Digital Signal Processing Reference
In-Depth Information
The coefficients and can be evaluated by the usual partial-fraction expansion.
These coefficients are complex valued, and is the conjugate of Thus, the in-
verse transform of (7.57) has two terms that are complex; however, the sum of these
two terms must be real. This sum is not a convenient form. We now present a differ-
ent procedure for finding the inverse transform, such that all terms are real.
In (7.57), we evaluate
k
1
k
2
k
2
k
1
.
k
1
and
k
2
from
`
s =a- jb
= ƒ k
1
ƒ e
ju
k
1
= (s - a + jb)F(s)
and
s =a+ jb
= ƒ k
1
ƒ e
-ju
.
`
k
2
= (s - a - jb)F(s)
(7.58)
Let
f
1
(t)
be the sum of the inverse transforms of the first two terms of (7.57):
f
1
(t) = ƒ k
1
ƒ e
ju
e
(a- jb)t
+ ƒ k
1
ƒ e
-ju
e
(a+ jb)t
.
Using Euler's relation, we can express this relation as
e
-j(bt -u)
+ e
j(bt -u)
2
f
1
(t)
= 2 ƒ k
1
ƒ e
at
B
R
= 2
ƒ
k
1
ƒ
e
at
cos
(bt - u).
(7.59)
Note that, in (7.58),
b
should be chosen positive, such that
b
in (7.59) is positive. The
sinusoidal expression in (7.59) is a more convenient form than the sum of complex
exponential functions. An example illustrating this procedure is given next.
Inverse Laplace transform involving complex poles
EXAMPLE 7.15
We now find the response of a system with the transfer function
3s + 1
H(s) =
s
2
+ 2s + 5
x(t) = e
-3t
.
to the input
We have
3s + 1
1
s + 3
Y(s) = H(s)X(s) =
+ 2s + 5
s
2
3s + 1
[(s + 1)
2
+ 2
2
](s + 3)
=
k
1
s + 1 + j2
+
k
2
s + 1 - j2
+
k
3
s + 3
,
=
