Frequency-Domain Analysis - Introduction to Digital Signal Processing and Filter Design - page 165

3

x(n)e − j( 2 π/N)( 1 .n) x( 0 )e − j( 2 π/ 4 )( 0 ) x( 1 )e − j( 2 π/ 4 )( 1 ) X p ( 1 ) = = +

n = 0

x( 2 )e − j( 2 π/ 4 )( 2 ) x( 3 )e − j( 2 π/ 4 )( 3 ) + + = 0 . 4 − j 0 . 4

3

x(n)e − j( 2 π/N)( 2 .n) x( 0 )e − j( 2 π/ 4 )( 0 ) x( 1 )e − j( 2 π/ 4 )( 2 ) X p ( 2 ) = = +

n = 0

x( 2 )e − j( 2 π/ 4 )( 4 ) x( 3 )e − j( 2 π/ 4 )( 6 ) + + = 0 . 0 + j 0 . 0

3

x(n)e − j( 2 π/N)( 3 .n) x( 0 )e − j( 2 π/ 4 )( 0 ) x( 1 )e − j( 2 π/ 4 )( 3 ) X p ( 3 ) = = +

n = 0

x( 2 )e − j( 2 π/ 4 )( 6 ) x( 3 )e − j( 2 π/ 4 )( 9 ) + + = 0 . 4 + j 0 . 4

Similarly, the DTFT of f(n) are computed as

3

f(n)e − j( 2 π/N)( 0 .n) f( 0 )e − j( 2 π/ 4 )( 0 ) f( 1 )e − j( 2 π/ 4 )( 0 )

F p ( 0 ) = = +

n = 0

f( 2 )e − j( 2 π/ 4 )( 0 ) f( 3 )e − j( 2 π/ 4 )( 0 )

+ + = 2 . 0

3

f(n)e − j( 2 π/N)( 1 .n) f( 0 )e − j( 2 π/ 4 )( 0 ) f( 1 )e − j( 2 π/ 4 )( 1 ) F p ( 1 ) = = +

n = 0

f( 2 )e − j( 2 π/ 4 )( 2 ) f( 3 )e − j( 2 π/ 4 )( 3 ) + + =

0 . 6

− j 0 . 6

3

f(n)e − j( 2 π/N)( 2 .n) f( 0 )e − j( 2 π/ 4 )( 0 ) f( 1 )e − j( 2 π/ 4 )( 2 ) F p ( 2 ) = = +

n = 0

f( 2 )e − j( 2 π/ 4 )( 4 ) f( 3 )e − j( 2 π/ 4 )( 6 )

+ + = 0 . 8 + j 0 . 0

3

f(n)e − j( 2 π/N)( 3 .n) f( 0 )e − j( 2 π/ 4 )( 0 ) f( 1 )e − j( 2 π/ 4 )( 3 )

F p ( 3 ) = = +

n = 0

f( 2 )e − j( 2 π/ 4 )( 6 ) f( 3 )e − j( 2 π/ 4 )( 9 ) + + = 0 . 6 + j 0 . 6

The term-by-term product of these vectors gives the DTFS of the output
Y p (k) = X p (k)F p (k) as

Y p (k) =

[6 . 40

− j 0 . 48 + j 00 . 0 + j 0 . 48]

0

and its inverse DTFS is computed from the formula

3

1
N

Y p (k)e j( 2 π/N)kn y p (n) =

k = 0

and we get y p (n) = [1 . 6 1.84 1.6 1.36].