Signals and Systems - Electrical Engineering

9.4 One-Sided Z-Transform 531

− 2 0 2

n

x[

n]

h

[n

]

y[

n]

n

n

4 6

(a) (b)

(c)

8

0

0.5

1

− 2 0 2 4 6 8

0

0.5

1

− 2 0 2 4 6 8

0

0.5

1

1.5

FIGURE 9.5
Convolution sum for an averager FIR: (a)x[n], (b)h[n], and (c)y[n]. The outputy[n]is of length 6 given thatx[n]
is of length 4 andh[n]is the impulse response of a second-order FIR filter of length 3.

and according to the convolution sum property,

Y(z)=X(z)H(z)=

1

2

( 1 + 2 z−^1 + 3 z−^2 + 3 z−^3 + 2 z−^4 +z−^5 )

Thus,y[0]=0.5,y[1]=1,y[2]=1.5,y[3]=1.5,y[4]=1, andy[5]=0.5, just as before.

In MATLAB the functionconvis used to compute the convolution sum giving the results shown

in Figure 9.5, which coincide with the ones obtained in the other approaches. n

nExample 9.8

Consider an FIR filter with impulse response

h[n]=δ[n]+δ[n−1]+δ[n−2]

Find the filter output for an inputx[n]=cos( 2 πn/ 3 )(u[n]−u[n−14]). Use the convolution sum

to find the output, and verify your results with MATLAB.