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CHAP. 61 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS AND SYSTEMS
(a) From Eq. (6.28)
1 sin Wn
IW
x[n] = - IT X(R) dfi = - elnn do = -
27 -, 27 -w rn
Thus, we obtain
sin Wn IRl s w
-x(n) =
7n WCI~IST
(b) The sequence x[n] is plotted in Fig. 6-13(b) for W = 7/4.
6.16. Verify the frequency-shifting property (6.44, that is,
eJnonx[n] tt X(n - 0,)
By Eq. (6.27)
m
~ ( ~ j f l t ~ = ~ [C ]e~fl~"X[n] e-~fl"
)
~
,,= -m
m
= C x[n] e-~(fl-flt~)n = X(fl -a,)
,,= -m
Hence,
e~~o"x[n] -X(R - a,)
6.17. Find the inverse Fourier transform x[n] of
-
x(n) = 2~qn no) WI, lfiol 5
From Eqs. (6.28) and (1.22) we have
Thus, we have
ejnon - 2rS(R - R,)
6.18. Find the Fourier transform of
x[n] = 1 all n
Setting R, = 0 in Eq. (6.1351, we get
x[n] = 1 o 2nS(R) In1 I
Equation (6.136) is depicted in Fig. 6-14.
Fig. 6-14 A constant sequence and its Fourier transform.