Signals and Systems - Electrical Engineering

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

According to the convolution property the Z-transform of the output of the noncausal

filter is

Y 1 (z)=X(z)H 1 (z)

=X(z)[zH(z)] (9.23)

where we let

H 1 (z)=Z[h 1 [n]]=

1

3

(z+ 1 +z−^1 )

=z

[

1

3

( 1 +z−^1 +z−^2 )

]

=zH(z)

whereH(z)=( 1 / 3 )Z[δ[n]+δ[n−1]+δ[n−2]] is the transfer function of a causal filter. Let

Y(z)=X(z)H(z)be the Z-transform of the convolution sumy[n]=[x∗h][n] ofx[n] andh[n], both

of which are causal and can be computed as before.

According to Equation (9.23), we then haveY 1 (z)=zY(z)ory 1 [n]=[x∗h 1 ][n]=y[n+1].

n

Letx 1 [n]be the input to a noncausal LTI system, with an impulse responseh 1 [n]such thath 1 [n]= 0 for

n<N 1 < 0. Assumex 1 [n]is also noncausal (i.e.,x 1 [n]= 0 forn<N 0 < 0 ). The outputy 1 [n]=[x 1 ∗h 1 ][n]

has a Z-transform of

Y 1 (z)=X 1 (z)H 1 (z)=[zN^0 X(z)][zN^1 H(z)]

whereX(z)andH(z)are the Z-transforms of a causal signalx[n]and of a causal impulse responseh[n]. If we

let

y[n]=[x∗h][n]=Z−^1 [X(z)H(z)]

then

y 1 [n]=[x 1 ∗h 1 ][n]=y[n+N 0 +N 1 ]

Remarks

n The impulse response of an IIR system, represented by a difference equation, is found by setting the initial

conditions to zero, therefore, the transfer function H(z)also requires a similar condition. If the initial

conditions are not zero, the Z-transform of the total response Y(z)is the sum of the Z-transforms of the

zero-state and the zero-input responses—that is, its Z-transform is of the form

Y(z)=

X(z)B(z)

A(z)

+

I 0 (z)

A(z)

(9.24)

and it does not permit us to compute the ratio Y(z)/X(z)unless the component due to the initial conditions

is I 0 (z)= 0.