Mathematical Foundations - Control of Color Imaging Systems: Analysis and Design - page 128

þ p p

1
j2

1
j2

1
2

X ( z ) z n 1 X ( j v) e j ( n 1 )v je j v dv ¼ X ( j v) e jn v dv

x ( n ) ¼ d z ¼

p p p

C p p

( 3 : 95 )

Example 3.26

Find the inverse DTFT of X( jv ) given by

¼ e jvj

X( jv ) for p < v < p

S OLUTION
Using the inversion integral, we have

p p

1
2

1
2

)e jnv d e jvj e jnv d x(n) ¼ X( jv v ¼ v

p p

p p

The above integral is decomposed into two integrals:

ð p

0

1
2

1
2

e v e jnv d e v e jnv d x(n) ¼ v þ v

p p

p 0

Hence,

ð p

0

0 p

1
2

1
2

1
2

e (1þjn)v
1

1
2

e (1jn)v
1

e (1þjn)v d e (1jn)v d

x(n) ¼

v þ v ¼

p p p þ jn p jn

p 0

p 0

e (1þjn)p
1

e (1jn)p e p e jnp

1

e p e jnp 1

2

1 1

2

1 1

2

1 1

2

1

¼ ¼

p þ jn p 1 jn p þ jn p 1 jn

Therefore,

1
2

2 2e p cos np

1

1 e p cos np

(1

x(n) ¼ ¼

p þ n 2 þ n 2 ) p

3.7 TWO-DIMENSIONAL z-TRANSFORM
The two-dimensional (2-D) z-transform of the 2-D sequence x ( n 1 , n 2 ) is de ned as

1 1

x ( n 1 , n 2 ) z n 1 z n 2

2

X ( z 1 , z 2 ) ¼ ( 3 : 96 )

1

n 1 ¼1 n 2 ¼1