Computer experiments with the Laplace Transform and transfer functions. Background: "Signals and Systems" by Ziemer, Tranter and Fannin, Chapter 5. (or an equivalent material from another textbook - a chapter on the Laplace Transform).
Before starting this exercise you need to review the material about the Laplace transform and transfer functions. You will need a table of Laplace transforms on your desk next to your computer or X-terminal.
Also, please use "help" command to review the description of the following commands in Matlab:
conv impulse ord2 poly rmodel residue roots step tf2zp zp2tf
Operations on polynomials:
Example:
multiply 
by 
.
We define 
In Matlab we just define the vectors of coefficients of 
and 
then 
where 
(analyze why)
that is 
Example:
find roots of a polynomial p whose vector of coefficients is
>> roots(
)
ans =
-7.0000
-5.0000
-4.0000
-2.0000
-1.0000
The above operations help greatly in manipulating the numerators and denominators of Laplace transforms.
To obtain an expression for the Laplace transform of a function, find Laplace transforms of each term from tables and add terms using Maltab operation "residue" to add partial fractions if any.
Example:
table lookup yields
We now form a vector of residues (the numbers in the numerators) and the vector of poles, and we define a constant k (here k=0). See "help residue" for more details.
>> r=[5 3 2 1];
>> p=[-2 1 0 -10];
>> k=0;
>> [b,a]=residue(r,p,k)
b =
11 104 24 -40
a =
1 11 8 -20 0
This gives us a rational function 
.
To add more complicated fractions you can use "conv" to find common denominators, or use the command "parallel" treating the Laplace transforms like if they were transfer functions of some systems.
More precisely, suppose we have two Laplace transforms in the form
and 
Then 
is given by the fraction
where
[n,d]=parallel(b1,a1,b2,a2);
Inverse Laplace transforms:
In the following problems, you need to use "zp2tf" and "residue" and the Laplace transform tables to obtain the inverse Laplace tranform of the given rational function of s. Once this is done, plot the numerical values of the inverse Laplace Transform in and interval from 0 to 10 sec and compare them with those obtained by using "impulse". The two should match (why?). If not, repeat the exercise more carefully.
Example:
by using partial fractions we get 
for 
.
This can be plotted using the time sequence vector 
;
We can get the result by Matlab in the following way:
>> [n,d]=zp2tf([ ],[-2 -8],10)
n =
0 0 10
In the instruction above, 
denotes an empty vector of zeros ( no zeros
in 
), 
is a vector of poles at 
and 
respectively, and
is the gain coefficient (scalar).
In more complicated problems it may be better to first define the vector of zeros as z and the vector of poles as p, and then use "zp2tf".
The next step is:
>> [r,p,g]=residue(n,d)
r =
-1.6667
1.6667
p =
-8
-2
g =
comment: 
is 
and 
. These are residues od poles at and
respectively.
Hence the partial fraction expansion is
and from the Laplace transform tables we get 
written above.