Digital Signal Processing Reference
In-Depth Information
h[n] = (
2
)
n
u[n + 1].
As a final example, consider
This system has memory. The system
is not causal, since
The system is stable, because, from earlier,
h[-1] = 2 Z 0.
n=-
q
ƒ h[n] ƒ =
q
q
n
1
2
1
1 - 1
¢
≤
= 2 +
2
= 4.
■
>
n=-1
We now relate the unit step response to the unit impulse response for an LTI sys-
tem. Suppose that the system input is the unit step function
u
[
n
]. We denote the unit
step response as
s
[
n
]. Then, from (10.30),
s[n] =
q
k=-
q
n
u[n - k]h[k] =
a
h[k],
(10.40)
k=-
q
since is zero for or for Hence, the unit step response
can be calculated directly from the unit impulse response.
From (10.40), we form the first difference for
s
[
n
]:
u[n - k]
(n - k) 6 0,
k 7 n.
n
n- 1
s[n] - s[n - 1] =
a
h[k] -
a
h[k] = h[n].
(10.41)
k=-
q
k=-
q
Thus, the impulse response can be obtained directly from the unit step response;
consequently, the unit step response also completely describes the input-output
characteristics of a system.
Step response from the impulse response
EXAMPLE 10.7
Consider again the system of Example 10.4, which has the impulse response
h[n] = 0.6
n
u[n].
(10.42)
This system is dynamic, causal, and stable. The unit step response is then, from (10.40),
n
n
0.6
k
.
s[n] =
a
h[k] =
a
k=-
q
k= 0
From Appendix C, we express this series in the closed form
n
1
-
0.6
n+ 1
1 - 0.6
0.6
k
u[n] = 2.5(1 - 0.6
n+ 1
)u[n].
s[n] =
a
=
k= 0
The factor
u
[
n
] is necessary, since for (causal system). This result verifies that
of Example 10.4, in which the step response was calculated with the use of the convolution
summation. Note that the impulse response is obtained from the step response by (10.41);
hence, for
s[n] = 0
n 6 0
n G 0,
h[n] = s[n] - s[n - 1]
= 2.5(1 - 0.6
n+ 1
)u[n] - 2.5(1 - 0.6
n
)u[n - 1].
