How to design a linear control system

transcript

In The Name Of God

University Of Zanjan

Electrical engineering department

Leanier systems control

Final project

Lecturer: Dr Farhad Bayat

Students :

Alireza Mirzaei 88442176

Mahyar Seyedan 88442141

Fall- 2011

Content :

Intruduction ………………………………………………………. (3)

Way of solving…………………………………………………...(3)

Simulink by matlab…………………………………………….(11)

Plots………………………………………………………………….(15)

Answers…………………………………………………………..(26)

Introduction :

In this project we are trying to show steps of designing of a system.in all steps we

have some problems and we want to solve all the problems in order to can say

this system is reliable.because of that in this project’s used Matlab to show the

result of each step and simulink parts of this system.

Way of solving :

In this part ,mathmatical way of solving each part has offered.In some of them

maybe you have another way to reach the result but these ways’re recommanded

and we use this ways in matlab.

Part(A)

Poles & Zeroes:

Example: Y(s)=𝐴(𝑠)

𝐵(𝑠)

Zeroes of Y(s) : roots of A(s)

Poles of Y(s) : roots of B(s)

When poles & zeroes of system sketched , we can write Y(s) in this form :

Y(s)= K s−z0 s−z1 s−z2 ….. s−Zn

s−P0 s−P1 s−P2 s−P3 ….. s−Pn

Notice : K is real number that sketched when we take apart Y(s) to some (s-Pn)

Approximating Y(s) :

We can decrease order of Y(s) by removing (s-Pn) or (s-Zn) forms of Y(s) to reach

your order. We can’t remove every poles or zeroes we want , but we choose

poles and zeroes which aren’t more effective on output of a system . These poles

are farther than other poles from Image axis and zeroes are near undeleted poles.

Notice : when a pole or zero got rid of a function ,we should put its effect in

steady state means for example effect of (s-P0) in steady state is (-P0).

Part (B) :

Step response means input of a system is a step signal and we want to know

output of system by this input . Impulse response means input of a system is

impulse .impulse is a signal that just have one infinite value in very short time and

then became zero.

This part simulinked by Matlab.

Part (C) :

Kp = lim𝑠→0

Ess means steady state error and when we use Kp for finding Ess when we want

to find steady state error for step input .Kp is called position error coefficient . Ess

Sketched by :

Ess = 1

1+𝐾𝑝

root locus is plot of roots when K(gain of Y(s)) changes from minus infinite to

positive infinite.

We find that by Matlab.

Part (D) :

Mp is overshoot of a system and is amount that is over than 1(according to this

plot below)

If we have this function :

Mp sketched by this equalization :

If Mp changes this change causes changing and by this change step response

or another features of system may change and we have different outputs by

different s.

First we find from Mp equalization because Mp is cleared in problem. Then we

input this new in function and sketch step response by Matlab .

Part (E) :

Nyquist :

If in general function (open loop) s=jw we can plot function by its size and angle

when w changes from minus infinite to positive infinite . in control we called this

plot Nyquist plot. By Nyquist plot we can find phase margin and gain margin and

some better information for analysis of a system.

for Example :

We sketch Nyquist plot of our aproximated function by Matlab.

Part (F) :

Frequency Analysis :

Sometimes we want to control a system and decrease steady state error or

increase stability condition of that. we should design a function which called

compensator to reach that condition. Compensator multiple to open-loop

function and changes stability condition or steady state error.

There are two kind of Compensator :

1.Lead Compensator : increase stability condition

2.Lag Compensator : decrease steady state error

Notice : sometimes both of them used for better result.

This are steps of designing :

1.Lead Compensator :

a) sketch the new poles with new condition

P1=- Wn+jWn 1 − ^2

P2=- Wn-jWn 1 − ^2

b) choose one of the new poles and find vector of old poles and that new poles :

a=Pold1-P1

b=Pold2-P1

c) find angel that in need to reach new condition

ph=180-(180-∟a)-(180-∟b)

d) finding 1

𝑇 & α

B=180-arctan(|𝑦𝑃1

𝑥𝑃1|);

M1=tan (𝑝ℎ+𝐵

M2=tan (𝐵−𝑝ℎ

𝑇 = 𝑥𝑃1-

𝑦𝑃1

𝛼𝑇 = 𝑥𝑃1-

𝑦𝑃1

e) check this equalization to find Kc :

if s=P1 |Kc 𝑠+

𝑠+1

𝛼𝑇

G(s)|=1

f) Lead Compensator : Gclead(s)= Kc 𝑠+

𝑠+1

𝛼𝑇

2.Lag Compensator :

a) A2 : Compensated gain & A1 : Uncompensated gain

𝐴1 = β

b) solve this equalization to find T :

if s=P1 | 𝑠+

𝑠+1

β𝑇

-5 ˂ ∟ 𝑠+

𝑠+1

β𝑇

e) check this equalization to find Kc :

if s=P1 | Kc 𝑠+

𝑠+1

β𝑇

G(s) |=1

c) Lag Compensator : Gclag(s)= Kc 𝑠+

𝑠+1

β𝑇

after designing both of them open-loop function will be :

Gc(s)= Gclead(s)G(s)Gclag(s)

Part (G) :

In state space a system defined by this matrixes :

X : state vector (n x 1)

U : input vector (m x 1)

Y : output vector (p x 1)

A : system matrix (n x n)

B : system input matrix (n x m)

C : output matrix (p x n)

D : input matrix ,usually for SISO systems is 0

When this matrixes go into state space :

sX(s)=AX(s)+BU(s)

Y(s)=CX(s)+DU(s)

If these equalization solved :

X(s)=(𝑠𝐼 − 𝐴)−1BU(s)

G(s)=C(𝑠𝐼 − 𝐴)−1B+D

Roots of det(𝑠𝐼 − 𝐴 ) will be Poles of G(s) .

We can check controllability of system by this way :

Q=[ B AB 𝐴2B …… 𝐴𝑛−1B]

If rank(Q)=n : system is controllable

Otherwise : system is uncontrollable

~ 10 ~

Pole replacement : sometimes we want to replace close-loop poles of system .

after checking controllability of system we can use this equalization to find K

matrix which added to system for replacement.

sX(s)=(A-BK)X(s)+BU(s)

by this equalization K matrix is found :

det(𝑠𝐼 − 𝐴+BK)=(s-P1)(s-P2)…..(s-Pn)

P1…Pn : replaced poles

In Matlab there is a code used for pole replacement.(see G.4 in Simulink by

matlab)

~ 11 ~

Simulink by Matlab :

%part A

N=[1 112/25 48/25];

D=[1 23/2 25 93/4 27/4];

G=tf(N,D);

p=pole(G);

z=zero(G);

g=zpk(G);

N1=[1 4];

D1=[1 2 1.5];

k1=polyval(N,0)/polyval(D,0);

k2=polyval(N1,0)/polyval(D1,0);

k3=(k1/k2);

N1=N1*k3;

G1=tf(N1,D1);

g1=zpk(G1);

figure

pzmap(G);

figure

pzmap(G1);

%----------------------------------------------------------

%PART B

figure

step(G1,G,'--');

figure

impulse(G1,G,'--');

%----------------------------------------------------------

%PART C

Kpgeneral=polyval(N,0)/polyval(D,0);

~ 12 ~

G2=Kpgeneral*G;

G2=feedback(G2,1);

Kpapx=polyval(N1,0)/polyval(D1,0);

G1apx=Kpapx*G1;

G1apx=feedback(G1apx,1);

figure

rlocus(G2);

figure

rlocus(G1apx);

[Gm,Pm,Wg,Wp] = margin(G1apx);

%----------------------------------------------------------

%PART D

[Nsys,Dsys]=feedback(N1,D1,1,1);

sysz1=tf(Nsys,Dsys);

Wncl=sqrt(Dsys(3));

zitasys=Dsys(2)/(2*Wncl);

zitasysmax=1/abs(9+(pi)^2);

zitasysmax=sqrt(zitasysmax);

Dsys(2)=2*zitasysmax*Wncl;

Nnew=Nsys;

Dnew=Dsys-Nsys;

G1new=tf(Nnew,Dnew);

Kpnew=polyval(Nnew,0)/polyval(Dnew,0);

figure

sysnew=feedback(G1new,1);

step(sysnew,sysz1,'--');

%----------------------------------------------------------

%PART E

figure

nyquist(G1,G,'--');

%----------------------------------------------------------

~ 13 ~

%PART F

%F.1(lead compensator)

[numT,denT]= feedback(N1,D1,1,1);

sys=tf(numT,denT);

Wn1=sqrt(denT(3));

Wn2=2*Wn1;

zita1=denT(2)/(2*Wn1);

zita2=zita1;

p1=-(zita2*Wn2)+(Wn2*(sqrt((zita2^2)-1)));

p2=-(zita2*Wn2)-(Wn2*(sqrt((zita2^2)-1)));

p3=pole(sys);

a=p1-p3(1);

b=p1-p3(2);

ph1=pi-angle(a);

ph2=pi-angle(b);

Cph=pi-ph1-ph2;

d=pi-atan(abs((imag(p1))/real(p1)));

m1=tan((Cph+d)/2);

m2=tan((d-Cph)/2);

T=-1/(real(p1)-((imag(p1))/m1));

alpha=-T/(real(p1)-((imag(p1))/m2));

N3=[1 1/T];

D3=[1 1/(alpha*T)];

Gc=tf(N3,D3);

zpk(Gc);

Kc=1/abs((polyval(N3,p1)/polyval(D3,p1))*(polyval(N1,p1)/po

lyval(D1,p1)));

Gc=Kc*Gc;

Gcc=G1*Gc;

sys2=feedback(Gcc,1);

figure

step(sys2,sys,'--');

%F.2(lag compensator)

Kp1=polyval(N1,0)/polyval(D1,0);

Kp2=100*Kp1;

beta=Kp2/Kp1;

T2=(1-(1/(beta^2))/(-2*real(p1)*(1-(1/beta))));

N4=[1 1/T2];

D4=[1 1/(beta*T2)];

~ 14 ~

Gc2=tf(N4,D4);

zpk(Gc2);

Kc2=1/abs((polyval(N4,p3(1))/polyval(D4,p3(1)))*(polyval(N1

,p3(1))/polyval(D1,p3(1))));

Gc2=Kc2*Gc2;

Gcclag=Gc2*G1;

figure

step(Gcclag,G1,'--');

sys3=feedback(Gcclag,1);

figure

step(sys3,sys,'--');

%F.3 Bode diagram

Gcomp=Gcc*G1*Gcclag;

Gcomp=feedback(Gcomp,1);

figure

bode(G1,Gcomp);

%----------------------------------------------------------

%PART G

%G FIND A,B,C,D STATE MATRIXS

[NF,DF]=feedback(N1,D1,1,1);

[A,B,C,D]=tf2ss(NF,DF);

%G.1 controlability check

Q=[B A*B ];

r=rank(Q);

disp(r);

s=roots(poly(A));

s0=s(1);

s1=s(2);

%G.3 simulinked by matlab (g3.mdl in directory)

~ 15 ~

sc1=2*s0;

sc2=4*s1;

J=[sc1 sc2];

K12=acker(A,B,J);

eig(A-B*K12);

disp(K12);

%G.5 simulinked by matlab (g5.mdl in directory)

sysg=ss(A-B*K12,eye(2),eye(2),eye(2));

t=0:0.01:4;

x=initial(sysg,[2;5],t);

x1=[2 5]*x';

x2=[5 2]*x';

figure

subplot(2,1,1);

plot(t,x1),grid;

subplot(2,1,2);

plot(t,x2),grid

Plots :

Pole zero map of G (general function)

-9 -8 -7 -6 -5 -4 -3 -2 -1 0

Pole-Zero Map

Real Axis

~ 16 ~

Pole and zero map of G1 (approximated function)

Step response for G(s) (general function) and G1(s) (approximated function)

-9 -8 -7 -6 -5 -4 -3 -2 -1 0-0.8

Pole-Zero Map

Real Axis

Imagin

0 1 2 3 4 5 6 70

Step Response

Time (sec)

~ 17 ~

Impuls e response for G(s) (general function) and G1(s) (approximated function)

root locus of closed loop system (with G (s)*Kp in open loop)

0 1 2 3 4 5 6 7 8 9 10-0.02

Impulse Response

Time (sec)

-10 -8 -6 -4 -2 0 2-25

Root Locus

Real Axis

Imagin

~ 18 ~

root locus of closed loop system (with G1 (s)*Kp in open loop)

Step response of sysnew and sysz1

(Notice : sysnew is new G1 with of maximum overshoot less than 5% and

sysz1 before changing )

When changes overshoot of system also will change.

-14 -12 -10 -8 -6 -4 -2 0 2-4

Root Locus

Real Axis

0 2 4 6 8 10 12 14 16 180

Step Response

Time (sec)

sysnew

~ 19 ~

Nyquist diagram of G(s) (general function) and G1(s) (approximated function)

according to nyquist diagrams in low frequencies both of them absolutely

the same , and in other frequencies there is a little difference between them.

Because in low frequencies they have the same dc value.

And in high frequencies because general function is order of 2 and went to zero

By ±180 but approximated function is order of 1 and went to zero by ±90.

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4-0.25

Nyquist Diagram

Real Axis

Imagin

~ 20 ~

F.1 : Step response of sys2 and sys close-loops.

(notice: sys is close-loop system of approximated function without lead

compensator(Gc), but sys2 is close-loop system of compensated of G1 with lead

compensator(Gc))

We can see in this below figure lead compensator go to steady state faster and

by this steady state changes and system stability increased.

0 1 2 3 4 5 60

Step Response

Time (sec)

~ 21 ~

I) step response of close-loop sys3 & sys

(Notice: sys3 is close-loop function of compensated with lag compensator and

Sys is close-loop function of G1)

We can see in this below figure steady state error decreased and output goes to

step input.

Step Response

Time (sec)

0 2 4 6 8 10 12 14 16 180

~ 22 ~

II) step response of open-loop Gcclag & G1

(Notice: Gcclag=G1Gclag)

Here’s zooming of pervious plot :

0 100 200 300 400 500 6000

Step Response

Time (sec)

Gcclag

0 5 10 15 20 25 30 350

Step Response

Time (sec)

Gcclag

~ 23 ~

F.3 : bode diagram of Gcomp & G1

(Notice : Gcomp is close-loop function of (Gclead x G1 x Gclag) and G1 in open-

Block diagram of G1 (approximated function)

Magnitu

Phase (

Bode Diagram

Frequency (rad/sec)

~ 24 ~

initial condition of [2 5]𝑇 (output of scope in simulink section of Matlab)

Block diagram of pole replaced system of G1 :

~ 25 ~

initial condition of [2 5]𝑇

I) outputs by M-file section in Matlab in different periods of time

II) output of scope in simulink section of Matlab

0 0.5 1 1.5 2 2.5 3 3.5 4-10

0 0.5 1 1.5 2 2.5 3 3.5 4-40

~ 26 ~

Answers:

Part (A) :

Sketch poles and zeros of system (G (s )) , then approximate G (s ) with a system of order 2 (G~(s) )).

Answer :

G~(s)=𝟎.𝟏𝟎𝟔𝟕 𝒔 + 𝟎.𝟒𝟐𝟔𝟕

𝐬^𝟐 + 𝟐 𝐬 + 𝟏.𝟓

Part (B) :

Sketch impulse and step response of open loop system before and after approximation (G (s ),G% (s )). Is your approximation acceptable?

Answer :

Yes, according to the plots of step & impulse response of G and G1 ,

approximation is acceptable because both of them are almost the same.

Part (C) :

Assume K (s) =Kp , then sketch the root locus of closed loop system for Kp and find the range of Kp for closed loop stability.

Answer :

Kp= 0.284444444444444 gain margin= infinite

Kp<gain margin === > Kp can amounted all positive real numbers!

~ 27 ~

Part (D) :

Assume H (s) =1 , find Kp such that the maximum overshoot be less than 5%. Then sketch the step response.

Answer :

Kpnew= 1.284444444444445

For step response see plots (D).

Part (E) :

(Nyquist Analysis) E.1- Sketch the nyquist diagram for main system (G (s )) and approximated system(G~ (s )) using MATLAB. Note: Writing all steps are necessary for G~ (s ) . E.2- Does the nyquist diagram of G (s ) and G~ (s ) confirm an accurate approximation? In which frequency two nyquist diagrams are similar and in which are different? Why? Explain your answer?

Answers :

E.1 . see plots (E.1)

E.2 . according to nyquist diagrams in low frequencies both of them absolutely

the same , and in other frequencies there is a little difference between them.

Because in low frequencies they have the same dc value.

And in high frequencies because general function is order of 2 and went to zero

By ±180 but approximated function is order of 1 and went to zero by ±90.

Part (F) :

~ 28 ~

(Frequency Analysis)

Answers :

Gclead(s)= 165.7 s + 786.8

s + 36.59

Gclag(s)=0.1626 𝑠2+ 0.8131 s + 0.6505

𝑠3+ 2.01 𝑠2+ 1.52 s + 0.015

Gctotal(s)=0.3066 𝑠5 + 5.441 𝑠4+ 37.32 𝑠3+ 121.7 𝑠2 + 182.7 s + 93.17

𝑠8+ 42.6 𝑠7 + 236.5 𝑠6+ 632.5 𝑠5+ 987.9 𝑠4+ 966.3 𝑠3 + 628.3 𝑠2 + 311.1 s+ 94.41

Part (G) :

Find a state space model (A,B ,C ,D ) for approximated system(G~ (s)

G.1- Check the controllability of system (A,B ,C ,D ) . G.2- Find all poles of open loop system (So1,So2 by solving det (sI -A) = 0 . G.3- Sketch the initial condition response of open loop system forX0= [ 2 5 ] 𝑇 G.4- Let U= K.x where K=[K1 K2] , design K1 K2 such that the poles of closed loop system place at Sc1=2So1 ,Sc2=4So2

~ 29 ~

G.5- Sketch the initial condition response of close loop system forX0= [2 5 ] 𝑇.

Answers :

A= −1.926666666666667 −1.926666666666667

C= [0.106666666666667 0.426666666666667]

Rank(Q)=2 and n=2 ==== > system is controllable!

So1= -1.053333333333333 + 0.903966567720044i

So2= -1.053333333333333 - 0.903966567720044i

~ 30 ~

See plots G.3

see and run g3.mdl in project directory

K=[K1 K2]= [4.213333333333333 13.486666666666668]

See plots G.5

see and run g5.mdl in project directory

THE END…………………………………………………………………………………………..

How to design a linear control system

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