Continuous and Discrete Time Signals and Systems (Mandal & Asif) solutions - chap04

Chapter 4: Signal Representation using Fourier Series Problem 4.1 (a)

Using Definition 4.4, the CT function x1(t) can be represented as x1(t) = c1φ1(t) + c2φ2(t) + c3φ3(t) with the coefficients cn, for n = 1,2, and 3, given by T

c1 =

1 2T

∫ x1 (t )φ1 (t )dt =

−T

0

1 2T

=

1 2T 1 2T

∫

∫ (− A)dt + ∫ Adt = 21T (− AT + AT ) = 0 ,

−T

0

−T / 2

T

c2 =

T

1 2T

x1 (t )φ 2 (t )dt =

−T

1 2T

∫

−T

1 2T

∫

(− A)(1)φ 2 (t )dt +

−T / 2

1 2T

∫

A(1)dt +

T 1 2T

0

∫ A(−1)dt

T /2

( AT2 − AT2 + AT2 − AT2 ) = 0, T

c3 =

and

T /2

0

(− A)(−1)dt +

1 2T

∫

0

x1 (t )φ3 (t )dt =

−T

1 2T

∫

T

∫ A(−1)dt = 21T (− AT − AT ) = − A .

1 2T

( − A)(1)dt +

−T

0

In other words, x1(t) = −Aφ3(t), which can also be proved by inspection. (b)

By inspection, x2(t) = −Aφ2(t), which can also be proven by evaluating the coefficients c1 = 0, c2 = −A, and c3 = 0.

(c)

Using Definition 4.4, the CT function x3(t) can be represented as x3(t) = c1φ1(t) + c2φ2(t) + c3φ3(t) with the coefficients cn, for n = 1,2, and 3, given by −T / 2

T

c1 =

∫ x3 (t )φ1 (t )dt =

1 2T

−T

c2 =

∫ x3 (t )φ2 (t )dt =

−T

1 2T

∫ ( A)(−1)dt +

−T

c3 =

1 2T

∫

x3 (t )φ3 (t )dt =

−T

1 2T

∫

∫ ( A)(1)dt = 21T ( AT2 + AT2 ) = A2 ,

T /2 T

1 2T

∫ ( A)(−1)dt = 21T (− AT2 − AT2 ) = − A2 ,

T /2

−T / 2

T

and

−T

T 1 2T

−T / 2

T 1 2T

∫ ( A)(1)dt +

1 2T

T

( A)(1)dt +

−T

1 2T

∫ ( A)(−1)dt = 21T ( AT − AT ) = 0 . 0

In other words, x3(t) = 0.5A(φ1(t) − φ2(t)), which can also be proved by inspection. Problem 4.2

Computing the integral ∞

∞

−∞

−∞

∫ φ1 (t )φ 2 (t )dt = ∫ e

−2 t

(1 − Ke )dt −4 t

Since the function inside the integral is even with respect to t, therefore,

▌

124

Chapter 4 ∞

∫

∞

(

∫

∞

)

∞

∫

∫

φ1 (t )φ 2 (t ) dt = 2 e − 2t 1 − Ke − 4t dt = 2 e − 2t dt − 2 K e −6t dt = 1 −

−∞

0

0

1−

For the functions to be orthogonal,

K 3

0

= 0 ⇒ K = 3.

K 3

▌

Problem 4.3

The following derivation shows that the individual functions {Pn(x), n = 0, 1, 2, 3} have nonzero finite energy. We use the notation Pm,n to represent the integral 1

∫ P ( x) P ( x)dx .

Pm,n =

m

n

−1

Computing the integrals 1

P0,0 = ∫ 1.1dx = [ x ]−1 = 2 , 1

−1

1

P1,1 =

∫ x dx =  2

1 3

−1 1

P2,2 =

1 4

2 2 ∫ (3x − 1) dx =

−1

and

1

P3,3 =

1 4

∫ ( 25 x

−1

6

1

1 4

∫ (9x

4

−1

1

x 3  = 23 , −1

)

1

− 6 x 2 + 1 dx = 14  59 x5 − 2 x3 + x  = 12 ( 59 − 2 + 1) = 52 , −1

)

1

− 30 x 4 + 9 x 2 dx = 14  257 x 7 − 6 x5 + 3 x3  = 12 ( 257 − 6 + 3) = 72 , −1

which shows that the functions Pn(x) have nonzero finite energy. To show that the functions are orthogonal with respect to each other, we determine the integrals 1

P0,1 = 1

P0,2 =

1 2

∫ (3x

2

−1

∫

x dx = 0 ,

−1 = odd 1

− 1)dx = 12  x 3 − x  = 0 , −1 1

P0,3 =

1 2

∫ (5 x

−1

3

= odd

1

P1,2 =

1 2

∫ (3x

P1,3 =

1 2

∫ (5 x

4

−1

and

1

P2,3 =

1 4



− x ) dx = 0 , 1

− 3 x 2 ) dx = 12  x 5 − x 3  = 0 , −1

∫ 15 x

−1

3

= odd

−1 1

− 3 x) dx = 0 ,

5

 − 14 x 3 + 3 x  dx = 0 . = odd 

▌

Solutions

125

Problem 4.4

The following derivation shows that the individual functions {Tn(x), n = 0, 1, 2, 3} have nonzero finite energy. We use the notation Tm,n to represent the integral 1

1

∫

Tm, n =

1− x

−1

2

Tm ( x)Tn ( x)dx .

Computing the integrals 1

T0,0 =

1

∫

1 − x2

−1

1

T1,1 =

∫

−1

1

∫

T2, 2 =

1 − x2

]

1 −1

=π,

x2

1 π dx = − 12 x 1 − x 2 + 12 sin −1 ( x) = ,   −1 2 2 1− x

4x4 − 4x2 + 1

−1

[

dx = sin −1 ( x)

1

dx = 4

x4

∫

1 − x2

−1

dx − 4T1,1 + T0,0 = 3π − 4(0.5π) + π = 2π ,

and similarly, the higher order Tm,n‘s can be proven to be nonzero for m = n. To show that the functions are orthogonal with respect to each other, we determine the integrals 1

dx = −  1 − x 2  = 0 ,   −1 2 1− x

−1

1

T0, 2 =

∫

−1

1

∫

T0,3 =

4 x 3 − 3x 1 − x2

−1

1

dx = 4

∫

−1

1

x

∫

T0,1 =

2x2 −1 1 − x2

dx = 2T1,1 − T0,0 = 0 ,

x3

1

dx − 3T0,1 = − 1 − x 2 + 13 (1 − x 2 )3  = 0 ,   −1 2 1− x

and similarly, the higher order Tm,n‘s can be proven to be zero for m ≠ n.

▌

Problem 4.5 1

Case I (m = p, n = q):

∫

H m, n (t ) H p , q (t )dt =

0

2 2 m ∫ [H m, n (t )] dt = ∫ [H 0,0 (2 t − n)] dt 1

1

0

0

x = ( 2 m t − n) ,

Substituting

2m −n

1

∫ H m,n (t ) H p,q (t )dt = ∫ [H 0,0 ( x)]

we get

2

dx = 2

−n

0

m

2 −m

−m

2m −n

∫ [H 0,0 ( x)]

2

dx .

0

m

Since (0 ≤ n ≤ 2 – 1), therefore, (2 – n) ≥ 1 and 1

∫ 0

H m, n (t ) H p , q (t ) dt = 2 − m

1

0.5

1

0

0

0.5

2 2 2 −m −m −m ∫ [H 0,0 ( x)] dx = 2 ∫ (1) dx + 2 ∫ (−1) dx = 2 .

126

Chapter 4

Case II (m ≠ p, n ≠ q):

1

1

0

0

∫ H m,n (t ) H p, q (t )dt = ∫ H 0,0 (2

t − n) H 0,0 ( 2 p t − q )dt

x = ( 2 m t − n) ,

Substituting 1

∫ H m,n (t ) H p, q (t )dt = 2

we get

m

−m

2m −n

∫ H 0, 0 ( x ) H 0, 0 ( 2

p −m

x + 2 p − m n − q)dx .

0

0

Since (0 ≤ n ≤ 2m – 1) or (2m – n) ≥ 1 and 1

1

∫ H m, n (t ) H p, q (t )dt = 2 ∫ H 0,0 ( x) H 0,0 (2 −m

or,

x + 2 p − m n − q) dx ,

0

0

1

p−m

0.5

∫ H m, n (t ) H p, q (t )dt = 2 ∫ H 0,0 (2 −m

p −m

x+2

p−m

n − q )dx − 2

1

∫ H 0, 0 ( 2

p−m

x + 2 p − m n − q) dx = 0 .

0.5

0

0

−m

▌ Problem 4.6

(a)

By inspection, we note that the time period T0 = 2π, which implies that the fundamental frequency ω0 = 1. Since the CTFS coefficient a0 represents the average value of the signal, therefore, a0 = 3/2. Using Eq. (4.31), the CTFS cosine coefficients an’s, for (n ≠ 0), are given by T0

an = T0 ∫ x1(t ) cos(nω0t )dt = 2

0

1 π

π

π

∫ 3cos(nω t )dt = ∫ 3cos(nt )dt 0

0

1 π

0

= n3π [sin(nt )]0 = n3π [sin(nπ ) − 0] = 0 π

Using Eq. (4.32), the CTFS sine coefficients bn’s are given by T0

bn = T0 ∫ x1(t )sin(nω0t )dt = 2

0

 n6π =  0

(b)

1 π

π

∫ 3sin(nt )dt = n3π [ − cos(nt )]

π 0

0

= n3π [ − cos(nπ ) + cos(0)] = n3π 1 − (−1)n 

n = odd n = even

By inspection, we note that the time period T0 = 2T, which implies that the fundamental frequency ω0 = π/T. Since the CTFS coefficient a0 represents the average value of the signal, therefore, a0 = 0.75. Using Eq. (4.31), the CTFS cosine coefficients an’s, for (n ≠ 0), are given by

Solutions

an =

T

2T

T

∫ x(t )cos(nω t ) dt = T ∫ x(t )cos(nω t )dt = T

2

2

0

−T

2

0

0

even function

127

T T / 2   ∫ 0.5cos( nω0t )dt + ∫ cos( nω0t )dt  T /2 0 

T /2 T = nω1 T [sin( nω0t )]0 + nω2 T [sin( nω0t )]T / 2 0

= =

1 nπ 2 nπ

0

[sin(nω0T / 2)] + [sin(nω0T ) − sin(nω0T / 2)] 2 nπ

[∵ ω0T = π ]

sin( nπ ) − n1π sin( nπ / 2) = − n1π sin( nπ / 2)

 0 n = even  = − n1π n = 4k + 1  1  nπ n = 4k + 3

Since x2(t) is even, therefore, the CTFS sine coefficients bn = 0. (c)

By inspection, we note that the time period T0 = T, which implies that the fundamental frequency ω0 = 2π/T. Since the CTFS coefficient a0 represents the average value of the signal, therefore, a0 = 1/2. Since the function [x3(t) − 0.5] is odd, therefore, the CTFS cosine coefficients an = 0, for (n ≠ 0). Using Eq. (4.32), the CTFS sine coefficients bn’s are given by T

bn =

2  t 1 −  sin(nω0t )dt T  T 0

∫

2  t  − cos(nω0t )  1  − sin(nω0t )  = 1 −  × − − ×  (nω0 ) T  T  (nω0 ) 2   T

(d)

T

0

=

2 −1  1  sin(0)   1  sin(nω0T ) + −  × − × 0 − (1) ×  2 (nω0 )  T  T  (nω0 )  T  (nω0 ) 2 

=

2 1 = nω0T nπ

By inspection, we note that the time period T0 = 2T, which implies that the fundamental frequency ω0 = π/T. Using Eq. (4.30), the CTFS coefficient T0 is given by a0 = 21T

T

∫

−T

T

x(t )dt = T1 ∫ x(t )dt = T1 × T2 = 12 . 0

Using Eq. (4.31), the CTFS cosine coefficients an’s, for (n ≠ 0), are given by

128

Chapter 4 an = 22T

T

∫

−T

T

T

T

0

0

0

x(t )cos( nω0t ) dt = T2 ∫ (1 − Tt ) cos( nω0t )dt = T2 ∫ cos( nω0t )dt − T22 ∫ t cos( nω0t )dt even function

T = nω2 T [sin( nω0t )]0 − ( 2)2

[cos(nω0t ) + nω0t sin(nω0t )]0

T

nω0 T 2

0

= n2π [sin( nω0T ) − 0] − 22 2 [cos( nω0T ) + nω0T sin( nω0T ) − 1]

[∵ ω0T = π ]

nπ

= n2π sin( nπ ) − n22π 2 [cos( nπ ) + nπ sin( nπ ) − 1] =0

 0 = n22 2 1 − ( −1)n  =  4 π  n2π 2

n = even n = odd

Since x4(t) is even, therefore, the CTFS sine coefficients bn = 0. (e)

By inspection, we note that the time period T0 = 2T, which implies that the fundamental frequency ω0 = π/T. Using Eq. (4.30), the CTFS coefficient T0 is given by a0 = 21T

2T

∫ 0

T

T

T

0

0

0

x (t )dt = 21T ∫ 1 − 0.5sin ( πTt )  dt = 21T ∫ dt − 41T ∫ sin ( πTt ) dt

[

]

T = 12 + 41T × π 1/ T cos( πTt ) 0 = 12 + 41π cos(π ) − cos(0) = 12 − 21π = π2−π1

Using Eq. (4.31), the CTFS cosine coefficients an’s, for (n ≠ 0), are given by T

T

0

0

T

an = 22T ∫ 1 − 0.5sin ( πTt )  cos( nω0t )dt = T1 ∫ cos( nω0t )dt − 21T ∫ sin ( πTt ) cos( nω0t )dt 0

=A

=B

where Integrals A and B are simplified as T A = nω1 T [sin( nω0t )]0 = n1π [sin( nω0T ) − 0] = n1π [sin( nπ ) − 0] = 0 0

and T

T

T

B = 21T ∫ sin ( πTt ) cos( nω0t )dt = 21T ∫ sin ( πTt ) cos ( nTπ t ) dt = 41T ∫ sin πTt ( n + 1) − sin ( πTt ( n − 1))  dt 0

1

0

−1

=

4T

=

4π ( n +1)

T

0

1

1

T

× π ( n+1)/ T cos T ( n + 1) 0 + 4T × π ( n−1)/ T cos T ( n − 1) 0 1

πt

πt

[for

[1 − cos π (n + 1)] − 4π (1n−1) [1 − cos π (n − 1)]

1 ≠ n = odd 0 1 ≠ n = odd  0  = 2 =  1 2 n = even n = even  4π ( n+1) − 4π ( n−1)  − π ( n2 −1)

For

T

T n = 1, B = 41T ∫ sin 2Tπ t dt = 41T × 2π−1/ T cos 2Tπ t  0 = 81π [1 − cos 2π ] = 0 . 0

In other words, which implies that

 0 B= 1 − π ( n2 −1)

n = odd n = even

n ≠ 1]

Solutions  0 an = A − B =  1  π ( n2 −1)

129

n = odd n = even

Using Eq. (4.32), the CTFS sine coefficients bn’s are given by bn =

2

2T

T

T

0

0

T

∫ 1 − 0.5sin ( πTt ) sin(nω0t )dt = T1 ∫ sin(nω0t )dt − 21T ∫ sin ( πTt ) sin(nω0t )dt 0

=C

=D

where Integrals C and D are simplified as  0 T C = nω1 T [ − cos( nω0t )]0 = n1π [ − cos( nω0T ) + cos(0)] = n1π [1 − cos( nπ )] =  2 0  nπ

n = even n = odd

and T

T

D = 21T ∫ sin ( πTt ) sin ( nTπ t ) dt = 41T ∫ cos πTt ( n − 1) − cos ( πTt (n + 1))  dt 0

0

T

T = 41T × π ( n−11)/ T sin πTt ( n − 1) 0 − 41T × π ( n+11)/ T sin πTt (n + 1) 0

=

1

4π ( n −1)

[for

n ≠ 1]

[sin π (n − 1) − sin(0)] − 4π (1n+1) [sin π (n + 1) − sin(0)] [ for

=0

n ≠ 1]

For (n = 1), T T   T D = 21T ∫ sin 2 ( πTt ) dt = 41T ∫ 1 − cos ( 2Tπ t )  dt =  14 − 4T ×12π / T sin 2Tπ t  0  = 14 .   0 0 =0  

In other words,

Therefore,

 14 n = 1 . D= 0 n > 1 0  bn = C − D =  π2 − 14  2  nπ

n = even n =1 1 ≠ n = odd .

Problem 4.7

By inspection, we note that the time period T0 = T, which implies that the fundamental frequency ω0 = 2π/T. Using Eq. (4.30), the CTFS coefficient a0 is given by

a0 =

1 T

T /2

1

∫ δ(t )dt = T .

−T / 2

Using Eq. (4.31), the CTFS cosine coefficients an’s are given by

▌

130

Chapter 4

an =

2 T

T /2

2

2

∫ δ(t ) cos(nω0t )dt = T cos(nω0t ) t =0 = T . .

−T / 2

Using Eq. (4.31), the CTFS sine coefficients bn’s are given by 2 bn = T

T /2

2

∫ δ(t ) sin(nω0t )dt = T sin(nω0t ) t =0 = 0.

−T / 2

The value for bn can also be derived by noting that x(t) is an even function. For such functions, the CTFS coefficient bn = 0.

▌

Problem 4.8

(i)

x1(t) = cos(7t) + sin(15t + π/2) = cos(7t) + cos(15t). The fundamental frequency of cos(7t) is given by ω1 = 7, which implies that the time period of this term is T1 = 2π/7. The fundamental frequency of cos(15t) is given by ω2 = 15, which implies that the time period of this term is T2 = 2π/15. T1 15 = T2 7

Since the ratio

is a rational number, x1(t) is periodic with the overall period T0 = mT1 = nT2 = 2π. The fundamental frequency is given by ω0 = 1. The CTFS expansion

x(t ) = a0 +

∞

∑ (an cos(nt ) + bn sin(nt )) n =1

we note that

a0 = 0, a7 = 1, and a15 = 1.

The remaining coefficients are all zero. In other words,

1 a0 = 0, an =  0

n = 7,15 and bn = 0 , otherwise

with the fundamental frequency ω0 = 1. (ii)

The fundamental frequency of sin(2t) is given by ω1 = 2, which implies that the time period of this term is T1 = π. The fundamental frequency of cos(4t + π/4) is given by ω2 = 4, which implies that the time period of this term is T2 = π/2. Since the ratio

T1 =2 T2

is a rational number, therefore, x2(t) is periodic with the overall period T0 = mT1 = nT2 = π. The fundamental frequency is given by ω0 = 2. Comparing x2(t) = 3 + sin(2t) + cos(4t + π/4) = 3 + sin(2t) + 0.707cos(4t) − 0.707sin(4t)

Solutions

with the CTFS expansion x(t ) = a0 +

131

∞

∑ (an cos(2nt ) + bn sin(2nt )) n =1

a0 = 3, b1 = 1, a2 = 0.707, and b2 = −0.707

we note that

The remaining coefficients are all zero.  1 n=2  and bn = − 0.707 otherwise  0 

0.707 In other words, a0 = 3, an =   0

n =1 n=2 otherwise,

with the fundamental frequency ω0 = 1. (iii) The fundamental frequency of exp(j2t) is given by ω1 = 2, which implies that the time period of this term is T1 = π. The fundamental frequency of exp(j5t) is given by ω2 = 5, which implies that the time period of this term is T2 = 2π/5. The fundamental frequency of exp(−j3t) is given by ω3 = 3, which implies that the time period of this term is T3 = 2π/3. Since the ratios

3 T1 5 T1 3 T = , = , and 2 = T2 2 T3 2 T3 5

are all rational numbers, therefore, x3(t) is periodic with the overall period T0 = mT1 = nT2 = pT3 = 2π. The fundamental frequency is given by ω0 = 1. Comparing x3 (t ) = 1.2 + e × e j 2t + e j 2 × e j 5t + e − j 2 × e − j 3t = 1.2 + e × cos( 2t ) + je × sin( 2t ) + e j 2 × cos(5t ) + je j 2 × sin(5t ) + e − j 2 × cos(3t ) − je − j 2 × sin(3t )

with the CTFS expansion x(t ) = a0 +

∞

∑ (an cos(2nt ) + bn sin(2nt )) n =1

we note that  e  − j 2 a0 = 1.2, an = e j 2 e  0

 je n=2  − je − j 2 n=3 and bn =  j2 n=5  je otherwise  0

n=2 n=3 n=5 otherwise

with the fundamental frequency ω0 = 1. (iv)

Because of the exp(t + 1) term, the signal x4(t) is not periodic. Therefore, the CTFS expansion cannot be obtained. ▌

132

Chapter 4

Problem 4.9

1 Dn = T0

By definition,

1 Dn = T0

which is expressed as

0

∫ x(t )e

T0 2

∫ x(t )e

− jnω 0 t

dt ,

−T0 2

− jnω 0 t

−T0 2

1 dt + T0

T0 2

∫ x(t )e

− jnω 0 t

dt .

0

B

A

Substituting t = −α in Integral A, we get 1 A= T0

0

∫ x ( − α )e

jnω 0 α

T0 2

1 ( − dα ) = T0

T0 2

∫ x ( −α )e

jnω 0 α

dα .

0

Since x(t) is an even function, therefore, x(−α) = x(α) and the above integral reduces to T0 2

1 A= T0

∫ x (α )e

jnω 0 α

dα .

0

Substituting the value of Integral A from the above expression, the exponential CTFS coefficients are given by T0 2

1 Dn = T0 or,

Dn =

1 T0

∫ x(t )e

jnω 0 t

0

1 dt + T0

T0 2

∫ [

T0 2

]

x(t ) e jnω0 t + e − jnω0 t dt =

0

∫ x(t )e

− jnω 0 t

dt

0

T0 2

2 T0

∫ x(t ) cos(nω0t )dt . 0

Problem 4.10

1 Dn = T0

By definition,

which is expressed as

1 Dn = T0

0

∫ x(t )e

T0 2

∫ x(t )e

− jnω 0 t

dt ,

−T0 2

− jnω 0 t

−T0 2

1 dt + T0

T0 2

∫ x(t )e

− jnω 0 t

dt .

0

B

A

Substituting t = −α in Integral A, we get 1 A= T0

0

∫ x ( − α )e

jnω 0 α

T0 2

1 ( − dα ) = T0

T0 2

∫ x ( −α )e

jnω 0 α

dα .

0

Since x(t) is an odd function, x(−α) = −x(α) and the above integral reduces to A=−

1 T0

T0 2

∫ x (α )e 0

jnω 0 α

dα .

▌

Solutions

133

Substituting the value of Integral A from the above expression, the exponential CTFS coefficients are given by 1 Dn = − T0

Dn = −

or,

1 T0

T0 2

∫ 0

T0 2

∫ x(t )e

jnω0 t

0

1 dt + T0

x(t )  e jnω0t − e − jnω0t  dt =

T0 2

∫ x(t )e

− jnω 0 t

dt

0

−2 j T0

T0 2

∫

x(t ) sin(nω0t )dt .

▌

0

Problem 4.11

(a)

By inspection, we note that the time period T0 = 2π, which implies that the fundamental frequency ω0 = 1. Using Eq. (4.44), the DTFS coefficients Dn’s are given by 1 Dn = T0

T0 2

∫

x(t )e

−T0 2

− jnω 0 t

3 π  , 1  2 − jnt dt = 3e dt =  3 − jnπ 2π  j 2 nπ 1 − e 0

∫

(

 3  2 3 Dn = 1 − ( −1) n =  0 j 2nπ  3  jnπ

(

or,

)

)

n=0 n ≠ 0.

n=0 even n, n ≠ 0. odd n

The magnitude and phase spectra are given by Magnitude Spectrum:

 3, n=0 2 Dn =  0, even n, n ≠ 0. 3 , odd n.  nπ

 0, even n  π ≺ Dn = − 2 , odd n, n > 0  π , odd n, n < 0.  2

Phase Spectrum:

The magnitude and phase spectra are shown in row 1 of the subplots included in Fig. S4.11. (b)

By inspection, we note that the time period T0 = 2Τ, which implies that the fundamental frequency ω0 = π/T. Since x(t) is an even function, therefore, the DTFS coefficients Dn’s are given by T  1 0.5T 1  0 . 5 dt dt = 0.25 + 0.5 = 0.75 n = 0 + T0 2 T T 0 2 0.5T Dn = x(t ) cos(nω0t ) dt =  0.5T T T0 1 1  0  T 0.5 cos(nω0t )dt + T cos(nω0t )dt , n ≠ 0. 0.5T  0

∫

∫

∫

For (n ≠ 0), the DTFS coefficients are given by

∫

∫

134

Chapter 4 or, Dn =

0.5 T

[

]

sin( nω0 t ) 0.5T nω 0 0

+

1 T

[

]

sin( nω0 t ) T nω 0 0.5T

=

0 .5 [sin(0.5nπ) − 2 sin(0.5nπ)] = − 0.5 sin(0.5nπ) . nπ nπ

Combining the above results, we get

 3  4  0  Dn =  1 − 2 n π  1  2 n π

n=0 even n, n ≠ 0. odd n, n = ( 4k + 1) odd n, n = (4k + 3).

The magnitude and phase spectra are given by  3, n=0  4  Dn =  0, even n, n ≠ 0.  1 odd n. 2 n π , 

Magnitude Spectrum:

even n 0,  ≺ Dn = π, odd n, n = (4k + 1) 0, odd n, n = (4k + 3).

Phase Spectrum:

The magnitude and phase spectra are shown in row 2 of the subplots included in Fig. S4.11. (c)

By inspection, we note that the time period T0 = Τ, which implies that the fundamental frequency ω0 = 2π/T. Using Eq. (4.44), the DTFS coefficients Dn’s are given by T

1 1 Dn = x(t )e − jnω0 t dt = T T

∫ 0

T

 1 T  × 2 = 12 , T  e − jnω0 t dt =  T 1 − jnω 0 t t dt T 1 − T e  0

∫ (1 − ) t T

0

∫(

n=0

)

n ≠ 0.

For (n ≠ 0), the DTFS coefficients are given by 1 Dn = T

T

∫ (1 − ) e t T

− jnω 0 t

0

T

 e − jnω0 t e − jnω0 t  dt =  1 − Tt − − T1  , (− jnω0 ) ( − jnω0 ) 2  0 

(

)

( )

which reduces to T

  1 e − jnω0T 1 1 . Dn = 0 − T1 + T1 − T1  = 2 2 (− jnω0 ) j 2 nπ (− jnω0 ) (− jnω0 )  0  Combining the two cases, we get  1 , n=0 Dn =  12  j 2 nπ , n ≠ 0. The magnitude and phase spectra are given by

Solutions  1,  2 Dn =  1  2 n π ,

Magnitude Spectrum:

135

n=0 n ≠ 0.

n=0  0,  n 0.

Phase Spectrum:

The magnitude and phase spectra are shown in row 3 of the subplots included in Fig. S4.11. (d)

By inspection, we note that the time period T0 = 2Τ, which implies that the fundamental frequency ω0 = π/T. Since x(t) is an even function, the DTFS coefficients Dn’s are given by  1T  1 − Tt dt = 0.50, n=0 T0 2  T 0 2 Dn = x(t ) cos(nω0t )dt =  T T0 1 t 0  T 1 − T cos(nω0t )dt , n ≠ 0.  0

∫(

∫

∫(

)

)

For (n ≠ 0), the DTFS coefficients are given by

1 Dn = T

T

0

T

sin(nω0t ) − cos(nω0t )  1 cos(nω0t )dt =  1 − Tt − − T1  , T  (nω0 ) (nω0 ) 2  0

∫ (1 − ) t T

(

)

( )

which reduces to  cos(nπ) 1  1 − (−1) n Dn = 0 − 0 − + . = ( nπ ) 2 (nπ) 2  ( nπ ) 2  Combining the two cases, we get

 1,  2  Dn =  0,  2  ( nπ) 2

n=0 even n, n ≠ 0 odd n, n ≠ 0.

Since Dn is always positive, its phase spectrum is 0. The magnitude and phase spectra are shown in row 4 of the subplots included in Fig. S4.11. (e)

By inspection, we note that the time period T0 = 2Τ, which implies that the fundamental frequency ω0 = π/T. For (n = 0), the exponential DTFS coefficients is given by D0 = 21T

2T

∫ 0

T

T

T

0

0

x(t )dt = 21T ∫ 1 − 0.5sin ( πTt )  dt = 21T ∫ dt − 41T ∫ sin ( πTt ) dt 0

= 12 + 41T × π 1/ T cos( T )  0 = 12 + 41π [ cos(π ) − cos(0)] = 12 − 21π πt

T

For (n = 0), the exponential DTFS coefficients is given by

136

Chapter 4 T

T

0

0

T

Dn = 21T ∫ 1 − 0.5sin ( πTt )  e− jnω0t dt = 21T ∫ e− jnω0t dt − 41T ∫ sin ( πTt ) e− jnω0t dt . 0

=A

=B

Solving for Integrals A and B, we get T

A = 21T ∫ e− jnω0t dt =

1

− j 2 nω0T

0

T

− jnω0t e  0 =

1

− j 2 nπ

e− jnπ − 1 = j 21nπ 1 − (−1)n 

and

CTFS coefficients for x1(t), Fig. P4.6(a)

CTFS coefficients for x1(t), Fig. P4.6(a)

1.5

2

0 1 1  2 ( a− n + jb− n ) = 2 a− n n < 0  12   0 = 1 4 2  2 n2π 2 = n2π 2 1 4 = 2  2 ( − n )2 π 2 n2π 2

(e)

n=0 a0  1 =  2 an n > 0 1  2 a− n n < 0

n=0 0 ≠ n = even n = odd , n > 0 n = odd , n < 0

 12  = 0  2  n2π 2

[∵

bn = b− n = 0]

.

n=0 0 ≠ n = even. n = odd

From the solution of Problem P4.6(e), we know that a0 = 12 − 21π

 0 , an =  1  π ( n2 −1)

0  , and b =  2 − 1 n nπ 4 n = even  2  nπ

n = odd

n = even n =1 1 ≠ n = odd

140

Chapter 4 Using Eq. (4.45), the exponential CTFS coefficients for x5(t) are given by

D0 = 12 − 21π

(n = 0): (n = 1):

D1 =

(n = −1):

D−1 =

(n > 1):

(n < −1):

1 2

( a1 − jb1 ) = − 2j ( π2 − 14 ) =

1 2

( a1 +

jb1 ) =

(

j 2 2 π

 − nπj Dn = ( an − jbn ) =  1  2π ( n2 −1)

(

)

(

)

j 18 − π1

)

− 14 = − j 18 − π1

1  jnπ = n = even  2π ( n12 −1)

n = odd

1 2

n = odd n = even

 2j ( − n2π ) n = odd  jn1π = 1 Dn = 12 ( a− n + jb− n ) =  1 = n even 2  2π ( n2 −1)  2π ( n −1)

n = odd n = even

Combining the above results, we obtain  12 (1 − π1 )   ± j 18 − π1 Dn =  1  2π ( n2 −1)  1  jnπ

(

n=0

)

n = ±1

▌

0 ≠ n = even ± 1 ≠ n = odd .

Problem 4.14

Problem 4.11(b) computes the exponential DTFS coefficients of x2(t) as CTFS

x 2(t ) ←→ Dnx = −

0.5 sin(0.5nπ) nπ

with fundamental frequency ω0 = π/T. Differentiating x2(t) with respect to t, we get ∞ ∞ dx 2(t ) = 0.5δ(t − 0.5T − 2kT ) − 0.5δ(t + 0.5T − 2kT ) , dt k = −∞ m = −∞

∑

∑

0.5 g (t +T )

0.5 g (t )

where the first term g(t) represents an impulse train with period T0 = 2T and with impulses located at (T/2 + 2kT). Using the time differentiation property, dx 2(t ) CTFS jnπ 0 .5 1 ←→ jnω0 Dnx = ×− sin(0.5nπ) = − j sin(0.5nπ) dt T nπ 2T

implying that

CTFS

0.5 g (t ) − 0.5 g (t + T ) ←→ − j

Using the time shifting property,

(

1 sin(0.5nπ) . 2T

)

(

g (t ) − g (t + T ) ←→ Dng 1 − e jnω0T = Dng 1 − e jnπ CTFS

with Dng representing the exponential CTFS coefficients of g(t). Hence,

(

)

Dng 1 − e jnπ = − j

1 sin(0.5nπ) T

)

Solutions

Dng = − j

or,

1 sin(0.5nπ) 1 sin(0.5nπ) e − jnπ / 2 j = − = . 2T 1 − e jnπ T e jnπ / 2 (−2 j sin(0.5nπ)) 2T

141

▌

Problem 4.15

(i)

As shown in Problem P4.8(i), x1(t) is periodic with the overall period T0 = 2π and fundamental frequency ω0 = 1. The function x1(t) can be expressed as follows:

x1 (t ) = cos(7t) + cos(15t) = 12 e j 7t + 12 e − j 7 t + 12 e j15t + 12 e− j15t . Comparing with the exponential CTFS expansion with ω0 = 1, x(t ) =

∞

∑ Dn exp( jnt ) , n =1

we note that

D7 = D− 7 = 0.5 and D15 = D−15 = 0.5.

The remaining coefficients are all zero. (ii)

As shown in Problem P4.8(ii), x2(t) is periodic with the overall period T0 = π and fundamental frequency ω0 = 2. Expanding x2(t) = 3 + sin(2t) + cos(4t + π/4) x2 (t ) = 3 +

as

1 j2

e j 2t − 12 e − j 2t + 12 e jπ / 4e j 4t + 12 e − jπ / 4e − j 4t

Comparing with the exponential CTFS expansion with ω0 = 2, x(t ) =

∞

∑ Dn exp( j 2nt ) , n =1

we note that D− 2 = 12 e − jπ / 4 , D−1 = j 12 , D0 = 3, D1 = − j 12

and D2 = 12 e jπ / 4 .

The remaining coefficients are all zero. (iii) As shown in Problem P4.8(iii), x3(t) is periodic with the overall period T0 = 2π and fundamental frequency ω0 = 1. Expanding x3 (t ) = 1.2 + e j 2t +1 + e j (5t + 2) + +e − j (3t + 2) as

x3 (t ) = 1.2 + e × e j 2t + e j 2 × e j 5t + e − j 2 × e − j 3t .

Comparing with the exponential CTFS expansion with ω0 = 1, x(t ) =

∞

∑ Dn exp( jnt ) , n =1

we note that

D− 3 = e − j 2 , D0 = 1.2, D2 = e, and D5 = e j 2 . The remaining coefficients are all zero.

142 (i)

Chapter 4 Since the signal is not periodic because of the exp(t + 1) term, the exponential CTFS expansion cannot be obtained. ▌

Problem 4.16

For the impulse train

p (t ) =

∞

1

CTFS → En = ∑ δ(t − 2kπ)← 2π

k = −∞

with period T0 = 2π and fundamental frequency ω0 = 1. Expressing

∞ ∞ dx(t ) = δ t + π4 − 2kπ − δ t − π4 − 2kπ , dt k = −∞ k = −∞

∑(

) ∑(

p (t + π / 4)

)

p (t − π / 4)

and using the time shifting property, we observe that dx(t ) CTFS ←→ e jnω0 π / 4 En − e − jnω0 π / 4 En . dt

Substituting ω0 = 1, we get

dx(t ) CTFS ←→ 2 j sin(0.25nπ) En . dt

Using the time differentiation property, jnω0 Dn = 2 j sin(0.25nπ) En , Dn = n2 sin(0.25nπ) En .

or, Substituting En = 1/2π, we get Dn =

1

πn

nπ ) sin(0.25nπ ) = 14 × sin(0.25 = 14 sinc(0.25n) . 0.25π n

▌

Problem 4.17

Example 4.14 derived the exponential DTFS coefficients of the square wave with the duty cycle (τ/T) as Dn = Tτ sinc( nTτ ) . (i)

For T = 5 ms, the fundamental frequency is f0 = 1/T = 1/5ms = 200 Hz, while the fundamental angular frequency is ω0 = 2πf0 = 400π radians/s. With τ = 1ms, the exponential CTFS coefficients are given by Dn = 15 sinc( n5 ) , which are plotted in Fig. S4.17(a) in terms of two scales: (a) number n of the CTFS coefficients; and (b) the corresponding frequency f = nf0 in Hz.

(ii)

For T = 10 ms, the fundamental frequency is f0 = 1/T = 1/10ms = 100 Hz, while the fundamental angular frequency is ω0 = 2πf0 = 200π radians/s. With τ = 2ms, the expression for the exponential CTFS coefficients stay the same as in part (i) and is given by Dn = 15 sinc( n5 ) , which are plotted in Fig. S4.17(b) in terms of two scales: (a) number n of the CTFS coefficients; and (b) the corresponding frequency f = nf0 in Hz.

Solutions

143

0.2 0.15 0.1 0.05 0 −15

−10

−5

−3000

−2000

0

−1000

5

0

1000

n

10

15

2000

3000

f

(a) 0.2

0.2

0.15

0.15

0.1

0.1

0.05

0.05 0

0

n

n

−30

−20

−10

0

10

20

30

−3000

−2000

−1000

0

1000

2000

3000

f

−60

−40

−20

0

20

−3000

−2000

−1000

0

1000

(b)

40

60

2000

3000

f

(c) Fig. S4.17: DTFS coefficients for Problem 4.17.

(iii) Finally, for T = 20 ms, the fundamental frequency is f0 = 1/T = 1/20ms = 50 Hz, while the fundamental angular frequency is ω0 = 2πf0 = 100π radians/s. With τ = 4ms, the expression for the exponential CTFS coefficients stay the same as in parts (i) and (ii) and is given by Dn = 15 sinc( n5 ) , which are plotted in Fig. S4.17(b) in terms of two scales: (a) number n of the CTFS coefficients; and (b) the corresponding frequency f = nf0 in Hz. From Fig. S4.17, we make the following observations. DC Coefficient: Keeping the duty cycle (τ/T) of the square wave constant maintains the same dc or average value of the signal. Therefore, the dc coefficient D0 stays the same for the three representations. Zero Crossings: Since the duty cycle (τ/T) is kept constant, the width of the main lobe and side lobes of the discrete sinc function stay the same in the discrete (n) domain. A change in the fundamental frequency causes the widths to be different in Hertz. ▌ Problem 4.18

(a)

In time domain, the average power of x1(t) is given by 1 Px1 = T

T

∫ 0

π

1 9 x1(t ) dt = 9dt = . 2π 2

∫

2

0

Using the Parseval’s theorem, the average power of x1(t) is given by

Px1 =

∞

∑

n = −∞

Dn

2

2

= D0 + 2

∞

∑

Dn

n =1,3,5,...

Using the results of Problem 4.21, we know that

2

= 2.25 +

18 π

2

∞

∑

1

n =1,3,5,... n

2

.

144

Chapter 4 π2 1 1 1 1 = 1 + 2 + 2 + 2 + 2 + ... , 8 3 5 7 9

Px1 = 2.25 +

which gives (b)

18 π2

×

π2 = 4.5 . 8

In time domain, the average power of x2(t) is given by 1 Px 2 = 2T

T

∫ x2(t )

2

dt =

0

1 [0.25T + T ] = 0.625. 2T

Using the Parseval’s theorem, the average power of x2(t) is given by ∞

∑ Dn

Px 2 =

2

∞

∑ Dn

2

= D0 + 2

n = −∞

2

=

n =1,3,5,...

9 2 + 2 16 4π

∞

1 . 2 n =1,3,5,... n

∑

Using the results of Problem 4.21, we know that π2 1 1 1 1 = 1 + 2 + 2 + 2 + 2 + ... 8 3 5 7 9 Px 2 =

which gives (c)

9 2 π 2 10 + 2× = = 0.625. 16 4π 8 16

In time domain, the average power of x3(t) is given by Px3 =

1 T

T

∫ x3(t )

2

dt = −

0

[

T (1 − t / T )3 3T

]

T 0

1 = . 3

Using the Parseval’s theorem, the average power of x3(t) is given by ∞

Px 3 =

∑

2

∞

2

Dn = D0 + 2

n = −∞

∑

2

Dn =

n =1, 2,3,...

1 2 + 2 4 4π

∞

1 . 2 n =1, 2,3,... n

∑

Using the results of Problem 4.23, we know that π2 1 1 1 1 = 1 + 2 + 2 + 2 + 2 + ... 6 2 3 4 5

Px3 =

which gives (d)

1 2 π2 1 + 2× = . 4 4π 6 3

In time domain, the average power of x4(t) is given by Px 4 =

1 2T

T

∫

2

x 4(t ) dt =

−T

[

T

1 T (1 − t / T ) 2 dt = − (1 − t / T )3 T 3T

∫ 0

]

T 0

1 = . 3

Using the Parseval’s theorem, the average power of x2(t) is given by Px 4 =

∞

∑

2

2

Dn = D0 + 2

n = −∞

∞

∑

2

Dn =

n =1,3,5,...

1 8 + 4 π4

∞

1 . 4 n =1,3,5,... n

∑

Solutions

Using the result which gives

1+

145

π2 1 1 1 1 + + + + = = ... 1 . 0147 96 34 5 4 7 4 9 4

Px 4 =

1 8 π2 1 + × = . 4 π 2 96 3

▌

Problem 4.19

(a)

Within one period t = [0, 2π], function x1(t) is absolutely integrable as 2π

π

0

0

∫ x1(t ) dt = ∫ 3dt = 3π.

Function x1(t) has only one maxima and one minima within one period, hence, has bounded variations. Finally, there are only two discontinuities within one period. Function x1(t) satisfies the Dirichlet conditions. (b)

Within one period t = [0, 2Τ], function x2(t) is absolutely integrable as 2T

∫ x2(t ) dt = 1.5T . 0

Function x2(t) has only one maxima at and two minimas within one period t = [0, Τ], hence, has bounded variations. Finally, there are only two discontinuities t = T/2 and 3T/2 within one period t = [0, Τ]. Function x2(t) satisfies the Dirichlet conditions. (c)

Within one period t = [0, Τ], function x3(t) is absolutely integrable as T

T

∫ x3(t ) dt = 2 . 0

Function x3(t) has only one minima and one maxima within one period t = [0, Τ], hence, has bounded variations. Finally, there are only one discontinuity at t = 0 within one period t = [0, Τ]. Function x3(t) satisfies the Dirichlet conditions. (d)

Within one period t = [0, 2Τ], function x4(t) is absolutely integrable as 2T

∫ x4(t ) dt = T . 0

Function x4(t) has only one minima and one maxima within one period t = [0, 2Τ], hence, has bounded variations. Finally, there is no discontinuity within one period t = [0, 2Τ]. Function x4(t) satisfies the Dirichlet conditions. (e)

Within one period t = [0, 2Τ], function x5(t) is absolutely integrable as

146

Chapter 4 2T

∫ x5(t ) dt = 0

(π − 1)T . 2π

Function x5(t) has only one minima and two maximas within one period t = [0, 2Τ], hence, has bounded variations. Finally, there is no discontinuity within one period t = [0, 2Τ]. Function x5(t) satisfies the Dirichlet conditions.

▌

Problem 4.20

Determine if the following functions satisfy the Dirichlet conditions and have CTFS representation. (i)

x(t ) = 1 / t , t = (0, 2] and x(t ) = x(t + 2) ;

(ii)

g (t ) = cos(π / 2t ) , t = (0, 1] and g (t ) = g (t + 1) ;

(iii)

h(t ) = sin(ln(t )) , t = (0, 1] and h(t ) = h(t + 1) .

Solution: 2

(i)

∫ 0

2

1 1 x(t ) dt = dt = − 2 t 2t 0

∫

2

=∞ 0

As the function x(t) is not absolutely integrable, x(t) does not satisfy the Dirichlet conditions. (ii)

As shown in Fig. S4.20 (top plot), function g(t) has an infinite number of maximas and minimas in one period. Therefore, g(t) does not satisfy the Dirichlet conditions.

(ii)

As shown in Fig. S4.20 (bottom plot), function h(t) appears to satisfy the Dirichlet conditions. However, Matlab is not able to plot all the peaks because of its limited resolution. When t = (0,1] , ln(t ) = (−∞, 0] and is a CT function. The function sin(ln(t )) will have a maxima every 2π interval of ln(t ) implying that the total number of maxima’s are infinite. The function h(t) therefore does not satisfy the Dirichlet conditions.

Problem 4.20, part (ii)

g(t) = cos(π/2t)

1 0.5 0 -0.5 -1

0

0.1

0.2

0.3

0

0.1

0.2

0.3

h(t) = sin(ln(t))

1

0.4

0.5 0.6 0.7 time (t) Problem 4.20, part (iii)

0.8

0.9

1

0.8

0.9

1

0.5 0 -0.5 -1

0.4

0.5 0.6 time (t)

0.7

Fig. S4.20: One period of the functions g(t) and h(t) in Problem 4.20(ii) and (iii).

Solutions

147

Problem 4.21

Example 4.9 derived the trigonometric CTFS coefficients of the triangular wave f(t), shown in Fig. S4.21, as follows f (t ) =

∞

24 24  1 1 1 cos(0.5nπt ) = 2 cos(0.5πt ) + 2 cos(1.5πt ) + 2 cos(2.5πt ) + 2 cos(3.5πt ) + 2 π  3 5 7 n =1,3,5,... ( nπ)

∑

  

Substituting (t = 0) on both sides, we get f ( 0) =

24 ∞ 1 24  1 1 1 1  = 2 1 + 2 + 2 + 2 + 2 + ... . 2 2 π n =1,3,5,... n π  3 5 7 9 

∑

f (t ) 3

−4

t

−2

0

2

4

−3

Figure S4.21: Periodic signal f(t) considered in Problem 4.21. f (0) = 3 .

From Fig. S4.21, we note that Equating the above two equations, we obtain

∞ π2 1 1 1 1 1 = = 1 + 2 + 2 + 2 + 2 + ... 2 8 n =1,3,5,... n 3 5 7 9

∑

▌

Problem 4.22

From the solution of Problem 4.6(c), we know that the trigonometric CTFS expansion of the half sawtooth wave is given by x3(t ) =

Substituting t = T/4, we get x3(T / 4) =

1 ∞ 1 + sin(2nπt / T ) 2 n =1 nπ

∑

1 ∞ 1 1 1 1 1 1 1 1  + sin(nπ / 2) = + 1 − + − + − + −... 2 n =1 nπ 2 π  3 5 7 9 11 

∑

Since x3(T/4) = (1 – (T/4)/T) = 0.75, therefore, 0.75 =

which implies that

1 1 1 1 1 1 1  + 1 − + − + − + −... 2 π  3 5 7 9 11 

π ∞ 1 1 1 1 1 1 = × (−1) n −1 = 1 − + − + − + −... 4 n =1 n 3 5 7 9 11

∑

▌

148

Chapter 4

Problem 4.23

From the solution of Problem 4.11(c), we know that the exponential CTFS expansion of the half sawtooth wave is given by  1 , n=0 Dn =  12  j 2 nπ , n ≠ 0. Computing the power from the exponential CTFS coefficients, we get

Px =

∞

∑

2

Dn =

n = −∞

∞ ∞ 1 1 1 1 1 1 2 + = + = + 2 2 2 2 2 4 n = −∞ 4n π 4 4 2π n =1 4n π

∑

∑

n≠0

∞

1

∑ n2 . n =1

Computing the power in the time domain, we obtain 1 Px = T

T

∫ 0

T

2

t t 1  1  x3(t ) dt = 1 −  dt = × (−T / 3) 1 −  T  T T  T 0 2

∫

3

T

= 0

1 1 × (−T / 3) × (0 − 1) = . T 3

Equating the two expressions for the power 1 1 1 = + 2 3 4 2π

∞

1

∑ n2 n =1

∞ π2 1 1 1 1 1 1 = = 1 + 2 + 2 + 2 + 2 + 2 + ... 6 n =1 n 2 2 3 4 5 6

∑

or,

▌

Problem 4.24

(i)

The transfer function H(ω) is given by ∞

H (ω) =

∫

e

−∞

− 2 t − jωt

e

∞

∫

dt = e − 2t e − jωt dt + 0

1 e − ( 2 + jω)t = (−2 − jω) =

(ii)

∞ 0

0

∫e

2t − jωt

e

−∞

1 e ( 2 − jω)t + (2 + jω)

dt

0 −∞

1 1 4 × [0 − 1] + × [1 − 0] = (−2 − jω) (2 + jω) 4 + ω2

Since the transfer function H(ω) is real valued, therefore, its magnitude spectrum H (ω) = H (ω) =

4 . 4 + ω2

Solutions

149

Magnitude Plot for h(t) = exp(−2|t|)

1

|H( ω)|

0.8 0.6 0.4 0.2 -8

-6

-4

-2

0 2 frequency ( ω)

4

6

8

Fig. S4.24: Magnitude spectrum for h(t) = exp(−2|t|) The magnitude spectrum |H(ω)| is shown in Fig. S4.24. (iii) The exponential CTFS coefficients En of the output signal y(t) are given by En = Dn H(ω0) where ω0 = 2π/T and Dn are the exponential CTFS coefficients for the input signal. As found in P4.12, the exponential CTFS coefficients for the input impulse train are given by: 1 Dn = T0

T0 2

∫ x(t )e

− jnω0 t

−T0 2

1 dt = T

T /2

∫ δ(t )e

− jnt

dt =

−T / 2

1 T

The exponential DTFS coefficients En are then given by   1 4  4T = 2 .   2 2 T  4 + ω  ω= 2 nπ / T  4T + (2nπ) 

En =

In the time domain, the output signal is expressed as ∞

y (t ) =

∑

En e jnω0 t =

n = −∞

∞

T e j 2nπt / T . 2 n = −∞ T + ( nπ)

∑

2

Problem 4.25

(i)

The transfer function H(ω) is given by ∞

H (ω) =

∫ (e

− 2t

−e

− 4t

)e

− jωt

0

−1 = e − ( 2 + jω)t ( 2 + jω) =

(ii)

∞

∫

dt = e

∞

0

− ( 2 + jω)t

∞

∫

dt − e − ( 4 + jω)t dt

0

0

−1 + e − ( 4 + jω)t (4 + jω)

∞

0

−1 −1 2 × [0 − 1] − × [0 − 1] = ( 2 + jω) (4 + jω) (2 + jω)(4 + jω)

The magnitude response is given by H (ω) =

2

(4 + ω )(16 + ω ) 2

The magnitude spectrum |H(ω)| is shown in Fig. S4.25.

2

.

▌

150

Chapter 4 Magnitude Plot for h(t) = [exp(−2t) − exp(−4t)] u(t)

0.25

|H( ω)|

0.2 0.15 0.1 0.05 −8

−6

−4

0 2 −2 frequency ( ω)

4

6

8

Fig. S4.25: Magnitude spectrum for h(t) = [exp(−2t) − exp(−4t)] u(t). (iii) The exponential CTFS coefficients En of the output signal y(t) are given by En = Dn H(ω0) with ω0 = π/T. For the raised cosine wave, the exponential CTFS coefficients Dn are given by 0.75 n=0  Dn =  0.5 . − sin(0.5nπ) n ≠ 0  nπ

Therefore, the CTFS coefficients En of the output signal y(t) are given by

0.75 n=0    2  ×  0.5 En =   sin(nπ / 2) n ≠ 0  (2 + jω )(4 + jω ) ω = nπ / T −  nπ 1/ 4 n=0   3/ 4 n=0   2 = ×  0.5 2T  (2T + jnπ )(4T + jnπ ) n ≠ 0  − nπ sin(nπ / 2) n ≠ 0  3 /16 n=0   = T 2 sin(nπ / 2) −  nπ (2T + jnπ )(4T + jnπ ) n ≠ 0.  In the time domain, the output signal is expressed as ∞

y (t ) =

∑

n =−∞

En e jnω0t =

∞ 3 T 2 sin(0.5nπ ) −∑ e jnπ t / T . 16 n =−∞ nπ (2T + jnπ )(4T + jnπ ) n≠0

Problem 4.26

(i)

The transfer function H(ω) is given by ∞

H (ω ) = ∫ te 0

= (ii)

−4 t

e

− jωt

∞

dt = ∫ te 0

∞

− (4 + jω ) t

e − (4+ jω ) t e − (4+ jω ) t − dt = t −(4 + jω ) 0 (4 + jω ) 2

−1 1 1 × [ 0 − 0] − × [ 0 − 1] = . 2 (4 + jω ) (4 + jω ) (4 + jω ) 2

The magnitude response is given by

∞

0

▌

Solutions H (ω ) =

1

(16 + ω )(16 + ω ) 2

2

=

151

1 . 16 + ω 2

The magnitude spectrum |H(ω)| is shown in Fig. S4.26. Magnitude Plot for h(t) = t exp(−4t) u(t) 0.06 0.04 0.02 −8

−6

−4

0 2 −2 frequency ( ω)

4

6

8

Fig. S4.26: Magnitude spectrum for h(t) = t exp(−4t) u(t). (iii) The exponential CTFS coefficients En of the output signal y(t) are given by En = Dn H(ω0) with ω0 = π/T. For the sawtooth wave, the exponential CTFS coefficients Dn are given by  1,  2  Dn =  0,  2  ( nπ) 2

n=0 even n, n ≠ 0 odd n, n ≠ 0.

Therefore, the CTFS coefficients En of the output signal y(t) are given by

 12 ,    1 × En =   0, 2  (4 + jω ) ω = 2 nπ / T  2  ( nπ )2

 1 32 ,  n=0   even n =  0,  odd n  2T 2  2 2  (nπ ) (4T + j 2nπ )

n=0 even n odd n.

In the time domain, the output signal is expressed as

y (t ) =

∞

∑

n =−∞

En e jnω0t =

∞ 1 2T 2 e jnπ t / T . + ∑ 2 2 32 n =−∞ (nπ ) (4T + j 2nπ ) n = odd

▌

152

Chapter 4

Problem 4.27

(i)

(a) Expressing x1 (t ) =

7 ∞ 1 7 1 1  sin[8π(2m + 1)t ] = sin(8πt ) + sin(24πt ) + sin(40πt ) + ... 3 5 π m = 0 2m + 1 π 

∑

we note that the signal x1(t) contains the fundamental component sin(8πt) and its harmonics. Therefore, the signal is periodic, and the fundamental frequency for x1(t) is given by ω0 = 8π radian/sec. The fundamental period is T0 = 2π/ω0 = 0.25 sec. (b) Since x1 ( −t ) =

7 ∞ 1 7 ∞ 1 sin[− 8π(2m + 1)t ] = − sin[8π(2m + 1)t ] = − x1 (t ) , π m = 0 2m + 1 π m = 0 2m + 1

∑

∑

the signal is odd. (c) The following MATLAB code is used to reconstruct the function in the time domain. The number n of harmonics is set to 4000. % initializing CTFS parameters nterms = 4000; w0 = 8*pi; t = -1:0.001:1; a0 = 0; an = zeros(1,nterms); nnz = 1:2:nterms; bn2d = zeros(2,nterms/2); bn2d(1,:) = 1./nnz; bn = reshape(bn2d,1,nterms); % calculating time-domain function x1 = (7/pi)* ictfs(w0,t, a0,an,bn); plot(t,x1); xlabel('t'); ylabel('x1(t)'); axis([-1 1 -3 3]), grid on; title ('Reconstruction from CTFS')

(d) The resulting waveform is shown in Fig. S4.27. Reconstruction from CTFS

3 2 x1(t)

1 0 −1 −2 −3 −1

−0.75

−0.5

−0.25

0 t

0.25

0.5

0.75

1

Fig. S4.27: Signal x1(t) reconstructed from the first 4000 trigonometric CTFS coefficients in Problem 4.27(a). (ii)

(a) Expressing x2 (t ) = 1.5 +

∞

1



1

1



∑ 4m + 1 cos[2π(4m + 1)t ] = 1.5 + cos(2πt ) + 5 cos(10πt ) + 9 sin(18πt ) + ...

m=0

Solutions

153

we note that the signal x2(t) contains the fundamental component cos(2πt) and its harmonics. Therefore, the signal is periodic, and the fundamental frequency for x2(t) is given by ω0 = 2π radian/sec. The fundamental period is T0 = 2π/ω0 = 1 sec. (b) Since x2 (−t ) = 1.5 +

∞

∞ 1 1 cos[− 2π(4m + 1)t ] = 1.5 + cos[2π(4m + 1)t ] = x2 (t ) , 4 1 m + 4 m + 1 m=0 m=0

∑

∑

the signal is even. (c) The following MATLAB code is used to reconstruct the function in the time domain. The number n of harmonics is set to 4000. % initializing CTFS parameters nterms = 4000 ; w0 = 2*pi ; t = -4:0.001:4 ; a0 = 1.5 ; nnz = 1:4:nterms; an2d = zeros(4,nterms/4); an2d(1,:) = 1./nnz ; an = reshape(an2d,1,nterms) ; bn = zeros(1,nterms) ; % calculating time-domain function x2 = ictfs(w0,t,a0,an,bn); plot(t,x2) xlabel('t'); ylabel('x2(t)'); axis([-2 2 -2 5]), grid on title ('Signal Reconstruction from CTFS')

(d) The resulting waveform is shown in Fig. S4.27.

▌

Signal Reconstruction from CTFS 4

x2(t)

3 2 1 0 −4

−3

−2

−1

0 t

1

2

3

4

Fig. S4.27: Signal x2(t) reconstructed from the first 4000 trigonometric CTFS coefficients in Problem 4.27(b). . Problem 4.28

From Example 4.8, the CTFS coefficients are given by a0 = 1.7079 , an = The periodic signal g(t) is, therefore, given by

3.4157 17.0787 n , and bn = . 2 1 + 25n 1 + 25n 2

154

Chapter 4

g (t ) = 1.7079 +

∞

∑

n =1

3.4157 1+ 25 n 2

cos(nt ) +

∞

∑ 171+.250787n n sin(nt ) 2

n =1

with the fundamental frequency ωo = 1 radians/s. The following MATLAB code is used to reconstruct the function in the time domain. The number n of harmonics is set to 4000. % initializing CTFS parameters nterms = 2000 ; n = 1:nterms; w0 = 1 ; t = -12:0.01:12 ; a0=1.7079 ; an = 3.4157./(1+25*n.*n) ; bn = 17.0787*n./(1+25*n.*n) ; % calculating time-domain function g = ictfs(w0,t, a0,an,bn) ; % plotting the function plot(t,g) xlabel('t'); ylabel('g(t)'); axis([-12 12 0 4]), grid on title ('Reconstruction of g(t) from CTFS')

The resulting waveform is shown in Fig. S4.28. It is observed that the plot is identical to that of Fig. 4.10. ▌ Reconstruction of g(t) from CTFS

4

g(t)

3 2 1 0 −12 −10

−8

−6

−4

−2

0 t

2

4

6

8

10

12

Fig. S4.28: Signal g(t) reconstructed from the first 2000 trigonometric CTFS coefficients in Problem 4.28. Problem 4.29

From Example 4.9, the CTFS coefficients are given by  0  a0 = 0 , an =  24  (nπ )2

n is even n is odd.

, and bn = 0 .

The periodic signal f(t) is, therefore, given by f (t ) =

∞

∑ (n24π) n =1

2

cos(0.5nπt ) =

24 π2

[cos(0.5πt ) + 19 cos(1.5πt ) + 251 cos(2.5πt ) + ]

with the fundamental frequency ωo = 0.5π radians/s.

Solutions

155

The following MATLAB code is used to reconstruct the function in the time domain. The number n of harmonics is set to 2000. % initializing CTFS parameters nterms = 2000 ; an = zeros(1,nterms); nnz = 1:2:nterms; w0 = 0.5*pi ; t = -8:0.01:8 ; a0=0 ; an2d = zeros(2,nterms/2); an2d(1,:) = 24./(pi*pi*nnz.*nnz) ; an=reshape(an2d,1,nterms) ; bn = zeros(1,nterms); % calculating time-domain function x = ictfs(w0,t, a0,an,bn) ; % plotting the function plot(t,x) xlabel('t'); ylabel('f(t)'); axis([-8 8 -4 4]), grid on title ('Reconstruction of f(t) from CTFS')

The resulting waveform is shown in Fig. S4.29. It is observed that the plot is identical to that of Fig. 4.11. ▌ Reconstruction of f(t) from CTFS

4

f(t)

2 0 −2 −4 −8

−6

−4

−2

0 t

2

4

6

8

Fig. S4.29: Signal f(t) reconstructed from the first 2000 trigonometric CTFS coefficients in Problem 4.29. Problem 4.30

From Example 4.12, the CTFS coefficients are given by Dn ≈

0.3416 0.2 + jn

.

The periodic signal g(t) is, therefore, given by g (t ) =

∞

exp( jnωοt ) ∑ 00..23416 + jn

n = −∞

with the fundamental frequency ωo = 1 radians/s. The following MATLAB code is used to reconstruct the function in the time domain. The number n of harmonics is set to 4000.

156

Chapter 4

% initializing CTFS parameters nterms = 4000 ; n=(-nterms/2):nterms/2; dn = 0.3416./(0.2+i*n); nnz = 1:2:nterms; w0 = 1; t = -12:0.01:12 ; % calculating time-domain function g = ictfs(w0,t,dn) ; % plotting the function plot(t,g) xlabel('t'); ylabel('g(t)'); axis([-12 12 0 4]), grid on title ('Reconstruction of g(t) from CTFS')

The resulting waveform is shown in Fig. S4.30. It is observed that the plot is identical to that of Fig. 4.10. ▌ Reconstruction of g(t) from CTFS

4

g(t)

3 2 1 0 −12 −10

−8

−6

−4

−2

0 t

2

4

6

8

10

12

Fig. S4.30: Signal g(t) reconstructed from the first 4000 exponential CTFS coefficients in Problem 4.30. Problem 4.31

From Example 4.13, the CTFS coefficients are given by

n = even

 0  Dn =  12  ( nπ )2

n = odd.

.

The periodic signal f(t) is, therefore, given by f (t ) =

∞

∑

n = −∞ n is odd

12

( nπ ) 2

exp( jnωοt )

with the fundamental frequency ωo = π/2 radians/s. The following MATLAB code is used to reconstruct the function in the time domain. The number n of harmonics is set to 4000.

Solutions

157

% initializing CTFS parameters nterms = 4000 ; w0 = 0.5*pi ; t = -8:0.01:8 ; nnz = 1:2:nterms; dn2d = zeros(2,nterms/2); dn2d(2,:) = 12./(pi*pi*nnz.*nnz) ; dn=reshape(dn2d,1,nterms) ; dn = [fliplr(dn(2:length(dn))), dn]; % calculating time-domain function f = ictfs(w0,t, dn) ; % plotting the function plot(t,f) xlabel('t'); ylabel('f(t)'); axis([-8 8 -4 4]), grid on title ('Signal Reconstruction from CTFS')

The resulting waveform is shown in Fig. S4.31. It is observed that the plot is identical to that of Fig. 4.11. ▌ Reconstruction of f(t) from CTFS

4

f(t)

2 0 −2 −4 −8

−6

−4

−2

0 t

2

4

6

8

Fig. S4.31: Signal f(t) reconstructed from the first 4000 exponential CTFS coefficients in Problem 4.31. Problem 4.32

From the solution of Problem 4.24, the exponential CTFS coefficients are given by

T   En =  2 2 2 T + n π  with the time domain representation

y (t ) =

∞

∑T

n =−∞

2

T e j 2 nπ t / T 2 2 +n π

where ωo = 2π/Τ = 2π radians/s The following MATLAB code is used to reconstruct the function in the time domain. The number n of harmonics is set to 4000.

158

Chapter 4

% initializing CTFS parameters nterms = 4000 ; T = 1; w0 = 2*pi/T; t = -6:0.01:6; nnz = 0:nterms; en = (4*T)./(4*T^2 + (nnz*pi).^2); en = [fliplr(en(2:length(en))), en]; % calculating time-domain function y = ictfs(w0,t, en) ; % plotting the function plot(t,y) xlabel('t'); ylabel('y(t)'); axis([-6 6 0 2.5]), grid on title ('Signal Reconstruction from CTFS') print -dtiff plot.tiff;

The resulting waveform is shown in Fig. S4.32.

Fig. S4.32: Signal y(t) reconstructed from the first 4000 exponential CTFS coefficients in Problem 4.32. Problem 4.33

From the solution of Problem 4.25, the exponential CTFS coefficients are given by

3/16   2 En =  T sin( nπ / 2)  − nπ (2T + jnπ )(4T + jnπ ) 

n=0 n ≠ 0.

with the time domain representation

y (t ) =

∞

∑Ee

n =−∞

n

jnω0t

∞ 3 T 2 sin( nπ / 2) = −∑ e jnπ t / T 16 n =−∞ nπ (2T + jnπ )(4T + jnπ ) n ≠0

where ωo = 2π/Τ = 2π radians/s The following MATLAB code is used to reconstruct the function in the time domain. The number n of harmonics is set to 4000.

Solutions

% initializing CTFS parameters nterms = 4000 ; T = 1; w0 = 2*pi/T; t = -6:0.01:6; n = -nterms:nterms; en = -(T^2*sin(0.5*n*pi))./((n+eps)*pi.*(2*T+j*n*pi).*(4*T+j*n*pi)); en(n == 0) = 3/16; % calculating time-domain function y = ictfs(w0,t, en) ; % plotting the function plot(t,real(y)) % imaginary part of y(t) is 0 xlabel('t'); ylabel('y(t)'); axis([-6 6 0.12 0.26]), grid on title ('Signal Reconstruction from CTFS') print -dtiff plot.tiff;

The resulting waveform is shown in Fig. S4.33.

Fig. S4.33: Signal y(t) reconstructed from the first 4000 exponential CTFS coefficients in Problem 4.33.

159