Frequency-Domain Analysis - Introduction to Digital Signal Processing and Filter Design - page 144

which we express in the form of

1

( 1 − 0 . 2 e − jω )( 1 − 0 . 5 e jω )

ae − jω ) or where each term in the denominator is of the general form (1 −

ae jω ) . The partial fraction expansion is now chosen to be in the form of (1 −

K 1
( 1 − 0 . 2 e − jω ) +

K 2
( 1 − 0 . 5 e jω )

Y(e jω )

=

The residue K 1 is determined by evaluating

e jω = 0 . 2 = 1 . 1111

Y(e jω )( 1 − 0 . 2 e − jω ) e jω = 0 . 2 =

1
( 1 − 0 . 5 e jω )

e jω = 2 = 1 . 111

Y(e jω )( 1 − 0 . 5 e jω ) e jω = 2 =

1
( 1 − 0 . 2 e − jω )

Similarly K 2 =

Example 3.13

e j( 0 . 3 πn) and h(n) ( 0 . 2 ) n u(n) . As in Example 3.12, we can find the Let x(n) = =

= 2 π δ(ω e j( 0 . 3 πn) as X(e jω ) DTFT of x(n) = − 0 . 3 π − 2 πk) and the DTFT

of h(n) as

1 − 0 . 2 e − jω

e jω
e jω

1

H(e jω ) = =

− 0 . 2

Thus

= 2 π δ(ω Y(e jω ) X(e jω )H (e jω ) − 2 πk)H (e jω ) = − 0 . 3 π

∞

− 2 πk)H (e j 0 . 3 π ) = 2 π δ(ω − 0 . 3 π

k =−∞

e j 0 . 3 π
(e j 0 . 3 π

2 π δ(ω

= − 0 . 3 π − 2 πk)

− 0 . 2 )

= 1 . 1146 e − j( 0 . 1813 ) 2 π δ(ω − 0 . 3 π − 2 πk)

= 1 . 1146 e − j( 0 . 1813 ) e j( 0 . 3 πn)

= 1 . 1146 e j( 0 . 3 πn − 0 . 1813 ) .
As an alternative method, we recollect that from the convolution of e jωn and
h(n) , we obtained y(n)

Therefore y(n)

e jωn H(e jω ) . In this example H(e jω ) = [ e jω /(e jω

= −

e j 0 . 3 πn H(e j 0 . 3 π ) :

0 . 2 ) ]and ω = 0 . 3 π . Therefore y(n) =

e j 0 . 3 πn e j 0 . 3 π
(e j 0 . 3 π

= 1 . 1146 e j( 0 . 3 πn − 0 . 1813 ) y(n) =

− 0 . 2 )