Digital Signal Processing Reference
In-Depth Information
real poles in the transfer function result in real exponential terms in the system
response.
We next illustrate the inverse transform of a function with a repeated pole.
Inverse transform involving repeated poles
EXAMPLE 7.16
The unit step response of a system with the third-order transfer function
Y(s)
X(s)
= H(s) =
4s
2
+ 4s + 4
s
3
+ 3s
2
+ 2s
will be found. Hence,
x(t) = u(t)
and
X(s) = 1/s.
The system output is then
4s
2
+
4s
+
4
s
3
+ 3s
2
+ 2s
1
s
¢
≤
Y(s) = H(s)X(s) =
.
Because this function is not in Table 7.2, we must find its partial-fraction expansion:
4s
2
k
1
s
2
+
k
2
s
+
k
3
s + 1
+
k
4
s + 2
.
+ 4s + 4
s
2
(s + 1)(s + 2)
Y(s) =
=
We solve first for
k
1
, k
3
,
and
k
4
:
4s
2
+ 4s + 4
(s + 1)(s + 2)
4
2
= 2;
k
1
=
s = 0
=
4s
2
+ 4s + 4
s
2
(s + 2)
4
-
4
+
4
(1)(1)
k
3
=
s =-1
=
= 4;
4s
2
+
4s
+
4
s
2
(s + 1)
16
-
8
+
4
(4)(-1)
k
4
=
s =-1
=
=-3.
We calculate
k
2
by Equation (F.8) of Appendix F:
4s
2
+
4s
+
4
s
2
d
ds
[s
2
Y(s)]
s = 0
=
d
ds
k
2
=
B
R
+ 3s + 2
s = 0
(s
2
+
3s
+
2)(8s
+
4)
-
(4s
2
+
4s
+
4)(2s
+
3)
[s
2
+ 3s + 2]
2
=
s = 0
(2)(4)
-
(4)(3)
4
=
=-1.
The partial-fraction expansion is then
4s
2
+ 4s + 4
s
2
(s
2
2
s
2
+
-1
s
+
4
s + 1
+
-3
s + 2
,
Y(s) =
=
+ 3s + 2)
