Signal representation using Fourier series - Continuous and Discrete Time Signals and Systems - page 170

and

b n
a n

− 1

< D − n =−< D n = tan , (4.51)

which illustrate that the magnitude spectrum is an even function and that the
phase spectrum is an odd function. Consider the magnitude and phase spectra
of the function g ( t ) in Example 4.11. The spectra are shown in Fig. 4.15. It
is observed that the amplitude spectrum is even, whereas the phase spectrum
is odd, which is expected as the function g ( t ) is real. Consider the rectangular
pulse train in Example 4.16, whose amplitude and phase spectra are shown
in Fig. 4.16. The amplitude function is again observed to be even symmetric.
However, the phase spectrum does not seem to be odd, although the time-domain
function is real-valued. Actually, the angle π r (i.e. 180 o ) is equivalent to −π r

(i.e. − 180 o ); the phase values π r

can be changed appropriately to satisfy the

odd property.

Parseval's theorem The power of a periodic signal x ( t ) can be calculated from
its exponential CTFS coefficients as follows (see Problem 1.9 in Chapter 1):

∞

= 1
T 0

x ( t ) 2 d t 2 . P x = D n (4.52)

n =−∞

T 0

For real-valued signals, D n = D − n , which results in the following simplified

formula:

∞ ∞

2 2 + 2 .

P x = D n = D 0 2 D n (4.53)

n =−∞ n = 1

Example 4.17
Calculate the power of the periodic signal f ( t ) shown in Fig. 4.11.

Solution
It was shown in Example 4.13 that the exponential CTFS coefficients of the
signal f ( t ) are given by

0

n is even

D n = 12

( n π ) 2

n is odd.

Since f ( t ) is real-valued, using Parseval's theorem (Eq. (4.53)) yields

∞ ∞ ∞ 2

12
n 2 π 2

2 2 + 2 2

P f = D n = D 0 D n = 2

n =−∞

n = 0 n = 1 , 3 , 5 ,...

∞

= 288
π 4

1
n 4

= 288
π 4

1 . 015 = 3 . (4.54)

n = 1 , 3 , 5 ,...