Signal representation using Fourier series - Continuous and Discrete Time Signals and Systems - page 180

Solution
To compute the sum S , we consider the periodic signal f ( t ) shown in Fig. 4.11.
As shown in Example 4.13, the exponential CTFS coefficients of f ( t ) are given
by

0

n is even

12
( n π ) 2

D n =

n is odd.

Using Parseval's theorem, the average power of f ( t )isgivenby

∞ ∞ ∞

144
π 4

1
n 4

= 288
π 4

2 2 + 2 2 P x = D n = D 0 D n = 2 S .

n =−∞

n = 1

n = 1
n = odd

(4.69)

Using the time-domain approach, it was shown in Example 4.17 that the average
power of f ( t )isgivenby

∞

= 1
T 0

x ( t ) 2 d t

P f = 3 . (4.70)

−∞

Combining Eqs. (4.69) and (4.70) gives (288 /π 4 ) S

=

3or

∞

= 3 π 4
288

= π 4
56

1
(2 n + 1) 4

S = ≈ 1 . 014 7 .

n = 0

4.8.2 Response of an LTIC system to periodic signals

As a second application of the exponential CTFS representation, we consider the
response y ( t ) of an LTIC system with the impulse response h ( t ) to an periodic
input x ( t ). The system is illustrated in Fig. 4.23. Assuming that the input signal
x ( t ) has the fundamental period T 0 , the exponential CTFS representation of
x ( t )isgivenby

∞

D n e j n ω 0 t , x ( t ) = (4.71)

m = 0

where the fundamental frequency ω 0 = 2 π/ T 0 . The steps involved in calculat-

ing the output y ( t ) are as follows.

Step 1 Based on Theorem 4.3.1, the output of an LTIC system y n ( t )toa
complex exponential x n ( t ) =

x ( t ) y ( t )

h ( t )

D n exp( j n ω 0 t )isgivenby

periodic
input

LTIC
system

periodic
output

D n H ( n ω 0 )e j n ω 0 t , y n ( t ) = (4.72)

Fig. 4.23. Response of an LTIC
system to a periodic input.

where H ( n ω 0 ) = H ( ω ), evaluated at ω = n ω 0 . The new term H ( ω ) is referred