =

Z1 ¡1

2

f (») 4 |

Z1

¡1

3

e¡j(»+!)t dt5 d» {z

}

recognize as 2¼±(»+!)

68

(6)

and so F[F (t)] = 2¼

Z1

f (»)±(» + !) d» = 2¼f (¡!):

(7)

¡1

Our desired duality property of the Fourier transform is here F[F (t)] = 2¼f (¡!) where F f (t) , F (!) is the notation for the original Fourier transform pair F[f (t)] = F (!). Example. Since we already know the Fourier transform pair μ ¶ t F ¦ , T sinc(!T =2); T

(8)

³ !´ F[T sinc(tT =2)] = 2¼¦ ¡ T

(9)

it follows from duality that

or

2¼ ³ ! ´ ¦ (10) T T by the linearity of the Fourier transform operator, and invoking the even symmetry of the pulse function ¦. Now T is just a remnant parameter from the original pair (8), so let's introduce a new (better!) constant ® = T =2 such that the above is F[sinc( T2 t)] =

¼ ³!´ F[sinc(®t)] = ¦ : ® 2®

(11)

Add this entry in your table of Fourier transforms! Another View. Consider a function ª(x) related to a given function Á(u) through the direct Fourier transform operator

ª(x) =

Z1

Á(u)e¡jux du:

(12)

¡1

Consider a second function Ã(x) related to our given function Á(u) through the inverse Fourier transform operator Z1 1 Ã(x) = Á(u)ejux du: (13) 2¼ ¡1

Observe that ª(x) = 2¼Ã(¡x). If we know the forward transform, then we also know the reverse transform of any function, by inspection.

69

Half-Power Bandwidth or Pulse width f1 (x) ... .... ......................... . . . ..... .... ..... .... .... .. .... ... ... .. ... . ... . . . .. . .... . . . . . . ... . . . .. ... . . . . .... . . . . .... .. . . . . . .... . . . . . .... . .. . . . . . ..... .. . . . . .... . .. . . ..... . . . . . . ...... . . . . . . .. ......... . . . . . . . . . . . ............................................................................................................................................................................................................................................................................................................................................................................................ ...

x

f2 (x) .. ... ......... .. ... .... ...... .. ... .. . .. .. . ... .. ... .. .. .. . .. . .. ... . . .. ... . . .. .. . .. . .. . .. . . ... .. . .. . .. . ................................ ...... .. . ... ....... ........ ................ . .. ... . ....... .... ........ ... .. ... ......................... .... . . . . . . . . . . . . . . . . . . . . .. .. ....... . ... .. . ... ....... . ....................... ............................................................................................................................................................................................................................................................................................................................................................ .... ........................... .... ..... .. ... .... . . . . . . . . ...... ..... ...... ............... ...

x

f3 (x) ..... .... .. ... .... ... ........ .... . . ..... .... .. .. .... .... .... ..... ... . . . . . . .. . . . . . . . . . .. .. ... ... .. ... . .. . . . . . . . . . . . . . . . . . . . .. ..... .. .. .. .. .. .. ... .. .. .. ... .. ...... .. .. .. ... .. .. .. .. ... ... .. .. .. .. .. .. .... .... .. .... .. .... .. .. .... .. .... ... .... . . . . . . . . . . . . . . . . . . . . . . ... .. . ..... . . . ... . .. . . .. . . ... ....... ....... .. ... .. .. .. .. ... .. ... ... .. .. ... .. .. ... ... ........... ....... .. .. ....... ... .. .. .... ... .... ... ... ..... ... .... ... ....... .... .... .. .. .. .... .. .... ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . .. . .. . .. .. . . . . . .. . ............................................................................................................................................................................................................................................................................................................................................................. ..... .. . .. .. .. .. .. .. .. .. .. .. .. .. .. .... .. .. ... ... .. . ... . . . . .. . . .. . .. ... ... .. .. .. .. .. ... .... ... ..... .. .... .. ... .. . . ... ... ...

x

Consider the three functions (or signals) graphed above. The function f1 (x) has a clear maximum and a \function width" or \pulse width" about the maximum where most of its energy is concentrated. The same observation applies to the signal f2 (x), but its \pulse width" is perhaps not as obvious. The oscillatory and somewhat wild signal f3 (x) possesses no obvious maximum that we see in its graph and it is unclear how to de¯ne any kind of a pulse width for it. Often our time-domain signals are one of the \linear" signals in an electric circuit, such as a voltage v(t) or a current i(t). In a mechanical system, the displacements x(t) or velocities v(t) are examples of one of the so-called \linear" signals that we might be interested in. The instantaneous power associated with one of the linear signals f (t) is proportional to jf (t)j2 . If the signal f (t) does possess a clear maximum and most of its energy or power 6 6 This

is an example of hazy (perhaps lazy, too) language. Power p(t) (in Watts) is not equal to energy E(t) (in Joules) but we are approximately thinking that p(t) / E(t) / jf (t)j2 . 70

is concentrated around the maximum (such as the functions f1 and f2 above), then we de¯ne the half-power pulse width as T1=2 = t2 ¡ t1 where

jf (t1;2 )j2 1 = 2 jfmax j 2

or

jf (t1;2 )j 1 =p : jfmax j 2

When the signal magnitude is given or graphed in units of decibels (dB), the half-power points t1;2 are also called the \3 (db)" points since p 10 log10 (1=2) = 20 log10 (1= 2) ¼ ¡3: The half-power pulse width is also then called the \3 (dB) pulse width." This measure of \signal duration" applies equally well in the frequency domain, where the term bandwidth is commonly used instead of pulse width. The symbol for bandwidth is typically BW in communications work, but a simple B1=2 seems ¯ne. Example: Gaussian pulse. Note in this case that jf (t)j = f (t) and jF (!)j = F (!). This will not be true in general. r ¼ F (!) = exp(¡! 2 =4®) f (t) = exp(¡®t2 ) ® .. .. ... ... ... ...................... ... ... ... . max . . ... ..... . . ... .... .... . . . . ... .. .... ..... ..... max . ... ........ ... ... ... ..... ... ... ............ . .... . . . .. .. . . . . .. . ... ... .... ... .. . .. .... .. ... .. . ... .... . . . .. .. .. . .. ... . . .. ... . .. .. .. . .. . . . ... . . . . . . . . ... . . . . . . . . ... .. . . . . . . ... . . .. . . . . . .... . . . . . . . . . .... .. . . . . . . . ...... . . . . . . . . . . . . . . . .................................................................................................................................................................................................................................................... . . . . ........................ ......................... 1=2 . .

.. .. ... ... ... ... ... ... ... ... ...................................... ... ... ... ... ... ... ... ... . ...... . .. max . .. . . ..... . ... . .. .... .. .... . . . . . . . ... ... .... .... .. .. ..... . max . ... ... ............... ... ... ... ... ... ... ............ ... ... ... ... ... ................. ... . . . . ...... . . . .. . . . . . . . . .. . ... ... ........ ... .. .... ... .... .. .. ..... .... .. . . . . .... . . . . . .. . . . . ... . . . . . .. .... . . . . . . . . .... .. . . . . . . . .... . . . . . . . .... . . . .. . . . . . . . ..... .. . . . . . . . ..... . . . . . . . . . ...... . . . . . ... .. .. .... .. .. .. . . . .............................................................................................................................................................................................................................. . .. ... .. ............................... 1=2 ............................ .

jf j

jF j

jf j p 2

T

t1

T1=2

jF j p 2

t

B

!1

t2

exp(¡®t22 ) 1 =p 1 2 r ln 2 t2 = 2® r 2 ln 2 = t2 ¡ t1 = 2t2 = ®

!2

exp(¡!22 =4®) 1 =p 1 2 !2 =

p 2® ln 2

B1=2 = !2 ¡ !1 = 2!2 =

The \BT product" B1=2 T1=2 = 4 ln 2 is independent of ®. 71

p 8® ln 2

!

Example: The sinc function. Examine the two graphs of j sinc(x)j = j sin(x)=xj, the one on the top using a linear ordinate scale and the one on the bottom using a logarithmic (dB) ordinate scale. The abscissa is x, which can serve as a scaled frequency ! for the spectrum of a time-domain rectangular pulse. 1.0 0.8 p 1= 2 0.6 jsinc(x)j

0.4 0.2

. . ............... ... ... .... ...... ... . ... . ... .. .. .... ... .. .. ... . . ... .. .. .. .. .. ... ... .. . .. . . .. .. .. .. .. .... ..... ..... . ... . . ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ..... ... ... ... ... ... ... ... ... ... ............ ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. . ... . . ....... ...... . ... .. .. ... .... ... .. ..... . ... ... ... ... . . .. ... . ... .. . ... ... .. .. .. . .. ... .... ... ... .. .. .. ... .. . ... ... . .. ... ... .. . .. . ... . . . . ... . .... . . ... . . . ... ... .... ... . ... . . . ............ ..... . . . . . . . . . . . ... . ........ ............. .... . . . ... . . . .... . . . ... . ... .... . . . . . . . . .... ... . ... ... . . . . . . . . . . . . ... . . ... ... . ....................... ............................ ... ... ... ... ... .... ... .... ..... ........ . . ...... ...... . ... . . . . . . . . . . .... ... .. . . ..... ..... . ... ... . . . . . . . . . . . .... .. .. . ..... ... .. ... ....... . . ...... .. . . . . . . . . . ..... .......... ... ... ... ... .... ..... ..... .... . . . . ..... ........ ... ....... ... .... .. ..

0.0 ¡10 0 ¡3 ¡5 ¡10 jsinc(x)j (dB) ¡15 ¡20 ¡25

¡6

¡4

¡2

0 x

x1

2

4

6

8

10

........................... ..... ... ... ............ ..... . . ..... .... .. ...... . . . . ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ........... ... ... ... ... ... ... ... ............. ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. .... ...... . . .. ... ... . . . . ... ... ... . .. ... . . . .. ... . . .. . . . . .. .. . . . .. . . ... . ... . . . . . ... .... .. .. .. .. .. .. .. .. .. .. .. .. .. . . .. ... . . . . . .. . . . . . . . . . ........... . ..... ......... .. . . . . ..... ........ . . . . .. . . . . . . . ... .. ... . .. . . . . . . . . . .. .. ... .. .. .. .. ... .. .. ... ... ... ... .. ... ... ... ... .. .. . .. .. . . . . . .. . .. . . . . . . . . . . . . . . . . . . .. . ... . . .. ................... . .... ....... . .. . . . . . . . . .. .. ... .... ... . ... . . . . .. . . . . . .. ... ... . . .. .. . . . ... . . . . . . . . . . ... .. ... .. ... .. .. . ... ... .. .. ... .. ... ... ... .. ... .. . . . . . . . .. . . ... .. .. .. ... . . ... . . . . . . ... .. . . . .. . . ... . . . . . . . . ... . . . .. .. . ... .. .. . . ... . .... .. . . . ... .. . ... . . .. .. . .. . . . .. . . . ... . .. . ... .. .. . .. . . . . . ... .. ... . .. . .. .. .. . ... . . . . . . . . . . . .. .. . . .. . ... ... .. .. .. . . .. . .. . . ... . . .. . . . ... .. ... .. .. .. ... ... . ... . . . .. ... . .. .. . .. .. . . . ... .. . . . . .. .. . .. . ... .. . . .. .. . . . . ... . . . . . .... . .. .. ... . .. .. .. .. .. .. ... ... ... . .. ... .. ... .. .. .. .. ... ... .. .. ... .. .. .. . .. . .. .. .. .. .................... ................. ....... ....... ............ . . ..........

¡30 ¡10 sin(x1 ) 1 =p x1 2

¡8

¡8 =)

¡6 x1 =

¡4

¡2

p 2 sin(x1 )

0 x

x1

=)

2

4

6

8

10

x1 = 1:391557378251510

If f (t) = ¦(t=T ) is the standard rectangular pulse of pulse width T , then since the spectrum is F (!) = T sinc(!T =2), we see that the half-power bandwidth is B1=2 = 2!1 ¼

2 £ 2 £ 1:39 5:56 ¼ : T T 72

Homework

1. Find the pulse width, bandwidth, and BT product of these two signals: f (t) = exp(¡®jtj)

and

g(t) = exp(¡®t)u(t):

Graph jf (t)j and its amplitude spectrum jF (!)j, and also jg(t)j and jG(!)j. Use at least two values for the constant ®.

2. Compare the half-power pulse width T1=2 described above with the (statistical class) of pulse width Dt that is used in the mathematical analysis of the Uncertainty Principle that appears on page 74. If the signal is normalized to have unit energy Z1 ¡1

jf (t)j2 dt = 1

then the (square of) pulse width is Dt2

=

Z1 ¡1

t2 jf (t)j2 dt:

In other words, simply take Dt as de¯ned here, to be yet another measure of a signal's pulse width. For a speci¯c signal, use the Gaussian pulse.

73

Uncertainty Principle for the Fourier Integral To simplify our notation, assume that the signals under consideration are normalized to have unit energy: Z1 Z1 1 jf (t)j2 dt = jF (!)j2 d! = 1: 2¼ ¡1

¡1

If we further assume that both f (t) and its Fourier transform F (!) are centered about t = 0 and ! = 0, respectively, then it is reasonable to de¯ne the t and !-domain durations by Z1 Z1 Dt2 = t2 jf (t)j2 dt and D!2 = ! 2 jF (!)j2 d!: ¡1

¡1

Our so-called \uncertainty principle" states: If

lim

t!§1

p

jtjf (t) = 0;

Then Dt D! ¸

r

¼ : 2

prf: Equality holds in Schwarz's inequality ¯2 ¯ b ¯Z ¯ Zb Zb ¯ ¯ ¯ g1 g2 dt¯ · jg1 j2 dt jg2 j2 dt ¯ ¯ ¯ ¯ a

a

a

only if g2 (t) = kg1 (t). Let g1 (t) = tf (t) and g2 (t) = f 0 (t): ¯ 1 ¯2 ¯Z ¯ ¯2 Z1 Z1 ¯ ¯ ¯ ¯ ¯ df (t) df (t) 2 ¯ ¯ ¯ · ¯ tf (t) dt jtf (t)j dt ¯ ¯ dt ¯ dt: ¯ dt ¯ ¯ ¡1

¡1

¡1

Integrate by parts

Z1 ¡1

¯1 Z1 df tf 2 (t) ¯¯ 1 1 tf dt = ¡ jf j2 dt = ¡ : ¯ dt 2 2 2 ¡1 | {z ¡1} | {z } =0 by above =1 by above

From the Fourier transform pair f 0 (t) () j!F (!) and Parseval's theorem, Z1 ¯ ¯2 Z1 ¯ df ¯ 1 ¯ ¯ dt = j!F (!)j2 d! ¯ dt ¯ 2¼

¡1

¡1

and so Schwarz's inequality gives 1 · 4

Z1 ¡1

1 jtf (t)j2 dt 2¼ 74

Z1 ¡1

j!F (!)j2 d!

or

¼ · Dt2 D!2 : 2

¥

Equality holds if df (t) = ktf (t): dt The solution of this di®erential equation is f (t) = Cekt

2

=2

:

Let k = ¡2®; this is the inescapable Gaussian function f (t) =

μ

2® ¼

¶1=4

¡®t2

e

() F (!) =

μ

2® ¼

75

¶1=4 r

¼ ¡!2 =4® = e ®

μ

2¼ ®

¶1=4

e¡!

2

=4®

:

Theorem (Riemann-Lebesgue Lemma). Let f (x) 2 L1 (¡1; 1). That is, f (x) is absolutely integrable Z1 jf (x)j dx < 1: ¡1

Then the integrals

Z1

Z1

f (x) cos ¸x dx;

¡1

f (x) sin ¸x dx;

¡1

tend to zero as ¸ ! 1. Proof. Consider the cosine integral. Let ² be a given positive number. Then we can choose X so large that ¡X Z1 Z jf (x)j dx < ²; jf (x)j dx < ²: ¡1

X

Hence

¯ ¯ 1 ¯Z ¯ ¯ ¯ ¯ f (x) cos ¸x dx¯ < ²; ¯ ¯ ¯ ¯

¯ ¯ ¡X ¯Z ¯ ¯ ¯ ¯ ¯ ¸0 . Then ¯ 1 ¯ ¯Z ¯ ¯ ¯ ¯ f (x) cos ¸x dx¯¯ < 4² (¸ > ¸0 ): ¯ ¯ ¯ ¡1

This proves the theorem for the cosine integral; a similar proof applies to the sine integral. ¥ 76

The above is the working-class proof of the Riemann-Lebesgue lemma, as given by E.C. Titchmarsh in Introduction to the Theory of Fourier Integrals, 1937. Perhaps you will prefer this one, from R.R. Goldberg, Fourier Transforms, Cambridge University Press, 1962. Definition (the Class Lp ). Suppose 1 · p < 1. The function f on (¡1; 1) is said R1 to be of class Lp (written f 2 Lp ) if jf (x)jp dx < 1. If f 2 Lp then kf kp is de¯ned to be

0

¡1

@

Z1

¡1

11=p

jf (x)jp dxA

:

The symbol kf kp is read as the Lp norm of f . Theorem (Riemann-Lebesgue). If f 2 L1 then Z1

lim F (!) = lim

!!§1

!!§1 ¡1

Proof. Since

Z1

F (!) =

e¡j!t f (t) dt = 0:

e¡j!t f (t) dt;

(1)

¡1

then ¡F (!) =

Z1

e¡j![t+(¼=!)] f (t) dt =

¡1

Z1 ¡1

e¡j!t f (t ¡ ¼=!) dt:

(2)

Subtracting (2) from (1) we obtain 2F (!) =

Z1 ¡1

Hence 2jF (!)j · But since f 2 L1 , lim

Z1

!!§1 ¡1

e¡j!t [f (t) ¡ f (t ¡ ¼=!)] dt: Z1

jf (t) ¡ f (t ¡ ¼=!)j dt:

(3)

jf (t) ¡ f (t ¡ ¼=!)j dt = 0

(4)

¡1

by the continuity-in-the-mean theorem. The theorem follows from (3) and (4). ¥ 77

Riemann-Lebesgue Lemma If f 0 (t) is absolutely integrable, then F (!) ! 0 as ! ! §1. PROOF: Integrate by parts F (!) =

Z1 ¡1

¡j!t

f (t)e

¯1 Z1 f (t)e¡j!t ¯¯ 1 dt = + f 0 (t)e¡j!t dt ¯ ¡j! j! t=¡1 ¡1

If f (§1) is ¯nite, then the boundary terms go to zero as ! ! §1. The integral is the Fourier transform of the derivative of f (t) Z1 ¡1

f 0 (t)e¡j!t dt = F [f 0 (t)]

and this Fourier integral exists (has ¯nite magnitude jF [f 0 (t)] j < 1) if the derivative is absolutely integrable Z1 jf 0 (t)j dt · A < 1: ¡1

If jF [f 0 (t)] j < 1, then the integral term F [f 0 (t)] ¡¡¡¡¡! 0: j! !!§1 Exercise. Examine the high frequency limiting behavior (! ! §1) of the spectra of these signal pairs and interpret in terms of the Riemann-Lebesgue Lemma: F

u(t) ()

1 + ¼±(!) j!

F

¦(t=T ) () T sinc(!T =2) ¼ F sinc(¯t) () ¦(!=2¯) ¯ 1 F e¡®t u(t) () ® + j! F

±(t) () 1 1 F ¼ () sgn(!) t j

78

Convergence of the Fourier Integral Representation

F (!) = F[f (t)] =

Z1

¡j!t

f (t)e

F

()

dt

¡1

f (t) = F

¡1

1 [F (!)] = 2¼

Z1

F (!)ej!t d!

¡1

Assume f 0 (t) is absolutely integrable so that the Riemann-Lebesgue lemma guarantees lim F (!) = 0:

!!1

Denote the inverse Fourier transform of the Fourier transform as 1 fe(t) = lim −!1 2¼ = lim

Z− Z1

f (¿ )e¡j!¿ d¿ ej!t d!

¡− ¡1

Z1

−!1 ¡1 Z1

Z−

1 f (¿ ) 2¼

¡−

sin[−(¿ ¡ t)] d¿ ¼(¿ ¡ t)

= lim

f (¿ )

= lim

f (x + t)

−!1 ¡1 Z1 −!1 ¡1

ej!(t¡¿ ) d! d¿

sin −x dx: ¼x

If f is continuous at t, then clearly fe(t) = f (t) if we accept the Dirac-delta representation sin −x = ±(x): −!1 ¼x lim

If f su®ers a jump discontinuity at t, in that lim f (t + x) = f (t¡ ) 6 = f (t+ ) = lim+ f (t + x);

x!0¡

x!0

then it is appropriate to break the integral above at x = 0 so that fe(t) = lim

Z0

−!1 ¡1

= lim

Z0

−!1 ¡1 Z1

sin −x f (x + t) dx + lim −!1 ¼x

Z1

f (x + t)

sin −x dx ¼x

0

f (x + t) ¡ f (t¡ ) sin −x dx + lim f (t¡ ) −!1 ¼x

−!1

sin −x dx ¼x

¡1

f (x + t) ¡ f (t+ ) sin −x dx + lim f (t+ ) −!1 ¼x

+ lim

Z0

0

Z1 0

79

sin −x dx ¼x

The function

f (x + t) ¡ f (t¡ ) ¼x

is continuous at x = 0¡ , and is well-behaved everywhere else by our initial assumption. (We are assuming that f is continuous everywhere except at the particular t of interest. If f actually has a ¯nite number of such jump discontinuities, then each one can be handled in a similar fashion.) By the Riemann-Lebesgue lemma, the ¯rst term on the right-hand side of the above integral vanishes as − ! 1. The same argument applies to the function f (x + t) ¡ f (t+ ) ¼x and its Fourier integral. From the famous integral Z1

sin u du = ¼ u

¡1

and its variation

Z1

sin ¯u du = ¼ sgn ¯; u

¡1

then for − > 0 we have

Z0 ¡1

sin −x dx = ¼x

Z1

sin −x dx = 12 : ¼x

0

Therefore, we have shown F ¡1 fF [f (t)]g = fe(t) =

80

1 2

£ ¡ ¤ f (t ) + f (t+ ) :

System Transfer Function

x(t)

.............................................................................................. ... ... ... ... ... .. ... .. ... . ........................................ ........................................... .. .. ... ... .. .. ... .. . . ................................................................................................

h(t)

y(t)

The zero-state response (in the time domain) of our linear, time-invariant (LTI) system is the convolution of the input signal with the system impulse response y(t) = x(t) ~ h(t):

(1)

The direct Fourier transform of this equation is Y (!) = X(!)H(!)

(2)

which exhibits the important role of the system transfer function Ffh(t)g = H(!):

(3)

By transforming to the frequency domain, the nontrivial convolution operator has been replaced by a simple multiplication! Besides algebraic simplicity, working in the frequency domain also gives us a whole another conceptual view of the input/output dynamics of our systems. Electrical engineers work so much in the frequency domain that we sometimes loose touch with the original time domain! For an arbitrary input signal x(t), the time-domain response y(t) now requires an inverse Fourier transform (4) y(t) = F ¡1 fX(!)H(!)g: However, the transfer function H(!) immediately gives the response of a LTI system to a time-harmonic signal. The simplest time-harmonic input signal is the complex exponential form (5) x(t) = ej!0 t : (The only other choice would be a real-valued cosine or sine, but as we have seen repeatedly this semester, the mathematics of signal analysis usually (always?) is much simpler in terms of the complex exponential form. And in deriving important concepts and principles, we always take the simpler path.) Note the use of !0 to denote some particular frequency, to avoid confusion with the Fourier transform frequency variable plain !. The response of our system to the excitation (5) is y(t) = h(t) ~ x(t) =

= ej!0 t

Z1

Z1 ¡1

h(¿ )x(t ¡ ¿ ) d¿ =

Z1 ¡1

h(¿ )e¡j!0 ¿ d¿ = ej!0 t H(!0 ):

¡1

81

h(¿ )ej!0 (t¡¿ ) d¿

(6)

Observe that the response (steady-state since the excitation has been on forever) of a LTI system to a time-harmonic signal is a time-harmonic signal of the same frequency as the excitation. Only the amplitude and phase of the response have been modi¯ed by the amplitude and phase of the system transfer function, evaluated (of course) at the excitation frequency. The amplitude and phase are apparent in the polar representation H(!) = jH(!)jej£(!) :

(7)

Note that if the system impulse response h(t) is a real-valued function of time, then we can recover the separate responses to the real and imaginary parts of a complex input by inspection:

x(t) = xr (t) + jxi (t)

.............................................................................................. ... .... ... .. ... ... ... .. .. . ......................................... ....................................... .. .. .. ... ... . .... .. . .. ..............................................................................................

y(t) = xr (t) ~ h(t) + jxi (t) ~ h(t)

Refx(t)g = xr (t)

.............................................................................................. .. ... .. ... ... ... .. .. .... ...................................... ......................................... .. ... .. ... ... .. .. .. ... . . ..............................................................................................

h(t) real

yr (t) = Refy(t)g

Imfx(t)g = xi (t)

.............................................................................................. ... ... ... ... .. ... .. ... ... . . . . . . ......................................... . ................................. ... .. .. ... ... .. .. .. ... . . .................................................................................................

yi (t) = Imfy(t)g

h(t)

h(t) real

F Exercise 1. Show that the time-harmonic response of a real LTI system to a cosine excitation x(t) = A cos(!0 t + Á) of arbitrary amplitude A and phase Á is y(t) = AjH(!0 )j cos[!0 t + Á + £(!0 )] where the amplitude and phase of the transfer function are de¯ned in (7). Hint: write x(t) = RefAejÁ ej!0 t g. In AC circuit analysis, we call AejÁ the complex phasor. F Exercise 2. Specialize the above result to the case where x(t) = sin(!0 t): F Exercise 3. Now consider the periodic excitation 1 X x(t) = cn ejn!0 t n=¡1

applied to a LTI system having arbitrary (not necessarily real) impulse response h(t). The transfer function is the generic H(!). What is the form of the response? Think about the linearity of the system. 82

Superposition Interpretation of System Transfer Function

x(t) =

1 2¼

Z1

j!t

X(!)e

¡1

d!

................................................. ... ... ... ... . ................................................. ..................................................... .. ... ... ... .... .. . ................................................

H(!)

1 2¼

Z1

H(!)X(!)ej!t d! = y(t)

¡1

................................................. ... ... .. .. H(! )X(! )ej!1 t ... 1 1 ... . . . . . ............................................................................................. . ............................................... ... .. ... . .... .... .... ... .. . .............................................. ... ... ... ... ................................................. ... . . ... .. j!2 t j!2 t .... . X(!2 )e . H(!2 )X(!2 )e .... .... . ... .......................................... ...................................................... ..... ... ... ... ..... ... .. ... . . . . . ... ... ..... . . . ..... .............................................. .... ..... ... ..... ..... .......... ..... ....... ................ ................................................. ...... .............. ... ... .... ... ... .. .. ... ... ... .. j!n t . . . .... . . . . . . . . . . . . . ....................................................................... .. ............................................ . ......................................... ... ... .... ... ... .. .. . . . ...... ... ... . ...... ... ..... ................................................ ....... .................... ..... ....... ..... . . . . . . ..... ... .... ... .................................................. ..... .. .. .. ..... . . . .. ... .. . . . . . . ... . . . . ..................................................... .. ............................................ .. .. .. ... ... ... .. .. .. .. ................................................ .. .. ... .. ................................................. ... . . . j!n t ... j!n t ... .. . . X(!n )e H(! )X(! )e . n n . . ... .............................................................................................. ..................................................... ... ... .. .. . ... ................................................

X(!1 )ej!1 t

H(!)

H(!)

x(t) =

P n

X(!n )e

H(!)

§

H(!)

H(!)

83

P n

H(!n )X(!n )ej!n t = y(t)

Transfer Function for Linear Electric Circuits Example.

+

. . . ...... ...... ...... ................................................ .... ..... .... ..... .... ................................................................................ ..... ..... ..... .... .... .... .. ......... .. . ........ .. . ......... ... ....... ... .. .......................................................................................................................................................................

²

R

x(t) ¡

+

y(t)

L

²

¡

Recall (look back in your notes; no one memorizes such speci¯cs) that the impulse response of the given RL circuit is μ ¶ R R h(t) = ±(t) ¡ exp ¡ t u(t): (8) L L Most of us would agree that the correct derivation of this time-domain result is a nontrivial exercise. We ¯rst found the step response s(t) and then di®erentiated to get h(t) = s0 (t). Fortunately, the Fourier transform allows us to circumvent the time-domain di®erential equations and work with purely algebraic equations. Temporarily set the input x(t) = vg (t) and the output y(t) = vL (t) for obvious dimensional consistency and to use signal labels for voltages that look like voltages. At least this helps my circuit analysis. . . . ...... ...... ...... .. . . .......................................................................................... ..... ..... ..... ..... ..... ............................................................................... .... .... .... .. .. .. .. .. . ..... ... . .. ...................... ......... .... ... . ... . ... . . .... ....... ... .. .. L . . ... ... ...... ......... .................. ... .... . ....... .. .. .. . ... ........................................................................................................................................................................................................

i(t)

vg (t)

²

R

+ ¡

+

v (t)

L

²

¡

Kirchho®'s voltage law is vg (t) = Ri(t) + L

di(t) dt

(9)

and the direct Fourier transform of this is Vg (!) = [R + j!L]I(!):

(10)

Note the natural occurrence of the familiar complex impedance. The ratio of output to input voltage spectra VL (!) j!L ZL = = (11) Vg (!) R + j!L ZR + ZL could have been written by inspection as a simple voltage divider. Our transfer function is Y (!) j!L j!L=R j!=!b H(!) = = = = (12) X(!) R + j!L 1 + j!L=R 1 + j!=!b where the constant !b = R=L (s¡1 ) is called the break frequency or half-power frequency or maybe even cut-o® frequency of this simple high-pass ¯lter. F Exercise 4. Verify that the Fourier transform of impulse response (8) gives transfer function (12). 84

0 ¡5 ¡10 ¡15 jHj (dB) ¡20 ¡25 ¡30 ¡35

..................................................................................................... ......................................................... ................ .......... ....... . . . . . . . ...... ..... .... .... . . . .... .... ... ... . . . .. .... .... ... . . ... ... ... ... . . ... ... ... ... . . . ... ... ... . . .. ... ... ... . . .. ... ... ... . . . ... ... ... . . .. ... ... ... . . . ... ... ... . . . ... ... ... . . .. ... ... ... . . .

¡40 10¡2

10¡1

100 !=!b

101

102

(a) Magnitude response.

¼=2

3¼=8 arg H ¼=4

¼=8

0

..................................................... ....................... ............... ............ ......... ........ ....... ...... ..... ..... .... .... .... .... ... ... ... ... ... .... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .... ... ... .... ... ... .... .... .... .... ..... ..... ...... ....... ....... ......... ........... ............... ...................... ............................................... .........

10¡2

10¡1

100 !=!b

101

(b) Phase response.

Fig. 1. Frequency behavior of single-pole high-pass ¯lter.

85

102

The magnitude and phase response of the transfer function (12) j!=!b (13) 1 + j!=!b are graphed versus a logarithmic and normalized frequency abscissa in Fig. 1. We also use a logarithmic (unnatural or base 10) scale to express the magnitude in decibels according to jHj (dB) = 20 log10 jHj: (14) H(!) =

Note that the low and high frequency approximations 8 j!=!b ; ! ¿ !b > > < j!=!b j H(!) = = (15) ; ! = !b 1 + j!=!b > 1+j > : 1; ! À !b clearly show that this ¯lter passes the high frequencies unscathed. In the low frequency regime (! ¿ !b ) where this ¯lter has the greatest e®ect between the input and output, the frequency-domain relationship is j!X(!) Y (!) ¼ : (16) !b Inverse Fourier transformation of (16) to the time domain gives 1 dx(t) (17) !b dt so we see that a high-pass ¯lter acts as a di®erentiator when it is having its biggest e®ect on the input. y(t) ¼

F Exercise 5. Consider the simple circuit having the positions of R and L reversed relative to the circuit above. Find its transfer function and show that this low-pass ¯lter acts as an integrator when it is having its greatest e®ect on the input signal. Homework Problem: Rectangular Pulse Response of Ideal Low-Pass Filter. Consider an ideal low-pass ¯lter of bandwidth B and subsequent transfer function ³ ! ´ H(!) = ¦ : 2B Find the transient response y(t) of this ¯lter to a rectangular pulse of pulse-width T x(t) = ¦(t=T ): Graphical results simplify with normalized time t=T as the independent variable, for various values of the dimensionless parameter BT . This problem is most readily solved in terms of a special function, the sine integral7 Zu sin » Si(u) = d» Si(¡u) = ¡ Si(u) Si(1) = ¼=2: » 0 7 M.

Abramowitz and I.A. Stegun, Handbook of Mathematical Functions. National Bureau of Standards, 1970, pp. 231{232, Table 5.1. (\Big Red"or \Citizen's Handbook"). MATLAB has it built-in, too! 86

Parseval's Theorem for the Fourier Integral The energy associated with the signal f (t) is E=

Z1 ¡1

jf (t)j2 dt =

Z1 ¡1

Z1

f (t)f ¤ (t) dt 2

32

Z1

Z1

3¤

1 4 1 F (!)ej!t d! 5 4 F (−)ej−t d−5 dt 2¼ 2¼ ¡1 ¡1 ¡1 2 32 3 1 1 Z Z Z1 1 4 1 = F (!)ej!t d! 5 4 F ¤ (−)e¡j−t d−5 dt 2¼ 2¼ ¡1 ¡1 ¡1 8 9 1 1 1 Z Z Z < = 1 1 = F (!) F ¤ (−) ej(!¡−)t dt d− d! : 2¼ ; 2¼ =

1 = 2¼ =

1 2¼

¡1 Z1 ¡1 Z1 ¡1

¡1 Z1

F (!)

¡1

¡1

1 F ¤ (−)±(− ¡ !) d− d! = 2¼

Z1

F (!)F ¤ (!) d!

¡1

jF (!)j2 d!

F Exercise 6. Derive the generalized Parseval theorem Z1

f (t)g ¤ (t) dt =

¡1

F Exercise 7. Find a simple evaluation for the integrals Z1

sin x dx x

¡1

Z1 ·

sin x x

¸2

dx:

¡1

We met the ¯rst one back on page 54. The similarity (more than similar! ) in the numerical values strikes me as a strange coincidence. What do you think? (You'll need the answers, ¯rst!) 87

Other Conventions for the Fourier Transform Pair Note that our forward Fourier transform could be scaled by any constant K as long as we divide by K in performing the inverse: Z1 Z1 1 F ¡j!t F (!) = K f (t)e dt () f (t) = F (!)ej!t d!: 2¼K ¡1 ¡1 p The common choice K = 1= 2¼ gives a pleasingly symmetric pair Z1 Z1 1 1 F ¡j!t F (!) = p f (t)e dt () f (t) = p F (!)ej!t d!: 2¼ 2¼ ¡1

¡1

One-sided (unilateral) trigonometric forms Fc (!) =

Z1

Fs (!) =

0 Z1

f (t) cos(!t) dt

f (t) sin(!t) dt

F

() F

()

2 f (t) = ¼ f (t) =

2 ¼

0

Z1

Fc (!) cos(!t) d!

0 Z1

Fs (!) sin(!t) d!

0

Note that the sine transform is a natural choice if f (0) = 0, while the cosine transform ¯ts the case when f 0 (0) = 0. The use of frequency f (Hz=cycle/s) instead of ! (1/s) Introduce the frequency f (Hz) such that ! = 2¼f , and our usual Fourier transform pair Z1 Z1 1 F ¡j!t X(!) = x(t)e dt () x(t) = X(!)ej!t d! 2¼ ¡1

becomes X(f ) =

Z1 ¡1

¡1

x(t)e¡j2¼f t dt

F

()

x(t) =

Z1 ¡1

X(f )ej2¼f t df:

This form has the advantage that both variables t and f appear symmetrically without any multiplying factors outside the integrals. Its disadvantage (to me) is the appearance of a 2¼ in every kernel function exp(§j2¼f t). Its supporters are often experimentalists, because of their natural preference to measure frequency f in (Hz), as opposed to the natural frequency !. The use of f can also pave the way for a slightly smoother transition to the discrete-time world. Either way you look at it, the Fourier transform pair has to have 2¼'s somewhere: The question is, do you place them as harmless scaling factors outside the integrals, or do you turn them loose as pesky irritants inside the arguments of the complex exponentials, so that every di®erential and integral operation results in a factor of 2¼ or (2¼)¡1 ? Seriously, the world is divided and you will inevitably need to translate between the two forms, but not in our course, where, at least for now, ! rules! 88

ECE 370

QUIZ 7

9 October 1940

Name

IgÄ or Reiruμ of, III

Take-Home: due next class, Wednesday October 31 12:00 noon Consider a real-valued function f (t) that possesses a de¯nite Fourier transform F (!). Assume that f (t) is absolutely integrable so that we are assured of the ¯niteness of F (!) for all frequencies. 1.) If f (t) = f (¡t) is even, what can you deduce about the symmetry (even/odd parity) of F (!)? Is F (!) pure real, pure imaginary, or arbitrarily complex? 2.) If f (t) = ¡f (¡t) is odd, what can you deduce about the symmetry (even/odd parity) of F (!)? Is F (!) pure real, pure imaginary, or arbitrarily complex? 3.) What is the physical (geometric) signi¯cance of F (0)? Z1

F (!) =

f (t)e¡j!t dt

¡1

************************** SOLUTION ************************** F (!) =

Z1 ¡1

f (t) cos(!t) dt ¡ j

Z1 ¡1

f (t) sin(!t) dt = Fr (!) ¡ jFi (!)

If f (t) = f (¡t) is even, then Fi (!) = 0 and Fr (!) = 2

Z1

f (t) cos(!t) dt = Fr (¡!):

0

That is, the spectrum of an even, real signal is pure real and is an even function of !.

If f (t) = ¡f (¡t) is odd, then Fr (!) = 0 and Fi (!) = 2

Z1

f (t) sin(!t) dt = ¡Fi (¡!):

0

That is, the spectrum of an odd, real signal is pure imaginary and is an odd function of !.

F (0) =

Z1

f (t) dt is the integral of f (t); the DC value, total area under the curve.

¡1

89

Cascade of Two LTI Systems

x(t)

............................................................................................. ............................................................................................. ... ... ... ... ... ... .. ... ... .. .. ... ... .. .. ... ... ... ... ... . . . . . ........................................ . ....................................... ...................................... . . .. ... ... 1 2 . . ... ..... .... .... .. .. .. .. .. ... . ... ... ............................................................................................... .................................................................................................

h (t)

h (t)

h(t) = h1 (t) ~ h2 (t) H(!) = H1 (!)H2 (!)

Linear Dispersionless Filter

x(t)

............................................................................................... ... .... ... .. ... ... ... .. . ...................................... ........................................... . .. 0 ..... ... .. . . .... .... .. . ............................................................................................

h(t) = ±(t ¡ t )

H(!) = e¡j!t0

y(t) = x(t ¡ t0 )

\linear phase"

90

y(t)

Homework Problem: Rectangular Pulse Response of Ideal Low-Pass Filter. Consider an ideal low-pass ¯lter of bandwidth B and subsequent transfer function ³ ! ´ : (1) H(!) = ¦ 2B Find the transient response y(t) of this ¯lter to a rectangular pulse of pulse-width T x(t) = ¦(t=T ):

(2)

Graphical results simplify with normalized time t=T as the independent variable, for various values of the dimensionless parameter BT . This problem is most readily solved in terms of a special function, the sine integral8 Si(z) =

Zz

sin » d» »

Si(¡z) = ¡ Si(z)

Si(1) = ¼=2:

(3)

0

x(t)

.............................................................................................. ... ... .. ..... .. ... . .... .. . ........................................ ......................................... ... ... .. .... .. ... . .... .. .. . ............................................................................................

h(t)

y(t)

The ¯lter impulse response is h(t) = F ¡1 [H(!)] =

B sinc(Bt) ¼

(4)

which can be found from your table of Fourier transforms or by direct evaluation of the inverse transform integral. Similarly, the spectrum of the input signal is X(!) = F[x(t)] = T sinc(!T =2):

(5)

At least three di®erent, but related, approaches to ¯nding the output y(t) are worth considering. Time-Domain Convolution. y(t) = x(t) ~ h(t) =

Z1 ¡1

B = ¼

T =2 Z ¡T =2

x(¿ )h(t ¡ ¿ ) d¿ =

sin[B(t ¡ ¿ )] B d¿ = B(t ¡ ¿ ) ¼

T =2 Z

h(t ¡ ¿ ) d¿

(6)

¡T =2 T =2 Z ¡T =2

8 M.

sin[B(¿ ¡ t)] 1 d¿ = B(¿ ¡ t) ¼

B(T Z=2¡t) B(¡T =2¡t)

sin » d» »

(7)

Abramowitz and I.A. Stegun, Handbook of Mathematical Functions. National Bureau of Standards, 1970, pp. 231{232, Table 5.1. (\Big Red"or \Citizen's Handbook"). MATLAB has it built-in, too! 91

ª 1© ª 1© Si[B(T =2 ¡ t)] ¡ Si[¡B(T =2 + t)] = Si[B(t + T =2)] ¡ Si[B(t ¡ T =2)] ¼ ¼ (8) ½ · μ ¶¸ · μ ¶¸¾ 1 t 1 t 1 = Si BT + ¡ Si BT ¡ (9) ¼ T 2 T 2

y(t) =

Note that y(t ! 1) = 0. Also note that time appears normalized to the incident pulse width T as in t=T , and that the single (dimensionless!) parameter that characterizes the ¯lter response to the rectangular pulse is the product BT .

Frequency-Domain Multiplication.

y(t) = F 1 = ¼

¡1

Z1

1 [X(!)H(!)] = 2¼

ZB

j!t

X(!)H(!)e

¡1

sin(!T =2) j!t e d! !

ZB

ZB

T sinc(!T =2)ej!t d!

¡B

only the even part of ej!t contributes

¡B

1 = ¼

1 d! = 2¼

sin(!T =2) 2 cos(!t) d! = ! ¼

¡B

ZB

sin(!T =2) cos(!t) d! !

(10) (11)

(12)

0

Use the trig identity 2 sin X cos Y = sin(X ¡ Y ) + sin(X + Y ) so that 1 y(t) = ¼

ZB © ª d! : sin[!(T =2 ¡ t)] + sin[!(T =2 + t)] !

(13)

0

Let » = !(T =2 ¨ t) so that ZB

d! sin[!(T =2 ¨ t)] = !

0

B(T Z=2¨t)

sin » d» = Si[B(T =2 ¨ t)] »

(14)

0

whereupon (13) is ª 1© Si[B(T =2 ¡ t)] + Si[B(T =2 + t)] ¼ 1© = Si[B(t + T =2)] ¡ Si[B(t ¡ T =2)]: ¼

y(t) =

92

(15)

Linear Combination of Step-Responses. Recognize that our particular input can be written as x(t) = ¦(t=T ) = u(t + T =2) ¡ u(t ¡ T =2) (16) so that the response is y(t) = s(t + T =2) ¡ s(t ¡ T =2)

(17)

where the step response is Zt

Zt ZBt B 1 sin » h(¿ ) d¿ = sinc(B¿ ) d¿ = s(t) = d» ¼ ¼ » ¡1 ¡1 ¡1 2 0 3 Bt Z Z i 1 sin » sin » 5 1 h ¼ 14 1 = d» + d» = + Si(Bt) = + Si(Bt): ¼ » » ¼ 2 2 ¼ ¡1

(18)

(19)

0

The linear combination (17) of shifted step responses now yields y(t) =

1© Si[B(t + T =2)] ¡ Si[B(t ¡ T =2)]: ¼

(20)

The MATLAB routine sinint.m calls the symbolic toolbox, built upon MAPLE, and does an extremely slow and ine±cient numerical integration to compute the function Si(x). It doesn't work at all on my computer, so here is an alternate computation that uses MATLAB's routine expint.m for the exponential integral, a di®erent but related special function. E1 (z) =

Z1

e¡t dt t

z

Si(x) =

¼ + ImfE1 (jx)g 2

The routine for E1 (jx) will not work for (real) values of x · 0, but that's ok since we know that Si(0) = 0 and it is an odd function Si(¡x) = ¡ Si(x). Therefore Si(x) = Si(jxj) sgn(x).

function f=sinintRWS(x) % f=sinintRWS(x) sine integral Si(x) for real x f=zeros(size(x)); k=find(x»=0); f(k)=(pi/2+imag(expint(i*abs(x(k))))).*sign(x(k));

93

1.2

1.0

.. ... ...... .... .. ... .. ... ... ... .. ... .... ............ .......... ... ... .......... .......... ... .... . . . . . .. .... . . . . . .. . . . .. ... ... ..... .. ... ......... ......... ... ... ....... .. ... .. .. ..... ... .. .. ... ............ ........ ........ ....... .... .... ... ........... .... .. ......... ... . .. .. .. . .. .. .... ... ... .. ... . .. .... .... ... . . .. ... .. .. .. ... .. . .. ... ... ... .. ... .. .. .. ... . .. ... ... ... .. .. .. .... ... . .. ... .. .. .. ... .. .. .. ... . ..... ...... .... ..... .... ...... .... .... .. ... ... .... ..... ..... ..... ...... ... ..... .... .... ... .. .. . ... .. .. .. ... .. ... ... ... .. . ... ... .. .. .. ... .. .. .. ... .. .. ... .... .. .. ... ... .. ... ... ... .. . .. ... .. .. .. .. .. ... . .. ... ... .. .. .. ... .. ... ... ... .. .. ... ... .. . . . . ... . . ... ... .. . . . . . . .. .... . . .... ....... ... . . . . . . .. ...... . . . . . . . . .. . . . .. ...... ... . ... .. ... .... .. . . . . . . . . . . . . . . . . . .. . ...... ........ ........ ........... ......... ........... .......... ......... ............... .......... ............... ..... .................................................................................................................................................................... ..... .... .... .... .. . . . . . . . . . ... . ...... ....... ...... ....... ........... ....... ......... ....... ....... ......... . . . . . . . . . . . . . . . . . . . . . ... . . . . . . .. .. .... . . .... .. .... ... .. . ..... .... ... .. .. ... ...... .. ... ...... .. .. ..... ..... .. .. .. . . . . . . . .. .. ... .. .. ... ........ ... ...

dashed curve: BT = 10 solid curve: BT = 50

0.8

0.6 y(t) 0.4

0.2

0.0

¡0:2 ¡2:0

¡1:5

¡1:0

¡0:5

0.0

0.5

1.0

t=T

Fig. 1. Rectangular pulse response of ideal low-pass ¯lter.

94

1.5

2.0

Ideal Band-Pass Filter H(!) .... .. .. .. ... ... ... ... ... ... .. .. .. .. ... ... ... ... .. .. .. . .. .. . . . . . .. . . . . . . . . . . . . . . . . . . . . ............................................ . . . . . . . . .................................... . .................................... ... ................................. .... ... .. .. .. .... ... .. .. .. .. ... ... ... ... .. .. . .. .. . .. .. . . . . . . . ............................................................................................................................................................................................................................................................................................................................................................................... . .

B

B

¡!0

!0

0

H(!) = ¦

μ

! ¡ !0 B

¶

+¦

μ

! + !0 B

!

¶

... .. .. .. ... .. ... ... .. .. . ....................................................................................................................................................................................... .... .. ... ... .. .. ... .. ... ... .. .. ... .. ... j!0 t .. 0 ... ... .. .. ..

f (t)

B sinc 2¼ e

μ

F (!)

Bt 2

¶

¦

³!´ B

F (! ¡ ! )

f (t)

¤ £ B sinc(Bt=2) ej!0 t + e¡j!0 t 2¼ B = sinc(Bt=2) cos(!0 t) ¼

h(t) =

h(t) ... ... ...... . .... ....... .... ........ ............ ......... ...... .......... ...... . . 0 .. ... ........ .. ... ..... ..... ...... .. ... ....... .. ... ..... ...... .. .. ........ .. .. ..... .... .... ..... .... .............. ..... ..... ..... .... . .. . .. ... .. . . .. ... .. ... ... ... ...... .. ... .. ... ... ...... .. .. ... ... ........ ... ... .. .. ...... ....... ... .... .... .... ......... ... .... ... .... ......... . .. . . . . ... .. .. . . . .. ... ... .. ... .. ... .. ... .. .. .. ... ... .. ... ... .... ... .... .... ... ... .... .... ... .... ..... ... .... .... .. . . . . . . . ... . .. .. ... .. ... .. .. .. .. ... .. ... .. ... ..... ..... ... ...... ...... ... ... .. ... ... ... .. .. .. .. .. .. .. ... .. ... .. ... ... .. .. .. ..... ... ... ... ... ...... ... .... .... ..... ... .... ... .... ... .... ... ... .... ..... .... .... ..... ..... ..... ........ .... ..... ..... ..... .... ..... ......... ... . . ....... ...... . . . . ...... ..... ..... ..... ..... .... ..... .... . ... . .. .. ... .. .. . . .. .. .. .. .. . .. .. .. ... .. .. .. ... ... ... .. ... ... . . . . . . . . . . . . . . . . . . . . ....... ... ... ... ... ....... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... . .. ... .. .. .. .. .. ..... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... .. ... .. .. .. .. . ...... ... .. .... ... .... .. ...... ..... ..... .......... . ...... .... .. ... .. ... .. .. .. .. .. .. .. .. .. .. ... ... ...... ..... ... ... ....... ...... ... ... ... ... .... . . ... ... .. .. ..... . .. .. .. . .. .. .. .. .. .. . ... ... .. .. ... .. .. ... ... ... ... .. ..... ... ... .. .. .. .. .. . .. .. .. .. .. .. .. .. ... ...... ....... ... . . . . . . . . . . . . . . . . . . . . . . . .. .. ... ... ... .. .. .. .. .. .. .. .. .. ... .. .. ... ... ... ... ... .. ... .. ... .. .. .. .. ... ... ... .. ... ... ...... ... .. ... ... .. .. ... .. .. .. ... .. ... ...... .... .. ... ... ... .. .. .. .. .. ... ... ... ... .... ... .. .. .. .. . . .. .. .. .. . .. ... ... ... ... ... .. ... ... .. .... ... .... ... .. ... ... ... ... ... ... .. ... ... .. ... .... ..... .. . ... .. .. .. .. .. .. .... ...... .. ... .. ... .. ... ... ... . ....... ... ... . ... ... .. ... .... .. .. ... .. ... .. ... ... .. .. .. ... ..... .. . . .. .. ..... ..... .. ... ... ... . ..... ...... ..... .. ..... ...... ... .... ....... ... ..... ...... .. ...... ..... ... ..... . .. .. .. ...

! = 100, B = 20

95

t

Complex Phasors for Time-Harmonic Circuit Analysis. Consider the sinusoidalsteady-state response of the given RLC circuit, that is ¯nd vo (t) if vi (t) = Vp cos(!0 t + Á). The time-domain integro-di®erential equation (Kirchho®'s voltage law) that is appropriate for this network is Zt di(t) 1 L + i(¿ ) d¿ + Ri(t) = vi (t) dt C ¡1

and the output voltage is simply a scaled version of the mesh current, that is vo (t) = Ri(t). ... ... .. .. ................................................. ................................................................................. ........................................................................ ... ... ... ................................... ..... ... .. ... ... .. ... ... ... .. .. ... ... ... ... . ... .. .......... ......... .. .......... ........ ................ . . . . . . . . ............. .. .. ... ... ........ .. ... ... ... .... .. .... ... .... ...... . ............. ... .. .. . i o .. ...... ... ... .. . . . .... ...... . ...... ... ....... ............ ..................... .. ... ... ... .. .. ... .. . ........................................................................................................................................................................................................................................................ ...... ......

+

²

L

i(t) +

C

v (t)

v (t)

R

¡

²

¡

(1) Solve for vo (t) by representing the time-domain signals in terms of their Fourier transforms. (2) Show that the Fourier transform analysis is equivalent to, and is therefore the motivation for, the (simple and compact!) complex phasor analysis of time-harmonic circuits such as this one. Recall that the time-harmonic signal i(t) is represented in terms of the complex phasor I via i(t) = RefIe+j!t g: The Fourier transform of the input or generator voltage is ½ ¾ ¤ Vp £ j(!0 t+Á) ¡j(!0 t+Á) Vi (!) = F[vi (t)] = F[Vp cos(!0 t + Á)] = F +e e 2 ¤ Vp £ = 2¼ejÁ ±(! ¡ !0 ) + 2¼e¡jÁ ±(! ¡ !0 ) : 2 The Fourier transform of the entire (KVL) equation above (both the LHS and RHS) is · ¸ £ ¤ 1 I(!) j!LI(!) + + ¼I(0)±(!) + RI(!) = ¼Vp ejÁ ±(! ¡ !0 ) + e¡jÁ ±(! + !0 ) : C j! Physically argue that I(0) = 0 here, since we expect zero steady-state DC current through the capacitor, whereupon μ ¶ £ ¤ 1 R + j!L + I(!) = ¼Vp ejÁ ±(! ¡ !0 ) + e¡jÁ ±(! + !0 ) : j!C | {z } Z(!)

96

Now solve this algebraic equation for the unknown spectrum ejÁ ±(! ¡ !0 ) + e¡jÁ ±(! + !0 ) : I(!) = ¼Vp Z(!) Fourier inversion proceeds as 1 i(t) = 2¼

Z1 ¡1

2

I(!)ej!t d! Z1

Z1

3

Vp 4 jÁ ±(! ¡ !0 ) j!t ±(! + !0 ) j!t 5 e e d! + e¡jÁ e d! 2 Z(!) Z(!) ¡1 ¡1 · ¸ j!0 t ¡j!0 t e Vp jÁ e = e + e¡jÁ note Z(¡!0 ) = Z ¤ (+!0 ) 2 Z(!0 ) Z(¡!0 ) · ¸ 1 Vp ejÁ j!0 t Vp e¡jÁ ¡j!0 t = e e + ¤ 2 Z(!0 ) Z (!0 ) · ¸ 1 Vp ejÁ j!0 t = e + c.c. an expression plus its complex conjugate 2 Z(!0 ) · ¸ £ ¤ Vp ejÁ j!0 t e = Re = Re Iej!0 t Z(!0 ) =

where the complex phasor I=

Vp ejÁ Vi = Z(!0 ) Z(!0 )

introduced itself! If you still want it, Vo = RI

and

vo (t) = Ri(t):

It looks like the two steps alluded to in the problem statement are more naturally done together.

97

Homework Problem on Uncertainty Principle Consider the signal 1 x(t) = p ¦(t=T ) T where T > 0 and the total energy in x(t) is

Ex =

Z1 ¡1

jx(t)j2 dt = 1:

(The multiplying factor T ¡1=2 ensures that Ex is independent of T .) If x(t) is the input voltage to the given electric circuit, with ¯xed values of R and L, consider the energy in the output voltage y(t) Z1 jy(t)j2 dt: Ey = ¡1

Again, note that changing T has no e®ect on the total input energy. If T is increased, will the output energy Ey increase, decrease, or remain the same? Explain. ........................................................................ ... .... .... .... ............................................................................ ............................................ .. .... ... .. ..... ...... .... ........ ............. ....... ....... ... ........ ............. ....... ...... .... ........ ............. ... .. ... ....................................................................................................................................................................................................

+

²

L

x(t)

y(t)

R

¡

²

98

+

¡

Approximate Analysis of Dispersion ................................................. ... ... .. ... .. ... ... . . ........................... ........................... .... .... .. .. ... .. . . ................................................

x(t)

y(t)

H(!)

Fig. 1. System Block Diagram

If the modulating signal or envelope of a high frequency sinusoid of frequency !0 is the baseband signal a(t), having bandwidth considerably less than !0 (say 10%), then the spectrum of the the modulated carrier x(t) = a(t) cos(!0 t)

(1)

X(!) = 12 [A(! ¡ !0 ) + A(! + !0 )]:

(2)

is An example spectrum is shown in Fig. 2. jX(!)j

... .... ........ .. .......... ... .... .. . .. ... ... ... .... .. .... . .. . . ... ... ... .. ... ... ... ... .. ... ... ... .. .. ... ... ... . ... . . ... ... .. . . .. . ... ... . .. ... ... ... . . . . . ... .. .. . .. . . . ... . . ... . . . . . . ... . . ... . . . . ... .. .. .. .. . . . .. .. . . . . . . ... ... . . . . . . .. ... . . . . . . ... . ... . .. . . . . ... . . .... . . . . . . .... . . . ..... . . . . . . . . . . . . . ................................................................................................................................................................................................................................................................................................................................................................................................................................................ . . .

¡!0

0

!

!0

Fig. 2. Typical Narrow-Band Spectrum For real signals x(t) and h(t), the spectra satisfy X(¡!) = X ¤ (!) and H(¡!) = H ¤ (!). The response of the system is the inverse Fourier transform of X(!)H(!), with the positive and negative frequency contributions written separately as

99

1 y(t) = 2¼

Z1

1 2¼

Z1 0

0

1 2¼

Z1

Z1

j!t

X(!)H(!)e

0

=

=

= =

1 2¼

0 Z1

Z0

1 d! + 2¼

X(!)H(!)ej!t d! +

X(!)H(!)ej!t d! +

1 2¼

X(!)H(!)ej!t d!

¡1 Z1

1 2¼

X(¡−)H(¡−)e¡j−t d− £ ¤¤ X(−)H(−)ej−t d−

0

X(!)H(!)ej!t d! + c.c.

0 1 2 [y+ (t)

¤ + y+ (t)]

(3)

The signal y+ (t) is called the pre-envelope of y(t), and consists of positive frequencies only Z1 1 y+ (t) = [A(! ¡ !0 ) + A(! + !0 )]H(!)ej!t d! 2¼ 0

1 ¼ 2¼

Z1 ¡1

1 A(! ¡ !0 )H(!)ej!t d! = 2¼

Z1

A(−)H(− + !0 )ej(−+!0 )t d−

(4)

¡1

where A(−) is the original, baseband or low-pass signal. The phase or argument of the transfer function H(!) in a neighborhood of the carrier frequency !0 is of most interest. Write H(!) = exp[j£(!)]:

(5)

A Taylor series about !0 of the phase function £(!0 + −) = £(!0 ) + −£0 (!0 ) + 12 −2 £00 (!0 ) + : : :

(6)

gives the approximate transfer function 0

H(− + !0 ) ¼ ej£(!0 ) ej−£ (!0 ) :

(7)

An approximation for the pre-envelope (4) is therefore Z1 0 1 j[!0 t+£(!0 )] y+ (t) ¼ e A(−)ej−[£ (!0 )+t] d− 2¼ ¡1

= ej[!0 t+£(!0 )] a[t + £0 (!0 )]

(8)

and so the ¯nal approximation for the output of the dispersive ¯lter (5) is y(t) ¼ a[t + £0 (!0 )] cos[!0 t + £(!0 )]:

(9)

The phase delay (of the carrier) is due to £(!0 ) and the group delay (of the envelope) is due to £0 (!0 ). 100

Numerical Example. The phase function of a length of waveguide transmission line is p £(!) = ¡ ! 2 ¡ !c2 t0 where !c is the cuto® frequency of the guide and t0 = d=c where d is the length and c is the free-space phase velocity. If !c = 0, then £0 (!0 ) = ¡t0 . If !c = 25, !0 = 30, t0 = 3, then (30)(3) £0 (!0 ) = ¡ p = ¡5:43 302 ¡ 252 The following ¯gures are the result of an inverse FFT approximation for the signal 2

x(t) = e¡t cos(!0 t) applied to the input of the above dispersive waveguide section. The 4096 point FFT uses ¢! = 0:3.

101

1.0

0.5

y(t)

0.0

¡0:5 ¡1:0

. . .. . .. .... .. .... ..... ..... ... ...... ...... ...... ..... .. ... ... ... .. ..... ..... ..... ..... ..... . .... .... .... .... .... . ..... ...... ....... ....... ....... ....... ...... .. .. .. .. .. .. .. c 0 0 ..... ..... ..... ..... ..... ..... ..... . .... .... .... .... .... .... .... . ...... ....... ...... ...... ...... ...... ...... ...... ...... ...... ..... .... ..... ..... ...... ..... ..... ..... .. .. ... .. .. .. .. .. .. ... ..... ..... .... ..... ..... ..... ..... ..... ..... ... ...... ....... ...... ....... ....... ....... ....... ....... ...... ....... ....... . . . . . ... .. ... ..... ..... ..... .... .... .... .... ... .... ... ..... ..... ..... ..... .... .... ...... ..... ..... ..... ..... .... ..... ..... ....... ......... ....... ....... ........ ........ ........ ....... ....... ....... ....... ........ ....... ....... . ... ... .. .. .. . . .. . .. . . . .. . . ... .. .... ... .. ... ......................................................................................... ... .... ... .... ... ... .... .... ... .... ... .... ... ... ........ .. ....................................................................................................................................................................................... . .... ..... ..... .... ..... .... .... ..... ..... .. .. .... ..... .... .... .... ..... .. .. .... ... .... .... .... .... ... ... .... .... ... .... .... .... . ..... .... ... .... ... .... .... .... ... ... ... ... .... .... .... ... ... ... .. ... ... .... .... .... .... .... . .... .... ... ... ... ... ... ... ... ... ... .... .... ..... ..... ...... ...... ...... ...... ...... ...... ...... ..... .... ..... .... ..... ..... ..... ..... ..... .... .... .... ..... ..... .... .... .... ..... ..... ..... ..... ...... ... ..... ..... ..... ..... ..... ..... ..... ..... ... ..... ...... ...... ...... ...... ..... ...... ...... ..... ..... ..... ..... ..... ..... .... ..... ... .... .... ..... .... .... .... .... . ..... .... ..... ..... ..... ..... .... .... ..... .... .... ..... .... .... ..... ..... ..... ..... ... ..... .... .... .... .. .... ..... .... ..... ... .. .. .. .... .... ..... ... . ..... ...... .. ... ...

! = 0, t = 3, ! = 30

0

1

2

3

4

5

6

7

8

9

10

t 1.0

0.5

y(t)

0.0

¡0:5 ¡1:0

. .. .. .. .... .. . . ..... .... ..... ... ...... ....... ....... ....... ........ .. . ... .. ... .. .. .. .... .... ..... ..... ..... ..... ..... .... ..... ..... .... ...... .... ..... ... c 0 ..... ....... ........ ....... ........ ....... ....... ........ ...... .. .. .. .. .. .. .. .. .. . . .... ..... ..... ..... ..... ..... ..... ..... ..... ..... ....... ....... ...... ...... ........ ....... ....... ........ ...... ........ ...... ... .. .. .. .. .. .. .. .. .. .. .. .. ... ..... ..... ..... ..... ..... ...... ..... ..... ...... ..... ..... .... . ..... .... ..... ...... ...... ..... ..... ..... ..... ..... .... ..... ..... ..... ...... ....... ....... ........ ....... ....... ........ ....... ........ ....... ....... ....... ...... ....... ...... ..... .. ... ... .... ... ... ... ... ... ... ... ... ... ... ... .... .... .. . ..... .... ..... .... .... ...... ..... ...... ..... ..... ..... ..... ..... ..... ..... .... ..... ..... .... .... ........ ........ ........ ........ ........ ........ ........ ........ ........ ......... ........ ........ ........ ........ ........ ........ ........ ....... ........ ........ ........ ...... .... .... .... ... ... . . . . . . . . . ........................................................................................................................ ..... .... ........... ..... .... ..... ..... ...... ..... ... .. ... .. ... ... ... .. .. .. ... .. ... .. ... ... .. .. ... .. .... ..... .... ............ ...... ..... ...... .......................... .... ... ... ... ... ... .... ... ... ... .... ... ... ... ... ... ... .... ... ..... .... .... .. . ...... .... .... .... .... .... .... ..... .... .... .... .... .... .... .... .... ... .... ... .. .... ..... ..... ...... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... .. ... ..... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .. .... ..... ..... ..... ...... ..... ...... ..... ..... ..... ..... ...... ..... ..... ..... .. ... ... .... ... ... ... .... .... ... ... ... ... ... ..... .... .... .... ..... ..... .... .... ..... ..... .... ...... ..... ... .... .... .... ... .... .... .... .... ... .... .... . .... .... .... ..... .... .... .... .... .... .... .... ..... .... ..... ..... .... .... ..... .... ..... .... .. .. ..... .... .... ...... ...... .... ..... ..... ...... .... ..... ..... .... .... .... ..... ... .. ... ... ... .... ... ... .. ... . ... .... ... ... ... ... ... .... .... ...... ...... ...... ...... ...... .... ... .... .... ... .... ..... . .... ..... .... ... .. .... ..... ...... ..... .. .. .. . ..

! = 25, t = 3, !0 = 30

0

1

2

3

4

5

6

7

t Fig. 3. Dispersive Filter Output

102

8

9

10

Fourier Integral Solution of Potential Problem Laplace's equation 2

r Ã(x; y) =

μ

@2 @2 + 2 @x2 @y

boundary conditions Ã(x; 0) =

½

¶

Ã(x; y) = 0

1; jxj < 1 0; jxj > 1

separate variables Ã(x; y) = X(x)Y (y) X 00 (x) Y 00 (y) + =0 X(x) Y (y) | {z } | {z } =¡k2

=+k2

solution form Ã(x; y) = cos(kx)e¡kjyj Ã(x; y) =

Z1

A(k) cos(kx)e¡kjyj dk

0

Ã(x; 0) =

Z1

A(k) cos(kx) dk

0

inverse Fourier transform 2 A(k) = ¼

Z1

2 Ã(x; 0) cos(kx) dx = ¼

0

Z1

cos(kx) dx =

2 sin k ¼ k

0

2 Ã(x; y) = ¼

Z1

sin k cos kx ¡kjyj e dk k

0

Z1

sin[k(x + 1)] ¡ sin[k(x ¡ 1)] ¡kjyj e dx k 0 · μ ¶ μ ¶¸ 1 x+1 x¡1 ¡1 ¡1 = tan ¡ tan ¼ jyj jyj

=

1 ¼

using G&R 3.941(1). 103

6

5

4

3

2

1

0 −3

−2

−1

0

Fig. 1. Equipotentials

104

1

2

3

Fourier Integral Solution of Potential Problem M.J. Ablowitz and A.S. Fokas, Complex Variables, Cambridge, 1997, page 60 prob 9. Laplace's equation r2 Ã(x; y) = 0 boundary conditions

Ã(x; 0) = sgn(x) = §1 (x ? 0)

separate variables Ã(x; y) = X(x)Y (y) X 00 (x) Y 00 (y) + =0 X(x) Y (y) | {z } | {z } =¡k2

=+k2

solution form Ã(x; y) = sin(kx)e¡kjyj Ã(x; y) =

Z1

B(k) sin(kx)e¡kjyj dk

0

Ã(x; 0) =

Z1

B(k) sin(kx) dk

0

inverse Fourier transform 2 B(k) = ¼

Z1

2 Ã(x; 0) sin(kx) dx = ¼

0

Z1

sin(kx) dx =

0

a bit tricky! recall ECE 370 and sgn(x) = u(x) ¡ u(¡x) 2 Ã(x; y) = ¼

Z1

sin kx ¡kjyj e dk k 0 μ ¶ μ ¶ 2 2 x jyj ¡1 ¡1 = tan = 1 ¡ tan ¼ jyj ¼ x

using G&R 3.941(1).

105

2 ¼k

The Fourier Transform of combT (t) is

2¼ T

comb2¼=T (!)

Recall our notation F (!) for the Fourier transform F (!) = F[f (t)] =

Z1

¡j!t

f (t)e

¡1

F

dt () f (t) = F

¡1

1 [F (!)] = 2¼

Z1

F (!)ej!t d!:

(1)

¡1

The Fourier transform of Woodward's comb function, or impulse train f (t) = combT (t) =

1 X

n=¡1

is therefore F[combT (t)] =

1 X

±(t ¡ nT )

e¡jn!T = F (!) = F (! + 2¼=T );

(2)

(3)

n=¡1

periodic in ! with period 2¼=T . Let's integrate F (!) over the range a < ! < b, where ¡¼=T < a < b < ¼=T . The minimum and maximum restrictions cover one period, centered about the origin (! = 0). We will make use of the discontinuous sum 1 X sin(nx) ¼ ¡ jxj = sgn(x) n 2 n=1

(¡2¼ < x < 2¼)

(4)

where, in this context, we de¯ne 8 > < ¡1; x < 0 sgn(x) = 0; x = 0 > : +1; x > 0

(5)

in order to ensure the vanishing of the sum for x = 0. The integral of interest is Zb X 1

a n=¡1

1 Z X

b

¡jn!T

e

d! = b ¡ a + 2

cos(n!T ) d!

n=1 a 1 X

2 1 [sin(nbT ) ¡ sin(naT )] T n=1 n ³¼ ´ ³¼ ´ =b¡a+ ¡ jbj sgn(b) ¡ ¡ jaj sgn(a) T T ½ 2¼=T if ¡ ¼=T < a < 0 < b < ¼=T = : 0 if a < b < 0 or 0 < a < b =b¡a+

(6)

Therefore, we see that F[combT (t)] = 2¼ T comb2¼=T (!): ¥ Now we are in a position to derive the Fourier series from the Fourier integral. And recall that we essentially derived the Fourier integral from the \spectral representation of the Dirac-delta" function. We are in excellent shape to now do some real damage. 106

()

... ... ... .. .. .......... .......... .......... .......... .......... ... ... ... ... ... .. .. .. .. .. .. .. .. .. .. .. . . ... . ... .... .... .... .. ... .. .. .. .. .. ... ... ... ... . . . . ................................................................................................................................................................................

:::

. . . . . ......... ......... ......... ......... ......... .. .. .. .. .. .. .. .. .. .. .. ... ... ... ... ... .. .. .. .. ... .. .. .. .. . . . . ...........................................................................................................................................................................................................................

:::

¡2T ¡T

0

2¼ comb 2¼ (!) T T

F

combT (t)

T

2T

::: t

:::

4¼ ¡ T

2¼ ¡ T

0

2¼ T

!

4¼ T

Fourier Transform of a Periodic Signal f (t) = f (t + T ) ........ ........ ........ ........ .. .. .. .. .. ... .. .. ... .... ... .... ... .... ... .... ..... ..... ..... . . . . . ....... . . . . . . . . . . . . . .. . . . .. .. .. ... ... ... ... .. . ... ........... ... ........... .... ... ........... .... ... ........... .... .... .... .... .... ........ .... .. ... .. ... .. ... .. ... .. .. .. .... .... .... .... .. ...... ...... ...... ...... .. .. .. .. ... ... ... ... . . . . . . . . . . . ... ... ... ... . . . . . . . . . . . . . . . . .. .. .. .. . . . . . . . . . . . . ... ... ... ... . . . . .. ...... .. ...... .. ...... .. ...... .. ... .. ... .. ... .. ... .................................................................................................................................................................................................................................................................................................................................................................................................... . . . .

:::

:::

¡T

0

T

t

2T

g(t) ... ... ... ... ... ... ... .... . ... ... .. .. ... .......... .... ... . .... . .. . . . . . ....... .. ... ... ... ... ... . . ... .. . . . .. . . .. .. . . .............................................................................................................................................................................................................................................................................................................................................................................................. . . . .

¡T

0

T

t

2T

Consider the arbitrary periodic signal f (t) = f (t + T ), and let's focus our attention on the particular period 0 · t · T . Any period would work, but since all periods are equivalent we might as well work with the selected one. (The function f does not have to go to zero at t = nT ; it was simply easier to draw one like this. And of course the function f does not have to be positive, or even real-valued.) Introduce the new aperiodic function g(t), also de¯ned for all t according to ½ f (t); 0 · t · T g(t) = (1) 0; t < 0 or t > T: We can easily reproduce the entire periodic signal f (t) by replicating g(t) all up and down the t-axis. That is, we write f (t) as the periodic extension of g(t) in the form f (t) =

1 X

n=¡1

107

g(t ¡ nT ):

(2)

Recognize that (2) can be written as f (t) = g(t) ~

1 X

n=¡1

±(t ¡ nT ) = g(t) ~ combT (t):

(3)

The Fourier transform of (3) is 2¼ comb 2¼ (!) T T 1 X 2¼ G(!) = ±(! ¡ n!0 ) T n=¡1

F (!) = G(!) ¢

where !0 ,

2¼ T

1 2¼ X = G(n!0 )±(! ¡ n!0 ): T n=¡1

(4)

The spectrum F (!) of the periodic signal f (t) consists of discrete spectral lines at integer multiples n!0 of the fundamental frequency !0 = 2¼=T . The nth multiple is called the nth harmonic. Note that negative n are just as important as positive n.

.... ...... ... ..... ... ........ ..... ... ... .. . . ... ... ...... .. .............. ......... . . .. . ... . .... .. . . . .. .. ... .. .... ... .. .. . .. .. . . . .. .. ... .. .. .... ... ... .. .. .. . . . . . .... . ... . . . . . . . ..... ............ .... .... ... .. .... . . . ... .. . . .. .. .. . . .. .. . .. . .... .... .... ... .. . .... . . . . .. . . . . . .. . . .. .. . . .... .... .... ... .... .. .. . . . . .. . .. . . . . . . . . . .. . .. . . . . . . . . . . . .. .... .... .... .... .... .. .. . . . . . . . .. .. .. .. .......... .... .... . . . ............ . . . .. . . . . . . . . . . . . ... .... .. . . . . . . .... .... .... .... .... .. .. .. ... .. .. .. .. .. .. .. .. ... ... .. . . . . . . . . . . . ... . . . . . . .. .. . . . . . .. .. . . . . . .. . . . . . .. .. .. .. .. .. .. .. . ... . . . . . .. . . . . . . .. . .. . . . . . . . . . . . . .. . . .. . . . . . .. .. .. .. .. .. . .. . . . . . . . . ... . . . . .. . . . ... . . . . . . . . . . . . . . . . . . . . . . .......... . . . . . . . . . . . . ...... . .. .. .. .. .. .. .. .. ... .. . . . . . . . ......... ...... . .... ........ . . .. . . . . . . . ... ... ... .. . . . . . . . . .. .. . . .. . . . . . . . . .. . . . . . . . . . . . ............................................................................................................................................................................................................................................................................................................................................................................................................................................................................. ... ... ... ... .. ... ... ... ... .. .. ... ..... . .. ... ... ... .... .... .. .... ... .... ........... .......... ... ... .. .. .............. ......... ..... . . . ... . ... . . .... ... ... . .. ... ... ... .. .. . ... .... .. .... . .. ... ... .... ...... ......... ..... .......... ...... .... ........ . ......... . . . . . . . . .... .. . ... ... . ... ... ... .... . ..

¡10

¡8

¡6

6

¡4

¡2

0

2

108

4

8

10

! !0

The inverse Fourier transform of F (!) should give us our original periodic f (t). It does. In fact, it gives us a whole new representation for a periodic signal 1 f (t) = 2¼

Z1 ¡1

j!t

F (!)e

Z1 1 1 X X G(n!0 ) G(n!0 ) jn!0 t d! = ±(! ¡ n!0 )ej!t d! = e T T n=¡1 n=¡1 ¡1

(5)

where ¯ G(n!0 ) = G(!)¯!=n!0 =

Z1

¡jn!0 t

g(t)e

dt =

¡1

ZT

¡jn!0 t

g(t)e

0

dt =

ZT

f (t)e¡jn!0 t dt:

(6)

0

We usually call the Fourier coe±cients cn instead of G(n!0 )=T and write the whole package as ZT 1 X 1 jn!0 t f (t) = cn e cn = f (t)e¡jn!0 t dt: (7) T n=¡1 0

Of course any period will work for the integration limits, since the total integrand is periodic with period T : tZ 0 +T 1 f (t)e¡jn!0 t dt: (8) cn = T t0

You now have the complete and uncluttered derivation of the Fourier series representation of a periodic signal, starting from the Fourier integral. The two key components are the correspondingly appropriate spectral representations for the aperiodic Dirac-delta function and for its periodic version, Woodward's comb function. All that was required on our part was a reasonable °uency with the Fourier transform properties and a few fundamental signals. F Exercise 8. Find an interested (or at least willing) fellow student or more who are at or above your present level with respect to Fourier analysis, and explain to them, in detail, the above derivation starting from the top of three pages back. If you explain it correctly and with some degree of coherence, and your entire audience has not deserted you, buy the survivor(s) a Coke. (They can split it to ¯t any ¯nancial boundary conditions, applied, of course, to your total solution.) You deserve one, too. Otherwise go back and do any of the homework exercises (including enough of the 16 Fourier transform pairs on page 59) that strike your fancy. If none ¯t that description, see me for more.

109

Periodic Signals and Fourier Series The period of a periodic signal is the smallest positive number T such that f (t) = f (t + T ) for all t. It will be frequently convenient to de¯ne !0 , 2¼=T to be the fundamental frequency of the periodic signal, with n!0 the nth harmonic frequency for integer n = 0; §1; §2; : : : . An important property of a periodic signal is that the integral tZ 0 +T t0

f (t) dt =

ZT

f (t) dt =

0

T =2 Z

f (t) dt =

¡T =2

3T Z =4

f (t) dt = : : :

¡T =4

is independent of the time t0 . That is, integration over any complete period yields the same result. Another useful observation is that if f (t) = f (t + T ), then the product f (t)ejn!0 t is also periodic with period T , so that the integral tZ 0 +T

f (t)ejn!0 t dt

t0

is also independent of the time t0 . F Be sure you can show/verify this important property for yourself. DEFN: Orthogonality. Two functions (generally complex) are said to be orthogonal on the domain a · t · b if the inner product ¯b Zb ¯ hf (t); g(t)i¯ , f (t)g ¤ (t) dt = 0: a

a

In any given situation, let's assume the interval of interest a · t · b is known and ¯xed, so that we can condense the notation on the inner product angle brackets. The inner product of a signal and itself is sometimes called the energy in the signal Zb

hf (t); f (t)i ,

a

jf (t)j2 dt = E

and the square-root of this necessarily non negative quantity is called the norm of the signal 2 b 31=2 Z p kf (t)k = + hf (t); f (t)i = 4 jf (t)j2 dt5 : a

When dealing with periodic signals so that the interval of interest is any period t0 · t · t0 + T , the e®ective or root-mean-square value is 2 t +T 31=2 Z0 1 frms = 4 jf (t)j2 dt5 : T t0

110

Kronecker delta. This simple function is a function of two integers m and n and is de¯ned as ½ 1; m = n ±m;n = 0; m 6 = n: It is unrelated to the Dirac-delta function ±(t) that is explicitly written as a function of the continuous time variable t. The operational advantage of introducing this notational convenience will be clear as we progress. Orthogonality of the Fourier basis. hejn!0 t ; ejm!0 t iT =

tZ 0 +T

hcos(n!0 t); cos(m!0 t)iT =

hsin(n!0 t); sin(m!0 t)iT =

ej(n¡m)!0 t dt = T ±n;m

t0

tZ 0 +T

(n; m = 0; §1; §2; : : : )

cos(n!0 t) cos(m!0 t) dt =

T ±n;m 2

(n; m = 1; 2; : : : )

sin(n!0 t) sin(m!0 t) dt =

T ±n;m 2

(n; m = 1; 2; : : : )

t0

tZ 0 +T t0

hcos(n!0 t); sin(m!0 t)iT =

tZ 0 +T

cos(n!0 t) sin(m!0 t) dt = 0

(n; m = 0; 1; 2; : : : )

t0

When n = 0 the nth cosine is unity, and we have

h1; cos(m!0 t)iT =

tZ 0 +T

cos(m!0 t) dt = T ±0;m

(m = 0; 1; 2; : : : )

t0

h1; sin(m!0 t)iT =

tZ 0 +T

sin(m!0 t) dt = 0

(m = 1; 2; : : : )

t0

Note that sin(m!0 t) is trivial, and therefore not used, when m = 0. F Homework: Verify/derive each of the above orthogonality relations. You will need to use trig identities to simplify the products in the integrands. 111

Derivation of Fourier Coe±cients by Minimizing the Mean Square Error The concept of the Fourier series arose naturally back on page 108 when we wrote a periodic signal as the inverse Fourier integral of its Fourier transform. In doing so, the Fourier coe±cients appeared naturally. Now let's start over by assuming that the periodic signal f (t) = f (t + T ) can be represented by a Fourier series, call it fe(t) =

1 X

cn ejn!0 t

n=¡1

to temporarily avoid any premature claims that fe(t) = f (t). One approach to derive the expression for the Fourier coe±cients cn is to ask the question: What values of the cn minimize the mean square error between the original signal f (t) and its proposed Fourier series representation fe(t) ? Firstly, write the mean square error 1 E = T 2

1 = T

tZ 0 +T t0

1 jf (t) ¡ fe(t)j2 dt = T

tZ 0 +T

£ ¤£ ¤ f (t) ¡ fe(t) f ¤ (t) ¡ fe¤ (t) dt

t0

tZ 0 +T t0

£ ¤ jf (t)j2 + jfe(t)j2 ¡ f (t)fe¤ (t) ¡ f ¤ (t)fe(t) dt:

We need the following component integrals.

1 T

tZ 0 +T t0

1 jfe(t)j2 dt = T

tZ 0 +T t0

1 X

cn e

n=¡1

|

{z

fe(t)

1 X

1 X

1 X

1 X

1 = cn c¤m T n=¡1 m=¡1 = =

n=¡1 1 X

cn

1 X

jn!0 t

m=¡1

cn c¤n

n=¡1

112

m=¡1

} |

tZ 0 +T t0

c¤m ±n;m

c¤m e¡jm!0 t dt {z

fe¤ (t)

ej(n¡m)!0 t dt

}

1 T

tZ 0 +T t0

1 T

tZ 0 +T

f (t)fe(t) dt = ¤

e¤

1 X

1 cn T n=¡1 1 X

f (t)f (t) dt =

t0

c¤n

n=¡1

1 T

tZ 0 +T

f ¤ (t)ejn!0 t dt

t0 tZ 0 +T

f (t)e¡jn!0 t dt

t0

Now the mean square error is

1 E = T 2

tZ 0 +T t0

2

jf (t)j dt +

1 X

cn c¤n

n=¡1

1 X

tZ 0 +T

1 X

tZ 0 +T

1 ¡ cn T n=¡1 ¡

c¤n

n=¡1

1 T

f ¤ (t)ejn!0 t dt

t0

f (t)e¡jn!0 t dt

t0

2

To minimize E with respect to the Fourier coe±cients cn , set to zero the partial derivative with respect to each of the ck (for all integer k) @ 2 1 E = c¤k ¡ @ck T or c¤k

1 = T

tZ 0 +T

f ¤ (t)ejk!0 t dt = 0

t0

tZ 0 +T

f ¤ (t)ejk!0 t dt

t0

the conjugate of which gives the required Fourier coe±cient formula 1 ck = T

tZ 0 +T

f (t)e¡jk!0 t dt

t0

where k can now be replaced by n 1 cn = T

tZ 0 +T

f (t)e¡jn!0 t dt:

t0

113

Fourier Coe±cients by a Direct Inner Product Write

1 X

f (t) =

cn ejn!0 t

n=¡1

and then take the inner product of both sides with ejm!0 t tZ 0 +T

1 X

f (t)e¡jm!0 t dt =

n=¡1

t0

=

1 X

cn

tZ 0 +T

ej(n¡m)!0 t dt

t0

cn T ±n;m = T cm

n=¡1

and therefore cm

1 = T

tZ 0 +T

f (t)e¡jm!0 t dt:

t0

Trigonometric Form of the Fourier Series 1 X ¤ £ f (t) = a0 + an cos(n!0 t) + bn sin(n!0 t) n=1

To ¯nd the trig coe±cients an and bn (n = 0; 1; 2; : : : ) in terms of the complex exponential coe±cients cn (n = : : : ; ¡1; 0; 1; 2; : : : ) set 1 X

cn e

n=¡1

c0 +

1 h X

c0 +

jn!0 t

cn e

n=1

jn!0 t

1 X ¤ £ = a0 + an cos(n!0 t) + bn sin(n!0 t) n=1

¡jn!0 t

+ c¡n e

i

1 X £ ¤ = a0 + an cos(n!0 t) + bn sin(n!0 t) n=1

1 h X © ª © ªi cn cos(n!0 t) + j sin(n!0 t) + c¡n cos(n!0 t) ¡ j sin(n!0 t)

n=1 1 X

= a0 +

n=1

¤ £ an cos(n!0 t) + bn sin(n!0 t)

1 h 1 i X X £ ¤ c0 + (cn +c¡n ) cos(n!0 t)+j(cn ¡c¡n ) sin(n!0 t) = a0 + an cos(n!0 t)+bn sin(n!0 t) n=1

n=1

114

Therefore 1 a0 = c0 = T

tZ 0 +T

f (t) dt

t0

an = cn + c¡n

2 = T

tZ 0 +T

f (t) cos(n!0 t) dt

(n = 1; 2; : : : )

t0

2 bn = j(cn ¡ c¡n ) = T

tZ 0 +T

f (t) sin(n!0 t) dt

(n = 1; 2; : : : )

t0

F Homework. Derive the trig Fourier coe±cients by taking the direct inner product of 1 X £ ¤ f (t) = a0 + an cos(n!0 t) + bn sin(n!0 t) n=1

with each of the trig Fourier basis functions cos(m!0 t), sin(m!0 t), and unity.

115

Two Standard Notations for the Trigonometric Fourier Series

f (t) = a0 +

1 X ¤ £ an cos(n!0 t) + bn sin(n!0 t)

n=1

1 a0 = T

tZ 0 +T

f (t) dt

t0

2 an = T

tZ 0 +T

f (t) cos(n!0 t) dt

(n = 1; 2; : : : )

f (t) sin(n!0 t) dt

(n = 1; 2; : : : )

t0

2 bn = T

tZ 0 +T t0

1 ¤ a0 X£ f (t) = + an cos(n!0 t) + bn sin(n!0 t) 2 n=1

2 an = T

tZ 0 +T

f (t) cos(n!0 t) dt

(n = 0; 1; 2; : : : )

t0

2 bn = T

tZ 0 +T

f (t) sin(n!0 t) dt

(n = 1; 2; : : : )

t0

The second form uses the general form for an to also calculate the DC term a0 , avoiding the need for a separate formula for a0 . It is silly, but observe that this captures the spirit: 1

¤ | X£ f (t) = + an cos(n!0 t) + bn sin(n!0 t) 7 n=1 7 |= T

tZ 0 +T

f (t) dt

t0

F Homework. In some applications, the Fourier series is written in the form 1 X f (t) = A0 + An cos(n!0 t + Án ): n=1

Find an expression for An and Án in terms of an and bn . 116

Riemann-Lebesgue Lemma - Fourier Coe±cients f (t) = f (t + T ) =

1 X

cn ejn!0 t ;

!0 = 2¼=T

n=¡1

T cn =

tZ 0 +T

f (t)e¡jn!0 t dt

t0

¯t0 +T tZ 0 +T f (t)e¡jn!0 t ¯¯ 1 = + f 0 (t)e¡jn!0 t dt ¯ ¡jn!0 ¯ jn!0 t=t0

t0

by integration-by-parts. The boundary terms are zero by the periodicity of the integrand. The remaining integral has ¯ ¯ t +T tZ 0 +T ¯ ¯ Z0 ¯ ¯ 0 ¡jn! t 0 ¯ f (t)e dt¯¯ · jf 0 (t)j dt · K < 1 ¯ ¯ ¯ t0

t0

if f 0 (t) is absolutlely integrable. Therefore jcn j ·

K : 2n¼

If f 0 (t) is absolutlely integrable, then cn ! 0 as n ! 1.

Corollary. If f (k) (t) is absolutely integrable, but f (k+1) (t) is not, then cn = O(n¡k ) as n ! 1.

117

Gibb's Phenomenon - Example Square wave

S5 (t)

1.0

0.5

f (t)

0.0

¡0:5 ¡1:0

... ..... ..... ... ... ... ... ... ... .. .. .. .... .. .... .... .......... ........... ......... . .. .... . . . . . . . .... ..... . . . ...... . . . . . . ................................................................................................................................................................................ ......................................................................................... .. .. ....... ....... ... .. ....... ... .. . ... ... . . . . .. ........ .. .. ...... . .... .. .. . ... .. .. .. ... ... .. ... ... .. ... .. ... .. .. .. .. .. ... ... .. .. .. .. .. .... ... .. ... .. ... .. .. .. .. .. ... ... ... .. ... .. .. ... .. .. .. .. .. .. .. .. .. .. .. .. ... .. ... .. ... .. ...... .... .... ....... .... .... ..... ..... ...... .... .... ..... .... .... .... .... ..... ..... .... ... ... ... ... ... .. ... .. . ... ... .. ..... ... ... .... .... ..... . .. .. ......... .... .... ..... ...... ...... . . ...... . ..... ... .. ........ ..... ..... .. ... ... ... ... ... . ... ... .. .. .. ... ... .. ... .. ... ... ... .. .. .. ... . . ... .... . . ... .. .. .. ... ..... .... .. .. .. .. ... .... ... .... ... ... . . . . ... ... .. .. .. .. .... .... .. .... .. .... ........... ........... ........... . . . . ....... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ............................................................................................................................................................................. ..................................................................................... .......... ........... ........... ........... .. .. .. .. .. ... ..... .. . .. . .. . ... .. ... .. ... ... .... .... ....

0.0

0.5

1.0

t=T Fourier series of the above square wave is f (t) =

1 4 X sin[(2k ¡ 1)!0 t] ; ¼ 2k ¡ 1

!0 = 2¼=T

k=1

Truncated Fourier series

N 4 X sin[(2k ¡ 1)!0 t] SN (t) = ¼ 2k ¡ 1 k=1

Di®erentiate to ¯nd its extrema 0 SN (t) =

N 8X cos[(2k ¡ 1)!0 t] T k=1

write cos[(2k ¡ 1)!0 t] =

sin(2k!0 t) ¡ sin[(2k ¡ 1)!0 t] 2 sin !0 t

All terms cancel except 0 SN (t) =

4 sin(2N !0 t) T sin !0 t 118

Smallest zero is

¼ 2N !0 h i h i (2k¡1)¼ N sin (2k¡1)¼ N sin X X 2N 2N 4 4 ¼ SN (t0 ) = = (2k¡1)¼ ¼ 2k ¡ 1 ¼ 2N t0 =

k=1

Riemann sum

Za

k=1

2N

£ ¤ N X sin 2k¡1 ¢x sin x 2 dx ¼ ¢x 2k¡1 x 2 ¢x k=1

0

where N ¢x = a. Here ¢x = ¼=N so a = ¼. 2 SN (t0 ) ¼ ¼

Z¼

sin x 2 dx = Si(¼) ¼ 1:18 x ¼

0

1:18 ¡ 1 £ 100% ¼ 9% total jump of 2

overshoot is

Pointwise Convergence of the Fourier Series The Dirichlet kernel is DN (x) =

N X

jnx

e

¡jN x

=e

2N X

ejnx = e¡jN x

n=0

n=¡N

sin[(N + 1=2)x] 1 ¡ ej(2N +1)x = ; jx 1¡e sin(x=2)

an even function DN (¡x) = DN (x). Consider a periodic function f (t) = f (t + T ) having a single jump discontinuity at the point ¡T =2 < t < T =2. De¯ne the right and left-hand limits and f (t¡ ) = lim+ f (t ¡ ²): f (t+ ) = lim+ f (t + ²) ²!0

²!0

With the Fourier coe±cients 1 cn = T

T =2 Z

f (¿ )e¡j2n¼¿ =T d¿;

¡T =2

the truncated Fourier series feN (t) =

N X

n=¡N

119

cn ej2n¼t=T

is feN (t) =

T =2 Z

T =2 Z N 1 X j2n¼(t¡¿ )=T 1 f (¿ ) e d¿ = f (¿ ) DN [2¼(t ¡ ¿ )=T ] d¿ T T n=¡N

¡T =2

Zt

=

1 f (¿ ) DN [2¼(¿ ¡ t)=T ] d¿ + T

¡T =2

|

Zt

I¡ =

¡T =2

{z

}

I¡

T =2 Z 1 f (¿ ) DN [2¼(¿ ¡ t)=T ] d¿ T t {z } | I+

1 [f (¿ ) ¡ f (t¡ )] DN [2¼(¿ ¡ t)=T ] d¿ + f (t¡ ) T

¡T =2

Zt

1 DN [2¼(¿ ¡ t)=T ] d¿ T

¡T =2

Introduce the change-of-integration variable x= such that

2¼ (N + 1=2)(¿ ¡ t) T

1 DN [2¼(¿ ¡ t)=T ] d¿ = T

sin x dx ¶ x (2N + 1)¼ sin 2N + 1 μ

¿ = ¡T =2 =) x = ¡(2N + 1)¼(1=2 + t=T ) = ¡X ¿ = t =) x = 0 T ¿ ¡t= (2N + 1)¼

1 I¡ = ¼

Z0 ½ · f ¡X

f (t¡ ) + ¼

Z0 ¡X

¾ ¸ T ¡ x + t ¡ f (t ) (2N + 1)¼

sin x μ (2N + 1) sin

x 2N + 1

Since ¡X < x < 0 , as N ! 1, ·

sin x μ (2N + 1) sin

¶ dx

¸ T f x + t ¡¡¡¡! f (t¡ ) (2N + 1)¼ N !1 120

x 2N + 1

¶ dx

so that the ¯rst integral above vanishes. The denominator becomes (2N + 1) sin

μ

x 2N + 1

¶

¡¡¡¡! x N !1

and with X ! 1, the second integral gives f (t¡ ) I¡ ¡¡¡¡! ¼ N !1

Z0

sin x f (t¡ ) dx = : x 2

¡1

The same philosophy yields f (t+ ) I+ ¡¡¡¡! ¼ N !1 so that

N X

j2n¼t=T

cn e

n=¡N

Z1

sin x f (t+ ) dx = x 2

0

f (t+ ) + f (t¡ ) ¡¡¡¡! : 2 N !1

¥

F Homework Consider the simpler case where f (t) is continuous everywhere, and therefore demonstrate that N X cn ej2n¼t=T ¡¡¡¡! f (t): N !1

n=¡N

121

ECE 370

16 January 1978

QUIZ 3

NAME

Take Home - Due Wednesday January 18 f (t) .

.. ... .. ... .. .. ... ... . . . . . . . ..... . . . . ..... . ......... . ........ . ... . .. ...... . ..... ......... ..... ........ . . . ... ..... ..... .. .. ..... ..... . . . . . . . . . . . . . . ..... ..... .... .... .... ..... ..... .... .... .... .... ..... ..... .. ..... ..... ..... ..... . . . . . . . . . . . . . . . . . . . . ..... ..... ..... . . . .. ..... ..... ..... ..... ..... ..... ..... ..... ... ......... ..... ..... ..... ..... ........ ..... .. .... ..... ........ ............ ............ ............. .. .. . . . ............................................................................................................................................................................................................................................................................................................. . . . . . .

1

:::

:::

¡T

¡T =2

T =2

0

T

t

3T =2

1.) Find a Fourier series representation of the signal f (t).

2.) Use MATLAB to graph the N th partial sum of this Fourier series. For example, if 1

a0 X f (t) = + an cos(n!0 t); 2 n=1 then let N

a0 X fN (t) = + an cos(n!0 t): 2 n=1 Pick several N to demonstrate (pictorially) the convergence of this Fourier series.

There are no sines (bn ´ 0) in this even function, and the DC or average value is a0 1 = 2 T

ZT

f (t) dt =

1 ; 2

(1)

0

as the graph shows (perhaps after a little thought). The cosine coe±cients (for n = 1; 2; : : : ) are 2 an = T

tZ 0 +T t0

2 f (t) cos(n!0 t) dt = T

T =2 Z

2 f (t) cos(n!0 t) dt = ¢ 2 T

T =2 Z f (t) cos(n!0 t) dt (2) 0

¡T =2

122

by the even symmetry of the product of f (¡t) = f (t) and the nth cosine. The last integral above requires an expression for f (t) in the range 0 · t · T =2, which is the line of slope 2=T through the origin: f (t) = 2t=T . Required now is 2 an = ¢ 2 T

T =2 Z

2t 8 cos(n!0 t) dt = 2 T T

T =2 Z t cos(n!0 t) dt:

0

(3)

0

In terms of di®erent variables, we need the integral

I=

Zd

x cos(®x) dx:

(4)

0

It is perfectly acceptable to use a table of integrals (perhaps in your calculus book, or a table such as the CRC). But here are two di®erent techniques to evaluate it. The ¯rst method is based on the observation that the parameter ® is independent from the integration variable x, so that we can write

I=

Zd

d x cos(®x) dx = d®

0

Zd

sin(®x) dx =

d 1 ¡ cos(®d) d® ®

0

d sin(®d) 1 ¡ cos(®d) = : ¡ ® ®2

(5)

The second method is integration by parts:

d(uv) = u dv + v du

=)

Zb a

The choice u=x dv = cos(®x) dx

=)

¯b Zb ¯ u dv = uv ¯ ¡ v du: a

a

du = dx 1 v = sin(®x) ®

gives

I=

Zd 0

Zd ¯d x 1 ¯ x cos(®x) dx = sin(®x)¯ ¡ sin(®x) dx ® ® x=0 0

¯d d 1 ¯ = sin(®d) + 2 cos(®x)¯ ® ® x=0 d cos(®d) ¡ 1 = sin(®d) + ; ® ®2 123

(6)

in agreement with the ¯rst result (5). The needed integral in (3) has upper limit d = T =2 and frequency (the factor multiplying the integration variable t in the argument of the cosine) ® = n!0 , whereupon ®d = n!0 T =2 = n¼ (recall !0 T = 2¼). Note that the values of the trig functions are thus sin(n¼) = 0 and cos(n¼) = (¡1)n . The Fourier coe±cients (3) are now 8 μ ¶2 1 2 < n 8 1 ¡ (¡1) 2 ¡ ; n odd n an = ¡ 2 =¡ [1 ¡ (¡1) ] = (7) ¼ n2 2 2 : T (n!0 ) (n¼) 0; n even: Together with the DC value (1), the entire trigonometric Fourier series for our even, periodic function f (t) is μ ¶2 X 1 2 1 1 f (t) = ¡ cos(2n¼t=T ): (8) 2 ¼ n2 n=1;3;5;::: The sum is reasonably clear (at least to me), but it would more commonly be written in the (probably nicer looking) completely equivalent form μ ¶2 X 1 1 1 2 cos[(2n ¡ 1)2¼t=T ]: (9) f (t) = ¡ 2 ¼ n=1 (2n ¡ 1)2 And it would be perfectly acceptable to write it in terms of the fundamental frequency as μ ¶2 X 1 1 2 1 f (t) = ¡ cos[(2n ¡ 1)!0 t]; 2 ¼ n=1 (2n ¡ 1)2 as long as it is perfectly clear that we have de¯ned !0 = 2¼=T . This series converges nicely, since jan j » n¡2 as n ! 1 (in fact, these an are 1=n2 for all odd n. In fact, Fourier series representations of known functions are used to sum certain series. In this case, note that f (T =2) = 1 and so (9) gives μ ¶2 X 1 2 1 1 cos[(2n ¡ 1)¼] 1= ¡ 2 2 ¼ (2n ¡ 1) n=1 or

1 X

1 ¼2 = : 2 (2n ¡ 1) 8 n=1 Note that the sum over both odd and even integers is 1 1 1 X X X 1 1 1 = + 2 2 n (2n ¡ 1) (2n)2 n=1 n=1 n=1 1 X

1 1 1X 1 = + 2 (2n ¡ 1) 4 n=1 n2 n=1

124

or 1 1 X 3X 1 1 = 2 4 n=1 n (2n ¡ 1)2 n=1

= and therefore

¼2 8

1 X 1 ¼2 = : n2 6 n=1

This is the Riemann zeta function 1 X 1 ³(x) = nx n=1

of argument x = 2.

125

On the Convergence and Summation of Some Series The basic form to consider is S=

1 X

an = a1 + a2 + a3 + : : :

n=1

where the indexing can just as easily start at n = 0 1 X

an = a0 + a1 + a2 + : : :

n=0

and negative indices are also commonly encountered 1 X

an = a0 +

n=¡1

If the series

1 P

n=1

1 X

an +

n=1

jan j converges, then we say that

1 X

n=1 1 P

a¡n :

an is absolutely convergent. Any series

n=1

that is absolutely convergent is convergent, since ¯1 ¯ 1 ¯X ¯ X ¯ ¯ an ¯ · jan j: ¯ ¯ ¯ n=1

n=1

A convergent series that is not absolutely convergent is called conditionally convergent.

Here is an important class of series: With p real 1 X 1 np n=1

converges i® p > 1.

This can be shown by using the integral argument (Do it!) on the next page.

126

If all of the terms of the original series are positive (true for exp(z) if z > 0), then a related integral can provide error bounds by the following reasoning. ..... .. .. .. ... ... ... ... ... .. ... .. ... ... ... .. ... ... .. ... ... ... ... ... .. ... .. ... ... ... 1 .. ... 1 ... ... ... ... .. ... .. ... ... ... .. ... ... ... n=N +1 .. ... N +1 ... ... ... ........................................ .. ...... ... .. . . ... .... ...... ... ... .. .. ...... ... ..... .. ... . ..... ... ... ..... ..... .. .. ...... .. .. . ........................................... .... ... .......... ... .. .. ... .......... . .. ... ........ .... ... ......... . ... ... ... ............................................... .. . ... ... ................ ... . .. . ................................................ ... ... .... ... ... ..................... .. ... ... .. .......................................................... .. ... ................................................................. ... ... .. ... ... ... ........................................... ... . . .. . ... ... ... .. ... .. . . . . ............................................................................................................................................................................................................................................................................................................................................................................................... ... .. ... .

g(x)

X

N +1 N +2 N +3

g(n) >

Z

g(x) dx

x

:::

... .. .. .. ... ... .. ... ... ... .. ... ... ... ... ... .. ... .. ... ... ... .. ... ... ... ... ... .. ... .. 1 ... 1 ... ... .. ... ... ... .. ... ... ... ... ... .. ... .. n=N +1 ... ... N ... .. ............................................. ... ... ... ... .. .. .. ... ..... .. ... ... ..... ..... ... .. ... ..... .. .. ... ..... ... ... ... ..... ... .. ... .. .. .. ................................................. .. . ... ........ ... ... ........ .... ... ... .. ....... . .. ... .... ... ........................................................ ... .. . . ... .................... ... ... ... ... ............................................................. ... ... ... ... ... ... .................................................................... ... .. . ... ....................................................................................................... ... . . .. . . . ... ... ... ......... . ....................... ... ... ... .... ... . . . . . . . . ........................................................................................................................................................................................................................................................................................................................................................................................... .. ... .. ..

g(x)

X

N

N +1 N +2 N +3

g(n) <

Z

g(x) dx

x

:::

Therefore, we have both a lower and an upper bound, so that the error or \tail" sum is Z1 N +1

g(x) dx <

1 X

n=N +1

127

g(n) <

Z1 N

g(x) dx:

If the series

1 P

n=1

jan j converges, then we must have lim an = 0: The converse is not true, n!1

as we demonstrate by example, since the series 1 X 1 np n=1

is divergent for p = 1=2 but 1 p ¡¡¡! 0: n n!1

Comparison Test. Let

1 P

an and

1 P

bn be two n=1 bn for all n > 1 P

n=1

exists an index N such that an · 1 P bn implies the convergence of the series n=1

implies the divergence of

1 P

n=1

series with nonnegative terms. If there

N , then the convergence of the series 1 P an , and the divergence of the series an n=1

bn .

n=1

Weierstrass M-Test for Absolute Convergence. Let

1 P

n=1

an and

1 P

bn be series.

n=1

Suppose there exists an index N such that jan j · bn for all n > N . Then a su±cient 1 1 P P an is that the series bn converges. condition for absolute convergence of the series n=1

n=1

¯ ¯ ¯ an+1 ¯ ¯ = ® exists for the Ratio (d'Alembert's) Test. Suppose that the limit lim ¯¯ n!1 an ¯ 1 P series an . n=1

(1) if ® < 1 then the series

1 P

an converges absolutely

1 P

an diverges

n=1

(2) if ® > 1 then the series

n=1

(3) there exist both absolutely convergent and divergent series for which ® = 1.

128

Acceleration of series convergence by subtracting out the asymptotic form (Kummer transform): Consider the series

1 P

f (n) where the large n form of f (n) is

n=1

f (n) ¡¡¡! fasy (n): n!1

If the original terms f (n) decay so slowly with n as to be numerically disagreeable, but the asymptotic form can be summed by some alternate method, then write 1 X

f (n) =

n=1

1 X

n=1

[f (n) ¡ fasy (n)] +

1 X

fasy (n):

n=1

The terms of the di®erence f (n) ¡ fasy (n) will approach zero faster than the original f (n), so that truncation of the di®erence sum can be made at a smaller n to achieve a given level of accuracy. Example. One series that can be summed exactly using the Poisson sum formula (on pages 150-151) is 1 X 1 ¼ = coth ¼b: 2 2 n +b b n=¡1 We choose an example series where we know the exact value of the sum simply to be able 1 P to check our result. Let's get this series into the form f (n), where f (¡n) = f (n) is an n=1

even function of the summation index, 1 X

1 X 1 1 1 = 2 + 2 2 2 2 2 n +b n +b b n=¡1 n=1

and therefore S=

1 X

n=1

n2

¼b coth ¼b ¡ 1 1 = : 2 +b 2b2

Now let's compare the brute-force summation of S as it stands, with a simple Kummer transform. Having the exact value of S allows us to gauge our accuracy. The asymptotic form of the summand f (n) is fasy (n) = n¡2 so that the Kummer transform is ¸ X 1 · 1 X 1 1 1 S= ¡ + : n2 + b 2 n2 n2 n=1 n=1 The asymptotic or \tail" sum is the Riemann zeta function of argument 2 (a well-known, actually famous sum) that we encountered back on page 124 1 X 1 ¼2 ³(2) = = : 2 n 6 n=1

129

The di®erence terms 1 n2 ¡ (n2 + b2 ) ¡b2 1 ¡ = = = O(n¡4 ) n2 + b2 n2 n2 (n2 + b2 ) n2 (n2 + b2 ) decay much faster than the original terms of O(n¡2 ). The Kummer transform of the original sum S is now 1 X 1 ¼2 S = ¡b2 + : 2 (n2 + b2 ) n 6 n=1 To do the comparison, label the truncated brute-force sum S0 (N; b) =

N X

n=1

n2

1 + b2

and the truncated Kummer-accelerated sum SK (N; b) = ¡b

2

N X

1 ¼2 + : n2 (n2 + b2 ) 6 n=1

With the exact value

¼b coth ¼b ¡ 1 2b2 let's compare the numerical performance of S(b) =

S0 (N; b) S(b)

with

SK (N; b) S(b)

as a function of the truncation index N for one or more values of the parameter b.

N

S0 (N; 10) S(10)

SK (N; 10) S(10)

101 102 103 104

0:499726021056655 0:934787298579118 0:993428003029169 0:999342482848786

1:125492496191760 1:000214644029801 1:000000218841745 1:000000000219156

Exercise. Using ³(4) =

1 P

n¡4 = ¼ 4 =90, perform a second Kummer transform on the

n=1

series in the example and compare its convergence to the data above.

130

Homework Problem: Convergence of Fourier Series. Consider the three periodic functions x(t), y(t), and z(t) as graphed below. Find appropriate Fourier series expansions for each of the three functions. Study and discuss the convergence of each series, including graphical display. For example, say the odd function y(t) is most conveniently represented by the Fourier sine series 1 X y(t) = bn sin(2n¼t=T ) n=1

where the coe±cients bn are known. Truncation of this Fourier series yields the approximation N X bn sin(2n¼t=T ): yN (t) = n=1

Compare these truncated series yN (t) with the original function y(t) for several (you decide which range of values has signi¯cance to your convergence study) values of the ¯nal or truncation index N . Certainly, the rate of decay of jbn j as n ! 1 is the important feature. Note: How are x(t), y(t), and z(t) related? How then are their Fourier series related? +1

x(t)

............. ....... ..... ..... .... ......... ..... ......... ..... ..... . ..... ..... ..... .... ..... . . . . . . . ..... ..... ..... .. .... ..... ..... ..... ..... . . . . . . . . ..... . . ..... ..... ..... ..... ..... ..... ..... ..... ..... .... ..... ..... ..... ..... ..... ........ . . . . . . . . ... ... .. ... ...................................................................................................................................................................................................................

¡T =2

T =2

0

t

T

+1

y(t)

..................... ...................................................... ......................................... ... .. ... .. ... .. .. ... ... . ... ... ... .. . .. ... ... . . .. ... ... .... ... ... ... .. . ...................................................................................................................................................................................................... ... ... ... ... ... ... .. .. .. .. .. .. ... ... ... ... .. .. .. .. ... .. ... .. . . . . ................................................... ................................................... ......... (+1) N ...... ...

z(t)

¡1

N

...... ......... ...... ............... ......... ...... ...... ...... ...... ......... ......... ...... ...... . . . ............................................................................................................................................................................................................ ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ......... ... ...

H

H (¡1)

131

t

t

Fourier Series of Full-Wave Recti¯ed Sine Wave sin( 12 !0 t) = sin(¼t=T )

................... .... ... ... ... . . ... . . ... . ... .. ... ... . . ... . ... . . .. .. . .. .. . . .. .. . .. . .. .. .. .. . .. .. . . .. ... . ..... . .... ... . . . ............................................................................................................................................................................................................................................ . . ...... . ...... . ... ... ... .. .. .. . . .. .. . .. .. . . .. .. .. .. .. . .. .. . . ... ... . . . ... ... . . . ... . . . ... . . . .... . . . .... . . . . ...............

¡T

0

t

T

f (t) = j sin( 12 !0 t)j = j sin(¼t=T )j

................... .................... ... .... ... ... ... .... ... ... . . . . . ... ... ... ... ... ... . . . . . ... . ... ... ... . . . . . ... ... ... ... . . . . .. .. .. . . . . . . . . ... .. .. . .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. . .. ... . . . . . . .. .. .. .. .. .. .. .. ...... ....... ... . ................................................................................................................................................................................................................................................ .. .. ..

¡T

0

t

T

The even function f (t) = f (¡t) is expanded in a Fourier series 1

a0 X f (t) = + an cos(n!0 t) 2 n=1

(1)

where its fundamental frequency is !0 = 2¼=T , in terms of its period T . Note that the period of the sine wave prior to recti¯cation is 2T . Equivalently, the frequency of the un-recti¯ed sine wave is half the frequency of the recti¯ed signal f (t). The DC or average value of f (t) is a0 1 = 2 T

ZT

1 f (t) dt = T

0

ZT 0

¯ ³! ´ ³ ! ´¯0 2 ¯ 0 0 sin t dt = cos t ¯ 2 !0 T 2 ¯

t=T

=

2 2 [1 ¡ cos ¼] = : (2) 2¼ ¼

For n = 1; 2; : : : 2 an = T

tZ 0 +T t0

2 f (t) cos(n!0 t) dt = ¢ 2 T

T =2 Z ³! ´ 0 sin t cos(n!0 t) dt 2 0

132

(3)

since

¯ ³ ! ´¯ ³! ´ ¯ ¯ 0 0 t ¯ = sin t f (t) = ¯sin 2 2 Use of the trig identity

for 0 · t · T =2:

(4)

2 sin A cos B = sin(A ¡ B) + sin(A + B) yields 2 an = T

T =2½ ¾ Z £ ¤ £ ¤ 1 1 sin (n + 2 )!0 t ¡ sin (n ¡ 2 )!0 t dt:

(5)

0

Let's economize the algebra and do both forms at once by considering ¯0 T =2 Z ¯ £ ¤ £ ¤ 2 2 1 ¯ 1 1 sin (n § 2 )!0 t dt = cos (n § 2 )!0 t ¯ 1 ¯ T T (n § 2 )!0 t=T =2 0 ¤ ¤ £ £ 1 1 ¡ cos (n § 2 )!0 T =2 1 ¡ cos (n § 12 )¼ 1 = = = : 1 1 (n § 2 )!0 T =2 (n § 2 )¼ (n § 12 )¼ (6) Insertion of (6) into (5) gives · ¸ 1 1 1 1 (n ¡ 12 ) ¡ (n + 12 ) 1 an = ¡ = =¡ : 1 1 1 1 ¼ n+ 2 ¼ (n + 2 )(n ¡ 2 ) n¡ 2 ¼(n2 ¡ 14 )

(7)

Together with the DC term (2), the full Fourier series for f (t) is now 1 2 1X 1 f (t) = ¡ 2 ¼ ¼ n=1 n ¡

1 4

cos(2n¼t=T );

(8)

and its truncated version (N th partial sum) is fN (t) =

N 1X 1 2 ¡ ¼ ¼ n=1 n2 ¡

1 4

cos(2n¼t=T ):

(9)

solid curve: N = 3 dashed curve: N = 20

......... .............. ...................... ........ ........ ........ ........ ....... ....... . . . . . ..... . . . . . . ..... . . . . . ..... .. . . ....... ... . ...... . ...... .... ..... ....... . . . ...... ..... ...... .... . . ..... ..... .. ...... . . . . . .... . .... .... .... ... ... ... ... ... .... ..... .... ..... ..... . . . . . . . . . ..... .. ..... ..... ..... ..... ..... ...... ..... ..... ..... ...... ..... ..... ........ . . . . ..... . . . . . . ..... . ...... ..... ..... ............ ................... .............. . .. .. . ......................................................................................................................................................................................................................................................................................... . .

0

1 133

t=T

Parseval's Theorem for the Fourier Series Consider a periodic signal f (t) = f (t + T ) and its Fourier series representation f (t) =

1 X

cn ejn!0 t

(!0 = 2¼=T ):

n=¡1

The average power (per period) \in" or \of" f (t) is

Pav

1 = T

tZ 0 +T t0

1 = T

tZ 0 +T

jf (t)j2 dt 1 f (t)f ¤ (t) dt = T

t0

=

1 X

n=¡1

=

1 X

n=¡1

tZ 0 +T

cn

m=¡1

c¤m

1 T

tZ 0 +T

jn!0 t

cn e

n=¡1

t0 1 X

1 X

j(n¡m)!0 t

e

1 X

c¤m e¡jm!0 t dt

m=¡1

dt =

1 X

n=¡1

t0

cn

1 X

m=¡1

c¤m ±mn

=

1 X

cn c¤n

n=¡1

jcn j2

F Homework: Find the comparable expression in terms of the trigonometric Fourier series coe±cients.

134

Homework Problem: Power Supply Performance. sin( 12 !0 t) = sin(¼t=T )

................... .... ... ... ... . . . ... ... ... ... . . ... ... . . ... ... . . .. .. . . .. .. . .. . .. . .. .. . . .. .. . .. . ... . .. . .. . .... .... ..... . . . . . . . . .......................................................................................................................................................................................................................................... . .... . .... .. . .. ... . . .. . ... . .. . .. .. . . .. .. . .. .. .. .. .. . .. . .. . ... ... . . ... ... . . . .. ... . . ... . . . ... . . . .... . . . ....... ....... ....

¡T

0

t

T

x(t) = j sin( 12 !0 t)j = j sin(¼t=T )j

................ ................. .... ... .... .... ... ... ... ... ... ... .. . . . . . . ... . ... . . ... . . . . ... . ... . . ... . . ... ... . ... .. . ... .. . . .. .. . . . . .. . . .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. . . . . . . . . . . .. . .. .. ... .. .. .. .. . . .. . .. ...... ...... .... . ............................................................................................................................................................................................................................................... .. .. ..

¡T

0

x(t) =

1 2 1X 1 ¡ ¼ ¼ n=1 n2 ¡

t

T

1 4

cos(2n¼t=T )

Consider a real power supply, consisting of a full-wave recti¯er followed by a simple lowpass ¯lter. Find an expression for, and graph, the time-domain output y(t) of the power supply for several values of the important parameter !b =!0 . De¯ne the percent ripple as the ratio of average power in the unwanted harmonics to the desired DC power. Graph percent ripple versus !b =!0 . Recall Parseval's theorem for Fourier series.

+

. . . ...... ...... ...... ................................................ ..... ..... ..... ..... ..... ................................................................................ .... .... .... ... ... ... .. . .........................

²

R

x(t) ¡

y(t)

C

........................ .... .. .. ... .. . . . . ...................................................................................................................................................................

²

135

+

¡

Solution. The transfer function of the low-pass ¯lter is 1 1 1 j!C H(!) = = = 1 1 + j!RC 1 + j!=!b R+ j!C

with !b ,

1 : RC

In polar form H(!) = jH(!)jej£(!) the magnitude of H(!) is 1 1 + (!=!b )2

jH(!)j = p and the phase of H(!) is

£(!) = ¡ tan¡1 (!=!b ): Note that the ¯lter impulse response is necessarily real, since the ¯lter is an actual, physical electric circuit. Also note that H(0) = 1 is real; that is, £(0) = 0. With the ¯lter input written as the particular Fourier series x(t) =

1 X

an cos(n!0 t);

n=0

the response of the ¯lter is y(t) =

1 X

n=0

an jH(n!0 )j cos[n!0 t + £(n!0 )]:

Note that the DC term is now called a0 , instead of a0 =2. Recall that !0 , 2¼=T is the fundamental frequency of the recti¯ed sine, which is twice the frequency of the pure sine wave that is the initial input to the full-wave recti¯er. Speci¯cally, the above expression is y(t) =

1 1X 2 1 ¡ ¼ ¼ n=1 n2 ¡

1 4

£ ¤ 1 p cos n!0 t ¡ tan¡1 (n!0 =!b ) : 1 + (n!0 =!b )2

Note that the single (dimensionless!) numerical parameter required to characterize the complete power supply output is !0 =!b . Time t is always multiplied by !0 = 2¼=T , so that in terms of normalized time t=T , we have μ ¶ 1 t 1 2 1X y = ¡ T ¼ ¼ n=1 n2 ¡

1 4

· ¸ 1 t ¡1 p cos 2n¼ ¡ tan (n!0 =!b ) : T 1 + (n!0 =!b )2

The form of the Fourier series for y(t) is, with An = an jH(n!0 )j and Án = £(n!0 ), y(t) =

1 X

An cos(n!0 t + Án )

n=0

136

(1)

and the total average power in the output y(t) is

Ptot

1 = T

tZ 0 +T t0

1 = T

jy(t)j2 dt

tZ 0 +T " 1 X

n=0

t0 1 1 X X

=

An cos(n!0 t + Án )

#"

Am cos(m!0 t + Ám )

#¤

dt

m=0

An A¤m

n=0 m=0

1 X

1 T

tZ 0 +T

cos(n!0 t + Án ) cos(m!0 t + Ám ) dt:

t0

The DC terms (n = 0 and/or m = 0) are a little di®erent, but I included them in the series (starting at n = 0 instead of starting at n = 1) for notational convenience. Recall or note that Á0 ´ 0. The inner products of interest are: 1 T

tZ 0 +T t0

1 T

tZ 0 +T t0

1 T

tZ 0 +T

1 ¢ 1 dt = 1

(n = m = 0)

1 ¢ cos(m!0 t + Ám ) dt = 0

cos(n!0 t + Án ) cos(m!0 t + Ám ) dt =

(n = 0; m = 1; 2; : : : )

1 ±nm 2

(n; m = 1; 2; : : : ):

t0

Therefore, the total average power in the signal y(t) is Ptot = jA0 j2 +

1 1X jAn j2 ; 2 n=1

which is the form of Parseval's theorem that applies for our particular representation of y(t). The DC (n = 0) term is our desired power supply output, while all of the higher harmonics (n = 1; 2; : : : ) give rise to the unwanted \ripple." Therefore, one de¯nition of percent ripple is unwanted power in the ripples £ 100% r, total power output or 1 P 1 jAn j2 2 n=1 r= £ 100%: (2) 1 P 1 2 2 jA0 j + 2 jAn j n=1

137

However, the problem statement evidently de¯ned percent ripple as 1 2

r=

1 P

n=1

jAn j2

jA0 j2

£ 100%:

These two r expressions will be nearly identical when the majority of the output power is in the DC component, but for a very poor power supply I suppose some marketing/sales people would prefer to use (2). The second form might be greater than 100%! 1.2 !i =!b = 0:1....................................................

1.0

..... . .... .... .... .... .... .... . . .... . .... . . . . . . .......... ... ...... . . . . . . . ... .. .......... ...... ... ...... . . . . . ... ... . ... .. .... ... ... ... ... ..... .... ... .. ... ... ... ... .... ... ... ... ... ... ... ... . . ... . . . ........... ......... . . . . . . . . . . . . . . . . . ......... .. ........ .. ....... ............. . . . . . . . . . . . . .... ... ......... ... ... .. ....... ......... .. ....... .. ........... ....... .. ....... .............. ......... ....... .... ....... .. ....... ......... .. ......... .. .. ... . .. ... ... ... ... ... ... .. .. ... ... ... ... ... . . .. . ... . . . . ... ... .. . . . .. . ... . . . . ... ... .. ... ... ... .... ... ... .... ... ... ... ... ..... .... ..... ... .... ... ... .. ... ... ............ . ... ... .... ..... ... ... . . ... ... .. ... ... ... .. ... ... .. .. ... . . . . . ... ... .. .. ... ... .. ... ... ... .. .. ... ... . ... . . . .. .. .. .. .. ... ... .. .. ... ... .. .. .. .... .. ... ... ... ... .... ....... .......

0.8

1

10

0.6 y(t) 0.4 0.2 0.0 ¡0:2

input frequency of unrecti¯ed sine is !i = !0 =2 ¡0:75

¡0:50

¡0:25

0.00

0.25

0.50

0.75

t=T 20 18 16 14 12 r (%) 10 8 6 4 2 0

..... ... .. ... .. .. .. ... ... ... ... ... ... ... ... .. .. .. .. ... ... ... ... .. .. ... ... ... ... ... ... .... .... ..... ..... ....... ........ ........ ........... .............. ................... .............................. .................................................... .............................................................................................................................................

0

1

2

3

4

!i =!b = freq of unrecti¯ed sine/¯lter break 138

5

Periodic Excitation of a Simple GLC Electric Circuit

v(t) ²

²

ig (t)

..................................................................................................................................................................................................................... ... .. .. .. .. .. .. ..... ... ... ... ... .. .. .. .. .. ...... ......... ... . .. .. . . . . . . . . . . . . .. ......... ............. ................ .......... ...... .. ......... . . . . . . . . . . . . . . . . ......... .. . ..... ........... ..... . .......... ......... . ... .. ............... .. ... ... .... ......... . . . . . . . . . . . . . . . . . . . . . . . . . ...... . ...... . ........ ... ............ .... .. ......... ... ... ............... ... .. .. ...... ... .. .. ... .. ... ... ... ... .. .. .. .. . . . .................................................................................................................................................................................................................... ... . .................... .......... ...

G

C

²

L

²

If the source signal ig (t) = ig (t + T ) is a periodic signal, then the steady-state solution for all of the linear signals (common voltage v(t), all three branch currents) will also be periodic with the same fundamental frequency !0 = 2¼=T of the excitation. If we know the waveform ig (t), then we can ¯nd its Fourier series representation ig (t) =

1 X

cn ejn!0 t :

(1)

n=¡1

Here is a case where it is more convenient and natural to use the complex exponential form of the Fourier series: The trigonometric form would su±ce in principle, but carrying around two sets of coe±cients an and bn is more awkward. Plus di®erentiating exp(jn!0 t) is much cleaner than di®erentiating cos(n!0 t) and sin(n!0 t). Kirchho®'s current law supplies the equation-of-motion for the unknown voltage signal v(t) in our circuit Zt 1 dv(t) + v(¿ ) d¿ = ig (t): (2) Gv(t) + C dt L ¡1

This integrodi®erential equation is converted to a pure di®erential equation by di®erentiating once to eliminate the integral operator in the inductive current C

d2 v(t) dv(t) 1 +G + v(t) = i0g (t): 2 dt dt L

(3)

The order of the terms was rearranged to put the di®erential equation in standard form, and the prime on the generator current denotes di®erentiation with respect to its argument, which is our independent variable t. The periodic voltage signal is written as a Fourier series 1 X v(t) = °n ejn!0 t (4) n=¡1

139

where the Fourier coe±cients °n are unknown, so far. Insertion of both series (1) and (4) into the di®erential equation (3) gives · ¸ 1 1 X X 1 jn!0 t 2 °n (jn!0 ) C + (jn!0 )G + e = (jn!0 )cn ejn!0 t : (5) L n=¡1 n=¡1 Since the above two Fourier series in (5) are equal for all t and they are written in terms of the same Fourier basis functions exp(jn!0 t), the corresponding coe±cients are equal ¸ · 1 2 = (jn!0 )cn : °n (jn!0 ) C + (jn!0 )G + (6) L Therefore the °n are given explicitly in terms of the known cn jn!0 jn!0 L °n = cn ; cn = 1 1 ¡ (n!0 )2 LC + jn!0 LG (jn!0 )2 C + (jn!0 )G + L and our particular solution (4) for the voltage is v(t) =

1 X

jn!0 L cn ejn!0 t : 2 1 ¡ (n!0 ) LC + jn!0 LG n=¡1

(7)

(8)

Note that the (SI or mksC) units of the numerator are those of !0 L, which are (H/s) or (−). All three terms in the denominator are dimensionless (check for yourself), and the Fourier coe±cients cn de¯ned in (1) are in (A). The units of the °n are therefore (V), as required. Problems 1. Find an expression for the average power dissipated as heat in this circuit. 2. Give a physical explanation for the absence (zero value) of the n = 0 term in the voltage. ¡1 3. If the lossy element is removed from the circuit (G p = 0 or G = R = 1), what is the physical signi¯cance of a harmonic having n!0 = 1= LC ?

4. Write down an approximation for the higher-order terms in the series for v(t), that is as n ! +1. (The terms with negative indices behave similarly.) Interpret this high frequency approximation in terms of the circuit. 5. If f (t) is a real function of time, show that the coe±cients cn in its complex exponential Fourier series 1 X cn ejn!0 t f (t) = n=¡1

must satisfy the requirement cn =

c¤¡n .

One way to see this is to examine f ¤ (t) = f (t).

6. If ig (t) is a real-valued source current, show that the expression (8) is also real. 7. Graph the signal v(t) when the current source is a square wave, say of amplitude §I0 . Show the e®ect of varying the circuit element values. Use normalized parameters, as much as possible, to simplify the results. 140

Simple Low-Pass and Hi-Pass Filter Response to a Periodic Input Problem: Find expressions for and graph the signals ya (t) and yb (t) for several values of the important (nondimensional) parameters that appear naturally. x(t)

A

.......... .......... ................. .......... ....... .... ....... .... ....... .... ....... .... ....... ....... ....... ....... ... ... ... ... ....... ....... ....... ....... . . . . . . . . . . . . . . . . . ... . . . . . . . . . . ... ... ... ..... ..... ..... ....... ... ... ... ....... ....... ....... ... . . . . . . ........ . . . . . . . . . .... . . . ... ... ... ... ... ... . . . ... . . . . . . ....... . . . . . . . . . ... ... ... ... . . . . . ....... . ... . . . . . . . . . . . . . . . . ... ... ... ....... ... ... ... . . . . . . . . ... . . . . . . . . . . . . . . . . ... ......... ... ......... ... ......... .. ........... .. .. .. ............................................................................................................................................................................................................................................................................................................................................................................................................

¡T

+

0

...... ...... ...... ................................................ ..... ..... ..... ..... ..... ............................................................................... ..... ..... ..... .... .... .. ... ........................

²

R

x(t) ¡

C

+

... .. ............................................................... .... .

... .. ...................................................................................... ..... .. .. ... .. . .. ....... .. ....... ............. ....... . ........ ............. ....... . . . ...... ............. ... .. . ......................................................................................................................................................................

²

C

y (t)

x(t)

¡

¡

a ........................ .. .. ... .. .. .. ......................................................................................................................................................................

²

2T

T

+

t

+

yb (t)

R

²

¡

Note that the excitation x(t) has both even and odd parts, but the even part is simply A=2. The graph of x e(t) = x(t) ¡A=2 is clearly an antisymmetric (odd) function, so that its Fourier series is most logically written in the trigonometric form with only sine functions. x e(t) = x(t) ¡ A=2

A=2

......... ........ ........ .......... .......... ....... ..... ....... .... ....... .... ....... .... ....... ....... ....... ....... ... ... ... ... ....... ....... ....... ....... . . . . . . . . . . . . . . . . . ... . . . . . . . . . . . .. .. .. ....... ....... ....... ....... ..................................................................................................................................................................................................................................................................................................................................................................................................................................... . . . . . . . . . . . . .... . . . ... . . ... . . . . . . ... ... ....... ..... ..... ..... .. .. ....... ....... ....... ....... ... ... .... .... ....... ....... ....... ....... ................ .. ............. ... ............. .. ............. .... .... ... ...........

¡T

0

¡A=2

x e(t) = ¡e x(¡t) x e(t) = T =2 Z

2 bn = T

¡T =2

4A = T

1 X

x e(t) = A bn sin(n!0 t)

2T

T

μ

1 t ¡ T 2

¶

(0 < t < T =2)

(!0 , 2¼=T )

n=1

4 x e(t) sin(n!0 t) dt = T

T =2μ Z

t 1 ¡ T 2

¶

sin(n!0 t) dt

0

141

T =2 Z x e(t) sin(n!0 t) dt 0

t

T =2 ¯0 Z cos(n!0 t) ¯¯ 1 ¡ cos(n!0 T =2) 1 ¡ cos(n¼) 1 ¡ (¡1)n sin(n!0 t) dt = = = = n!0 ¯t=T =2 n!0 n!0 n!0 0 8 < 2 n odd n!0 = (but this form is not as useful as it might appear) : 0 n even

Z

Z u dv = uv¡ v du

u=t

du = dt

dv = sin(n!0 t) dt

v=¡

cos(n!0 t) n!0

T =2 T =2 ¯0 Z Z t cos(n!0 t) ¯¯ 1 t sin(n!0 t) dt = + cos(n!0 t) dt ¯ n!0 n!0 t=T =2 0

0

¯T =2 sin(n!0 t) ¯¯ T sin(n¼) T =¡ cos(n!0 T =2) + = ¡ cos(n¼) + 2n!0 (n!0 )2 ¯t=0 2n!0 (n!0 )2 ¡T (¡1)n = 2n!0

Now go back and assemble the required Fourier coe±cient · ¸ 4A ¡(¡1)n 1 ¡ (¡1)n ¡A ¡4A = : bn = ¡ = T 2n!0 2n!0 2n!0 T n¼ The Fourier series representation of the input signal is therefore 1 A A X sin(n!0 t) : x(t) = ¡ 2 ¼ n=1 n

The transfer functions of our two ¯lters, from a simple frequency domain voltage divider, are 1 1 j!C Ha (!) = = = jHa (!)jej£a (!) 1 1 + j!RC R+ j!C Hb (!) =

R 1 R+ j!C

=

j!RC = jHb (!)jej£b (!) : 1 + j!RC

Let's not write out the polar representations explicitly: MATLAB will compute those for us. That is, assume the magnitude function jHa (!)j and £a (!) are known (they are!), and obviously the same for the b circuit. Note that Ha (0) = 1 and Hb (0) = 0. 142

The Fourier series for the two output signals are therefore 1 £ ¤ A A X jHa (n!0 )j sin n!0 t + £a (n!0 ) ya (t) = ¡ 2 ¼ n=1 n 1 ¤ £ A X jHb (n!0 )j yb (t) = ¡ sin n!0 t + £b (n!0 ) ¼ n=1 n

Let's use normalized time t=T and normalize by the amplitude A of the input to write 1 £ ¤ 1 1 X jHa (n!0 )j ya (t=T ) = ¡ sin 2n¼t=T + £a (n!0 ) A 2 ¼ n=1 n 1 £ ¤ yb (t=T ) 1 X jHb (n!0 )j =¡ sin 2n¼t=T + £b (n!0 ) A ¼ n=1 n

We need Ha (n!0 ) =

1 !b 1 = ¡¡¡! 1 + jn!0 RC 1 + jn!0 =!b n!1 jn!0

Hb (n!0 ) =

jn!0 RC jn!0 =!b = ¡¡¡! 1 1 + jn!0 RC 1 + jn!0 =!b n!1

where !b ,

1 RC

is the break frequency (or cut-o® frequency or half-power frequency) of (both of) the ¯lters. Therefore, the important nondimensional parameter of our problem is !0 =!b = !0 RC. Recall that !0 = 2¼=T is the fundamental frequency of the periodic excitation. As n ! 1, the terms in the series for the low-pass ¯lter output ya (t) vary like n¡2 while the higher-order terms in the high-pass ¯lter output yb (t) vary like n¡1 . Therefore, the numerical convergence of the signal ya (t) is much faster. That is, truncation of the ya (t) series at some ¯nite value of n will give a much better approximation for the in¯nite series than will the same truncation of the series for yb (t). Note that the outputs ya (t) and yb (t) are intimately related since Ha (!) + Hb (!) = 1

=)

ya (t) + yb (t) = x(t):

Therefore, we can numerically evaluate the series for ya (t) (because its series converges faster), and then avoid computing the poorly convergent series for yb (t) simply by using the known relationship yb (t) = x(t) ¡ ya (t):

143

!0 =!b = 0:1 y (t)

....... ........... a ............ .. . .......... .. ................. ..... .................. ..... . . . . . . . . . . . . . . ... ........ .. ........ ............. ............ ... ... .............. ............... .... ... ................. ................ . . . . . . . . . . . . . . ..... . . . . ... ... . . . . . . . ................. . .... . . . ..... ......... . . . . . . . ........ .... ..... . . . . . . . ...... . ..... . . ...... . . . . ........... .... . ...... . . . . . .... .... ..... ................. . . . . . .... . . ..... ......... . . . . . . . . .... ..... ... . . . . . . . . . . . . .... ......... ...... . . . . . . ........ . ...... . .. ... . . . . . ....... .. ... .. ... .................. . .. ... . . . . .. ... ......... . . . . . .. ... . ....... .. ... . . . . . . . ......... . .. ... .. ... . . . . . ... . . . . . . .. .... . . . ... ..... . . ...... .... .. ... .. .. ................. ............... . . . . . . . . .. ... ...... . ................. . . . . . . . . . . . . ... .... . . . .. ... ............ .............. .. .... .. ... ............ ............. .. ............................. .. ............................ .. ...... .. ... ..... ............................................................................................................................................................................................................................................................................................................................................................................

t=T

y (t)

..... ...... b .... .. .... ... ..... .. . ..... . . . . . . .. . .. ... ..... .. .... .. .... ..... .. .. ..... .... . . . . . . . . . .. . . . . ..... ... .. ... . . . . .. .. ..... .. . .. . . . . . .. . .. . . . . . .. . .. .. . . . .. .. . . . . . .. .. .. .. ..... .. . . . . . .. . .. . .. . . .. . .. . . . . .. . . .. . . . . .. .. .. . . . . .. .. ..... .. . . . .. ... ... .. . . .. . . ..... . . .. . .. .. .... . . . .. ..... .. ... . . . . . . . .. .. .... ..... .. .. ..... .... .. ...... .. ........ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ............................................................................................................................................................................................................................................................................................................................................................................................................................. . . . .... .... .... .... .. ... .. ... ... .. ... .. .. .. .. .. ... .. ... .. .. ... .. ... ... .. ... .. . . .. .... ... .... ... .. .. .. .. ... . ... . .. .... .. .... ... ... ... ... .... .... ..... ..... .... .... ..... ..... .... .... ..... ..... ..... ..... ..... ..... .... .... ..... ..... . . .. ....... ....... .. ... ... ... .. .. ... ... .. ..

144

t=T

!0 =!b = 1 y (t)

..... ...... a .... .. . ... ... ..... .. . ..... . . . . . . .. .. ... ... . .... .... ..... ..... ........... ........... .... ..... ........ ........ ........ ....... . . . . . . . . . . . . . . . . . . . . . . . . . ... . .... .. .... ........ ........ .. ... ..... ........ ........ .. ... .... .. ... ........ ........ ..... ....... ........ . ..... . . . . .. .... . . . . . . . . . . . . . . . ... ... ... .. ... . . . . . . . .. .... . . . . . . . . . . ... . ... . . . . ... ...... . . . .. . .. . . . . ... ... . . . . ... . . . . . .. . . ... .. . .. . . . . . ... . . . . . . . . . ... .. .. ... .... . . . . . . . . . . . . . .... .. ... .... . . . . .. . . . . . . . . . . .... .... . ... . . .. . . . . . . . . . ..... . . . ..... . . .... . . . . . . . . . .. ..... . . ..... . . . ... ... . . . . . ...... ...... . . .. . . . . . . .. .... . . . ........ ........ . . . . . . . . . . . . .. . . . .................. ........... ........... . . . . . . . .................................................... .......... .. .. . . . . . ... . . .. .. . . . . . . . ..... . . . . .. . . . . .. . . . . . . . . .. . . . .. ..... .. .... .. ..... ..... .. ... ....... .... .. ........ . ... ... ...........................................................................................................................................................................................................................................................................................................................................................................

t=T

y (t)

..... ...... b .... .. .... ... ..... .. . ..... . . . . . . .. . .. ... ..... .. .... .. .... ..... .. .. ..... .... . . . . . . . . . .. . . . . ..... ... .. ... . . . . .. .. ..... .. . .. . . . . . .. . .. . . . . . .. . .. .. . . . .. .. . . . . . .. .. .. .. ..... .. . . . . . .. . .. . .. . . .. . .. . . . . .. . . .. . . . . .. .. .. . . . . .. .. ..... .. . . . .. .. ... .. . . .. . .... . . . .. . ..... . .. . . . ..... .......................................................................... . . . . . . . . . . . . . . .................................................................................... .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... ... ........... . ... .......................... ... ... ..... .... .................. ... ........ ... ........ ............ . .......... . . . . . . . . . . . . . . . . . . ................................................................................................................................................................................................................................................................................................................................................................................ .. . .... ....... . ... ... ...... ..... . ..... ..... . . ... .... . ..... .. . . . . .. . . . . ... .... .... .... ... .. .. .... ... ... ... ... .... ... . .. . . . . . ... .... ... ... .. .. ... ... ... ... ... ... . . .. . . . . ... .... ... ... .. .. ... ... ... ... ... ... . . .. . . . ... ..... .... ..... .. ... .. ... ... . ... . .. ... .. ... ... ... ... ... .. .... .. .... ... .. ... .. .... .... .... .... .... ....

145

t=T

!0 =!b = 10 y (t)

..... ...... a .... .. . ... ... ..... .. . ..... . . . . . . .. .. ... ... .. .... .... .. ..... ..... .. .. .... ..... . . . . . . . .. . . . . . .... ... . .. .. . . . . .. .. . . . . .. .. .. ..... .. . .. . . . . . .. . .. . . . . . . . ............................. ............................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ................................. .................................. .. ......... .......... . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ....................... . . . . . . . . . . . . . . . . . . . ......................................... . . . ................................................................................................................................................................................. .. .. . .. .. ..... . . . .. . . .... .. .. ..... .. .. .... ..... .. .. . . . .. . . . .. . .. ... . . .. .. . . ..... . . .. . . . . .... . . . . . . . .. . . . .. ..... .... .. ... ..... ..... .. .... . ....... .. .. ...... .. ..... .............................................................................................................................................................................................................................................................................................................................................................................

t=T

y (t)

..... ...... b .... .. .... ... ..... .. . ..... . . . . . . .. . .. ... ..... .. .... .. .... ..... .. .. ..... .... . . . . . . . . . .. . . . . ..... ... .. ... . . . . .. .. ..... .. . .. . . . . . .. . .. . . . . . .. . .. .. . . . .. .. . . . . . .. .. .. .. ..... .. . . . . . . . . . . . . . . . . . .. .... ... ..... ... . . . . . . . . . . . . . . . . . . . . . ... . . ... ... . . . . . . . . . . . . . . . . . . . . . . . . .. ... ... .. .......... .......... ..... ... ... .......... .......... .... .......... ... ... .......... ..... .......... ......... . . ... . . ... . . . . . . . .......... . .... ... . . ... . ... ... . . . . . . . . ... . . ..... . . . . ... . ... . . . . . .. .... . .... . . . . ... . . .. . . . . ... ..... . ... . . . ... . . . . . . . . . . . ... ... ... ......... .... ... .... ......... .. ........ ... ........ ........ ... ... ........ . . . . . . . . . . . . . . . . . . ................................................................................................................................................................................................................................................................................................................................................................................. . .... .... ........ ........ ... ... ....... . . . . ... ... . . ... . . . . ... ... . . ... . . . . . ... ... . .. . . . . . . . ... ... ... . . . . . . ... . . . ....... ..... ..... . . .... ...... .. .. ...... ...... . . . . . . . . ... . . . . . . ... ....... .. ....... ...... ...... ... .... ...... ...... .. .. ....... ....... . . . . . . . ... . . . . ... ....... .. ........... . .... .......... ... ..... ..... .....

146

t=T

The Vibrating String This is one of the most celebrated problems of mathematical physics. And since the early 1950's, the unchallenged lead instrument in the sound track of Western culture has been constructed around a set of six such steel strings. Let's see what our Fourier analysis can tell us about this simple and important wave equation. y(x; t0 ) .. ... ................... ..... .... .. .... ... ... .... . ... . .. . .. . ... .. . . . . ... . ... . .. . .. ... . . . ... . ... ... . .. ... . . . . ... . ... ........ . . . ... . ... .... ...... . . . ... .. . ... ... . . .. ........ ... . ... . .. .. ... ... ......................................................................................................................................................................................................................................................................................................................................................... . ... ... . ... . ... ... . . . . ... . . ... ... ..... ... ... ..... .... ...... ... ........... ................... ... ..... . . . .. ... .... ... ... ... ... ... . . .. ... ... ... .. ... .. . ... . ..... ....... ...

x

a

0

Fig. 1. Snapshot of string displacement at time t = t0 .

Fig. 1 shows the pro¯le (at some particular time t0 ) of the vertical de°ection of a string, of total length a, about its equilibrium position y = 0. The string is ¯xed at each end; hence y(0; t) = y(a; t) = 0

(1)

are the boundary conditions imposed on our unknown function y(x; t), for all t. ............. ............... .......... . ........... . . . . . 0 . . . . .. ........... ............................................................................

μ

........................................................ .. ...... ...... ...... . . . . . . ...... . ........ .......... ............

μ

T

.. ......... ........ .. ......... .. . . . . . . . .. ....... ....... .. ........ . .. . . . .. . . . . . .. . ... . . . . . .. .. . . . . . .. .. . . . . . .. ... ...... . . .. . . . .. . . . . .. . .. . . . . . .. .. . . . . . . .. .... . . . . .. .... ....... .. . . . . .. . . . .. . . ........ .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .

x

x + ¢x

Fig. 2. Free-body diagram of elemental string length. 147

T

Consider the free-body diagram (Fig. 2) of a di®erential length of string. If the entire string is subject to a tensile force T (N), then the sum of the forces in the y-direction, according to Newton's second law, must be T sin μ0 ¡ T sin μ = ½ ¢x

@2y @t2

(2)

where ½ is the lineal mass density (kg/m) of the string. For small displacements about equilibrium sin μ0 ¼ tan μ0 =

@ y(x + ¢x) @x

and

sin μ ¼ tan μ =

@ y(x) @x

(3)

resulting in @ @ y(x + ¢x; t) ¡ y(x; t) ½ @2 @x @x = y(x; t): ¢x T @t2

(4)

In the limit as ¢x ! 0, we have the wave equation ·

where c =

¸ 1 @2 @2 ¡ 2 2 y(x; t) = 0 @x2 c @t

(5)

p T =½ will be the wave speed in (m/s). Á(x) ... .. .................... .. ... ... .. ... ... ... ... . .. ... . . . ... ... . . . ... ... . . . .. . ... . . .. . ... . . ... . ... . .. .... . ... . . . . ......................................................................................................................................................................................................................................................... ... ...

x

a

0

Fig. 3. Initial displacement of string.

Let's excite our string by holding it in the initial rest position y(x; 0) = Á(x) @ y(x; 0) = 0 @t

(one possibility shown in Fig. 3) (initially at rest)

(6) (7)

and releasing it. We can solve our partial di®erential equation (5) by expanding y(x; t) into a Fourier series in x, which is equivalent to periodically extending the function y(x) beyond our original physical domain 0 · x · a. Since y vanishes at the end points x = 0 148

and x = a, we can use a sine series, which is the odd periodic extension of period 2a. So the form of our sought-after solution is 1 X

y(x; t) =

Bn (t) sin(n¼x=a);

(8)

n=1

where the functions Bn (t) (Fourier coe±cients for the x-expansion) are unknown at this point. We need the appropriate Fourier series representation of our initial displacement (6); write it as 1 X Á(x) = bn sin(n¼x=a) (9) n=1

where the bn are known, in principle and in reality, since we know the initial displacement Á(x): Za 2 Á(x) sin(n¼x=a) dx: (10) bn = a 0

Insertion of our solution form (8) into the wave equation (5) yields ¸ 1 · ³ X 1 d2 n¼ ´2 ¡ 2 2 Bn (t) sin(n¼x=a) = 0: ¡ a c dt n=1

(11)

This is itself a Fourier series representation of zero, so each coe±cient must be zero; that is · 2 ³ n¼c ´2 ¸ d + (12) Bn (t) = 0: dt2 a The solution of this homogeneous di®erential equation is a linear combination of ½

cos sin

¾μ

n¼ct a

¶

:

We select the cosine (discard the sine) in accordance with the initial condition (7). Now we have Bn (t) = ¯n cos(n¼ct=a) in terms of (unknown) constants ¯n , which gives y(x; t) =

1 X

¯n cos(n¼ct=a) sin(n¼x=a):

(13)

n=1

Evaluation at t = 0, according to the initial condition (6), provides ¯n = bn , and so y(x; t) =

1 X

bn cos(n¼ct=a) sin(n¼x=a):

n=1

149

(14)

The physical interpretation of our solution(!) (this is it!!!) is much clearer upon rewriting it (via a trig identity) as 1 i h n¼ io 1 X n h n¼ bn sin (x ¡ ct) + sin (x + ct) y(x; t) = 2 n=1 a a

or, ¯nally, y(x; t) =

i 1 h~ ~ + ct) Á(x ¡ ct) + Á(x 2

(15)

(16)

~ means the odd periodic extension (Fig. 4) of the initial displacement Á(x) as where Á(¢) given in the sine series (9): ~ Á(x) =

1 X

n=¡1

Á(x ¡ 2na) ¡

1 X

n=¡1

Á[¡x ¡ (2n + 1)a]:

(17)

In (17) we are taking Á(») to be zero outside of the original domain 0 · » · a. Also note ~ ¨ ct) is a wave propagating with speed c in the §x-direction. that Á(x

~ Á(x)

.............. .............. ... ... ... ... ... ... ... ... . . . . ... .. . .. . . . . ... ... ... ... ... ... . . . . ... ... .. .. . . ... . . . . . . ... ... .. .. . . . ... .... ... . . ........................................................................... ........................ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .................................................. ................................................... . . ...................................................................................... .. ... ... . ... . ... . ... . ... . . . . . ... ... .. .. ... ... ... ... ... ... ... ... ... . . . . . . ... ... .. .. ... ... ... ... ... ... .. .. ............. ..............

¡a

a

0

2a

x

Fig. 4. Odd periodic extension of initial displacement of string.

Problems 1.) Verify that any functions f (x + ct) and g(x ¡ ct) satisfy the wave equation (5). Certainly, then, (16) is a solution. Such forms are called the d'Alembert solution. 2.) Verify that (16) satis¯es the initial conditions (6) and (7), and the homogeneous boundary conditions (1). 3.) Why do we say that f (x + ct) and g(x ¡ ct) are propagating in the ¨x-directions, respectively, with speed c?

150

Poisson Sum Formula Let f (t) be any function that is Fourier transformable. Then its periodic extension can be represented by a Fourier series: 1 X

g(t) =

f (t + nT ) =

n=¡1

1 X

ck ej2k¼t=T

k=¡1

with Fourier coe±cients 1 ck = T

ZT

g(t)e¡j2k¼t=T dt

0 1 Z 1 X = f (t + nT )e¡j2k¼t=T dt T n=¡1 T

0

with ¿ = t + nT (n+1)T Z

1 1 X = T n=¡1

1 = T

Z1

f (¿ )e¡j2k¼¿ =T ej2kn¼ d¿

nT ¡j2k¼¿ =T

f (¿ )e

1 d¿ = F T

μ

2k¼ T

¶

:

¡1

The resulting Poisson sum formula μ ¶ 1 1 X 2k¼ f (t + nT ) = F ej2k¼t=T T T n=¡1 1 X

k=¡1

is often written in the common form, with t = 0 and T = 1 such that 1 X

f (n) =

n=¡1

1 X

F (2k¼):

k=¡1

F Homework. Sum the slowly converging series 1 X

n=¡1

n2

1 ¼ = coth ¼b: 2 +b b

151

Solution. Recall that the Fourier transform of g(t) = exp(¡®jtj) with ® > 0 is G(!) =

Z0

e(®¡j!)t dt +

¡1

Z1

e¡(®+j!)t dt =

0

1 2® 1 + = 2 : ® ¡ j! ® + j! ® + !2

Duality says that the Fourier transform pair F

g(t) () G(!)

F

G(t) () 2¼g(¡!):

also gives

so that the Fourier transform of f (t) = is F (!) =

t2

1 + ®2

¼ ¡®j!j e : ®

We will also use the geometric series 1 X

zk =

k=1

The Poisson sum formula

1 X

z 1¡z

f (n) =

n=¡1

(jzj < 1):

1 X

F (2k¼)

k=¡1

therefore gives (assume b > 0 and note that the desired sum is even in b) " # · ¸ 1 1 1 X X 1 ¼ X ¡2jkj¼b ¼ 2e¡2¼b ¼ ¡2¼bk 1+ = e = 1+2 e = 2 + b2 n b b b 1 ¡ e¡2¼b n=¡1 k=¡1

=

k=1

¡2¼b

¡¼b

¼b

¼1+e ¼e +e ¼ = = coth(¼b): ¡2¼b ¼b ¡¼b b 1¡e b e ¡e b

~

It is interesting to look at the limiting case of very small b. If jbj ¿ 1, then 1 X

1 X 1 1 1 1 1 ¼2 : ¡¡¡! +2 = 2 + 2³(2) = 2 + n2 + b2 b!0 b2 n2 b b 3 n=¡1 n=1

Formula 4.5.67 of Big Red shows that, for jzj < ¼, coth z = so that

1 z z3 + ¡ + ::: z 3 45

· ¸ ¼ ¼b (¼b)3 1 ¼2 ¼ 1 coth(¼b) ¡¡¡! + ¡ +::: = 2 + + O(b2 ): b b ¼b 3 45 b 3 b!0

How can you see the behavior of the sum for very large b, that is for b ! 1 ? 152

Sampling and Reconstruction .......................................................... ... .... .... ... ........ ........ .. ........... ............. ... s ... . . . . . . . . . . . . . . . . . . . ................................................ ......................................... ........... .............................................................. .. .. ... ... ...... ...... ... ... ...... ....... ... . .. ............... ... . ..... ......................................................... . ......... ... .. ...

x (t)

x(t)..

h(t)

y(t)

combT (t) Consider a continuous-time (analog) signal x(t) that is uniformly sampled at the times tn = nT (n = : : : ; ¡2; ¡1; 0; 1; 2; 3; : : : ) so that the sampling frequency is

2¼ (1) T in terms of the sampling period or interval T . Once x(t) has been sampled, all we have is a table (list) of numbers x(nT ). But a useful representation of the sampled signal xs (t) in continuous-time notation is 1 X xs (t) = x(t) combT (t) = x(nT )±(t ¡ nT ): (2) !s =

n=¡1

From our table of Fourier transform properties we need £ ¤ 1 F (!) ~ G(!): F f (t)g(t) = 2¼

(3)

The easiest way to get this is to take the inverse Fourier transform of the frequency-domain convolution Z1 Z1 £ ¤ 1 ¡1 F F (−)G(! ¡ −) d− ej!t d! F (!) ~ G(!) = 2¼ ¡1 ¡1

=

Z1

1 F (−) 2¼

¡1

Z1 ¡1

G(! ¡ −)ej!t d! d−:

The inner integral (in !) is 1 2¼

Z1 ¡1

1 G(! ¡ −)ej!t d! = 2¼

Z1

G(¸)ej(¸+−)t d¸ = ej−t g(t)

¡1

and therefore £ ¤ F ¡1 F (!) ~ G(!) = g(t)

Z1

F (−)ej−t d− = 2¼f (t)g(t)

¡1

which is identical to (3). Typeset by AMS-TEX 153

The Fourier fransform of the time-domain comb function is £ ¤ 2¼ comb 2¼ (!) F combT (t) = T T

(4)

and thus the spectrum of the sampled signal xs (t) is Xs (!) =

1 X 1 1 2¼ ±(! ¡ k!s ) X(!) ~ comb 2¼ (!) = X(!) ~ 2¼ T T T k=¡1

=

1 T

1 X

k=¡1

X(! ¡ k!s ):

(5)

If x(t) is a \low-pass" signal, then X(!) = 0 for j!j > !h :

(6)

This means that the spectrum of x(t) has a highest frequency !h . An example low-pass spectrum appears in the ¯rst row of the ¯gure below. X(!)

........... ..... .......... ..... ..... . . . . . ...... ..... ...... ..... ...... ..... . . ..... . . ... ...... . . . . . ... . . . . . . . .......................................................................................................................................................................................................................................................................................................................................................................................................................

¡!h

!

!h

0

Xs (!) when !s =2 > !h

.... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. .. . .. . ... .. ....... ............... . ..... ..... .. . . ..... .......... . ... . ...... ..... ........... . ... . ...... . ...... .. . . . . . . . . ...... ...... .. .. ..... . ...... . . . . . . . . . . . . . . . . . ...... ...... . ...... ... ... ... .. .. ..... ..... ..... ..... ..... ..... . . . . . . . . . . . . . . ...... . . ...... ...... .. .. .. . ... . . . . . . . . . . . ...... . . . . . . ...... ...... . . .. . . . . . . . . . . . .... . . . . . . . ...... ...... . . . . . . . . . . . . . . . . . . . . . . . . . . . . ........................... ........................................................................................ .................................. ........................................................................................ .................................. ........................................................................................................................ .. .. . .

¡!s

Xs (!) when !s =2 < !h

0

¡!s =2

!s =2

.... ... ... ... ... ... ... ... ... ... ... ... ... ... .... . . .. .. ........ .......... ...... . . . ........ ........ . . . . .. . . .. ...... .. ....... ...... .......... . ...... .......... ... . ...... .......... . . . . . . . . . . . . .. ...... ...... ...... . ..... ...... ..... . ...... . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . ...... ...... ...... ..... ...... . . . ... .... .... ...... ..... ..... ... ......... ..... ... ......... ..... ..... ..... ..... ...... ...... . ...... ...... ...... ...... ...... .......... ..... .... ..... . . . . . . . . . . . . . . . . ...... ......... . . . . . . . . . . . . . . . . . . . .......... ........... .......... ........... ........... ..... . . . ...... ......... .. ....... ........... ........... . . . . . . . . . . . . . . . ........... . . . . . . . . . . . ..................................................................................................................................................................................................................................................................................................................................................................................................................................................... ... ... . .

¡2!s

¡!s

¡!s =2

0

!s =2

!s

!

!s

!

2!s

The middle curve shows Xs (!) for the case !s > 2!h and the bottom curve is for !s < 2!h . If we pass xs (t) through an ideal low-pass ¯lter having transfer function H(!) = T ¦ 154

μ

! !s

¶

;

(7)

then we see that, for all !, Y (!) = X(!) if !s > 2!h :

(8)

In this case, y(t) = x(t) and we have exactly recovered the low-pass signal x(t) from its sample values! The minimum sampling frequency is called the Nyquist rate or Nyquist frequency 2!h . If we sample at too low a frequency, as shown in the bottom curve of the previous ¯gure, then Y (!) 6 = X(!) if !s < 2!h (9)

and of course then y(t) 6 = x(t). The error or distortion that results from the overlap in the adjacent shifted transforms X(! ¡ k!s ) is called aliasing. The whole concept is usually called: Shannon's sampling theorem. If the low-pass signal x(t) having spectrum X(!) = 0 for j!j > !h is sampled uniformly at the rate !s > 2!h (the Nyquist rate), then x(t) is exactly recoverable by passing the sampled signal through the ideal reconstruction ¯lter having transfer function H(!) = T ¦(!=!s ). The frequency domain renders this derivation clear and simple. Now let's examine the output y(t) in the time-domain to get another perspective. The impulse response of the ideal reconstruction ¯lter is ¤ £ h(t) = F ¡1 T ¦(!=!s ) = sinc(!s t=2) = sinc(¼t=T ); (10)

and this ideal low-pass ¯lter is also called the sinc interpolation ¯lter. Independent of x(t) being a low-pass signal or not, and for an arbitrary sampling frequency (that is, ignoring both Nyquist and Shannon), the output of the ¯lter in response to the sampled signal (2) is always y(t) = xs (t) ~ h(t) =

1 X

n=¡1

x(nT )±(t ¡ nT ) ~ h(t) =

· μ ¶¸ t = x(nT ) sinc ¼ ¡n : T n=¡1

1 X

n=¡1

x(nT )h(t ¡ nT )

1 X

(11)

If k is an integer, observe that sin(k¼) sinc(k¼) = = k¼

½

1; k = 0 = ±k0 ; 0; k = 60

(12)

in terms of the Kronecker-delta. Therefore, we see that at the sample times t = mT , the output is always equal to the input in that y(mT ) =

1 X

n=¡1

x(nT ) sinc [¼ (m ¡ n)] =

155

1 X

n=¡1

x(nT )±mn = x(mT ):

(13)

Dissection of the Time-Domain Output of the sinc Interpolation Filter: Consider a small number (four) of samples x[n] = x(nT ) for n = 0; 1; 2; 3. If all other (n = 4; 5; : : : and n = ¡1; ¡2; : : : ) of the x[n] are zero (or absent), then the output (11) of the sinc interpolation ¯lter is ¶ · μ ¶¸ μ t t + x[1] sinc ¼ ¡1 y(t) = x[0] sinc ¼ T T · μ ¶¸ · μ ¶¸ t t + x[2] sinc ¼ ¡ 2 + x[3] sinc ¼ ¡3 : T T Each of these individual terms are graphed below for the example data x[0] = 0:5

x[1] = 1:0

x[2] = 0:8

x[3] = 0:6

along with the sum of all four terms. Do you see which curve is which?

1.0

0.8

0.6

²

........... . ........ ................ .... . . ....... .... .... ... .... . . .. . .. ..... .... ... ... . . ... ..... .. . . . .. ..... .. .. . . ..... . .. . . . ... ...... . . . . . ...... . . . . . . . ... .. .. ......... . . .. . .. ............... . . . . . ........ . . . . ... . ....... . .. . . . .. . ...... . . ..... ... ... . . . . . . ... . .... . . . . . . . ... . . . . . . . . ... . ... .. . . . . . . .. . ... . . . . . .. . . . . . . . .... . ........... . . . . . ... ........... . . . . . . . . . . ... .. ... . . .. . . . ... .. . . .... ... ... ... . ... . ... ... ... .. . .. ...... . . . . . . . . . . ... ... .. ... .... ..... . . . . . ... .. . .. .. . .. . . . . . . ... ... .. . ... . ... . . . . . .... ... .. .. . ... . . . . . . ... ... . . .. .. . . . . . .. ... .. .. . .. .. ... . . . . . ... ... . . . . . .. . . . . ... ... .. . .. . .. . . .. . . . . ... ... .. .. .. . . . . . . . . . ... .. . . . .. . . . . . . . . . . ... ... ... . . . . . . . . . .. . ... .. . . . . . . . . . . . .. . . . . ... ... . . . . . . . . . . . . ... .. . . .. . . .. . . . . ... .. . . . . . . . .. . . . . . . ... .. . . .. ... .. . .. . . . ... ... . . .. .. . . . . . . .. .. .. . . . .. . .. . . . . ... .. . . . . . . ... . . .. .. .. . .... .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. ..... ......... . . .. . . ... ..... . . . . . . . . . . . . . . ... .... . . .. . . ... .......... . ....... ..... . . . . . . . . . . . . ... .. . . . . ............ ... ... . ....... . . . . . . . . . . . . . . . . . . ... . . . . . . . ... .. . ... ... ... ... .... . . .. . ....... . . .... . .... .... .. . ..... .. .. ...... ... .. ... ... .... .. .. ..... ... ... ...... ... ......................... .... ............. . . . . . . . . ... ... . .. . . . . ..... . ..... .. ...... . .... . ... .. ..... . ......... . .. .. . . . . . . . . . . . . . . . . . . . . . . . . . .......................................................................................................................................................................................................................................................................................................................................................................................................................................................................... .................. ..... . .... .. ..... . . . ........ ... . ........ ......... .. .. ....... ........ ........... .... .... .. ............ . . ....... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... ... .......... ... .... .. ................... . . .. .. ..... . . . . ....... . .. .. ... .... .. ...... .... .. ......... ......................... ....... .. . . . . . . . . . . . . . . . .... . . . . . . ... ... ............... .... ... ........ . . .. . . . ..... ... ...... .. ... .. . ........... ... ....... . . .. .. ...... .......... . . ... ............. ... .. ... .. .... . .. . .. ... . . . . . . . . . . . ... ... . .... .... . . ... .. . . . . . .. ... . .. . . . .... .... ........ ..... .....

²

²

²

0.4

0.2

0.0

¡0:2 ¡0:4

¡2

¡1

0

1

2 t=T

156

3

4

5

DISCRETE-TIME SIGNALS, SYSTEMS, AND TRANSFORMS A common source for a discrete-time signal is to uniformly sample a continuous-time signal, as in x[n] = x(nT ) n = : : : ; ¡2; ¡1; 0; 1; 2; 3; : : : The discrete-time signal x[n] is also called a sequence; it is a function of the integers n. Usually we are not concerned with any underlying sample period T that was used to obtain our sequence from some original x(t). Once we're in the discrete-time world, the independent time variable is simply the integers n. Our sequence x[n] is simply a list or table of ordered numbers. Note that the sequence x[n] can take on real or complex values.

Many of our familiar continuous-time signals have discrete-time analogues or cousins, but be aware that their behavior is often (usually? always?) quite di®erent. For starters, there is no such thing as continuity in discrete-time. The continuous-time calculus operators of di®erentiation and integration have analogous operators of di®erencing and accumulation in discrete-time. Sometimes the discrete-time version of a common signal, for example the Heaviside unit-step function u[n], goes by the same name and similar symbol as its continuous-time counterpart. In other cases, di®erent names are used. I suppose the Heaviside unit-step sequence is a better name that draws attention to its di®erence with the continuous-time signal u(t). Heaviside unit-step sequence. ½ 0; n = ¡1; ¡2; : : : u[n] , 1; n = 0; 1; 2; : : :

u[n]

²......

²

²

²

²

0

1

2

3

4

... ... ... ... ... ... ... ... .. .. .. .. .. .. .. .. .. .. ... ... ... ... .. .. .. . . . ... ... . ... ... . . . . .............................................................................................................................................................................................................................................................

:::

² ¡3

² ¡2

² ¡1

::: n

Unlike u(t) that is not well-de¯ned at t = 0, the unit-step sequence clearly takes on the value u[0] = 1. Note, however that u[0:5] is complete notational nonsense. Kronecker delta sequence. (also called unit impulse sequence) ±[n] ²..... .... ::: ::: .. ½ ... .. 0; n = §1; §2; §3; : : : . ..............² ................................² ................................² ...............................................................² ...............................² ................................² ...............................² .................. n ±[n] , ¡3 ¡2 ¡1 1; n = 0 0 1 2 3 4 Unlike the continuous-time Dirac-delta function ±(t) that is illegal at t = 0, the discretetime Kronecker-delta sequence clearly takes on the value ±[0] = 1. The Kronecker-delta possesses the sampling property x[n] =

1 X

k=¡1

x[k]±[k ¡ n]:

Relationship. ±[n] = u[n] ¡ u[n ¡ 1]

u[n] =

n X

k=¡1

157

±[k]

Complex exponential. First, recall that for any positive real number ! (called the frequency), the continuous-time complex exponential x(t) = ej!t is always periodic in time t with period T = 2¼=!, since x(t + T ) = ej!(t+T ) = ej!T ej!t = ej!t = x(t) whenever !T = 2¼, since e2¼j = 1.

The discrete-time complex exponential is substantially di®erent. Consider the discretetime sequence x[n] = ej−n where the frequency − is any real number: Take it to be positive for starters. Is x[n] periodic in the independent time variable n ? Let's denote its candidate period by the speci¯c integer N , and examine x[n + N ] = ej−(n+N ) = ej−N ej−n : If x[n] is periodic with the period N , then x[n + N ] = x[n] which requires that −N = 2k¼ where k is any integer. This only happens if −=

k 2¼ = a rational multiple of 2¼: N

However, the discrete-time complex exponential is a periodic function of the frequency variable −, since ejn(−+2¼) = e2n¼j ejn− = ejn− : Therefore, in discrete-time work, we only have to consider frequencies − in the primary range ¡¼ < − · ¼, for example. The highest frequency possible is − = ¼ (or equivalently, ¡¼), and the lowest frequency is − = 0 (which is identical with − = 2¼).

158

N ²......

Exercises.

... .. .. .. ... .. .... ... .. .. .. ... ... ... .. .. .. .. ... ... .. .. .. ... .. ... ... .. ... .. .. .. ... .. .. ... .. ... .. .. .. . . .... .... ... .... ... .. .. .. .. ... .. .. .. ... ... .. .. ... .. .. .. ... .. .. ... ... ... .. .. . . . ... ... .... .... ... ... ... ... ... .. ... .. .. .. .. .. .. .. .. .. ... ... ... ... .. .. ... .. .. .. .. .. .. .. ... ... ... ... ... ... .. ... .. .. .. .. .. .. .. .. .. .. .. ... ... ... ... .. ... .. ... .. .. .. .. .. .. .. .. .. ... .. ... ... ... ... ... . . . . . .........................................................................................................................................................................................................................................................................................................................................................................................

N ¡1 ²

²

x[n]

1. Write an expression, in terms of n, N , and u[¢], for the given sequence x[n].

3²

2²

1²

:::

² ¡3

² ¡2

² ¡1

² 0

1

2

3

: : : N ¡1 N

² N +1

::: ² n

2. Graph the exponential (or geometric) sequence x[n] = an u[n]. Look at the following speci¯c cases: a = 1=2;

a = ¡1=2;

a = 2;

a = rej−

where r = 1=2 and − = 1

Discrete-Time Systems

x[n]

.............................................................................................. ... .... ... .. ... .. ... ... .. . . . . . . ......................................... . . ................................. .. ... . .... .. .. .. ... ... .. . ..............................................................................................

y[n]

The equation-of-motion, that describes the relationship between the input x[n] and the output y[n], for a discrete-time system is often a di®erence equation. An example of a second-order, constant-coe±cient di®erence equation is c0 y[n] + c1 y[n ¡ 1] + c2 y[n ¡ 2] = b0 x[n] + b1 x[n ¡ 1]: When accompanied by two (the di®erence equation is second order) initial conditions of the form y[¡1] = y¡1 and y[¡2] = y¡2 ; we have a discrete-time initial value problem. Classical solution techniques are analogous to the classical methods used to solve ordinary di®erential equations in continuous-time. Di®erence equations can sometimes be solved recursively: y[n] = ¡

c2 b0 b1 c1 y[n ¡ 1] ¡ y[n ¡ 2] + x[n] + x[n ¡ 1]: c0 c0 c0 c0

The discrete-time version of the Laplace transform (or Fourier transform) is called the Z-transform, and it, too, provides an attractive path to solve discrete-time di®erence equations by converting them into purely algebraic equations in the transform (z) domain. The inverse Z-transform gets our solution back to the time [n] domain. 159

Linear Time-Invariant Discrete-Time Systems

x[n]

.............................................................................................. ... ... .. .. ... ... ... .. ... .. ......................................... . ....................................... .. ... .. ..... ... .. .. ... .. ...............................................................................................

y[n] = T fx[n]g

Assume that the zero-state response sequence y[n] to the exciting sequence x[n] is given by the operator or system transformation rule y[n] = T fx[n]g. If the operator T satis¯es the superposition property T fc1 x1 [n] + c2 x2 [n]g = c1 T fx1 [n]g + c2 T fx2 [n]g for any values of the constants c1 and c2 and for arbitrary sequences x1 [n] and x2 [n], then the discrete-time system is linear. Given that the zero-state response to the input sequence x[n] is y[n], if the response to the delayed input x[n ¡ n0 ] is the delayed response y[n ¡ n0 ], then the system is time-invariant.

x[n]

x[n ¡ n0 ]

.............................................................................................. ... .... ... .. ... .. ... ... . .. ......................................... ...................................... .. ... ... .. .. .. ... ... . .. .............................................................................................. .............................................................................................. ... ... ... ... .. ... .. ... ... . . . . . . ......................................... ..................................... .. .. ... ... .. .. .. ... . . .................................................................................................

y[n] = T fx[n]g

y[n ¡ n0 ] = T fx[n ¡ n0 ]g

Input/Output relationship. Using the sampling property of the Kronecker-delta sequence, write the input sequence as x[n] =

1 X

k=¡1

x[k]±[n ¡ k]:

The system operator T operates only on functions of the running time variable n, treating all other variables as constants, so that the T operator can be moved inside the summation in ( 1 ) 1 X X y[n] = T fx[n]g = T x[k]±[n ¡ k] = x[k]T f±[n ¡ k]g: k=¡1

k=¡1

The response to a Kronecker-delta or unit impulse sequence is the system impulse response h[n] = T f±[n]g. By the (assumed or de¯ned) time-invariance of the linear system, the 160

response due to a delayed unit impulse ±[n ¡ k] is the delayed h[n ¡ k] = T f±[n ¡ k]g, and therefore the zero-state response is y[n] =

1 X

k=¡1

x[k]h[n ¡ k] = x[n] ~ h[n]:

This is the discrete-time convolution sum between the two signals x[n] and h[n].

x[n]

.............................................................................................. ... ... .. ..... .. ... . .... .. . ....................................... ....................................... ... ... .. .... .. ... . .... .. .. . ............................................................................................

h[n]

y[n] = x[n] ~ h[n]

Example of a LTI Discrete-Time System. Let's look at one particular discrete-time system in detail and study its input/output dynamics from several perspectives. One common way to specify a discrete-time system is via a system block diagram. We need to ¯rst introduce several basic components of such system block diagrams.

summer

f [n]

..................... .... ... ... ................................... . . . . . . ................................... .... .. ... ... . ..... ................... ... . ...... .. ... .. ... ... .

P

f [n] + g[n]

g[n]

ampli¯er/attenuator

unit delay

f [n]

.......... ... .......... ........ . . ........................................ . . . . ..................................... .. ........ ..... ............. ........

®

®f [n]

............................... ... ... .. .

f [n] ............................................. z ¡1 ...........................................f [n ¡ 1] ... . .............................

The reason that the unit delay is represented by the symbol z ¡1 will be clear when we study the Z-transform.

161

x[n]

......................... ....... ..... ..... .... ... ... . . . ... . . ... .. . .......................................................................................................................................................................................................... . . . . . . . ......................................................... ... .... ... ... ... . ... .. . ... . ... . . . .... . .... ...... .... ......... ..... .... ....................... ... ........ ... .. ... ... ..... .. .. ... ... ... .. .... .. ... ... ... ...... .... . ... . . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... . . .. . .. .. . . . . . . . ... ... . . . . .. . . . . . . . . . . . . .. . . . ....................................... .................................... ..... ................................................ ¡1 . . . ...... . . . . . ... . . . .. . ...... . . ...... ..... ... ... ...... .. . .. ...... ....................................... ...... ... . ....... ...

P

²

®

y[n]

z

The output of the summer is y[n] = x[n] + ®y[n ¡ 1] and so the discrete-time system depicted by the block diagram above is governed by an equation-of-motion that is a linear, constant coe±cient, ¯rst-order di®erence equation. The impulse response h[n] is the response y[n] to the input sequence x[n] = ±[n], and being a zero-state response, its initial condition is h[¡1] = 0. Sometimes di®erence equations can be solved recursively, and this is one h[n] = ®h[n ¡ 1] + ±[n]

h[¡1] = 0

h[0] = ®h[¡1] + 1 = 1 h[1] = ®h[0] + 0 = ® h[2] = ®h[1] = ®2 h[3] = ®h[2] = ®3 .. . h[n] = ®n u[n] With the unit impulse response h[n] in hand, the zero-state response due to any input sequence is readily computed via the convolution sum. For example, the unit step response is s[n] = u[n] ~ h[n] =

1 X

k=¡1

h[k]u[n ¡ k] =

n X

k=¡1

h[k] =

n X

k=¡1

n+1

=

1¡® 1¡®

u[n]:

The required ¯nite geometric series is detailed two pages ahead.

162

k

® u[k] =

n X

k=0

®k

The Z-Transform In order to derive the Z-transform from the Laplace transform, we need to represent our sequence x[n] in continuous-time notation: Take a step backwards, so to speak, and write a sampled (discretized) continuous-time signal as ......... .. .... 1 1 .. ..... .... .... .... .. ... . ... .. ... d T . .. ... ... ... ... .. .. ... n=¡1 n=¡1 ... .. ... . ... ... ... .. ... ... .. ... . ... ... ... .. ... .. .. .. ... ....... .. ...... .. ... .. ... .. .. .. ... ... .. . ... .. ... ........ .. ... .. .... ... .. .. ..... . . ... . .. ... .. ... ...... .. ... ... .. .... .. ... .. ... ... .. ... ... . .. .. ... . .. .... .. .. ... .... ... ... .. .. .. ... ...... ........................................... .......................... ......................... .. .. .. .... .... ... ............... .............. . . . . . . . . . . . . . . .. .. .. . . . . ............... .. . . . ..... . . . . . . .... . . . . . . . . . . . . . . . . . . . . . . . ........................................................................................................................................................................................................................................................................................................................................................................................................... . . ... . . ... ... ... ... ... . ... .. .. .. . .. ........ ... ... ... ... ... ... ..... ... ... .. .. .. .. .... ... ... ... ... ... ..... . . .. .. ... .. . .. ... ....... ... ... ... ... .. ...... .. .. ..... ...... ... ... .... .. .. ... ..... . . . . . . ... ... .... .... .... . ..... .. ............ ........ ................

²

x (t) = x(t) comb (t) =

X

x(nT )±(t ¡ nT ) =

X

x[n]±(t ¡ nT ):

²

²

T

²

²

² ²

²

²

² ² ² ² ² ² ² ² ² ²

t

The bilateral Laplace transform of a continuous-time signal f (t) is de¯ned as the integral Lff (t)g = F (s) =

Z1

f (t)e¡st dt

¡1

where s = ¾ +j! is a complex frequency variable. The Laplace-transform of the discretized signal xd (t) is therefore Xd (s) =

1 X

n=¡1

x[n]

Z1

¡1

¡st

±(t ¡ nT )e

dt =

1 X

x[n]e¡nsT :

n=¡1

Introduce the transformed variable z = esT so that the above Laplace-transform is now called the Z-transform Zfx[n]g = X(z) =

1 X

x[n]z ¡n :

n=¡1

The Z-transform converts a function of the integers n (a sequence) to a function of a complex variable z. 163

Geometric series. Firstly, consider the ¯nite geometric series N X

S=

an

n=0

where a is an arbitrary complex number. There is no question about the convergence of this series, since it is simply the sum of a ¯nite number of terms. We can obtain a closed-form expression for the sum S by this neat trick: S = 1+a + a2 + a3 + ¢ ¢ ¢ + aN ¡1 + aN

a + a2 + a3 + ¢ ¢ ¢ + aN ¡1 + aN + aN +1

aS =

S ¡ aS = 1¡aN +1

N +1

S(1 ¡ a) = 1 ¡ a

=)

S=

N X

an =

n=0

1 ¡ aN +1 1¡a

Note that when a = 1, we have S = N + 1, which is the limit as a ! 1 of the above result. Similarly, the expression for the in¯nite geometric series 1 X

an =

n=0

1 1¡a

(jaj < 1)

explicitly displays its region-of-convergence jaj < 1. Z-transform example: causal exponential sequence. Consider the sequence x[n] = an u[n] where a is an arbitrary complex number. Then the Z-transform of x[n] is X(z) =

1 ³ ´n X a

n=0

z

=

1

z a = z¡a 1¡ z

(jzj > jaj):

The region-of-convergence in the complex z-plane for X(z) to be a valid transform of x[n] is the exterior of the circle of radius jzj = jaj.

164

Fibonacci Sequence - Di®erence Equation and Z-Transform f [n] = f [n ¡ 1] + f [n ¡ 2]

initial conditions: f [¡2] = 0;

Zff [n]g = F (z) = Zff [n ¡ 1]g = f [¡1] + z ¡1 F (z);

1 X

f [¡1] = 1

f [n]z ¡n

n=0

Zff [n ¡ 2]g = f [¡2] + z ¡1 f [¡1] + z ¡2 F (z)

F (z) = 1 + z ¡1 F (z) + z ¡1 + z ¡2 F (z) F (z) =

z(z + 1) 1 + z ¡1 = 2 ¡1 ¡2 1¡z ¡z z ¡z¡1

F (z) z+1 z+1 = 2 =³ p ´³ p ´ z z ¡z¡1 z ¡ 1+2 5 z ¡ 1¡2 5 ³ ´ ³ ´ 1 1 p3 p3 1 + 1 ¡ 2 2 5 5 p p = + via partial fraction expansion z ¡ 1+2 5 z ¡ 1¡2 5 1 f [n] = 2

p !n p !n μ μ ¶Ã ¶Ã 3 1 3 1+ 5 1¡ 5 1+ p 1¡ p + 2 2 2 5 5

for n = ¡2; ¡1; 0; 1; 2; : : : p f [n + 1] 1+ 5 ¡¡¡! n!1 f [n] 2

f = 0; 1; 1; 2; 3; 5; 8; 13; 21; 34; 55; 89; 144; : : :

165

Convolution Theorem of the Z-Transform Zff [n] ~ g[n]g = =

(

1 X

n=¡1 1 X

1 X

k=¡1

f [k]

"

f [k]

"

k=¡1

(let m = n ¡ k)

=

=

1 X

k=¡1 1 X

)

f [k]z

f [k]g[n ¡ k] z ¡n

1 X

n=¡1 1 X

g[n ¡ k]z ¡n

#

g[m]z ¡(m+k)

m=¡1 1 X ¡k

#

g[m]z ¡m

m=¡1

k=¡1

= F (z)G(z)

Z-Transform of a Delayed Sequence. Zff [n ¡ n0 ]g =

1 X

n=¡1 ¡n0

=z

f [n ¡ n0 ]z ¡n =

1 X

f [m]z ¡(m+n0 ) = z ¡n0

m=¡1

1 X

f [m]z ¡m

m=¡1

F (z)

System Transfer Function. y[n] = x[n] ~ h[n]

Z

() H(z) =

Y (z) = X(z)H(z)

Y (z) X(z)

Recall the system block diagram for the example four pages back. The Z-transform of the governing di®erence equation y[n] ¡ ®y[n ¡ 1] = x[n] is (1 ¡ ®z ¡1 )Y (z) = X(z) and therefore the system transfer function is Zfh[n]g = H(z) =

1 z = : 1 ¡ ®z ¡1 z¡®

166

Discrete-Time Fourier Transform. With z written in polar form as z = rej− , if we restrict z to lie on the unit-circle (jzj = r = 1), then we have the discrete-time Fourier transform 1 ¯ X ¯ j− = X(e ) = x[n]e¡jn− : X(z)¯ jzj=1

n=¡1

The discrete-time Fourier transform (DTFT) is a mapping from a function of the integers n to a periodic function of the continuous real variable −, with period 2¼. Observe that the DTFT is essentially a Fourier series representation of the periodic function X(ej− ), where x[n] are the Fourier coe±cients! Therefore, the inverse DTFT is simply obtained by applying the operator Z¼ 1 d− ejm− 2¼ ¡¼

to the DTFT above 1 2¼

Z¼

j−

jm−

X(e )e

¡¼

Z¼ 1 d− = x[n] ej(m¡n)− d− = x[m]: 2¼ n=¡1 ¡¼ | {z } 1 X

±mn

The discrete-time Fourier transform pair is therefore j−

DTFTfx[n]g = X(e ) =

1 X

x[n]e¡jn−

n=¡1

¡1

DTFT

1 fX(e )g = x[n] = 2¼ j−

Z¼

X(ej− )ejn− d−:

¡¼

Discrete Fourier Transform. Consider a ¯nite or partial sequence x[n], where our attention is centered on the indices n = 0; 1; 2; : : : ; N ¡ 1. Its DTFT is j−

X(e ) =

N ¡1 X

x[n]e¡jn− :

n=0

If we now sample the DTFT at N values of − −k =

2k¼ N

(k = 0; 1; 2; : : : ; N ¡ 1)

that are uniformly distributed around the unit circle, we have j−k

X(e

)=

N ¡1 X

x[n]e¡j2nk¼=N :

n=0

167

This is a mapping from a function x[n] of the integers n = 0; 1; 2; : : : ; N ¡ 1 to another function of the same integers k = 0; 1; 2; : : : ; N ¡1. Rather than write it as the cumbersome X(ej−k ), we focus our attention on the discrete frequency parameter k and write it as the discrete Fourier transform or DFT X[k] =

N ¡1 X

x[n]e¡j2nk¼=N

(k = 0; 1; 2; : : : ; N ¡ 1):

n=0

The inverse DFT is most easily derived by ¯rst looking at the comparable discrete-time spectral representation of the Kronecker-delta sequence ±[n] =

N ¡1 1 X §j2nk¼=N e : N k=0

The truth of this last equation can be established in a number of ways. Observe that the sum is a ¯nite geometric series (page 163): The result follows at once. Therefore, application of the operator N ¡1 1 X j2mk¼=N e N k=0

(m = 0; 1; 2; : : : ; N ¡ 1)

to the DFT above gives N ¡1 N ¡1 N ¡1 X 1 X 1 X j2(m¡n)k=N j2mk¼=N X[k]e = x[n] e = x[m]: N N n=0 k=0 k=0 | {z } ±mn

Hence, the DFT transform pair is DFTfx[n]g = X[k] =

N ¡1 X

x[n]e¡j2nk¼=N

n=0

¡1

DFT

N ¡1 1 X fX[k]g = x[n] = X[k]ej2nk¼=N N k=0

(k = 0; 1; 2; : : : ; N ¡ 1) (n = 0; 1; 2; : : : ; N ¡ 1):

The DFT (FFT is the fast algorithm that e±ciently computes the DFT) is therefore a transformation that converts N numbers into N other numbers. But each of the sets of N numbers, x[n] and X[k], are a single period of periodic sequences, each with period N . To see this, observe that x[n + N ] =

N ¡1 N ¡1 1 X 1 X X[k]ej2(n+N )k¼=N = X[k]ej2nk¼=N ei2k¼ N N k=0

=

1 N

N ¡1 X

k=0

X[k]ej2nk¼=N = x[n]

k=0

168

and X[k + N ] =

N ¡1 X

x[n]e

N ¡1 X

x[n]e¡j2nk¼=N = X[k]:

¡j2(k+N )n¼=N

=

n=0

=

N ¡1 X

x[n]e¡j2nk¼=N e¡j2n¼

n=0

n=0

Exercises. 1. Find the Z-transform and its region-of-convergence for the ¯nite, rectangular pulse sequence x[n] = u[n] ¡ u[n ¡ N ], where N > 0. 2. Given the sequence x[n] = ±[n] (n = 0; 1; 2; : : : ; N ¡ 1), ¯nd its DFT X[k]. 3. Given the sequence x[n] = 1 (n = 0; 1; 2; : : : ; N ¡ 1), ¯nd its DFT X[k]. 4. The Bessel function of order n and argument t is de¯ned as the integral 1 Jn (t) = 2¼

Z2¼

ej(nÁ¡t sin Á) dÁ:

0

Approximate this integral using the DFT. Use the fft algorithm in MATLAB to generate values of Jn (1) and Jn (50) for n = 0; 1; 2; : : : ; 20. Compare with Table 9.4 of Big Red: M. Abramowitz and I.A. Stegun, Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. National Bureau of Standards (also reprinted by Dover). 5. Recall from page 165 that the Z-domain transfer function of the system having block diagram on page 161 is 1 : H(z) = 1 ¡ ®z ¡1

Graph the magnitude and phase response H(ej− ) as a function of − for these values of the constant ®: (a) ® = 1 and ® = ¡1 (graph on the same axes) (b) ® = 1=2 and ® = ¡1=2 (also graph on common axes)

Graph 20 log10 fjH(j−)jg (dB) for the magnitude characteristic. Explain the frequency behavior with respect to the sign of ®.

169

DFT APPROXIMATION OF THE CONTINUOUS-TIME FOURIER TRANSFORM

F (!) =

Z1

¡j!t

f (t)e

¡1

dt ¼

1 X

N=2 ¡j!n ¢t

f (n ¢t)e

n=¡1

¢t ¼ ¢t

X

f (n ¢t)e¡jn! ¢t ;

n=¡N=2+1

subject to appropriate restrictions. Evaluation at speci¯c frequencies gives

F

μ

k2¼ N ¢t

¶

N=2

¼ ¢t

X

f (n ¢t)e¡j2¼nk=N :

n=¡N=2+1

De¯ne the periodic sequence x[n] =

½

f (n ¢t); n = 0; 1; : : : ; N=2 f ((n ¡ N )¢t); n = N=2 + 1; : : : ; N ¡ 1

where x[n + N ] = x[n]. Also note that exp(¡j2¼nk=N ) is periodic in both n and k with period N . The approximate Fourier transform above is now F

μ

k2¼ N ¢t

¶

¼ ¢t

N ¡1 X n=0

x[n]e¡j2¼nk=N ¼ ¢t X[k]

in terms of the DFT X[k] of the sequence x[n]. If F (!k ) is desired for negative k, employ the periodicity of X[k] = X[k + N ]. Also note that the range of available frequencies is ¼ ¡ ¢t

μ ¶ 2 ¼ 1¡ ·!· : N ¢t

.......................................................... ....... .... .................................. ..... ... . . . .. . ............. .. .. .. ...... .................................. . . ..... .... ... . . .... . . . . . . . . ..... ... . .. . . . . . . . ................................ . . ... ... . .... .. ........ .... .... ................................................................... ... ................ ..... ... .. .. .... . ................................................................... ......................... .................... ..... . . . . . .. . . .. ........ . . . . . . .... . . . . ... . . . . . ................................... . . . . . . . . . . . .. . . . . . ................................ ... ..... ... .... .... .... .... ........... .... . ....... . . . ...... .... . . . . ... . . . . . . ... . . . . . . . . . ...... ... . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . .............................. .... ... .... .... ............... . .... .... .... .... .... .... ..... ....... .... .. .. .. .. .. .. .. .. .. ...................................... . ......... .. . . . . . . . . . . . . . . . ........................................ . . . ... .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .... ..... .. .. .. .. .. .. .. .. .. ... .. ...................... ... ........................................... ....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... ... . .................................. ... .

170

¢t

... . .................................. ... .

t

Numerical Example of Using the FFT to Evaluate the Fourier Integral: Time-Domain Response of an RC Filter to a Gaussian Pulse .... .. .. ... ......................................................................................................................... .. .. ... ... .. ...... ...... . ........ . . . . . . ........ ........ ...... ... ........ ............ ........ ...... ... ....... ............. .... .. ... ....................................................................................................................................................................................................

.... .. .. . ................................................................ ... .. .. ..

²

+

+

C

x(t)

y(t)

R

¡

²

F

y(t) = x(t) ~ h(t)

()

¡

Y (!) = X(!)H(!)

(1)

The time-domain input to the simple RC ¯lter is a Gaussian pulse 2

x(t) = e¡®t

(2)

r

(3)

having spectrum X(!) =

¼ ¡!2 =4® : e ®

The circuit transfer function is H(!) =

Y (!) = X(!)

R 1 R+ j!C

j!RC j!=!b = 1 + j!RC 1 + j!=!b

=

(4)

where the ¯lter break frequency is de¯ned to be !b =

1 : RC

(5)

The time-domain response is the inverse Fourier transform 1 y(t) = 2¼

Z1

j!t

H(!)X(!)e

1 d! = 2¼

¡1

Z1 ¡1

j!=!b 1 + j!=!b

r

¼ ¡!2 =4® j!t e e d!: ®

(6)

Let ® = ¿ ¡2 so that (6) becomes a little clearer 1 y(t) = p 2 ¼

Z1 ¡1

j!¿ · ¸ ¤ £ 1 t !b ¿ 2 ¿ d!: exp ¡ 4 (!¿ ) exp j!¿ j!¿ ¿ 1+ !b ¿ 171

(7)

Now introduce the normalized variables z = !¿;

zb = !b ¿;

v=

t ¿

(8)

so that the circuit input and output signals are x(v) = e¡v

2

1 y(v) = p 2 ¼

(9) Z1 ¡1

jz=zb ¡z2 =4 jzv e e dz: 1 + jz=zb

(10)

Normalized time is v = t=¿ and the single parameter requiring a numerical value is zb = !b ¿ =

¿ 1 =p : RC ®RC

(11)

This parameter is essentially the ratio of the input \pulse width" to the circuit timeconstant. Note that as the cut-o® frequency of the high-pass ¯lter approaces zero, the output approaches the input 1 y(v) ¡¡¡! p zb !0 2 ¼

1 x(v) = p 2 ¼ Let

Z1

¡z 2 =4 jzv

e

e

¡1

Z1

e¡z

2

=4 jzv

e

dz = x(v):

(12)

¡1

1 dz ¼ p 2 ¼

N=2

X

2

e¡(n¢z)

=4 jn¢z v

n=¡N=2+1

2¼k (k = ¡N=2 + 1; : : : ; N=2) N ¢z μ ¶ N=2 X 2 2¼k ¢z ¼ p x e¡(n¢z) =4 ej2¼nk=N N ¢z 2 ¼ vk =

n=¡N=2+1

x[k] =

N ¢z p FFT¡1 fX[n]g 2 ¼

with

2

x[n] = e¡(n¢z)

172

=4

:

e

¢z

1.00

...... . .. . . .. . . . . . ............ .. . .. ... . . ... . . ... .. .. ... ... . . ... ... . . .... ... . . ... .. . ..... ... . . .. ... . ...... ... . .. ... . . .... ... .. . .. ... . . .. .... ... . ... ... . ... .. ..... ... . .... ... . ... ... ... . ... ... .. .... .... ...... . . ... . ... . . . . ... . .. ...... .. ... . .. .. .. ... . ..... ... .. ... .. .. . .. ... . .... .. .. ... . ... . .. ... . ...... ... .. ... .. ... .. .. ... . .... .. ... . .. ... . ... . ... . . . .... .. ... .. .. ... . ......... .. ..... ... . .. ... . ... . .. ..... ... .. ...... . . ... . .. ..... ... . ........ .. . ... . .. .. .. ... . . . . ....... ... .. ............ . . ... . . . . . . . . . . ... ..... . ... . . . . . . . . . . . . . . . . .... ... ... .. .... .... .. .... ... ..................... ........... ..... . . . . . . . . .. . .. .... ..... .. .. .... . .. ... . .............. .............. ........................................... ...................................... ..... .. ... .............. .. ..... .............................................................. . . . . . ... ... .................. .. ..... .............................................................................. . .... ...... . . ..... ... . .. ... ... . ... ... . .......... ............... ..... ... ... .... ....... .... ..... . . .. . . ...... .... .. ....... ............................................................................................. .. .. .... . . .. ... .. ... ... .. ... . . .. . .. ... ... .. ... .. ... . . .. ...... ... ......

x(t)

0.75

0.50 y(t) 0.25

0.00

¡0:25

¡0:50

0:1

1

¿ =RC = 10

¡4

¡3

¡2

¡1

0

1

2

3

4

t=¿

Fig. 1. Response of RC High-Pass Filter to a Gaussian Pulse x(t) = exp[¡(t=¿ )2 ] Cases: ¿ =RC = 0:1; 1; 10

173

Example: DFT Approximation of a Fourier Sine Integral 8 0; > > > 2t > > < ; T f (t) = 2 > > > (T ¡ t); > ........... . . . . . . > T ........ ....... : ............ ................ . . . ......... .... ......... ....... 0; .......... ......... .. ........

f (t)

t 0 application of L¡1 d=dt to eliminate the integral yields the homogeneous second-order di®erential equation d2 i(t) R di(t) 1 + i(t) = 0 + dt2 L dt LC

(t > 0)

A second initial condition i0 (0) = V0 =L comes from the ¯rst integro-di®erential equation evaluated at t = 0+ with the capacitor initially uncharged. A nice, clean statement of the pertinent initial value problem is now di(t) d2 i(t) + 2® + !02 i(t) = 0 2 dt dt with initial conditions i(0) = 0

and

i0 (0) = V0 =L;

where the introduction of ®,

R 2L

and

!02 ,

1 LC

simpli¯es the mathematics by clarifying the physically-important combinations of the original R, L, and C circuit values. The exact solution to this well-posed initial value problem is V0 ¡®t i(t) = e sin ¯t ¯L where q ¯ , !02 ¡ ®2 : p This form is most natural for the subject underdamped case where R < 2 L=C. 177

Discrete-time approximation: The discretization of the derivative derives from the de¯ning limit as f (nT ) ¡ f ((n ¡ 1)T ) f [n] ¡ f [n ¡ 1] df (t) f (t) ¡ f (t ¡ ¢t) , lim =) =) : ¢t!0 dt ¢t T T A successive application of this gives the second-derivative d df (t) =) dt dt

f [n] ¡ f [n ¡ 1] f [n ¡ 1] ¡ f [n ¡ 2] ¡ f [n] ¡ 2f [n ¡ 1] + f [n ¡ 2] T T = : T T2

The discretized forms of the initial conditions i(0) = 0 and i0 (0) = V0 =L are thus i[0] = 0

i[¡1] = ¡V0 T =L:

and

The di®erence equation that approximates the continuous-time di®erential equation is similarly ª © 1 + 2®T + (!0 T )2 i[n] ¡ 2(1 + ®T )i[n ¡ 1] + i[n ¡ 2] = 0: Using the two initial conditions at n = ¡1 and n = 0, a recursive solution can be constructed for n = 1; 2; : : : according to i[n] =

2(1 + ®T ) 1 i[n ¡ 1] ¡ i[n ¡ 2]: 2 1 + 2®T + (!0 T ) 1 + 2®T + (!0 T )2

Numerical Results: The circuit values R = 200 (−)

L = 20 (H)

C = 40 (¹F)

give ® = 5 (s¡1 )

and

¯ = 35 (s¡1 ):

The MATLAB code discreteRLC.m compares the normalized exact solution ¯L i(t) = e¡®t sin ¯t V0 to the recursive solution of the discrete-time problem.

178

Normalized Current 1.0 solid curve: exact solution dashed curve: sampling period T = 0:010 (s)

0.5

0.0

¡0:5

............ ... .... .. .. . .. . . .. .. ... ..... . . ..... ... .. . .. ... .... ... .. .. . .. ... .. .. .. ... .. .. .. .. ...... .. .... .. . ..... ..... .... ..... .... ..... ........ ... ..... ...... ........ .... ... ... ... .... ... ... ... ... . ... . .... .. ... ..... ... . ...... .. . ..... . ... .. . ...................... ..... . ... .. . ........ ...... ... .. . .. ... ... ... ... ... ..... ..... ..... . . . . . . .. . ... .. ...... . . ..... .... . . . . . . . .. . . . . .. . . .............................................................................................................................................................................................................................................................................................................................................................................................................................................. . . . . . ...... . . . . . . . . . . . ... ... ... ... ....... ... ... .. . ... .. ....... . . . ... ... .. ........................ .. .. ......... ... ... .. . . . . . . . ... . ... . ... . . . . . . . . . . . . .. .. .... ..... . . . . . . .. .. .. .. ..... ....................... .. ..... .... ... .. .. ... ... ... .. .... .. .. ... .. . ... ... .. . ... .. .. ... .. .. .. . ... .. .. .. ... ..... ....... .....

0.0

0.1

0.2

0.3

0.4

0.5

t (s)

Normalized Current 1.0 solid curve: exact solution dashed curve: sampling period T = 0:001 (s)

0.5

0.0

¡0:5

.... .............. .... ........ ..... ..... . . ..... ... ..... ... ..... .... .. .... . ..... .... ... .. ... .. .... . .. .... ... ... ... .. .. .......... .. ... ..... ...... .. ... ... ... ... ............ .. ........ ... ...... .. . .. ...... ... . .... ..... .. .. .... ... ..... ... . ... .... .............. ... .. . ........ . ... ......... ... ... ... .. . ...... ... .. ... .................... ... ... ... ........... ... . ....... . . . . . . . . . ... . . . . . ......................................................................................................................................................................................................................................................................................................................................................................................................... .......... .. .. .... ... .. ... ......... ... ... ... ... . . . . . . . .. ..... . . . ...................... . . . . . ... ...... ... . . . . . . . . ... ...... .... . .. . . ... . . . . .......... . ... ........ . .... . . . ..... .... ...................... ... .... ... ... .... .... . . . ..... ... ..... ... ..... .... .... ..... . .... ...... ......... ... ... ... .. .... ..... .......

0.0

0.1

0.2

0.3 t (s)

179

0.4

0.5