Mathematical Foundations - Control of Color Imaging Systems: Analysis and Design - page 116

Example 3.15

Convolve the following two sequences:

x(n) ¼ ½ 231 and h(n) ¼ ½ 1 232

S OLUTION
The z-transforms of x(n) and h(n) are

3z 1 þ z 2 X(z) ¼ 2 þ

2z 1 3z 2 2z 3 H(z) ¼ 1 þ þ

Using the convolution property of z-transform, we have

3z 1 þ z 2 )(1 2z 1 3z 2 2z 3 )

Y(z) ¼ X(z)H(z) ¼ (2 þ þ þ

z 1 þ z 2 11z 3 9z 4 2z 5

¼ 2 þ þ þ

Therefore,

y(n) ¼ x(n) * h(n) ¼ ½ 2 111192

Example 3.16

Convolve the following two sequences using z-transform:

5) n u(n) x(n) ¼ (0 :

7) n u(n)

h(n) ¼ (0 :

S OLUTION
The z-transform of y(n), the convolution of x(n) and h(n)is

z
z

z
z

z 2

Y(z) ¼ X(z)H(z) ¼

7 ¼

0 : 5 0 : (z 0 : 5)(z 0 : 7)

Performing partial fraction, we have

5z
z

3 : 5z

z

2 :

Y(z) ¼

7

0 : 0 : 5

Taking the inverse z-transform yields

7) n u(n) 5) n u(n)

y(n) ¼ 3 : 5(0 : 2 : 5(0 :