The z-transform - Continuous and Discrete Time Signals and Systems - page 605

Solution
(i) Taking the z-transform of the input sequence, we obtain
X 1 ( z ) = z

z − 1 .
Applying the convolution property, the z-transform Y 1 ( z ) of the output response
is given by

− 2
(1 − z − 1 )(1 − z − 1 + z − 2 ) .
Taking the partial fraction expansion of Y 1 ( z ) yields

z
( z − 1)( z 2 − z + 1)

z

Y 1 ( z ) = H ( z ) X 1 ( z ) = =

1
1 − z − 1

− 1

1 − z − 1 + z − 2 .
Using entries (3) and (12) of Table 13.1 (see Problem 13.5), we obtain

Y 1 ( z ) =

1
1 − z − 1

←→

u [ k ]

and

√ π k

3

+ π
6

←→ 1

1 − z − 1 + z − 2 .
Using the linearity property, the output y 1 [ k ]isgivenby

3 sin

u [ k ]

π k
3

+ π
6

√

y 1 [ k ] = 1 − u [ k ] .

3 sin

Note that the output response contains a unit step function and a sinusoidal term
and is, therefore, bounded.
(ii) Taking the z-transform of the input sequence, we obtain

√

− 1

( 3 / 2) z

X 2 ( z ) =

− 2 .

1 − z

− 1 + z

Applying the convolution property, the z-transform Y 2 ( z ) of the output response
is given by

√ √

− 3
(1 − z − 1 + z − 2 ) 2 .
Using the frequency-differentiation property (see Problem 13.6), it can be
shown that the following is a z-transform pair:

− 1 − 2

( 3 / 2) z z ( 3 / 2) z

Y 2 ( z ) = H ( z ) X 2 ( z ) = =

1 − z − 1 + z − 2 1 − z − 1 + z − 2

√

− 3
(1 − z − 1 + z − 2 ) 2 .

2
3

π
3 k

√ π

3 k + π

( 3 / 2) z ←→ =

sin − sin u [ k ]

6

3

Therefore, the output response is given by

2
3

π
3 k

√ π

3 k + π

y 2 [ k ] = sin − sin u [ k ] .

6

3

√

Note that the output is a growing sinusoid function because of the k / 3 scaling

factor. Therefore, as k increases, the y 2 [ k ] increases without bound, leading
to an unstable situation.