Continuous and Discrete Time Signals and Systems (Mandal & Asif) Solutions - Chap05

Chapter 5: Continuous-time Fourier Transform Problem 5.1 (a)

The CTFT for x1(t) is given by ∞

∫

X 1 (ω) =

− jωt − jωt ∫ [ τt ] e dt + ∫ [1 − τt ] e dt

−∞

−τ

 = 1+ 

[ ] t τ

− jωt

2 ω2 τ

=τ (b)

−

[]

e × − (− jω)

 1 = −  (− jω) =

τ

0

x1 (t )e − jωt dt = 1 +

[1τ ]×

(− jω)

ω2 τ

(ωτ / 2)

2

  + × 1−  (− jω) 2  −τ  e

1

2 cos(ωτ)

sin 2 (ωτ / 2)

1 τ

0

0

2

= 2×

+

− jωt

[1τ ]×

[ ] t τ

τ

e − jωt e − jωt  × − − 1τ ×  (− jω) (− jω) 2  0

[ ]

e jωτ   1 e − jωτ 1 + × − −   τ 2 2 (− jω) (− jω)   (− jω)

[]

1 − cos(ωτ) ω2 τ

= 2×

[1τ ]×

  (− jω)  1

2

2 sin 2 (ωτ / 2) ω2 τ

 ωτ  = τ sinc 2   .  2π 

The CTFT for x2(t) is given by

X 2 (ω ) =

∞

∫

x2 (t )e− jω t dt =

−∞

= t 4 

e− ( a+ jω ) t ( − ( a + jω ))

∞

∞

−∞

0

4 − at 4 − ( a + jω ) t − jω t dt ∫ t e u(t )e dt = ∫ t e

− ( a + jω ) t

+ 4t 3 (( −e( a + jω ))2 + 12t 2

e− ( a+ jω ) t ( − ( a + jω ))3

∞

+ 24t (( −e( a + jω ))4 + 24 ( −e( a + jω ))5  0 − ( a+ jω ) t

− ( a + jω ) t

= [0 + 0 + 0 + 0 + 0] − 0 + 0 + 0 + 0 + 24 ( − ( a +1 jω ))5  = ( a +24jω )5 .   (c)

The CTFT for x3(t) is given by

X 3 (ω ) =

∞

∫ x (t )e 3

−∞

=

1 2

− jω t

∞

dt =

∫e

− at

−∞

cos(ω0t )u(t )e

− jω t

∞

dt =

1 2

∫e

− ( a + jω ) t

0

∞

∞

0

0

∞

 e jω0t + e− jω0t  dt ∞

0 0 − ( a + jω − jω0 ) t dt + 12 ∫ e− ( a + jω + jω0 )t dt = 12  ( −e( a + jω − jω0 ))  + 12  ( −e( a + jω + jω0 ))  ∫e 0 0 − ( a + jω − jω ) t

− ( a + jω + jω ) t

= 12 0 − ( − ( a + jω1 − jω0 ))  + 12 0 − ( − ( a + jω1 + jω0 ))  = 12  ( a + jω1− jω0 ) + ( a + jω1+ jω0 )  = ( a +ajω+ )jω2 +ω 2 . 0 (d)

The CTFT for x4(t) is given by

160

Chapter 5

X 4 (ω ) =

∞

∫

x4 (t )e− jω t dt =

−∞

∞

∫

2 2

∫

2

−∞

)  = exp  ( jωσ ⋅  2σ 2 

∞

exp( − 2tσ 2 )e− jω t dt =

∞

∫ exp −

( t + jωσ 2 )2 2σ 2

−∞

exp[ − t

2

+ 2 jωσ 2t 2σ

−∞

2

∞

]dt =

∫ exp −

t 2 + 2 jωσ 2t + ( jωσ 2 )2

−∞

2σ 2

 exp  ( jωσ 2 )  dt   2σ  2 2

 dt = exp  ( jωσ 2 )  2πσ = 2πσ exp  − ω 22σ 2  .     2σ  2 2

▌ Problem 5.2 (a)

By definition,

[ ]

π

∫

X 1 (ω) = 3e − jωt dt = 3 (e− jω) − jωt

0

3 e − jωπ / 2 jω

=−

π 0

=−

3 jω

[e

− jωπ

]

−1 = −

3 − jωπ / 2 e jω

[e

− jωπ / 2

By definition,

X 2 (ω ) =

0.5T

∫

0.5e

− jωt

1.5T

dt +

−0.5T

∫

0.5T

e− jωt dt = 0.5  (e− jω )  − jωt

 − j 0.5ωT − e j 0.5ωT  − = − 0.5 jω  e = − 0.5 jω [ −2 j sin(0.5ωT ) ] −

1 jω

1 jω

0.5T − 0.5T

+  (e− jω )  − jωt

e− j1.5ωT − e− j 0.5ωT 

e− jωT [ −2 j sin(0.5ωT )]

sin(0.5ωT ) sin(0.5ωT ) T T  + 2e− jωT  0.5  =  0.5 0.5T × ω 0.5T × ω = 0.5Tsinc ( 0.5πωT ) + Te− jωT sinc ( 0.5πωT ) .

(c)

By definition, T

∫(

)

(

X 3 (ω) = 1 − Tt e − jωt dt =  1 − Tt  0

) (e− jω) − (− T1 ) (−e jω)  − jωt

− jωt

2

T 0

( ) (−e jω)  −  (−1jω) − (− T1 ) (− j1ω) 

= 0 − − T1  =−

1 ω2T

− jωT

e − jωT +

2

1 jω T

For ω = 0,

∫(

2

+

1 ω2T

)

=

1 jω

+

1 ω2T

(

(1 − e

X 3 (ω) = 1 − Tt dt = − T2 1 − Tt  0

]

/ 2) [− 2 j sin(ωπ / 2)] = 6e − jωπ / 2 [sin(ωπ ]= 6e − jωπ / 2 [21/ π × sin(ωπωπ/ 2/ 2) ] ω

= 3πe − jωπ / 2 sinc(ω / 2). (b)

− e jωπ / 2

− jωT

)2 

T 0

).

= 0 + T2 = T2 .

1.5T 0.5T

Solutions (d)

By definition, 0

X 4 (ω) =

∫ (1 + )e t T

− jωt

T

∫(

−T

(

=  1 + Tt 

(e)

2 ω2T

0

) (e− jω) − (T1 ) (−e jω)  − jωt

− jωt

2

=  ( − 1jω) −  =

)

dt + 1 − Tt e − jωt dt

(T1 ) (− j1ω)

2

−0+

0 −T

(

+  1 − Tt 

) (e− jω) − (− T1 ) (−e jω)  − jωt

− jωt

2

(T1 ) (−e jω)  + 0 − (− T1 ) (−e jω) jωT

− jωT

2

2

T 0

( ) (− j1ω) 

− ( − 1jω) + − T1

[1 − cos(ωT )] = 2×2 sinω (T0.5ωT ) = 1/(0.45 T ) × sin(0.(50ω.5Tω)T ) = Tsinc 2 ( 0.5πωT ) . 2

2

2

2

2

By definition, T

T

0

0

T

X 5 (ω ) = ∫ 1 − 0.5sin ( πTt )  e− jωt dt = ∫ e− jωt dt − 0.5∫ sin ( πTt ) e− jωt dt 0

=A

=B

We consider different cases for the above integral. Case I: ( ω = 0) −∞

X 5 (0) =

∫

∞

x(t )dt =

∞

=T

∫

−∞

+ π0.5 /T

T

T

0

0

1 − 0.5sin ( πTt )  dt = ∫ dt − 0.5∫ sin ( πTt ) dt T

cos( T )  0 = T + πt

T 2π

[cos(π ) − cos(0)] = T − Tπ = T (1 − π1 )

Case II: ( ω ≠ 0, ω ≠ π/T): T

T A = ∫ e− jωt dt = −1jω e− jωt  = −1jω e− jωT − 1 = j1ω 1 − e− jωT  0

[ω ≠ 0]

0

T

 − jωt  B = 0.5  πe2 2 {− jω sin ( πTt ) − πT cos ( πTt )} −ω 2 T 0

for ω ≠ 0, ± πT T

     = π 20.5−ωT2T 2 e− jωt  jω sin ( πTt ) + πT cos ( πTt )    =0 at t =0,T   0 2

= π 20.5−ωT2T 2  − πT e− jωT − πT  = π 20.5−ωπ2TT 2 1 + e− jωT  2

Case III: ( ω = π/T):

2

161

162

Chapter 5

T

B = 0.5∫ sin ( πTt ) e− jωt dt = 0

T 0.5 2j

T

−j − j (ω − ) t − j (ω + ) t j − jωt ∫ e T − e T  e dt = 0.52 j ∫ e T − e T  dt πt

πt

0

 0.5 − j 2Tπ t  dt ω = π  2 j ∫ 1 − e T   0 = T  0.5  j 2Tπ t π  −  2 j ∫ 1 − e  dt ω = − T 0 

π

π

0

T

 ± j 2Tπ t is periodic with period T ,  As e 

T

∫e 0

± j 2Tπ t

 dt = 0 

0.5T = ± 0.5 2 j [ t ]0 = ± 2 j T

Combining, the above results, the CTFT can be expressed as  T (1 − π1 )  X 5 (ω ) =  j1ω 1 − e− jωT  ∓ 0.52 jT 1 − jωT  − π 20.5−ωπ2TT 2 1 + e− jωT   jω 1 − e  T (1 − π1 )  =  ± 2jπT ∓ 4Tj 1 − jωT  − π 20.5−ωπ2TT 2 1 + e− jωT   jω 1 − e

ω =0 ω = ± πT otherwise

ω =0 ω = ± πT otherwise

▌ Problem P5.3

From magnitude and phase spectra shown in Fig. P5.3, the individual CTFT’s can be expressed as follows Fig. P5.3(b):

j 0.5ω  X 1 (ω) = 1 × e 0 

Fig. P5.3(c):

1× e− j 0.5ω  X 2 (ω ) =  1× e j 0.5ω  0 

Fig. P5.3(d):

1× e− jπ / 3  X 3 (ω ) =  1× e jπ / 3  0 

−W ≤ ω≤W otherwise

−W ≤ ω ≤ 0 0 ≤ω ≤W otherwise −W ≤ ω ≤ 0 0 ≤ω ≤W otherwise

Using the CTFT synthesis Eq. (5.9), the function x1 (t ) is calculated as follows.

Solutions

x1 (t ) =

∞

1 2π

∫

X (ω )e jωt dω =

−∞

W

1 2π

∫

e j 0.5ω e jωt dω =

−W

W

1 2π

∫e

=

 1 2 j sin [ (0.5 + t )W ] = j (0.5 + t )  2π

1  e j (0.5+t )ω  1  e j (0.5+t )W − e− j (0.5+t )W =    2π  j (0.5 + t )  −W 2π  j (0.5 + t ) W

π

×

sin [ (0.5 + t )W ] (0.5 + t )W

=

W

dω

−W

W

=

j (0.5 + t )ω

sinc Wπ (t + 0.5) .

π

Using the CTFT synthesis Eq. (5.9), the function x2 (t ) is calculated as follows.

1 x2 (t ) = 2π

∞

0

1 ∫−∞ X (ω )e dω = 2π

∫e

jωt

j ( −0.5+ t )ω

−W

1 dω + 2π

0

W

∫e

j (0.5+ t )ω

dω

0

W

1  e j ( −0.5+t )ω  1  e j (0.5+t )ω  1 1 − e− j ( −0.5+t )W e j (0.5+t )W − 1  = + = + j (0.5 + t )  2π  j (−0.5 + t )  −W 2π  j (0.5 + t )  0 2π  j (−0.5 + t ) =

2 jt sin(tW ) − cos(tW )  1  1 + e j 0.5W   2 j (t 2 − 0.25) 2π  j (t − 0.25) 

=

1 1 + e j 0.5W ( 2 jt sin(tW ) − cos(tW ) )  . j 2π (t − 0.25)  2

Clearly, at (t = ±0.5), x2(t) is undefined in the above expression. Computing directly, we obtain W

At t = 0.5:

x2 (0.5) =

1 2π

j 0.5ω j 0.5ω ∫ e e dω =

W 1 2π

0

0

At t = −0.5: x2 ( −0.5) =

1 2π

∫

e − j 0.5ω e − j 0.5ω d ω =

−W

∫e

jω

dω =

1 2 jπ

0

0

1 2π

∫e

− jω

W

 e jω  = 0

dω =

1 −2 jπ

−W

1 2 jπ

 e jW − 1 . 0

 e − jω  = −W

1 2 jπ

 e jW − 1 .

Using the CTFT synthesis Eq. (5.9), the function x3 (t ) is calculated as follows.

x3 (t ) =

1 2π

∞

∫

X (ω )e jωt d ω =

−∞ 0

1 2π

0

∫

e − jπ / 3e jωt dω +

−W

1 2π

0

∫e

jπ / 3

e jωt dω

−W

W

jWt 1 − jπ / 3  e jωt  1 jπ / 3  e jωt  1  − jπ / 3 1 − e − jWt − 1 jπ / 3 e e e e = + = +e      jt jt  2π  jt  −W 2π  jt  0 2π 

=

1 sin(Wt + π / 3) − sin(π / 3)  [ 2 j sin(Wt + π / 3) − 2 j sin(π / 3)] =   πt j 2π t 

Clearly, at (t = 0), x3(t) is undefined in the above expression. Computing directly, we get

163

164

Chapter 5

x3 (0) =

1 4π

0 W    (1 − j 3) ∫ d ω + (1 + j 3) ∫ d ω  = −W 0  

W 4π

1 − j 3 + 1 + j 3  =  

W 2π

.

Although the functions x1 (t ) , x2 (t ) , and x3 (t ) have the same magnitude spectra, their phase spectra are different. As a result, the time domain representations of these functions are different.

abs(x2 (t))

x1 (t)

For the special case W = π, the three functions are plotted in Fig. S5.3. Since x2(t) is a complex function, its magnitude is plotted in Fig. S5.3. The Matlab code is also included below. ▌

Problem 5.3

1 0.5 0 -0.5 -1 -5

-4

-3

-2

-1

0 t

1

2

3

4

5

1 0.75 0.5 0.25 0 -5

-4

-3

-2

-1

0 t

1

2

3

4

5

-4

-3

-2

-1

0 t

1

2

3

4

5

x3 (t)

1 0.5 0 -0.5 -1 -5

Fig. S5.3. Plots of functions in Problem P5.3. % MATLAB code to plot the functions in Problem 5.3 del = 0.01; t = -5:del:5; W = pi ; x1 = (W/pi)*sinc((W/pi)*(t+0.5)) ; x2 = 1./(j*2*pi*(t.^2-0.25)).*(1+exp(j*0.5*W)*(2*j*t.*sin(t*W)-cos(t*W))); x2(t==0.5) = 1./(j*2*pi)*(exp(j*W)-1); x2(t==-0.5) = 1./(j*2*pi)*(exp(j*W)-1); x3 = (sin(W*t+pi/3)-sin(pi/3))./(pi*t); x3(t==0) = W/(2*pi) ; subplot(3,1,1), plot(t, x1), grid on title('Problem 5.3');

Solutions

xlabel('t') % Label of ylabel('x_1(t)') % Label of % subplot(3,1,2), plot(t, abs(x2)), grid xlabel('t') % Label of ylabel('abs(x_2(t))') % Label of

X-axis Y-axis on X-axis Y-axis

subplot(3,1,3), plot(t, x3), grid xlabel('t') % Label of X-axis ylabel('x_3(t)') % Label of Y-axis

Problem 5.4

(a)

The partial fraction expansion is given by X 1 (ω) =

(1 + jω) −1 2 ≡ + (2 + jω)(3 + jω) (2 + jω) (3 + jω)

Calculating the inverse CTFT, we obtain

x1 (t ) = − e −2t u (t ) + 2e −3t u (t ) . (b)

The partial fraction expansion is given by X 2 (ω) =

−1 1 0 .5 0.5 ≡ + + (1 + jω)(2 + jω)(3 + jω) (1 + jω) (2 + jω) (3 + jω)

Calculating the inverse CTFT, we obtain

x 2 (t ) = 0.5e −t u (t ) − e −2t u (t ) + 0.5e −3t u (t ) . (c)

The partial fraction expansion is given by

X 3 (ω) =

1 2

(1 + jω)(2 + jω) (3 + jω)

≡

−1 0 .5 0 − 0 .5 + + + (1 + jω) (2 + jω) (2 + jω) 2 (3 + jω)

Calculating the inverse CTFT, we obtain

x3 (t ) = 0.5e −t u (t ) − te −2t u (t ) + 0.5e −3t u (t ) . (d)

The partial fraction expansion is given by X 4 (ω) =

1 2

(1 + jω)(2 + 2 jω + ( jω) ) X 4 (ω) =

or,

≡

1 + jω 1 − (1 + jω) (2 + 2 jω + ( jω) 2 )

1 + jω 1 − (1 + jω) 1 + (1 + jω) 2

Calculating the inverse CTFT, we obtain

x 4 (t ) = e −t u (t ) − e −t cos t u (t ) . (e)

The partial fraction expansion is given by X 5 (ω) =

1 (1 + jω) 2 (2 + 2 jω + ( jω) 2 ) 2

≡

1 (1 + jω) 2

−

1.50 (2 + 2 jω + ( jω) 2 )

+

0.25(4 jω + ( jω) 2 ) (2 + 2 jω + ( jω) 2 ) 2

165

166

Chapter 5

or,

X 5 (ω) =

1 (1 + jω) 2

−

1.50 1 + (1 + jω) 2

+

0.25(4 jω + ( jω) 2 ) (1 + (1 + jω) 2 ) 2

Calculating the inverse CTFT, we obtain  0.25(4 jω + ( jω) x5 (t ) = te −t u (t ) − 1.50e −t sin t u (t ) + ℑ −1   (1 + (1 + jω) 2 ) 2

2

)  . 

▌

Problem 5.5

Consider an arbitrary function φ(t), and assume that ∞

p (t ) =

∫e

j ωt

dt .

−∞

Now, consider the integral ∞

∞

∞



∞

−∞

−∞

 −∞



-∞

∞



 −∞



jω ( t −T ) jω t − jωT ∫ φ (t ) p(t − T )dt = ∫ φ (t )  ∫ e dω  dt = ∫ e  ∫ φ (t )e dt  dω = Changing the order of integration −∞

=

∫e

jω ′T

∞

∫e

Φ(ω ′)( −dω ′) =

jω ′T

∞

∫e

− jωT

Φ(−ω )dω

−∞

Φ(ω )=ℑ{φ ( t )}

Φ(ω ′)dω ′

−∞

∞

ω =−ω ′, dω =− dω ′ ∞

=

∫ Φ(ω )e

jωT

dω

................................................(1)

−∞

Note that the right hand side of Eq. (1) is the inverse CTFT of Φ(ω) computed at t = T, i.e., φ(T). Hence, ∞

∫

∞

φ(t ) p(t − T )dt =

−∞

∫ Φ(ω)e

jωT

dω = 2πφ(T ) .

−∞

The above equation is valid for any arbitrary φ(t) if and only if p(t) = 2πδ(t) as can be seen from the following property of the impulse response ∞

∫

2π φ(t )δ(t − T )dt = 2πφ(T ) . −∞

In other words, ∞

∫e

j ωt

dt = 2πδ(t ) .

−∞

Interchanging the variables, t and ω, we obtain the required identity ∞

∫e

−∞

j ωt

dω = 2πδ(ω) .

▌

Solutions

167

Alternate Proof:

Note that the above result can be proved directly from the CTFT pair CTFT 1 ← → 2πδ(ω) .

∞

∫

2πδ(ω) = 1 × e − jωt dt

By definition,

−∞ ∞

Since δ(−ω) = δ(ω),

∫e

2πδ(ω) = 2πδ(−ω) =

jωt

dt .

▌

−∞

Problem 5.6

Using Eq. (5.40), the CTFT for a real-valued even function x(t) can be expressed as ∞

∫

X (ω) =

∞

∫

x(t )e − jωt dt = 2 x(t ) cos(ωt )dt .

−∞

0

Since there is no complex value in the above equation, X(ω) is real valued, i.e., Im{X(ω)} = 0. ∞

∞

∫

∫

Also, X (−ω) = 2 x(t ) cos(−ωt )dt = 2 x(t ) cos(ωt )dt = X (ω) . 0

0

Therefore, X(ω) is also an even function with respect to ω. Since, X(ω) is real valued, Re{X(ω)} = Re{X(−ω)}. ▌ Problem 5.7

Using Eq. (5.40), the CTFT for a real-valued odd function x(t) can be expressed as ∞

X (ω) =

∫

∞

∫

x(t )e − jωt dt = − j 2 x(t ) sin(ωt )dt .

−∞

0

Since x(t) is real, the product x(t)sin(ωt) is also real and so is the integral. Therefore, X(ω) is pure imaginary, i.e., Re{X(ω)} = 0. ∞

Also,

∫

∞

∫

X (−ω) = 2 x(t ) sin( −ωt )dt = −2 x(t ) sin(ωt )dt = − X (ω) . 0

0

Therefore, X(ω) is also an odd function with respect to ω. Since, X(ω) is imaginary-valued, Im{X(ω)} = −Im{X(−ω)}. ▌ Problem 5.8

(a)

Since is not equal to

X 1 (− ω) =

5 2 + j ( − ω− 5 )

X 1∗ (ω) =

=

5 2 − j ( ω+ 5 )

5 2 − j (ω−5 )

,

168

Chapter 5 X1(ω) does not satisfy the Hermitian property. Its inverse CTFT x1(t) is not real valued and is complex. Nothing can be stated about the odd and even property of x1(t) from the Hermitian property.

(b)

(

X 2 (− ω) = cos − 2ω +

Since

is not equal to

(

X 2∗ (ω) = cos 2ω +

π 6

π 6

)=

)=

3 2

3 2

cos(2ω) + 12 sin( 2ω)

cos(2ω) − 12 sin(2ω) ,

X2(ω) does not satisfy the Hermitian property. Its inverse CTFT x2(t) is not real valued and is complex. Nothing can be stated about the odd and even property of x2(t) from the Hermitian property.

(c)

X 3 (− ω) =

Since

5 sin [4(− ω− π )] ( − ω− π )

X 3∗ (ω) =

is not equal to

=

5 sin [4(ω+ π )] ( ω+ π )

5 sin [4 (ω− π )] (ω − π )

[4ω] = 5 (sin ω+ π )

[4ω] , = 5 (sin ω− π )

X3(ω) does not satisfy the Hermitian property. Its inverse CTFT x3(t) is not real valued and is complex. Nothing can be stated about the odd and even property of x3(t) from the Hermitian property.

(d)

Since X 4 (− ω) = (3 + 2 j ) δ(− ω − 10) + (1 − 2 j )δ(− ω + 10) = (3 + 2 j )δ(ω + 10) + (1 − 2 j )δ(ω − 10)

is not equal to

X 4∗ (ω) = (3 + 2 j ) δ(ω − 10) + (1 − 2 j )δ(ω + 10) ,

X4(ω) does not satisfy the Hermitian property. Its inverse CTFT x4(t) is not real valued and is complex. Nothing can be stated about the odd and even property of x4(t) from the Hermitian property. (e)

Since

X 5 (− ω) =

is equal to

X 5∗ (ω) =

1

(1− jω)(3− jω)2 (5+ ω2 ) 1

(1− jω)(3− jω)2 (5+ ω2 )

,

X4(ω) satisfies the Hermitian property. Its inverse CTFT x4(t) is real valued. Since X4(ω) is complex (neither pure real-valued or pure imaginary), x4(t) is neither even nor odd with respect to t. ▌ Problem 5.9

(a)

Applying the linearity property,

{

}

{

}

X 1 (ω) = ℑ 5 + 3 cos(10t ) − 7e −2t sin(3t )u (t ) = 5ℑ{1} + 3ℑ{cos(10t )} − 7 ℑ e −2t sin(3t )u (t ) . By selecting the appropriate CTFT pairs from Table 5.2, we get X 1 (ω) = 10δ(ω)ℑ{1} + 3πδ(ω − 10) + 3πδ(ω − 10) − (b)

Entry (8) of Table 5.2 provides the CTFT pair CTFT

sgn(t ) ← → Using the duality property,

2 jt

CTFT

2 jω

.

← → 2π sgn( −ω) ,

21 ( 2 + jω) 2 + 3 2

.

Solutions

1 πt

or, (c)

CTFT

← → − j sgn(ω) .

Entry (7) of Table 5.2 provides the CTFT pair

e Using the time shifting property, e

−4 t

−4 t −5

CTFT

← → 4+8jω .

← → 4+8jω e − j 5ω . CTFT

Using the frequency differentiation property,

t2 e t2 e

or, (d)

169

−4 t − 5

−4 t −5

CTFT

← →( j ) 2

← → 200e − j 5ω CTFT

d2 dω 2 1 4+ jω

{e

− j 5ω

8 4 + jω

+ 16e − j 5ω

}

1 ( 4 + jω) 3

.

Entry (17) of Table 5.2 provides the CTFT pair

(6ωπ ) sin(5 πt ) CTFT 5 sinc(5t ) = 5 5πt ← → rect (10ωπ ) sin(3πt ) 3πt

3 sinc(3t ) = 3

and

CTFT

← → rect

Using the multiplication property

[ (6ωπ )∗ rect(10ωπ )] sin( 3πt ) sin( 5 πt ) CTFT ← → π2 [rect (6ωπ ) ∗ rect (10ωπ )] , t

π2 × or, or,

sin( 3πt ) πt

×

sin( 5 πt ) πt

CTFT

← → 2ππ rect 2

2

πt) CTFT ← → 52π  rect ( 6ωπ ) ∗ rect ( 10ωπ )  , 5 sin(3π t t)sin(5 2

where * is the convolution operation. (e)

Entry (17) of Table 5.2 provides the CTFT pair

(6ωπ ) sin( 4 πt ) CTFT 4 sinc(4t ) = 4 4 πt ← → rect (8ωπ ). sin(3πt ) 3πt

3 sinc(3t ) = 3

and

CTFT

← → rect

Using the time differentiation property, 1 d sin( 4 πt ) π dt t

CTFT

← →( jω) rect

(8ωπ ).

Using the convolution property

[ (6ωπ )× jωrect(8ωπ )] sin( 3πt ) sin( 4 πt ) CTFT ∗ dtd ← → π2 [rect (6ωπ ) × jωrect (8ωπ )], t t sin( 3πt ) sin( 4 πt ) CTFT 4 t ∗ dtd ← → j 2π rect (6ωπ ). t

π2 × or, or,

sin(3πt ) πt

∗ π1

d sin( 4 πt ) dt t

CTFT

← → 2ππ rect 2

▌

170

Chapter 5

Problem 5.10

Using the linearity property,

{

}

X (ω ) = ℑ  136 e−2t − 136 cos(3t ) + 134 sin(3t )  u (t )

{

}

= 136 ℑ e−2t u (t ) − 136 ℑ{cos(3t )u (t )} + 134 ℑ{sin(3t )u (t )} = 136 2+1jω − 136  π2 δ (ω − 3) + π2 δ (ω + 3)  − 136 9−jωω 2 + 134  − 2jπ δ (ω − 3) +

jπ 2

δ (ω + 3)  + 134 9−3ω

2

π = 136  2+1jω − 9−jωω 2 + 9−2ω 2  − 26 [6δ (ω − 3) + 6δ (ω + 3) + 4 jδ (ω − 3) − 4 jδ (ω + 3)] π = 136  (9−ω(2)++j(2ω −)(9jω−ω)(22 )+ jω )  − 26 [(6 + j 4)δ (ω − 3) + (6 − j 4)δ (ω + 3)]   6 = (2+ jω )(9 − 13π [ (3 + j 2)δ (ω − 3) + (3 − j 2)δ (ω + 3)] −ω 2 ) 2

which is the required result.

▌

Problem 5.11 ∞

F {x(at )} =

From the definition of CTFT,

∫ x(at )e

− jω t

dt .

−∞

We consider two different cases (a > 0) and (a < 0) Case 1:

Assume a > 0. Substitute r = at in the above expression. The upper and lower limits of integration stay the same and dr = a dt. The final result is F {x(at )}

Case 2:

∞

∞

− j (ω ) r = x(r )e a dra −∞

∫

=

1 a

∫ x ( r )e

− j ( ωa ) r

−∞

dr = 1a X

(ωa ) .

Assume a < 0. Substitute r = at in the above expression. The upper limit of integration is r → −∞ and the lower limit of integration is r → ∞ , and dr = a dt. The final result is F {x(at )} =

∞

∫

x ( r )e

−∞

− j ( ωa ) r dr a

−∞

=

1 a

∫

x ( r )e

− j ( ωa ) r

∞

dr = −

∞

Combining the two cases, yields F {x(at )} =

1 a

X

1 a

∫ x ( r )e

−∞

(ωa ) .

− j ( ωa ) r

dr = −

1 a

X

(ωa ) . ▌

Problem 5.12

Comparing with Fig. 5.9(a), we observe that h(t ) = x1

H (ω) = 2 X 1 (2ω)

Using the scaling property, or, which simplifies to

(2t ) .

H (ω) =

2 ω2

[2ω sin( 4ω) + cos(2ω) − 1] ,

H (ω) = 16sinc

(4πω ) − 4sinc 2 (ωπ ) .

▌

Solutions

171

Problem 5.13

Using the definition of CTFT, we obtain

{

Fe

jω 0 t

∞

} ∫

x(t ) = e

jω 0 t

x(t )e

− jω t

−∞

∞

dt =

∫ x(t )e

− j ( ω − ω0 ) t

dt = X (ω − ω 0 ) .

▌

−∞

Problem 5.14

Using the convolution property,

[

x(t ) ∗ u (t ) ←  → X (ω) 2πδ(ω) + CTFT

1 jω

],

∞

∫ x(τ )u(t − τ )dτ ←→ 2π X (0)δ (ω ) + CTFT

or,

X (ω ) jω

,

−∞

∞

∫ x(τ )u(−(τ − t ))dτ ←→ 2π X (0)δ (ω ) + CTFT

or,

X (ω ) jω

,

−∞

t

∫ x(τ )dτ ←→ 2π X (0)δ (ω ) + CTFT

or,

X (ω ) jω

.

−∞

Problem 5.15

(a)

→ Using the time scaling property, x(2t ) ←  CTFT

1 2

X

(ω2 ).

CTFT → Using the frequency shifting property, e − j 5t x(2t ) ←

1 2

X ( ω2+ 5 ) .

Substituting the value of X(ω), we obtain

1 − 3 ℑ{e − j 5t x(2t )} = 12   0  ω12+11  −ω =  112  0 

ω +5

(b)

ω +5 ≤ 3 elsewhere −11 ≤ ω ≤ −5 −5 ≤ ω ≤ −1 elsewhere.

Using the frequency differentiation property, CTFT ( jt ) 2 x(t ) ← →

d2X dω2

,

CTFT t 2 x(t ) ← → − ddωX2 . 2

or, The CTFT of t2 x(t) is given by

{

}

F t 2 x(t ) = −

d2 dω 2

[∆(ω3 )] = − ddω [rect(ω3 )] = −[δ(ω + 3) − δ(ω − 3)] = [δ(ω − 3) − δ(ω + 3)] .

▌

172

(c)

Chapter 5 =t Express (t + 5) dx dt

dx dt

+ 5 dx . dt

Using the time differentiation property, the CTFT of dx dt

dx dt

is given by

CTFT ← → jωX (ω) .

Applying the frequency differentiation property to the above CTFT pair, gives

t

dx dt

CTFT ←  → j ddω [ jωX (ω)] = − X (ω) − ω dX dω .

The CTFT of ( t + 5 ) dx dt is given by

{

}

ℑ (t + 5) dx = − X (ω) − ω dX + 5 jωX (ω) . dt dω

Substituting the value of X(ω), we obtain

{

ℑ (t + 5)

(d)

dx dt

}

( (

) ( ) (

 j 5ω 1 − ω3 − 1 −  =  j 5ω 1 + ω3 − 1 +  0 

2ω 3 2ω 3

) )

0≤ω≤3 −3≤ω≤0 elsewhere.

Using the time multiplication property, x(t ) ⋅ x(t ) ← → 21π [ X (ω) ∗ X (ω)] , CTFT

which implies that F {x(t ) ⋅ x(t )} =

(e)

1 2π

[∆(ω3 )∗ ∆(ω3 )].

Using the time convolution property, CTFT x(t ) * x(t ) ← → X (ω) ⋅ X (ω) ,

which reduces to 2  ω  1 − 3   F {x(t ) ∗ x(t )} =    0 

(f)

ω ≤3

1 +  = elsewhere 

ω2 9

x(t ) ⋅ cos ω 0 t ← → 21π X (ω) ∗ πδ(ω − ω 0 ) +

1 2π

−

2ω 3

0

ω ≤3 elsewhere.

Using the time multiplication property, CTFT

or,

x(t ) ⋅ cos ω 0 t ← → 12 X (ω − ω 0 ) + CTFT

1 2

X (ω) ∗ πδ(ω + ω 0 ) , X (ω + ω 0 )

Case I: For ω0 = 3/2, we obtain

(

x(t ) ⋅ cos(3t / 2) ← → 12 X ω − CTFT

3 2

) + 12 X (ω + 32 ).

The two replicas overlap over (−3/2 ≤ ω < 3/2), therefore,

Solutions

 12 + ω+63 / 2  1  F {x(t ) cos(3t / 2)} =  1 ω−3 / 2 2 + 6  0 

173

− 92 ≤ ω ≤ − 32 − 32 ≤ ω ≤ 32 3 ≤ω≤ 9 2 2 elsewhere.

Case II: For ω0 = 3, we obtain x(t ) ⋅ cos 3t ← → 12 X (ω − 3) + CTFT

1 2

X (ω + 3) .

Since there is no overlap between the two shifted replicas, 1 − ω+3 3  ω−3  1 F {x(t ) cos 3t} = 2 1 − 3   0

or,

 1 − ω+ 3 6 2  1 ω− 3 F {x(t ) cos 3t} =  2 − 6   0

ω+3 ≤3 ω−3 ≤3 elsewhere. −6≤ω1

The above output y (t ) is plotted in the last row of Fig. S5.22. The output response matches our expectation from our circuit theory knowledge. At t = −T / 2 , the input voltage becomes 1 volt, and the capacitor starts charging resulting in an increase in the output voltage. The increase continues until t = T / 2 at which the input becomes zero. After t = T / 2 , the capacitor starts discharging resulting in an ▌ exponential decrease of the output voltage. The output voltage becomes zero at t = ∞ .

186

Chapter 5

h(τ) 1 CR

1 e − τ / CR CR

h(τ)

τ

0

1

v(τ)

v(τ) 0

− T2

τ

T 2

v(t − τ)

1

v(t − τ) t − T2

τ

0

t + T2

h(τ) v(t − τ) 1 CR

Case I: (t < −T/2) t − T2

1 e − τ / CR CR

τ

0

t + T2

h(τ) v(t − τ) 1 CR

Case II: (−T/2 < t < T/2)

1 e − τ / CR CR

τ

0 t+T 2

t − T2

h(τ) v(t − τ)

Case III: (t > T/2)

1 CR

1 e − τ / CR CR

0

y (t ) ( for T = 2, R = 1 MΩ, C = 1 µ F)

t − T2

t + T2

τ

Solutions Figure S5.22: Convolution of the input signal v(t) with the impulse response h(t) in Problem 5.22.

Program: The MATLAB code to plot y(t) in Problem 5.22 t = -2:0.001:3; % P5.20(a) y = 0*(t-1).*(t1); plot(t,y); grid on; xlabel('t'); ylabel('y(t)');

Problem 5.23 (i)

As determined in Problem 5.22, the transfer function of the RC series circuit is given by H (ω) =

1 1 × . CR 1 /(CR ) + jω

Calculating the CTFT of the input, we obtain

X 1 (ω) = π[δ(ω − ω 0 ) + δ(ω + ω 0 )] . Using the modulation property, the CTFT of the output signal is obtained as Y (ω) = H (ω) X 1 (ω) =

π π 1 1 × δ(ω − ω 0 ) + × δ( ω + ω 0 ) CR 1 /(CR ) + jω CR 1 /(CR ) + jω

which reduces to

or,

Y (ω) =

π 1 π 1 × δ( ω − ω 0 ) + × δ( ω + ω 0 ) , CR 1 /(CR ) + jω 0 CR 1 /(CR ) − jω 0

Y (ω) =

π 1 /(CR ) − jω 0 π 1 /(CR ) + jω 0 × δ( ω − ω 0 ) + × δ( ω + ω 0 ) , 2 2 CR 1 /(CR ) + ω 0 CR 1 /(CR ) 2 + ω 02 1 /(CR ) π [δ(ω − ω0 ) + δ(ω + ω0 )] × CR 1 /(CR ) 2 + ω 02 − jω 0 π [δ(ω − ω0 ) − δ(ω + ω0 )] . + × CR 1 /(CR ) 2 + ω 02

Y (ω) =

or,

Calculating the inverse CTFT, we obtain Y (ω) =

− jω 0 1 /(CR ) 1 1 × cos( ω t ) + × j sin(ω 0 t ) , 0 CR 1 /(CR ) 2 + ω 02 CR 1 /(CR ) 2 + ω 02

or, y (t ) = which can be expressed as

1 2

1 + C R 2 ω 02

[cos(ω0 t ) + ω0 CR sin(ω0 t )] ,

187

188

Chapter 5

y (t ) = (b)

1 2

1+ C R

2

ω 02

[

]

cos ω 0 t − tan −1 (ω 0 CR ) .

As determined in Problem 5.22, the transfer function of the RC series circuit is given by H (ω) =

1 1 × . CR 1 /(CR ) + jω

Calculating the CTFT of the input, we obtain X 1 (ω) = jπ[δ(ω − ω 0 ) − δ(ω + ω 0 )] . Using the modulation property, the CTFT of the output signal is given by Y (ω) = H (ω) X 1 (ω) =

π 1 π 1 × δ( ω − ω 0 ) − × δ( ω + ω 0 ) jCR 1 /(CR ) + jω jCR 1 /(CR ) + jω

which reduces to

or,

Y (ω) =

π π 1 1 × δ( ω − ω 0 ) − × δ( ω + ω 0 ) , jCR 1 /(CR ) + jω 0 jCR 1 /(CR ) − jω 0

Y (ω) =

1 /(CR ) + jω 0 1 /(CR ) − jω 0 π π δ( ω − ω 0 ) − × δ( ω + ω 0 ) , × 2 2 jCR 1 /(CR ) + ω 0 jCR 1 /(CR ) 2 + ω 02 1 /(CR ) π [δ(ω − ω0 ) − δ(ω + ω0 )] × jCR 1 /(CR ) 2 + ω 02 − jω 0 π [δ(ω − ω0 ) + δ(ω + ω0 )] . + × jCR 1 /(CR ) 2 + ω 02

Y (ω) = or,

Calculating the inverse CTFT, we obtain Y (ω) =

ω0 1 /(CR ) 1 1 sin( ) × ω t − × cos(ω 0 t ) , 0 CR 1 /(CR ) 2 + ω 02 CR 1 /(CR ) 2 + ω 02

or, y (t ) =

1 2

1 + C R 2 ω 02

[sin(ω0 t ) − ω0 CR cos(ω0 t )] ,

which can be expressed as y (t ) =

1 1 + C 2 R 2 ω 02

[

Problem 5.24

The transfer functions obtained in Problem 5.20 are as follows: (a) H (ω ) = (b) H (ω ) =

1

( jω )

3

+ 6 ( jω ) + 11( jω ) + 6 2

1

( jω )

2

+ 3 ( jω ) + 2

]

sin ω 0 t − tan −1 (ω 0 CR ) .

▌

Solutions

1

(c) H (ω ) = (d)

( jω ) + 2 ( jω ) + 1 ( jω ) + 4 1 = . H (ω ) = 2 ( jω ) + 6 ( jω ) + 8 2 + jω

(e) H (ω ) =

2

1

( jω )

3

+ 8 ( jω ) + 19 ( jω ) + 12 2

The MATLAB code to plot the magnitude and phase spectra is given below: w = -5:0.001:5; % % P5.20(a) H = 1./((j*w).^3+6*(j*w).^2+11*(j*w)+6); subplot(5,2,1) plot(w,abs(H)); grid on; xlabel('\omega (radians/s)'); ylabel('P5.20(a): |H_1(\omega)|'); axis tight subplot(5,2,2) plot(w,angle(H)); grid on; xlabel('\omega (radians/s)'); ylabel('P5.20(a):