Signals and Systems - Electrical Engineering

9.4 One-Sided Z-Transform 529

nExample 9.7

Consider computing the output of an FIR filter,

y[n]=

1

2

(

x[n]+x[n−1]+x[n−2]

)

for an inputx[n]=u[n]−u[n−4] using the convolution sum, analytically and graphically, and

the Z-transform.

Solution

The impulse response ish[n]=0.5(δ[n]+δ[n−1]+δ[n−2]), so that h[0], h[1], h[2] are,

respectively, 0.5, 0.5, and 0.5, andh[n] is zero otherwise.

Convolution sum formula:The equation

y[n]=

∑n

k= 0

h[k]x[n−k]

=x[0]h[n]+x[1]h[n−1]+···+x[n]h[0] n≥ 0

with the condition that in each entry the arguments ofx[.] andh[.] add ton≥0, gives

y[0]=x[0]h[0]=0.5

y[1]=x[0]h[1]+x[1]h[0]= 1

y[2]=x[0]h[2]+x[1]h[1]+x[2]h[0]=1.5

y[3]=x[0]h[3]+x[1]h[2]+x[2]h[1]+x[3]h[0]=x[1]h[2]+x[2]h[1]+x[3]h[0]=1.5

y[4]=x[0]h[4]+x[1]h[3]+x[2]h[2]+x[3]h[1]+x[4]h[0]=x[2]h[2]+x[3]h[1]= 1

y[5]=x[0]h[5]+x[1]h[4]+x[2]h[3]+x[3]h[2]+x[4]h[1]+x[5]h[0]=x[3]h[2]=0.5

and the rest are zero. In the above computations, we notice that the length ofy[n] is 4+ 3 − 1 =

6 since the length ofx[n] is 4 and that ofh[n] is 3.

Graphical approach:The convolution sum is given by either

y[n]=

∑n

k= 0

x[k]h[n−k]

=

∑n

k= 0

h[k]x[n−k]

Choosing one of these equations, let’s say the first one, we needx[k] andh[n−k], as functions

ofk, for different values ofn. Multiply them and then add the nonzero values. For instance, for

n=0 the sequenceh[−k] is the reflection ofh[k]; multiplyingx[k] byh[−k] gives only one value

different from zero atk=0, ory[0]= 1 /2. Forn=1, the sequenceh[1−k], as a function ofk,