Digital Signal Processing Systems, Basic Filtering Types, and Digital Filter Realizations - Digital Signal Processing: Fundamentals and Applications - page 168

In DSP applications, the given transfer function of a digital system can converted into a difference
equation for DSP implementation. The following example illustrates the procedure.

EXAMPLE 6.5
Convert each of the following transfer functions into its difference equation.

z 2 1
z 2 þ 1:3z þ 0:36

a. HðzÞ¼

z 2 0:5 z þ 0:36
z 2

b. HðzÞ¼

Solution:
a. Dividing the numerator and denominator by z 2 to obtain the transfer function whose numerator and denominator
polynomials have the negative power of z, it follows that

ðz 2 1Þ=z 2
ðz 2 þ 1:3z þ 0:36Þ=z 2 ¼

1 z 2
1 þ 1:3z 1 þ 0:36z 2

HðzÞ¼

We write the transfer function using the ratio of Y ðzÞ to X ðzÞ :

1 z 2
1 þ 1:3z 1 þ 0:36z 2

Y ð z Þ
X ðzÞ ¼

Then we have

Y ðzÞð1 þ 1:3z 1 þ 0:36z 2 Þ¼X ðzÞð1 z 2 Þ

By distributing Y ðzÞ and X ðzÞ, we yield
Y ðzÞþ1:3z 1 Y ðzÞþ0:36z 2 Y ðzÞ¼X ðzÞz 2 X ðzÞ

Applying the inverse z-transform and using the shift property in Equation (5.3) of Chapter 5, we get

yðnÞþ1:3yðn 1Þþ0:36yðn 2Þ¼xðnÞxðn 2Þ

Writing the output yðnÞ in terms of inputs and past outputs leads to

yðnÞ¼xðnÞxðn 2Þ1:3yðn 1Þ0:36yðn 2Þ

b. Similarly, dividing the numerator and denominator by z 2 , we obtain

X ðzÞ ¼ ð z 2 0:5 z þ 0:36Þ= z 2

Y ð z Þ

¼ 1 0:5z 1 þ 0:36z 2

HðzÞ¼

z 2 =z 2

Thus

Y ðzÞ¼X ðzÞð1 0:5z 1 þ 0:36z 2 Þ

By distributing X ðzÞ, we yield

Y ðzÞ¼X ðzÞ0:5z 1 X ðzÞþ0:36z 2 X ðzÞ

Applying the inverse z-transform with using the shift property in Equation (5.3), we obtain

yðnÞ¼xðnÞ0:5xðn 1Þþ0:36xðn 2Þ