Signals and Systems - Electrical Engineering

11.6 Realization of Discrete Filters 695

because a 2 =b 2 − 0 and the lower delay because it is not needed once these constant multiplier are
deleted).
n Obtaining the transfer function from a direct form II realization is not as obvious as it is from a direct
form I realization. For a given realization direct form II, we need to use the auxiliary variable w[n]to
obtain the transfer function from the realization. Instead of using the time-domain equations one should
use the Z-transform equations to obtain the overall transfer function. For instance, consider the direct form
II realization in Figure 11.28 for the second-order system. From the realization we obtain

w[n]=x[n]−a 1 w[n−1]−a 2 w[n−2]⇒( 1 +a 1 z−^1 +a 2 z−^2 )W(z)=X(z)

y[n]=b 0 w[n]+b 1 w[n−1]+b 2 w[n−2]⇒Y(z)=(b 0 +b 1 z−^1 +b 2 z−^2 )W(z)

Solving for W(z)in the top equation and replacing it in the bottom equation will give H 2 (z).

Cascade Realization
Thecascade realizationis obtained by representing the given transfer functionH(z)=B(z)/A(z)as a
product of first- and second-order filtersHi(z)with real coefficients:

H(z)=

∏

i

Hi(z) (11.73)

Each transfer functionHi(z)is realized by direct form II and cascaded. Different from the analog case,
this cascade realization is not constrained by loading.

nExample 11.16

Obtain a cascade realization of the filter with transfer function

H(z)=

3 +3.6z−^1 +0.6z−^2

1 +0.1z−^1 −0.2z−^2

Solution

The poles ofH(z)arez=−0.5 andz=0.4, and the zeros arez=−1 andz=−0.2, all of which

are real. One way of obtaining the cascade realization is to expressH(z)as

H(z)=

[

3 ( 1 +z−^1 )

1 +0.5z−^1

][

1 +0.2z−^1

1 −0.4z−^1

]

If we let

H 1 (z)=

3 ( 1 +z−^1 )

1 +0.5z−^1

H 2 (z)=

1 +0.2z−^1

1 −0.4z−^1

RealizingH 1 (z)andH 2 (z)separately and then cascading them we obtain the realization forH(z)

shown in Figure 11.29.