The z-Transform - Digital Signal Processing: Fundamentals and Applications - page 155

Solution:
a. Applying the z-transform on both sides of the difference equation and using Equation (5.3) or Equation (5.14) ,
we obtain

Y ðzÞþ0:1Y ðzÞz 1 0:2Y ðzÞz 2 ¼ X ðzÞþX ðzÞz 1

(5.15)

Factoring out YðzÞ on the left side and substituting X ðzÞ¼ZðdðnÞÞ ¼ 1 to the right side in the Equation (5.15)
we get

Y ðzÞð1 þ 0:1z 1 0:2z 2 Þ¼1ð1 þ z 1 Þ

Then YðzÞ can be expressed as

1 þ z 1
1 þ 0:1z 1 0:2z 2

Y ðzÞ¼

To obtain the impulse response, which is the inverse z-transform of the transfer function, we multiply the
numerator and denominator by z 2 .
Thus

z 2 þ z
z 2 þ 0:1z 0:2 ¼

zðz þ 1Þ
ðz 0:4Þðz þ 0:5Þ

Y ðzÞ¼

Using the partial fraction expansion method leads to
Y ð z z ¼

z þ 1
ðz 0:4Þðz þ 0:5Þ ¼

A
z 0:4 þ

B
z þ 0:5

where

z¼0:4 ¼ z¼0:4 ¼

Y ð z Þ
z

z þ 1
z þ 0:5

0:4 þ 1
0:4 þ 0:5 ¼ 1:5556

A ¼ðz 0:4Þ

z¼0:5 ¼ z¼0:5 ¼

Y ð z Þ
z

z þ 1
z 0:4

0:5 þ 1
0:5 0:4 ¼0:5556

B ¼ðz þ 0:5Þ

Thus

Y ðzÞ¼ 1:5556z

ðz 0:4Þ þ 0:5556 z ðz þ 0:5Þ

which gives the impulse response
y n ¼ 1:5556ð0:4Þ

n u n 0:5556ð0:5Þ n u n

b. To obtain the response due to a unit step function, the input sequence is set to be

xðnÞ¼uðnÞ

and the corresponding z-transform is given by

z
z 1

X ðzÞ¼