Signals and Systems - Electrical Engineering

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

Remarks

n As before, multiplication in one domain causes convolution in the other domain.

n In computing a periodic convolution we need to remember that: (1) the convolving sequences must have

the same period, and (2) the Fourier series coefficients of a periodic signal share the same period with the

signal.

nExample 10.18

To understand how the periodic convolution sum results, consider the product of two periodic

signalsx[n] andy[n] of periodN=2. Find the Fourier series of their productv[n]=x[n]y[n].

Solution

The multiplication of the Fourier series

x[n]=X[0]+X[1]ejω^0 n

y[n]=Y[0]+Y[1]ejω^0 n ω 0 = 2 π/N=π

can be seen as a product of two polynomials in complex exponentialsζ[n]=ejω^0 nso that

x[n]y[n]=(X[0]+X[1]ζ[n])(Y[0]+Y[1]ζ[n])

=X[0]Y[0]+

(

X[0]Y[1]+X[1]Y[0]

)

ζ[n]+X[1]Y[1]ζ^2 [n]

Nowζ^2 [n]=ej^2 ω^0 n=ej^2 πn=1 so that

x[n]y[n]=

(

X[0]Y[0]+X[1]Y[1]

)

︸ ︷︷ ︸

V[0]

+

(

X[0]Y[1]+X[1]Y[0]

)

︸ ︷︷ ︸

V[1]

ejω^0 n=v[n]

after replacingζ[n]. When using the periodic convolution formula, we have

V[0]=

∑^1

k= 0

X[k]Y[−k]=X[0]Y[0]+X[1]Y[−1]=X[0]Y[0]+X[1]Y[2−1]

V[1]=

∑^1

k= 0

X[k]Y[1−k]=X[0]Y[1]+X[1]Y[0]

where in the upper equation we used the periodicity ofY[k] so thatY[−1]=Y[− 1 +N]=Y[− 1 +

2]=Y[1]. Thus, the multiplication of periodic signals can be seen as the product of polynomials

inζ[n]=ejω^0 n=ej^2 πn/Nsuch that

ζpN+m[n]=ej

2 π

N(pN+m)=ej

2 π

Nm=ζm[n] p=±1,±2,..., 0≤m≤N− 1

which ensures that the resulting polynomial is always of orderN−1.

Graphically, we proceed in a manner analogous to the convolution sum (see Figure 10.12) by

representingX[k] andY[m−k] circularly, and shiftingY[m−k] clockwise bym=0 andm=1,