Although Laplace transforms are rarely solved using integration
(
tables and
computers (
e.g. Matlab) are much more
common), we will provide the
bilateral Laplace transform
pair here. These define the forward and inverse Laplace
transformations. Notice the similarities between the forward
and inverse transforms. This will give rise to many of the same
symmetries found in
Fourier
analysis.
Laplace Transform
Fs=∫-∞∞ftⅇ-stdt
F
s
t
f
t
s
t
(1)
Inverse Laplace Transform
ft=12πⅈ∫c-ⅈ∞c+ⅈ∞Fsⅇstds
f
t
1
2
s
c
c
F
s
s
t
(2)
Finding the Laplace and Inverse Laplace Transforms
Solving the Integral
Probably the most difficult and least used method for
finding the Laplace transform of a signal is solving the
integral. Although it is technically possible, it is
extremely time consuming. Given how easy the next two
methods are for finding it, we will not provide any more
than this. The integrals are primarily there in order to
understand where the following methods originate from.
Using a Computer
Using a computer to find Laplace transforms is relatively
painless. Matlab has two functions,
laplace and
ilaplace, that are both part of the
symbolic toolbox, and will find the Laplace and inverse
Laplace transforms respectively. This method is generally
preferred for more complicated functions. Simpler and more
contrived functions are usually found easily enough by
using tables.
Using Tables
When first learning about the Laplace transform, tables are
the most common means for finding it. With enough practice,
the tables themselves may become unnecessary, as the common
transforms can become second nature. For the purpose of
this section, we will focus on the inverse Laplace
transform, since most design applications will begin in the
Laplace domain and give rise to a result in the time domain.
The method is as follows:
-
Write the function you wish to transform,
Hs
H
s
,
as a sum of other functions,
Hs=∑i=1m
H
i
s
H
s
i
1
m
H
i
s
where each of the
H
i
H
i
is known from a table.
-
Invert each
H
i
s
H
i
s
to get its
h
i
t
h
i
t
.
-
Sum up the
h
i
t
h
i
t
to get
ht=∑i=1m
h
i
t
h
t
i
1
m
h
i
t
Example 1
Compute
ht
h
t
for
∀s,ℜs>-5:Hs=1s+5
s
s
-5
H
s
1
s
5
This can be solved directly from the
table to be
ht=ⅇ-5t
h
t
5
t
Example 2
Find the time domain representation,
ht
h
t
,
of
∀s,ℜs>-10:Hs=25s+10
s
s
-10
H
s
25
s
10
To solve this, we first notice that
Hs
H
s
can also be written as
251s+10
25
1
s
10
. We can then go to the
table to find
ht=25ⅇ-10t
h
t
25
10
t
Example 3
We can now extend the two previous examples by finding
ht
h
t
for
∀s,ℜs>-5:Hs=1s+5+25s+10
s
s
5
H
s
1
s
5
25
s
10
To do this, we take advantage of the additive property of
linearity and the three-step method described above to
yield the result
ht=ⅇ-5t+25ⅇ-10t
h
t
5
t
25
10
t
For more complicated examples, it may be more difficult to
break up the transfer function into parts that exist in a
table. In this case, it is often necessary to use
partial fraction
expansion to get the transfer function into a more
usable form.
Visualizing the Laplace Transform
With the Fourier transform, we had a complex-valued
function of a purely imaginary
variable,
Fⅈω
F
ω
. This was something we could envision with two
2-dimensional plots (real and imaginary parts or magnitude and
phase). However, with Laplace, we have a complex-valued
function of a complex variable.
In order to examine the magnitude and phase or real and
imaginary parts of this function, we must examine
3-dimensional surface plots of each component.
While these are legitimate ways of looking at a signal in the
Laplace domain, it is quite difficult to draw and/or analyze.
For this reason, a simpler method has been developed.
Although it will not be discussed in detail here, the method
of
Poles and Zeros
is much easier to understand and is the way both the Laplace
transform and its discrete-time counterpart the
Z-transform are
represented graphically.
"My introduction to signal processing course at Rice University."