Signals and Systems - Electrical Engineering

612 C H A P T E R 10: Fourier Analysis of Discrete-Time Signals and Systems

nExample 10.19

A periodic signalx 1 [n] of periodN=4 has a period

x 1 [n]=

{

1 n=0, 1

0 n=2, 3

Suppose that we want to find the periodic convolution sum ofx[n] with itself—call itv[n]. Let then

y[n]=x[n−2], and find the periodic convolution sum ofx[n] andy[n]—call itz[n]. How doesv[n]

compare withz[n]?

Solution

The circular representation ofx[n] is shown in Figure 10.13. To find the periodic convolution

sum we consider a periodx 1 [n] and represent the stationary signal by the internal circle, and the

circularly shifted signal by the outside circle. Multiplying the values in each of the spokes and

adding them we find the values of a period ofv[n], which is given by

v 1 [n]=











1 n= 0

2 n= 1

1 n= 2

0 n= 3

Analytically, the Fourier series coefficients ofv[n] areV[k]=N(X[k])^2 = 4 (X[k])^2. Using the

Z-transform,X 1 (z)= 1 +z−^1 so thatX^21 (z)= 1 + 2 z−^1 +z−^2 and

V[k]= 4

X^21 (z)

4 × 4

|z=ej 2 πk/ 4

=

1

4

( 1 + 2 e−j^2 πk/^4 +e−j^2 π^2 k/^4 )=

1

4

( 1 + 2 e−jπk/^2 +e−jπk)

This can be verified by using the period obtained from the periodic convolution sum so that

V[k]=

1

N

N∑− 1

n= 0

v[n]e−j^2 πnk/N=

1

4

( 1 + 2 e−j^2 πk/^4 +e−j^2 π^2 k/^4 )

which equals the above expression.

Graphically, the periodic convolution ofx[n] andy[n] is shown in Figure 10.13(c) where the sta-

tionary signal is chosen asx[m], represented by the inner circle, and the circularly shifted signal

is chosen asy[n−m], represented by the outer circle. The result of the convolution is a periodic

signalz[n] of period

z 1 [n]=











1 n= 0

0 n= 1

1 n= 2

2 n= 3