Signal representation using Fourier series - Continuous and Discrete Time Signals and Systems - page 175

yields

= ( jn ) 1

n
4

= j n
4

n
4

G n

4 sinc

sinc .

The function r ( t ) can now be represented as an exponential CTFS as follows:

n =−∞ G n e j n ω 0 t n =−∞ (j n ) sinc

∞ ∞

= 1
4

n
2

e j nt ,

r ( t ) =

where the fundamental frequency ω 0 is set to 1.

4.6.1 CTFS with different periods

In this section, we consider the variation of the CTFS when the period of a
function is changed. We use the rectangular pulse train for simplicity as its
CTFS coefficients are real-valued.

Example 4.22
Consider the periodic function x ( t ) in Fig. 4.13 (in Example 4.14) for the
following three cases:

(a) τ = 1msand T = 5 ms;

(b) τ = 1msand T = 10 ms;

(c) τ = 1msand T = 20 ms.

In each of the above cases, (i) determine the fundamental frequency, (ii) plot
the CTFS coefficients, and (iii) determine the higher-order harmonics absent in
the function.

Solution
It was shown in Example 4.14 that the exponential DTFS coefficients are given
by

= τ
T

n τ
T

.

D n sinc

(a) With T = 5 ms, the fundamental frequency is f 0 = 1 / T = 1 / 5ms =

200 Hz, while the fundamental angular frequency is ω 0 = 400 π

radians/s. The corresponding exponential CTFS coefficients are given by

= 2 π f 0

= 1

n
5

D n

5 sinc

,

which are plotted in Fig. 4.21(a) using two scales on the horizontal axis. The
first scale represents the number n of the CTFS coefficients and the second scale