control lab #3

7/29/2019 Control Lab #3

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Experiment 3 System Interconnections and Response Analysis Page 1 of 12

NATIONAL UNIVERSITY

OF

SCIENCES AND TECHNOLOGY

COLLEGE OF EME

Control Systems Lab Experiment #3

System Inter Connections and Response

Analysis

Submitted By: NS Saad IftikharDe 32 EE A

Nust 039

Objectives:

Practice on developing different possible system inter connections

Developing Transfer functions for MISO Systems

Overshoot and Steady State Error Calculations

System Reduction to Dominant Poles


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New MATLAB commands to learn in this experiment

Command Description

Feedback feedback(sys, K)implements a feedback path across sys with gain K.

Enterhelp feedback in the Matlab Command Window for a detailed description.Series Series (sys1, sys2) implements a series connection of sys1 and sys2.

Pzmap Pzmap(sys) gives the sketch of poles and zeros on S-plane

Procedure

Section-1

An open-loop Transfer Function of a separately excited DC motor is shown below,

with the system parameters as,

For the open-loop speed control system of a dc motor (shown above), what is the followingtransfer function

G(s)Td=0 = (s)/Va(s) = =

roots= -200,-0.74

Ra L a Km J b Kb

1 0.005 10 2 0.5 0.1

(s) (Output)Va(s) (Input) 1

Las+Ra1

Js+BKm

Kb

Td(s)Disturbance Torque


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Write MATLAB commands to determine the above mentioned transfer function.

Matlab Code:km=10;Ra=1;La=0.005;J=2;b=0.5;

Kb=0.1;H1=tf([1],[ La Ra]);H2=Kb;H3=tf([1],[ J b]);

H_1=series(H1,H2);H=series(H_1,H3);sys=feedback(H,Kb);pzmap(sys)subplot(3,1,1),impulse(sys)subplot(3,1,2),step(sys)subplot(3,1,3),pzmap(sys)

Attach the open-loop step Response. Label the speed and time axes properly.Open-loop Step Response


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Determine the transfer T(s)Va=0 = (s)/Td(s) manually.

T(s)Va=0 = (s)/Td(s) = Zeroes= -200

=

roots= -200, 0.749

Write MATLAB commands to determine the above mentioned transfer function.

MATLAB CODE:

km=10;Ra=1;

La=0.005;J=2;b=0.5;Kb=0.1;H1=tf([1],[ La Ra]);H2=Kb;H3=tf([1],[ J b]);sys1=series(H1,H2);sys2=series(-Kb,sys1);feed=feedback(H3,sys2);feed1=feed*-1;pzmap(feed1)step(feed1)subplot(3,1,1),impulse(feed1)subplot(3,1,2),step(feed1)subplot(3,1,3),pzmap(feed1)


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Attach the open-loop step Response. Label the speed and time axes properly.

Write your comments on the step response due to Va (Td=0) and due to Td(Va=0).

In the first case when the disturbance torque td=0 there is no zero in the response and first

pole is at -200 which makes jw axis present in ROC so the response of the system is stable.

In second case when there is Va=0, In this case a zero is present at -200 which cancels the pole

at -200, so Only one pole is present at 0 which is at the jw axis so this makes the response

unstable , as pole is at the origin so this response is of an integrator circuit.


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For a unity-feedback control system shown in Fig. below, the errorfunction is determined by the equation and can directly be foundfrom MATLAB.

To be able to determine the error function in MATLAB, we need firstto determine the vector containing the input signal r(t), and the

vector containing the actual output signal y(t). The next step issubtract y(t) from r(t) to obtain the error function. MATLAB codebelow is an example that determines and plots the error of thesystem as a function of time, from an input unit-step function. Thesystem is defined by the following model

>> % Generate unity-feedback system

>> % Determine output and time vectors>> % Determine error function

Matlab code:g=tf([ 2 3],[1 3 6])Ys=feedback(g,1)

[y t]=step(Ys);error=1-y;plot(error)subplot(3,1,1), pzmap(Ys);subplot(3,1,2) ,step(Ys);subplot(3,1,3), plot(t,error);

ERROR PLOT:


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Section-2:

Dominant Poles and Second Order App roximat ion:

Both Zeros and Poles may affect the system response, but the dominant poles are theone who have significant amount of impact on system response.


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For the following transfer function

Write MATLAB commands to find poles and zeros.

G=62.5*tf([1 2.5], [1 12.25 62.5 156.25] )p=[1 12.25 62.5 156.35]poles=roots(p)p1=[1 2.5]zeros=roots(p1)

Plot polesand zerosin s-plane. What is the command that will be used?


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Now find and sketch the step response.

What is %Overshoot and Settling Time from the response?

% overshoot38%

Settling time 1.04 sec

As an approximation, we neglect the 3rd(real) pole, and again find theresponse of the system.


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Again find step response and values of %Overshoot and Settling Time for theapproximated system. (You can use step(sys1, sys2) for comparison of responses)

MATLAB COMMANDS:

G=62.5*tf([1 2.5], [1 12.25 62.5 156.25] );p=[1 12.25 62.5 156.35];poles=roots(p);p1=[1 2.5];zeros=roots(p1);%subplot(4,1,1) ,pzmap(G)%subplot(4,1,2), step(G)G1=10*tf([1 2.5], [1 6 25] );%subplot(4,1,3), step(G1)%subplot(4,1,4),

step(G,G1)


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What is the change in performance parameters (settling time, damping,

%overshoot) that has been observed after ignoring the 3rd (real) pole of

the system?

The settling time has increased to 1.05 sec The overshoot percentage has increased to 54%. Anddamping effect has increased . so as we have removed the real pole now we only have a zero anda complex pole so this complex pole effects the system more and as a result of this complex poleour response is more under damped as compared to the previous case.

Similarly for the following system, using second order approximation, comparethe performance parameters (%O.S, settling time, damping) of the actual andapproximated system. (You can use step(sys1, sys2)for comparison ofresponses)

Approximated system is (ignoring pole at50)

MATLAB CODE:

G=62.5*tf([1 2.5], [1 12.25 62.5 156.25] );

p=[1 12.25 62.5 156.35];poles=roots(p);p1=[1 2.5];zeros=roots(p1);%subplot(4,1,1) ,pzmap(G)%subplot(4,1,2), step(G)G1=10*tf([1 2.5], [1 6 25] );%subplot(4,1,3), step(G1)%subplot(4,1,4),

step(G,G1)


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What is the change in performance parameters (settling time, damping,

%overshoot) that has been observed after ignoring the 3rd (real) pole of the

system?

Can this pole be ignored?

What criteria do you observe for the valid approximation?

There is no change in the performance of both response as the response is criticallydamped response so there is no percentage overshoot in both cases similary nosettling time and damping in the response.

As no complex poles or zeroes are present in the response only real poles arepresent hence as a result the response is c critically damped and similar in both thecases.Yes the pole can be ignored. The reason is above mentioned this pole is furthest awayfrom the jw axis so the other pole acts as dominant pole compared to this pole so this

pole can be easily ignored without any significant change in the response and dampingof the circuit

control lab #3

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