Page 338 - Schaum's Outline of Theory and Problems of Signals and Systems

CHAP. 61 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS AND SYSTEMS

Noting that (1 - e-jn) = 0 for R = 0, X(R) must be of the form

where A is a constant. To determine A we proceed as follows. From Eq. (1.5) the even
component of u[nl is given by
u,[n] = $ + f6[n]
Then the odd component of u[n] is given by

uo[n] = u[n] - u,[n] = u[n] - f - $[n]
1 1
and y{uo[nl] =A + - e-jn - aS(R) - -
2
From Eq. (6.63b) the Fourier transform of an odd real sequence must be purely imaginary.
Thus, we must have A = a, and

6.29. Verify the accumulation property (6.571, that is,

From Eq. (2.132)

Thus, by the convolution theorem (6.58) and Eq. (6.142) we get

6.30. Using the accumulation property (6.57) and Eq. (1.501, find the Fourier transform of
4nI.
From Eq. (1.50)

Now, from Eq. (6.36) we have

s[n] H 1
Setting x[kl= 6[k] in Eq. (6.571, we have
x[n] = 6[n] HX(R) = 1 and X(0) = 1