Signals and Systems - Electrical Engineering

11.6 Realization of Discrete Filters 693

By realizing the all-pole filter given in Equation (11.67), and using its outputw[n] in the realization

of Equation (11.68), we minimize the number of delays used. The realization of Equation (11.68)

does not require new delays, as the delayedw[n]’s are already available from the realization of the

all-pole filter. Thus, the number of delays used corresponds to the order of the denominatorA(z),

which is the order of the filter.

nExample 11.15

Consider the same transfer function as in Example 11.14 to obtain a direct form II realization of it.

Solution

We have that

W(z)=

X(z)

1 +0.1z−^1

gives

w[n]=x[n]−0.1w[n−1]

Now, according to Equation (11.68),

Y(z)=( 1 +1.5z−^1 )W(z)

which corresponds to the difference equation

y[n]=w[n]+1.5w[n−1] (11.69)

Notice that in Equation (11.69)w[n] andw[n−1] are already available, and thus there is no

need for new delays in this step. The direct form II realization using only one delay is shown

in Figure 11.27.

Obtaining the transfer function from the direct form II realization is not as obvious as from the

direct form I. In this case, we need to obtain the transfer function corresponding to the all-pole

filter first and use it to obtain the overall transfer function. From the above realization, we have

w[n]=x[n]−0.1w[n−1]

y[n]=w[n]+1.5w[n−1]

FIGURE 11.27
Direct form II realization of
H(z)=( 1 +1.5z−^1 )/( 1 +0.1z−^1 ). This is a
minimal realization ofH(z), which is a first-order
system, as only one delay is used. 0.1

x[n] 1.5

y[n]

w[n]

z−^1

+

−

+ +