ass2 simulation final report thurai

8/8/2019 ASS2 Simulation Final Report Thurai

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Experiment 1: DC Motor Speed Modeling

Exercises:

1. Table below shows the values of undamped natural frequency and damping ratio for a general

second order control system. For the parameter values shown in the table, sketch the step

response of the system. Compare and discuss the results.

Undamped NaturalFrequency (n)

Damping Factor ()

2 0.4

2 0.1

1 0.3

1 0.2

2 0.5

1 0.8

(15 marks)

Classified the table to which systemTable 1

Undamped NaturalFrequency (n)

Damping Factor () Sys no

2 0.4 sys1

2 0.1 sys2

1 0.3 sys3

1 0.2 sys4

2 0.5 sys5

1 0.8 sys6

Using this formula for the general second order

Use the value in table 1 and substitute inside the above equation [1]

The m-file script;a=2^2;

b=[1,(0.4*a),a];

c=[1,(0.1*a),a];

d=1^2;

e=[1,(0.3*1*2),d];

f=[1,(0.2*2*1),d];

g=[1,(0.5*a),a];

h=[1,(0.8*2*1),d];

sys1=tf(a,b);

sys2=tf(a,c);

sys3=tf(d,e);

sys4=tf(d,f);

sys5=tf(a,g);

sys6=tf(d,h);


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From the Figure1 we obtained Table 2

Table 2

SysPeak

AmplitudeOvershooting

(%)Rise time

(s)Settingtime (s)

Steady

state finalvalue

sys1 1.25 25.4 0.736 4.25 1

sys2 1.73 72.9 0.556 19.2 1

sys3 1.37 37.2 1.33 11.2 1

sys4 1.53 52.7 1.21 19.6 1

sys5 1.16 16.3 0.824 4.04 1

sys6 1.02 1.52 2.48 3.76 1

Discussion

From the Table 2 we can assume

y Went the damping ratio is about is increase the overshoot percentage when down.y Went the damping ratio is decrease the settling time increase.y Went the undamped natural frequency is less the settling time increase.

All the system can be classified as under-damped system because their damping ratio are less

than 1, which is under-damped. The oscillates within a decaying envelope is called under-damped response. to reach steady-state the oscillation will take longer in a high underdamped

system .[2]


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0 5 10 15 20 25 300

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

Step Response For 6 System

Time (sec)

Amplitude

sys1

sys2

sys3

sys4

sys5

sys6

Figure 1: The step response of six systems Ex. (1)


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2. The motor transfer function is modified by adding compensators and a unity gain negativefeedback. The resultant open loop transfer function is

)21)(1(

)2()(

sss

sKsG

!

X

There are two parameters that can be changed (K and X ). Choose at least 5 different values ofK

and X ,

a) check for stability of the system by pole zero plot in each case(7 marks)

b) sketch the step response of the system in each case and calculate the transient parameters(8 marks)

Five values that has been choose

Table 3

System K Sys1 5 4

Sys2 4 5

Sys3 3 2

Sys4 2 3

Sys5 1 1

The value above will be substitute in the equation given .Part (a)

The m-file script;

an=[5 2*5];ad=[2*4 (4+2) 1 0];

sys1t=tf(an,ad)

sys1=feedback(sys1t,1)

subplot(3,3,1); %multiple plot in one figure%

pzmap(sys1); %pzmap funtion%

title('sys1');

bn=[4 2*4];

bd=[2*5 (5+2) 1 0];

sys2t=tf(bn,bd)


subplot(3,3,2);

pzmap(sys2);title('sys2');

cn=[3 2*3];

cd=[2*2 (2+2) 1 0];

sys3t=tf(cn,cd)


subplot(3,3,3);

pzmap(sys3);

title('sys3');


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dn=[2 2*2];

dd=[2*3 (3+2) 1 0];

sys4t=tf(dn,dd)


subplot(3,3,4);

pzmap(sys4);

title('sys4');

sys4=tf(dn,dd);

en=[1 2*1];

ed=[2*1 (1+2) 1 0];

sys5t=tf(en,ed)


subplot(3,3,5);

pzmap(sys5);

title('sys5');

Screen shot for 5 different value that been shown in the Table3.

Screen Shot 2: The closed loop transfer function for the five values. Ex 2(a)


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The plot;

Figure 2: The pzmap Ex. 2(a)

Discussion

y only the sys that have poles have the left side of axis are stabley from the pzmap we can find the sys5 are stabley The value choose for sys5 are small value compare to other valuey This show the sys only will be stable in under-damper or critically-damped systemy

In the pzmap we can find that only the poles that are in left side and in the imaginary axis stablesystem.

Part (b)The m-file

an=[5 2*5]

bn=[4 2*4]

cn=[3 2*3]

dn=[2 2*2]

en=[1 2*1]

ad=[2*4 (4+2) 1 0]

bd=[2*5 (5+2) 1 0]

cd=[2*2 (2+2) 1 0]

dd=[2*3 (3+2) 1 0]

ed=[2*1 (1+2) 1 0]

sys1t=tf(an,ad)

sys2t=tf(bn,bd)

sys3t=tf(cn,cd)

sys4t=tf(dn,dd)

sys5t=tf(en,ed)

-2 0 2-2

0

2

sys1

Real Axis

Imagina

ryAxis

-2 -1 0 1-1

0

1

sys2

Real Axis

Imagina

ryAxis

-2 -1 0 1-2

0

2

sys3

Real Axis

Imagina

ryAxis

-2 0 2-1

0

1

sys4

Real Axis

ImaginaryAxis

-2 -1 0-1

0

1

sys5

Real Axis

ImaginaryAxis


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step(sys1,sys2,sys3,sys4,sys5)

The transient parameters table 4 form the figure 3 at below

Table 4

System Peak amplitude Rise time (s)Settling time

(s)Overshoot %

Steady state final

value

sys1 >10.3 - - - Infinitysys2 >0.524 - - - Infinity

sys3 >9.19 - - - Infinity

sys4 >-4.26 - - - Infinity

sys5 1.65 1.41 40.7 64.6 1

Discussion

y In the step response of the figure 3 we could only have the sys5 value of rise time,settling time, overshoot percent and steady state final value. This is because only sys5 are

stable other are not stable.

y From what we get is pzmap we can conculude that only sys5 are stable and other notstable.

y This motor transfer function will only will stable in under-damped or critically-dampedsystem, which mean that damping ratio must be 1 or less than 1.


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The plot;

0 10 20 30 40 50 60 70-1

-0 .5

0

0.5

1

1.5x 1 0

5 Step Response

Time (sec )

Amplitude

sys1

sys2

sys3

sys4

sys5

Figure 3: The step response Ex. 2(b)


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Exercises

1. Simulate the response of the system for five different values of Kp if proportionalcontroller is used. Discuss the results obtained and propose a suitable proportionalcontroller to improve the performance of the system and satisfy the design specifications.

(15 marks)

2. Simulate the response of the system for five different values of Kp and KD if PDcontroller is used. Discuss the results obtained and propose a suitable PD controller to

improve the performance of the system and satisfy the design specifications.(15 marks)

3. Simulate the response of the system for five different values of Kp and KI if PI controlleris used. Discuss the results obtained and propose a suitable PI controller to improve theperformance of the system and satisfy the design specifications.

(15 marks)

4. Simulate the response of the system for at least five different values of Kp,KI, and Kd ifPID controller is used. Discuss the results obtained and propose a suitable PID controller

to improve the performance of the system and satisfy the design specifications.(15 marks)

5. Based on the obtained results compare the performance of various controllers.(10 marks)


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The table 4 below is the 5 value for that will be used for question 1, 2 3

and 4. Table 5

System

sys 0.05 10 3

sys1 0.5 100 9

sys2 5 0.1 8

sys3 50 0.01 6

sys4 13.9 27.5 5

For Question 1

The m-file;J=0.01; %rotor inertia%b=0.1; %damping ratio of mechanical system%K=0.01; %Electromotive force constant%R=1; %electric resistance%L=0.5; %electric inductance%num=K;den=[(J*L) ((J*R)+(L*b)) ((b*R)+K^2)];M1=tf(num,den);

Kp=0.05;Gc=Kp;sysP=Gc*M1;sys=feedback(sysP,1)t=[0:0.01:5];hold on;step(sys,t);

Kp1=0.5;Gc1=Kp1;sysP1=Gc1*M1;sys1=feedback(sysP1,1)step(sys1,t);

Kp2=5;Gc2=Kp2;sysP2=Gc2*M1;sys2=feedback(sysP2,1)step(sys2,t);

Kp3=50;Gc3=Kp3;sysP3=Gc3*M1;sys3=feedback(sysP3,1)


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step(sys3,t);

Kp4=13.9;Gc4=Kp4;sysP4=Gc4*M1;sys4=feedback(sysP4,1)step(sys4,t);

Screen shot

Screen Shot 3: The closed loop function Ex. 1


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The plot

Figure 3: The step response of Kp. Ex. 1

Step Response

Time (sec)

Amp

litude

0 0 .5 1 1 .5 2 2.5 3 3 .5 4 4 .5 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

sy s

sys1

sys2

sys3

sys4


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Dissusion

Table 6: Kp Controller Ex.1

system

Peak

amplitute

Rise time

(s)

Settling

Time (s)

Overshooting

%

Steady

state finalvalue

Steadystate error

%

sys 0.00497 1.13 2.05 0 0.00497 99.5

sys1 0.0476 1.07 1.95 0 0.0476 95.2

sys2 0.333 0.7 1.25 0 0.333 66.7

sys3 0.94 0.159 0.532 12.8 0.94 6

sys4 0.532 0.396 0.631 0.146 0.532 46

From the table we can assume

y sys2, sys3 and sys4 have fulfill two criteria of the design requirement than needy

For Kp we can say that almost fulfill the settling time with almost less than 2secondy For overshooting we hat 4 system that fulfill the requirementy For steady state error

For Question 2

The m-fileJ=0.01;

b=0.1;

K=0.01;

R=1;

L=0.5;num=K;

den=[(J*L) ((J*R)+(L*b)) ((b*R)+K^2)];

M1=tf(num,den);

Kp=0.05;

Kdsn=[10 0];

Kdsd=[0 1];

Kds=tf(Kdsn,Kdsd);

Gc=Kp+Kds;

sysPD=Gc*M1;

sys=feedback(sysPD,1)

t=[0:0.01:5];

hold on;

step(sys,t);

Kp1=0.5;

Kdsn1=[100 0];

Kdsd1=[0 1];

Kds1=tf(Kdsn1,Kdsd1);

Gc1=Kp1+Kds1;

sysPD1=Gc1*M1;

sys1=feedback(sysPD1,1)


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step(sys1,t);

Kp2=5;

Kdsn2=[01 0];

Kdsd2=[0 1];


Gc2=Kp2+Kds2;

sysPD2=Gc2*M1;


step(sys2,t);

Kp3=50;

Kdsn3=[0.01 0];

Kdsd3=[0 1];


Gc3=Kp3+Kds3;

sysPD3=Gc3*M1;


step(sys3,t);

Kp4=13.9;

Kdsn4=[27.5 0];Kdsd4=[0 1];


Gc4=Kp4+Kds4;

sysPD4=Gc4*M1;


step(sys4,t);

Screen shot

Screen Shot 4: The closed loop function. Ex. 2


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The plot

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Step Response

Time (se c)

Amp

litu

de

sys

sys1

sys2sys3

sys4

Figure 4: The step response of PD. Ex. 2


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Discussion

Table 7: PD controller Ex.2

systemPeak

amplitute

Rise time

(s)

Settling

Time (s)

Overshooting

%

Steadystate final

value

Steadystate error

sys 0.588 0.000232 - 1.17e+004 0.00497 99.5

sys1 0.941 0.000456 - 1.88e+003 0.0476 95.2

sys2 0.333 0.699 1.3 0 0.333 66.7

sys3 0.939 0.159 0.532 12.7 0.833 16.7

sys4 0.815 0.0144 4.26 40.1 0.581 42

y from the Table 7 above we find that PD controller that only 2 system fulfill thesettling time

y there no system that fulfill the steady state error requirementy only one system fulfill the overshooting requirementy PD controller have high steady state error

For Question 3

The m-fileJ=0.01;

b=0.1;

K=0.01;

R=1;

L=0.5;

num=K;den=[(J*L) ((J*R)+(L*b)) ((b*R)+K^2)];

M1=tf(num,den);

Kp=0.05;

Kisn=[0 3];

Kisd=[1 0];

Kis=tf(Kisn,Kisd);

Gc=Kp+Kis;

sysPI=Gc*M1;

sys=feedback(sysPI,1);

t=[0:0.01:40];

hold on;

step(sys,t);

Kp1=0.5;

Kisn1=[0 9];

Kisd1=[1 0];

Kis1=tf(Kisn1,Kisd1);

Gc1=Kp1+Kis1;

sysPI1=Gc1*M1;

sys1=feedback(sysPI1,1);

step(sys1,t);


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Kp2=5;

Kisn2=[0 8];

Kisd2=[1 0];


Gc2=Kp2+Kis2;

sysPI2=Gc2*M1;


step(sys2,t);

Kp3=50;

Kisn3=[0 6];

Kisd3=[1 0];


Gc3=Kp3+Kis3;

sysPI3=Gc3*M1;


step(sys3,t);

Kp4=13.9;

Kisn4=[0 5];Kisd4=[1 0];


Gc4=Kp4+Kis4;

sysPI4=Gc4*M1;


step(sys4,t);

Screen shot

Screen Shot 5: The closed loop function of PI. Ex. 3


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Plot;

Step Response

Time (sec)

Amplitude

0 5 10 15 20 25 30 35 400

0.2

0.4

0.6

0.8

1

1.2

1.4

sys

sys1

sys2

sys3

sys4

Figure 5: The step response of PI. Ex.3


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Discussion

Table 8: PI controller Ex.3

systemPeak

amplitute

Rise time

(s)

Settling

Time (s)

Overshooting

%

Steadystate final

value

Steadystate error

sys 1 6.03 11 0 1 0

sys1 1.04 1.63 4.5 4.05 1 0

sys2 1 2.7 5.07 0 1 0

sys3 0.997 0.216 20.6 0 1 0

sys4 1 6.26 13.6 0 1 0

y From the above table we fulfill the overshooting and steady state error requirement.y For settling time we did not fulfill any requirementy PI controller can be used as steady state error controller and over shooting controller

For Question 4

The m-file

J=0.01;

b=0.1;

K=0.01;

R=1;

L=0.5;

num=K;

den=[(J*L) ((J*R)+(L*b)) ((b*R)+K^2)];

M1=tf(num,den);

Kp=0.05;

Kisn=[0 3];

Kisd=[1 0];

Kis=tf(Kisn,Kisd);

Kdsn=[10 0];

Kdsd=[0 1];Kds=tf(Kdsn,Kdsd);

Gc=Kp+Kis+Kds;

sysPID=Gc*M1;

sys=feedback(sysPID,1)

t=[0:0.01:40];

holdon;

step(sys,t);

Kp1=0.5;


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Kisn1=[0 9];

Kisd1=[1 0];


Kdsn1=[100 0];

Kdsd1=[0 1];


Gc1=Kp1+Kis1+Kds1;

sysPID1=Gc1*M1;

sys1=feedback(sysPID1,1)

step(sys1,t);

Kp2=5;

Kisn2=[0 8];

Kisd2=[1 0];


Kdsn2=[0.1 0];

Kdsd2=[0 1];


Gc2=Kp2+Kis2+Kds2;

sysPID2=Gc2*M1;

sys2=feedback(sysPID2,1)step(sys2,t);

Kp3=50;

Kisn3=[0 6];

Kisd3=[1 0];


Kdsn3=[0.01 0];

Kdsd3=[0 1];


Gc3=Kp3+Kis3+Kds3;

sysPID3=Gc3*M1;


step(sys3,t);

Kp4=13.9;

Kisn4=[0 5];

Kisd4=[1 0];


Kdsn4=[27.5 0];

Kdsd4=[0 1];


Gc4=Kp4+Kis4+Kds4;

sysPID4=Gc4*M1;


step(sys4,t);


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Screen shot;

Screen Shot 6: The closed loop function of PID. Ex. 4

Discussion

Table 9: PID controller Ex. 4

systemPeak

amplitute

Rise time

(s)

Settling

Time (s)

Overshooting

%

Steady

state finalvalue

Steady

state error

sys 1.02 7.55 14 2.44 1 0

sys1 1.19 0.0159 - 15 1 0

sys2 1 2.72 5.05 0 1 0

sys3 0.997 0.216 20.6 0 1 0

sys4 1 7.61 12.1 0.0141 1 0

y From the table above we can determine that steady state error requirement has beenfulfils.

y For the overshooting 4 of 5 requirements are fulfill.y The settling time requirement did not been fulfill in PID controller.


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Plot;

Figure 6: The step response of PID. Ex. 4

Step Response

Time (se c)

Amplitude

0 5 10 15 20 25 30 35 400

0.2

0.4

0.6

0.8

1

1.2

1.4

sy s

sys1

sys2

sys3

sys4


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Question 5

Based on the results that obtained on , PD, PI and PID controller. PI and PID almost

fulfill the requirement of the design of motor output speed based on the value that been selected

from Table 3 above.

PI controller has two system that almost fulfilling the requirement but the settling timedid not fulfill the requirement. PID controller have 1 system almost fulfill the requirement butdid not fulfill the settling time.

For and PD controller can fulfill the settling time and overshooting but cannot fulfill

the steady state error from the result been obtained. and PD controller are best to use in

controlling settling time and overshoot. For PI and PID controller is best to use to controlovershooting and steady state error.

Based on the results obtained PID controller is the best controller to use in this system.


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Reference[1] Richard , C.D., Robert, H.B 2008., Modern ControlSystem, 11

thedn, Pearson

International Edition, Singapore.

[2] Wikipedia 2010, The Transient Response, online retrieved 2 May 2010, fromhttp://en.wikipedia.org/wiki/Transient_response#Underdamped.

ass2 simulation final report thurai

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