Example H4.4
MATLAB code for example 4.4 from the book "Regeltechniek voor het HBO"
Determine the pole and zero plot from a unit step response
- Date : 23/02/2022
- Revision : 1.0
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Table of Contents
Assignment
The unit step response of a spring-mass-damper system (section 2.5.2) has been measured and is shown in figure 4.9a. How does the pole zero plot look like?
Solution
The step response suggests a second order process and as we know from section 2.5.2 this is really the case. Moreover the horizontal start of the response underlines this. Because of the overshoot this has to be a second order process with β < 1.
First of all it can be seen from the final value of 6 as a response to step of amplitude 1, that 
The overshoot is determined as:
This value we can associate with the well known formulas you might remember of a (pure) second order process (4.18):
= 
. From this we get: 
and 
and There is still more to notice from the step response: 
or (equivalent): 
. From: 
or 
we get 
rad/s and 
. So the poles equal 
.
or (equivalent): . From: or we get rad/s and . So the poles equal .To complete the parameters as used in formula 4.11 we still have to calculate 
and 
. With 
(see 4.13) we find: 
and 
(4.14) we find: and 
.
and . With (see 4.13) we find: and (4.14) we find: and .In figure 4.9b you see the pole and zero plot. You should check the step response of this process with transfer function: 
.
. Maybe you would prefer another sequence to calculate the parameters, but the starting point in each case is 
ω (from 
and D. From this you can detect the parameterset (
) or (
). It is not a big problem to get the formulas you need by table lookup.
ω (from and D. From this you can detect the parameterset () or (). It is not a big problem to get the formulas you need by table lookup.MATLAB code for this example:
% clear all variables from Workspace and close all figures
clear variables;
close all;
H=tf(18.6,[1 1.6 3.1]);
figure(101);
step(H);
grid on;
Higher order systems can be seen as dominant first or second order
Maybe you wonder until now why we pay so much attention to the first and second order processes. The answer is that in many cases higher order systems can be seen as DOMINANT first or second order. To be able to manipulate these two types of systems means that you have got already a lot of knowledge of many processes.
Experiment 1
Examine for instance the step response of our system when you add a pole in 
(third order; multiply 
with 
)
(third order; multiply with ) s=tf('s');
H=18.6/(s^2+1.6*s+3.1);
% Add an extra pole to the system and compare the stepresponses
p1=8;
H_tot=H*(p1/(s+p1));
figure(102);
step(H,H_tot);
grid on;
legend("Orginal","Extra Pole");
Experiment 2
Or examine for instance the step response of our system when you add two poles in 
(fourth order; multiply with 
).
(fourth order; multiply with ).s=tf('s');
H=18.6/(s^2+1.6*s+3.1);
% Add an extra pair of complex conjugated poles to the system and compare the stepresponses
Lambda_1=33;
Omega_1=5;
H_extra=(Lambda_1^2+Omega_1^2)/((s+Lambda_1+1i*Omega_1)*(s+Lambda_1-1i*Omega_1));
H_tot=H*H_extra;
figure(103);
step(H,H_tot);
grid on;
legend("Orginal","Extra Pole Pair");
Experiment 3
In the same way you could examine the second order process of example 4.3a with poles in 
and 
When you skip the pole in 
(by multiplying equation 4.19 with 
; then 
will keep the same) you will see that the step response hardly differs from the second order one.
and When you skip the pole in (by multiplying equation 4.19 with ; then will keep the same) you will see that the step response hardly differs from the second order one.Try this out your self using the code examples above.
% Your code