Discrete-time Fourier series and transform - Continuous and Discrete Time Signals and Systems - page 487

Fig. 11.9. DTFT of the periodic
sequence x 1 [ k ] = k ,
0 ≤ k ≤ 3, with fundamental
period K 0 = 4. (a) Magnitude
spectrum; (b) phase spectrum.

X 1 ( W )

3 p 3 p 3 p

2 p 2 p

2 p

2 p

p p

W

−2 p −1.5 p − p −0.5 p 0 0.5 p p 1.5 p 2 p

(a)

< X 1 ( W )

3 p 3 p

p p

4 4

−0.5 p 1.5 p

W

−1.5 p −2 p − p 0 0.5 p p 2 p

3 p 3 p

− −

4 4

(b)

Solution
(i) Using Eq. (11.36a), the DTFT of x 1 [ k ]isgivenby

n =−∞ D n δ

∞

n =−∞ D n δ

∞

− 2 n π
4

− n π
2

X 1 ( Ω ) = 2 π = 2 π .

Ω Ω

Substituting Ω 0 = 2 π/ K 0 = π/ 2 in Eq. (11.36b), the DTFS coefficients D n for

x 1 [ k ] are given by

3

= 1
4

= 1

− j n π k / 2 − j n π/ 2 + 2e − j n π − j3 n π/ 2 ] .

+ 3e

D n k e

4 [e

k = 0

For 0 ≤ n ≤ 3, the values of the DTFS coefficients are as follows:

= 1

4 [1 + 2 1 + 3 1] = 3

n = 0 D 0

2 ;

= 1

− j π/ 2 + 2 e − j π − j3 π/ 2 ]

n = 1 D 1

4 [e

+ 3 e

= 1

4 [ − j + 2( − 1) + 3( j )] =− 1

2 [1 − j];

= 1

− j π − j2 π − j3 π ]

n = 2 D 2

4 [e

+ 2 e + 3 e

= 1

4 [ − 1 + 2(1) + 3( − 1)] =− 1 2 ;

= 1

− j3 π/ 2 + 2 e − j3 π − j9 π/ 2 ]

n = 3 D 3

4 [e

+ 3 e

= 1 + 2( − 1) + 3( − j)] =− 1

4 [ j 2 [1 +

j ] .

The values of the DTFS coefficients that lie outside the range 0 ≤ n

≤ 3 can

= D n .
Since X 1 ( Ω ) is a complex-valued function, its magnitude and phase spectra
are plotted separately in Figs. 11.9(a) and (b). The area enclosed by the impulse

be obtained by using the periodicity property D n + 4