Discrete-time Fourier series and transform - Continuous and Discrete Time Signals and Systems - page 471

x [ k ]

1 1 1

0.5
0.25 0.13

0.5 0.25 0.13 0.5

0.25 0.13

k

−20 −18 −16 −14 −12 −10 −8 −6 −4 −2 0 16 18 20

24 6 8 0 2 4

Fig. 11.2. Periodic DT sequence
defined in Example 11.3.

Solution
The DT sequence x [ k ] is plotted in Fig. 11.2. Since its period K 0

= 15, the

fundamental frequency Ω 0 = 2 π/ 15. The DTFS coefficients D n are given by

14 14

1
15

1
15

0 . 5 k e

− j n Ω 0 k − j n Ω 0 ) k ,

= =

(0 . 5e

D n

k = 0 k = 0

which is a GP series that simplifies to

15

− j n Ω 0

1 −

0 . 5e

1 − 0 . 5 15 e

− j15 n Ω 0
1 − 0 . 5e − j n Ω 0

1
15

1
15

D n = = .

1 − 0 . 5e − j n Ω 0

Since Ω 0 = 2 π/ 15, the exponential term in the numerator, exp( − j15 n Ω 0 ) =
exp( − j2 n π ) = 1. Expanding the exponential term in the denominator as
exp( − j n Ω 0 ) = cos( n Ω 0 ) − j sin( n Ω 0 ), the DTFS coefficients are given by

1 − 0 . 5 15
1 − 0 . 5 cos( n Ω 0 ) + j0 . 5 sin( n Ω 0 )

1
15

D n =

1
15

1 ≈

1 − 0 . 5 cos( n Ω 0 ) + j0 . 5 sin( n Ω 0 ) .

(11.11)

As the DTFS coefficients are complex, we determine the magnitude and phase
of the coefficients as follows:

= 1
15

1

D n

magnitude

(1 − 0 . 5 cos( n Ω 0 )) 2 + (0 . 5 sin( n Ω 0 )) 2

1
15

1

= √

1 . 25 − cos( n Ω 0 ) ;

(11.12)

0 . 5 sin( n Ω 0 )
1 − 0 . 5 cos( n Ω 0 )

− 1

< D n =− tan ,

phase (11.13)

where Ω 0 = 2 π/ 15. The magnitude and phase spectra of the DTFS coefficients

are plotted in Figs. 11.3(a) and (b), in which one period of D n is highlighted by
a shaded region.

Example 11.4
Determine the DTFS coefficients of the following periodic function:

A e j((2 π m / N ) k +θ ) , x [ k ] = (11.14)