Frequency-Domain Analysis - Introduction to Digital Signal Processing and Filter Design - page 143

filters. However, as mentioned earlier, they can also be used in time-domain
analysis, particularly when the z -transform technique cannot be used. To illustrate
this point, we provide the following examples.
Let the unit impulse response of a discrete-time system be given as h(n) =

( 0 . 2 ) n u(n) and the input sequence be given as x(n) ( 0 . 5 ) − n u(

= − n) . Therefore

e jω
e jω

1
1 − 0 . 2 e − jω =

H(e jω ) =

− 0 . 2

e − jω
e − jω

1

X(e jω ) =

0 . 5 e jω = 1 − − 0 . 5

e jω
e jω

e − jω
e − jω

Y(e jω ) H(e jω )X(e jω )

= =

− 0 . 2 − 0 . 5

Now let

k 1 e jω
e jω

k 2 e − jω
e − jω

Y(e jω ) =

− 0 . 2 +

− 0 . 5

so that we can easily obtain the inverse DTFT of each term. Note the difference
in the two terms.
Then we compute k 1 from the following method, which is slightly different
from the partial fraction method we have used earlier:

e − jω
e − jω

k 2 (e jω − 0 . 2 )

Y(e jω )e − jω (e jω

− 0 . 5 ) e − j 2 ω

−

0 . 2 )

= = k 1 +

(e − jω − 0 . 5

Evaluating both terms at e jω = 0 . 2 yields

e jω = 0 . 2 =

e − jω
e − jω

5
4 . 5 = 1 . 111

k 1 =

− 0 . 5

Similarly, the constant k 2 is evaluated as follows (again, this is slightly different
from the partial fraction expansion method we have used earlier):

e jω
e jω

k 1 e j 2 ω (e − jω − 0 . 5 )

Y(e jω )e jω (e − jω −

0 . 5 )

= + k 2 =

e jω − 0 . 2 − 0 . 2

Evaluating the two terms at e jω

2
1 . 8

= 2 . 0 yields k 2 = = 1 . 111. Therefore we

have Y(e jω ) = 1 . 111 e jω /(e jω + 1 . 111 e − jω /(e − jω − 0 . 2 ) − 0 . 5 ) , and the output

= 1 . 111 ( 0 . 2 ) n u(n) + 1 . 111 ( 0 . 5 ) − n u( is y(n) n) .

Now we describe a preferred method of finding the inverse DTFT of

−

e jω
e jω

e − jω
e − jω

Y(e jω ) H(e jω )X(e jω ) = =

− 0 . 2 − 0 . 5