Signals and Systems - Electrical Engineering

528 C H A P T E R 9: The Z-Transform

and

X 2 (z)= 1 +b 1 z−^1

Their product is

X 1 (z)X 2 (z)= 1 +b 1 z−^1 +a 1 z−^1 +a 1 b 1 z−^2 +a 2 z−^2 +a 2 b 1 z−^3

= 1 +(b 1 +a 1 )z−^1 +(a 1 b 1 +a 2 )z−^2 +a 2 b 1 z−^3

The convolution sum of the two sequences[1a 1 a 2 ]and[1b 1 ], formed by the coefficients of X 1 (z)and

X 2 (z), is[1(a 1 +b 1 ) (a 2 +b 1 a 1 )a 2 ], which corresponds to the coefficients of the product of the polyno-

mials X 1 (z)X 2 (z). Also notice that the sequence of length 3 (corresponding to the first-order polynomial

X 1 (z))and the sequence of length 2 (corresponding to the second-order polynomial X 2 (z)) when convolved

give a sequence of length 3 + 2 − 1 = 4 (corresponding to the third-order polynomial X 1 (z)X 2 (z)).

n A finite-impulse response or FIR filter is implemented by means of the convolution sum. Consider an FIR

with an input–output equation

y[n]=

N∑− 1

k= 0

bkx[n−k] (9.22)

where x[n]is the input and y[n]is the output. The impulse response of this filter is (let x[n]=δ[n]and

set initial conditions to zero, so that y[n]=h[n])

h[n]=

N∑− 1

k= 0

bkδ[n−k]

giving h[n]=bnfor n=0,...,N− 1 , and accordingly, we can write Equation (9.22) as

y[n]=

N∑− 1

k= 0

h[k]x[n−k]

which is the convolution of the input x[n]and the impulse response h[n]of the FIR filter. Thus, if X(z)=

Z(x[n])and H(z)=Z(h[n]), then

Y(z)=H(z)X(z) and y[n]=Z−^1 [Y(z)]

n The length of the convolution of two sequences of lengthsMandNisM+N− 1.

n If one of the sequences is of infinite length, the length of the convolution is infinite. Thus, for an infinite-

impulse response (IIR) or recursive filters the output is always of infinite length for any input signal, given

that the impulse response of these filters is of infinite length.