9.4 One-Sided Z-Transform 533− (^1051015)
−0.5
0
0.5
1
1.5
n
x[
n]
0 5 10 15
0
0.5
1
n
h
[n
]
− 2 0 2 4 6 8 10 12 14 16 18
− 1
−0.5
0
0.5
1
n
y[
n]
(a) (b)
(c)
FIGURE 9.6
Convolution sum for FIR filter: (a)x[n], (b)h[n], and (c)y[n].
n The length of the convolution sum=length ofx[n]+length ofh[n]− 1 = 14 + 3 − 1 =
16—that is,Y(z)is a polynomial of order 15. n
nExample 9.9
The convolution sum of noncausal signals is more complicated graphically than that of the causal
signals we showed in the previous examples. Let
h 1 [n]=
1
3
(
δ[n+1]+δ[n]+δ[n−1])
be the impulse response of a noncausal averager FIR filter, andx[n]=u[n]−u[n−4] be the input.
Compute the filter output using the convolution sum.Solution
Graphically, it is a bit confusing to ploth 1 [n−k], as a function ofk, to do the convolution
sum. Using the convolution and the time-shifting properties of the Z-transform we can view the
computation more clearly.