Page 307 - Schaum's Outline of Theory and Problems of Signals and Systems

294 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS AND SYSTEMS [CHAP. 6

and X(ejn) mean the same thing whenever we connect the Fourier transform with the
z-transform. Because the Fourier transform is the z-transform with z = ein, it should not
be assumed automatically that the Fourier transform of a sequence x[n] is the z-transform
with z replaced by eiR. If x[n] is absolutely summable, that is, if x[n] satisfies condition
(6.311, the Fourier transform of x[n] can be obtained from the z-transform of x[n] with
= eifl since the ROC of X(z) will contain the unit circle; that is, leinJ = 1. This is not
generally true of sequences which are not absolutely summable. The following examples
illustrate the above statements.

EXAMPLE 6.1. Consider the unit impulse sequence 6[nl.
From Eq. (4.14) the z-transform of 6[n] is

By definitions (6.27) and (1.45) the Fourier transform of 6[n] is

Thus, the z-transform and the Fourier transform of 6[n] are the same. Note that 6[n] is absolutely
summable and that the ROC of the z-transform of 6[nl contains the unit circle.

EXAMPLE 6.2. Consider the causal exponential sequence
x[n] = anu[n] a real

From Eq. (4.9) the z-transform of x[n] is given by

Thus, X(ei") exists for la1 < 1 because the ROC of X(z) then contains the unit circle. That is,

Next, by definition (6.27) and Eq. (1.91) the Fourier transform of x[n] is

Thus, comparing Eqs. (6.37) and (6.38), we have

X(R) = X(z)l,=p
Note that x[n] is absolutely summable.

EXAMPLE 6.3. Consider the unit step sequence u[nl.
From Eq. (4.16) the z-transform of u[nl is

The Fourier transform of u[n] cannot be obtained from its z-transform because the ROC of the