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Control System Designer Toolbox

August 21, 2020

By Manoj sai

CONTROL SYSTEM

Have you ever thought of how control systems are involved in our day to day life? An Air conditioner functions depending on the temperature of the room and the desired temperature, which is set by us using a remote. The temperature of the room is being sensed by a sensor, and the error will be given as an input to the Air conditioner. According to the received error signal, it will work. Thus the room temperature is being controlled by the Air conditioner.

To control means to regulate, to direct, or to command. Hence a control system is an arrangement of different physical elements connected in such a manner to regulate, direct or command itself or some other system. In other words, the Control system is a system of devices, whichmanages, commands, directs, or regulates the behavior of the system to achieve desired results.Now let us have a look at the various systems involved in Control system.

Open-loop system :

A system in which output is dependent on input but controlling action is entirely independent of the output of the systemis called an open-loop system. A traffic control system that operates througha signal on a time basis is one of the examples of an open-loop control system.

open loop system

Consider a bread toaster with a pre-selected timer knob setting. The control objective is to toast the bread to the desired color (usually light brown). This is also an example of an open control system.

Advantages

Such systems are simple in construction
These are easy from the maintenance point of view
These are simple to design and hence economical

Disadvantages

The open-loop system give inaccurate results if there are variable in the external environment
Similarly, they cannot sense internal disturbances in the system, after the controller stage

To overcome all the above disadvantages, generally, in practice, closed-loop systems are used.

Closed-loop system

A system in which the controlling action is somehow dependent on the output is called a closed-loop system.To have a dependence of input, on the output, such a system uses the feedback property. Feedback is a property of a system by which it permits the output to be compared with the reference input to generate an error signal based on this appropriate controlling action will be decided.In these systems, a part of the output is fedback to the input for comparison with reference input applied to it.

The best example of a closed-loop control system is a human being. If a person wants to reach a book on the table, the position of the book is given as the reference. The feedback signal from the eyes compares the actual position of hands with the reference position. An error signal is given to the brain. Brain manipulates this error and givesa signal to the hands. This process continues until the positions of the hands get achieved appropriately

Close loop system

Advantages

Accuracy of the system is always very high because the controller modifies and manipulates the actuating signal such that error in the system will be zero.
This system sense environmental changes, as well as internal disturbances and accordingly, modifies the error.

Disadvantages

Systems are complicated and time-consuming from the design point of view and hence costlier.
Due to feedback, the system may try to overcorrect the error cause oscillations. This leads to system instability and must be taken care of while designing the system.

Transfer function

The transfer function is a ratio of the output of a system to the input of the system in the Laplace domain with zero initial conditions. This function models the system.

Frequency response

The frequency response of a control system can be carried by various graphical techniques like bode plot, polar plot, and Nicholas plot. The frequency response plots are used to determine the frequency specifications to study the stability of the systems.

Bode plot

Bode plot in MATLAB

Take numerator and denominator coefficients from the user and pass it in tf function to produce the transfer function

numerator=input('enter the coefficients in numerator');

denominator=input('enter the coefficients in denominator');

freq_domain=tf(numerator,denominator);

Converting the transfer function to symbolic expression:

syms s;

poly_Num = cell2mat(freq_domain.num);

poly_Den = cell2mat(freq_domain.den);

%Create a symbolic polynomial using poly2sym

sym_num = poly2sym(poly_Num, s);

sym_den = poly2sym(poly_Den, s);

%Divide the numerator and denominator expressions

sym_TF = sym_num/sym_den;

Find corner frequency(wc), maximum frequency (wh), and minimum frequency(wl)

%Divide the numerator and denominator coefficients by their greatest common divisor

num=numerator/(gcd(sym(numerator)));

den=denominator/(gcd(sym(denominator)));

%Divide the least coefficient in num and den with 10

wl=min([num,den])/10;

%Multiply the highest coefficient in num and den with 10

wh=max([num,den])*10;

%Find the unique elements in num and den

wc=unique([num,den]);

%Find the unique elements in wl,wc, and wh

w=unique([wl wc wh]);

Remove 0 frequencies

w=w(w~=0);

Substitute jw in the transfer function and finding magnitude using abs function

Mag=abs(subs(sym_TF,s,w*1i));

Convert into dB scale

Mag_dB=double(20*log10(Mag));

Substitute jw in the transfer function and finding phase using angle function

phase_rad=angle(subs(sym_TF,s,w*1i));

Convert into degrees

phase_deg=double(phase_rad*(180/pi));

Removing positive phase angle, by subtracting 360. There is no effect by adding -360 because tan(-360) is 0

for i=1:length(phase_deg)

if(phase_deg(i)>0)

phase_deg(i)=phase_deg(i)-360;

end

end

Plotting magnitude and phase plots

figure(1)

subplot(2,1,1);

plot(w,Mag_dB);

title('Bode Plot');

ylabel('Magnitude in dB');

xlim([0 10]);

subplot(2,1,2);

plot(w,phase_deg);

xlabel('frequency in rad/sec');

ylabel('phase in deg');

Plotting bode plot using predefined bode function in the control system toolbox

figure(2)

bode(freq_domain)

Figure 1 and figure 2 are shown below

Bode plot in MATLAB

Polar plot

The polar plot of a transfer function G(s) is a plot of the magnitude of G(jw) versus the phase angle of G(jw) on polar coordinates as w varies from 0 to infinity. This polar plot is also called a Nyquist plot.

Nicholas plot

The Nicholas plot is a frequency response plot of the given transfer function. The Nicholas plot is a graph between the magnitude of G(jw) in decibels and phase of G(jw) in degrees.

Root locus

The path taken by the roots of the system when any system parameters are varied is called root locus

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Time response

The time response of a system is defined as the output of a system as a function of time.

Step Response

Step response in MATLAB

Compensator

It is necessary to introduce additional devices or components in the system to alter the behavior and to meet the desired specifications. Such a device inserted into the system to satisfy the specification is called a compensator

Why do we need compensator?

The main reason for designing a control system is for obtaining the desired output. So, the control systems are designed to perform specific tasks. The requirements are usually related to accuracy, stability, and response speed.

In the time domain, the specifications are given in terms of rise time, maximum overshoot, settling time. In the frequency domain, the specifications are given in terms of Gain margin and Phase margin.

The first step in design is the adjustment of gain to meet the desired specifications. In some cases, adjustment of gain alone is insufficient to meet the given specifications. In many cases, increasing the gain may lead to instability.

In such cases, it is necessary to introduce additional devices or components in the system. Such an addition of a device or component is called compensation. The compensator introduces pole/zero in the open-loop transfer function to modify the system to meet the design requirements.

CONTROL SYSTEM DESIGNER TOOLBOX

Using this toolbox, we can directly get the plots like Bode, Nyquist, Nicholas, root locus, pole-zero map, and also response for unit step and impulse inputs directly without writing any program as mentioned above for bode plot and unit step response.

CONTROL SYSTEM DESIGNER TOOLBOX

We can design a compensator for the given transfer function. Here compensator is used for manipulating the system transfer function for meeting the required design parameters.

We should add these requirements in step response by right-clicking on the step response window. After adding, the graph should be in the white color space in the window for meeting the required parameters.

CONTROL SYSTEM DESIGNER TOOLBOX

So we should tune the compensator for getting the step response within the white-colored space, which will ultimately meet the required output.

We can tune the compensator graphically by tuning the poles and zeros on design plots, such as Bode and root locus.

This can be done by right-clicking on the root locus plot and selecting the add pole/zero or delete pole/zero

CONTROL SYSTEM DESIGNER TOOLBOX

We can tune the compensator using automated design methods, such as PID tuning, IMC, and LQG

CONTROL SYSTEM DESIGNER TOOLBOX

We can visualize and validate the step response which will dynamically update with respect to the changes made.

Tuning compensator graphically

Example: The DC motor

A Simple Model of a DC Motor Driving an Inertial Load

The given example aims to control the angular rate by varying the applied voltage.

CONTROL SYSTEM DESIGNER TOOLBOX

The transfer function of the system is

$G(s)=\frac{1}{s^2+14s+40.02}$

For the following parameters:

R= 2.0 Ω

L= 0.5 H

Km = .015 (Torque constant)

Kb = .015 (EMF constant)

Kf = 0.2 Nms

J= 0.02 kg.m^2

For this example, the design requirements are:

Rise time of less than 0.5 seconds
Steady-state error of less than 5%
Overshoot of less than 10%
Gain margin greater than 20 dB
Phase margin greater than 40 degrees

Open the Control System Designer Toolbox by the following command

controlSystemDesigner(tf(1,[1 14 40.02]))

This will open the toolbox with bode editor, root locus plot, and step response.

CONTROL SYSTEM DESIGNER TOOLBOX

Initially, the values of F, C, and H will be unity. So there will be no effect on step response (r to y). The uncompensated or plant transfer function is G.

CONTROL SYSTEM DESIGNER TOOLBOX

Select the grid option by right-clicking on the bode editor.

CONTROL SYSTEM DESIGNER TOOLBOX

The design requirements are given by right-clicking on the step response plot and selecting Design Requirements > New

CONTROL SYSTEM DESIGNER TOOLBOX

The rise time is 0.5 seconds, the settling time is 1 second, and the percentage of overshoot is 10%. After setting these requirements, the step response will look like, as shown below.

CONTROL SYSTEM DESIGNER TOOLBOX

The plot should be in the white region to meet the given requirements.

Step - I :

Since the rise time should be less than 0.5 sec. So, the gain crossover frequencyshould be higher than 2 rad/sec. Drag the magnitude plot upwards till the crossover frequency is nearly 3 rad/sec. Doing this will automatically update the step response of the system.

CONTROL SYSTEM DESIGNER TOOLBOX

Now the rise time became less than 0.5 seconds.

Step-II :

To eliminate the steady-state error value, add an integrator that means adding a pole at the origin. In the root locus editor right-click on the plot area and select Add Pole/Zero > Integrator

After adding an Integrator, the system step response is

CONTROL SYSTEM DESIGNER TOOLBOX

Step-III :

The rise time should be decreased to 0.5 seconds. To achieve this, again change the gain crossover frequency to 3 rad/sec.

CONTROL SYSTEM DESIGNER TOOLBOX

Step-IV :

The peak overshoot is in the Yellow region, The Gain margin and the phase margin are lower than the required range. Compensator containing gain and Integrator is not sufficient to meet the design requirements. Therefore compensator needs a lead network.

To add a lead network, in the root locus editor right-click on the plot area and select Add Pole/Zero >Lead

Now place a pole (red x) and zero (red o) in root locus at -2 and -10 respectively

CONTROL SYSTEM DESIGNER TOOLBOX

Step-V :

Now the GM is 11.3 dB, and PM is 50.6 degrees.

For changing the lead’s pole and lead’s zero:

In the data browser tab

CONTROL SYSTEM DESIGNER TOOLBOX

Right-click on C. It will open the compensator editor window.

The method for finding Lead network is as follows:

$\phi =PM_{desired}+\epsilon -PM_{actual}$

Where,

$\epsilon$ = Random multiple of 5 (25)

Substitute,

Desired Phase Margin = 60

Actual Phase Margin = 31

$\phi =60+25-31=54^{\circ}$

Substitute $\phi$ in the below equation for getting $\alpha$

$\alpha =\frac{1-\sin \phi }{1+\sin \phi }=0.105$

Substitute $\alpha$ in the below equation

$A=-20\times \log (\frac{1}{\sqrt{\alpha }})=-10$

At -10 dB the frequency (w) is 9 rad/sec

$T=\frac{1}{w\sqrt{\alpha }}=0.35$

Using T and $\alpha$ , we will find pole and zero location in the lead network

$P=\frac{-1}{\alpha T}=-27.2$

$Z=\frac{-1}{ T}=-2.857$

CONTROL SYSTEM DESIGNER TOOLBOX

At zero = -2.857 and pole = -27.2, and changing the gain value to 110. The GM = 21 dB and the PM = 76.3 degrees, the step response is also in the white region. So, this satisfied all design requirements.

CONTROL SYSTEM DESIGNER TOOLBOX

To view step response characters:

To add the characteristics to the step response plot, right-click the plot area, and select Characteristics > Rise Time, and peak Response

CONTROL SYSTEM DESIGNER TOOLBOX

The system performance characteristics are:

Rise time is 0.4seconds
Steady-state error is zero.
Overshoot is 0%.
The gain margin is 21dB
Phase margin is 76.3 degrees

The system response meets all of the design requirements.

Example: RC circuit

An electronic circuit consists of a voltage source, resistance, and a capacitor. The voltage across the capacitor is controlled by the input voltage.

RC Circuit

The transfer function of the system is

RC Circuit

For the following parameters:

R= 10 MΩ

C= 1µF

For this example, the design requirements are:

The rise time of less than 10seconds
A steady-state error of less than 5%
An overshoot of less than 20%
Gain margin greater than 20 dB
Phase margin greater than 40 degrees

Open the Control System Designer Toolbox by the following command

controlSystemDesigner(tf(1,[10 1]))

This will open the toolbox with bode editor, root locus plot and step response.

RC Circuit

Initially, the values of F, C, and H will be unity. So there will be no effect on step response (r to y). The uncompensated or plant transfer function is G.

RC Circuit

Select the grid option by right-clicking on the bode editor

RC Circuit

The design requirements are given by right-clicking on the step response plot and selecting Design Requirements > New

RC Circuit

The rise time is 10 seconds, the settling time is 20 seconds, and the percentage overshoot is 20%. After setting these requirements, the step response will look like, as shown below.

RC Circuit

Step-I :

To eliminate the steady-state error value, add an integrator that means adding a pole at the origin. In the root locus editor right-click on the plot area and select Add Pole/Zero > Integrator

After adding an Integrator, the system step response is

RC Circuit

Step-II:

To reduce the overshoot, decrease the gain by dragging down the magnitude plot. Drag until the response is in the white region. This is shown below.

RC Circuit

Step-III :

The PM is 29 degrees. In order to meet 60 degrees, design a lead network using the following method:

$\phi =PM_{desired}+\epsilon -PM_{actual}$

Where,

$\epsilon$ = Random multiple of 5 (10)

Substitute,

Desired Phase Margin = 40

Actual Phase Margin = 28

$\phi =40+10-28=27$

Substitute $\phi$ in the below equation and get $\alpha$

$\alpha =\frac{1-\sin \phi }{1+\sin \phi }=0.375$

Substitute $\alpha$ in the below equation

$A=-20\times \log (\frac{1}{\sqrt{\alpha }})=-4.26$

At -2.464 dB the frequency (w) is 0.25 rad/sec

$T=\frac{1}{w\sqrt{\alpha }}=6.177$

Using T and $\alpha$ , we will find pole and zero location in the lead network

$P=\frac{-1}{\alpha T}=-0.408$

$Z=\frac{-1}{ T}=-0.153$

RC Circuit

After editing the compensator, the step response changes automatically. The updated step response is in the white region.

RC Circuit

To view step response characters:

To add the characteristics to the step response plot, right-click the plot area, and select Characteristics > Rise Time, and peak Response

RC Circuit

The system performance characteristics are:

Rise time is 5.71 seconds
Steady-state error is zero
Overshoot is 14.7%
The gain margin is infinity
Phase margin is 56.5 degrees

The system response meets all of the design requirements.

Tuning compensator using automated tuning methods

Internal Model Control (IMC) Tuning

Conclusion :

For designing a control system, the Bode plot and the step response are crucial. We saw the program for drawing the Bode plot and step response for the given transfer function. The Control System Designer Toolbox made it so simple. By opening the toolbox itself, the bode plot, step response, and root locus are coming. This is just like a drop in the ocean. There are a lot more that can be done using this toolbox. We can design a control system with a tunable compensator for the given design requirements. This tuning can be done graphically and automatically (using PID, LQG, and IMC). After designing the complete system, we can either export the complete system to Simulink or export the values of the blocks in the system to MATLAB workspace. This is the complete overview of the Control System Designer Toolbox.

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About the author

Manoj sai

I am currently pursuing b.tech, always interested to learn new things, and a good listener. I strongly believe that " Dream, Dream Dream, Dreams transform into thoughts and thoughts result in action." - Abdul Kalam

Excellent very useful session.
Thanks

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