Report Exam II Linear Systems

8/8/2019 Report Exam II Linear Systems

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UNIVERSIDAD AUTNOMA DE AGUASCALIENTES

CENTRO DE CIENCIAS BSICASDEPARTAMENTO DE SISTEMAS ELECTRNICOS.SECOND PARTIAL EXAM LINEAR SYSTEMS NOVEMBER 22 ND, 2010

STUDENT: JUVENTINO SAUCEDO NAVARRO

Exercise 1

Here is the root locus for a simple control system. The block diagram is shown first, and the root locusfollows below.

From the root locus, you should be able to determine the following.

The transfers function for the system being controlled. The transfer function can be formed justby knowing the poles and zeroes, which are shown on the root locus plot.

Determine an adjusted system that has a DC gain of 1.0 Using your adjusted system as G(s), determine the gain, K, which will produce poles at -2.5 +/- j

2.5.

Solution:

We can see that the root locus plot only has 2 poles (at -1 and -4) and the zeros are located at the infinity


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Simplifying it we have the following:

Using Matlab to obtain the transfer function

>> GS = zpk ([], [-1 -4], [1])

Zero/pole/gain:

1

-----------(s+1) (s+4)

Taking into account the value of k = 1 and adjusting the system with a gain of 1 we have the followingresponse>> k = 1;

>> t = feedback (k*GS, 1);

>> t

Zero/pole/gain:1

--------------------------

(s+1.382) (s+3.618)

>> rltool (t)

Finally, the gain k obtained is 7.48 when the poles are in 2.5 +/- 2.5j

C(s)

R(s)=

kG(s)

1+ kG(s)=

k1

(s+1)(s+ 4)

1+ k1

(s+1)(s+ 4)

=

k

(s+1)(s+ 4)

(s+1)(s+ 4) + k

(s+1)(s+ 4)

C(s)

R(s)=

k(s+1)(s+ 4)

(s+1)(s+ 4)[(s+1)(s+ 4) + k]=

k

(s+1)(s+ 4) + k=

k

s2+ 5s+ 4 + k


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Exercise 3

Assume that you have a system with three poles at s = -1, -2 and -3. The G(s) for the system is:

The system is being controlled in a feedback system as shown below.

Predict step response performance for K = 1, 2, 5, 10, 20, 50

Solution:

First we need to obtain the transfer function:

With the help of matlab we obtain the following

G(s) =5

(s+1)(s+ 2)(s+ 3)

C(s)

R(s)=

kG(s)

1+ kG(s)=

5k

(s+1)(s+ 2)(s+ 3)

1+5k

(s+1)(s+ 2)(s+ 3)

=

5k

(s+1)(s+ 2)(s+ 3)

(s+1)(s+ 2)(s+ 3)+ 5k

(s+1)(s+ 2)(s+ 3)

=5k

(s+1)(s+ 2)(s+ 3)+ 5k

C(s)

R(s)=

5k

(s2+ 3s+ 2)(s+ 3)+ 5k

=5k

(s3+ 6s

2+11s+ 6) + 5k


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>> GS = zpk ([], [-1 -2 -3], [5])

Zero/pole/gain:5

-----------------(s+1) (s+2) (s+3)

For k = 1

>> k = 1;>> t = feedback( k*GS, 1 );>> t

Zero/pole/gain:5

-------------------------------------------(s+3.904) (s^2 + 2.096s + 2.818)

>> rltool()

In the Architecture Menu and the System Data submenu, importing G(s), and changing the compensatorvalue in the Compensator Editor C=1 in the SISO we have the following graph

By the previous plot we can see that the answer is Under-damped.


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For k = 2

>> k = 2;>> t = feedback( k*GS, 1 );>> t

Zero/pole/gain:10

--------------------------------(s+4.309) (s^2 + 1.691s + 3.713)

>> rltool()



For k = 5


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>> k = 5;>> t = feedback( k*GS, 1 );>> t

Zero/pole/gain:25

-------------------------------------------

(s+5.038) (s^2 + 0.962s + 6.153)

>> rltoolIn the Architecture Menu and the System Data submenu, importing G(s), and changing the compensatorvalue in the Compensator Editor C=5 in the SISO we have the following graph


For k = 10

>> k = 10;>> t = feedback( k*GS, 1 );>> t

Zero/pole/gain:50

---------------------------------------------


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(s+5.774) (s^2 + 0.2255s + 9.698)

>> rltool


In the plot we can see that the answer is Under-damped.

For k = 20

>> k = 20;>> t = feedback( k*GS, 1 );>> t

Zero/pole/gain:100

--------------------------------------------(s+6.713) (s^2 - 0.7134s + 15.79)

>> rltool



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In the plot we can see that the systems answer is unstable.

For k = 50

>> k = 50;>> t = feedback( k*GS, 1 );>> t

Zero/pole/gain:250

------------------------------------------(s+8.353) (s^2 - 2.353s + 30.65)

>> rltool



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In the plot we can see that the systems answer is unstable.

Exercise 5

Assume that you have a system with three poles at s = -1, -3, and -5, and a single zero at s = -2. The G(s)for the system is:


G(s) =5(s+ 2)

(s+1)(s+ 3)(s+ 5)


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You need to determine how the poles move in the s-plane as the gain, K, is varied. (i.e. plot the rootlocus)

Using analysis techniques, obtain an expression for the closed loop poles. Sketch the root locus on the plot below

Solution:

Using the Matlabs Workspace and the SISO tool we have the following:>> GS = zpk ([-2], [-1 -3 -5], [5])

Zero/pole/gain:5 (s+2)

----------------------(s+1) (s+3) (s+5)

>> feedback( GS, 1 )

Zero/pole/gain:5 (s+2)

-------------------------------------------(s+1.484) (s^2 + 7.516s + 16.85)>> rlocus(ans)>> rltool


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In the Architecture Menu and the System Data submenu, importing G(s),we have the following graphs:

In the plot we can see that the systems response is critically damped.

Exercise 7

Dr. Abner Mallity has been working for a small electronics design company. They need to produce a chipfor a precise amplifier. Here the problems they have.

They have a design for an amplifier stage.o When they implement the design the gain changes due to component variability,temperature (and this is for a wide temperature range application).


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They think that they can use a feedback configuration to reduce the variability.o In the feedback configuration they would use three stages.o Each stage has a nominal transfer function of 100/(10-5s + 1)o The propose using a feedback configuration like this one, but instead of unity gain in the

feedback path, they propose using a gain of .01.

Using root locus analysis, determine if they can use this configuration.

>> GS=tf([100],[10E-5 1])

Transfer function:100

------------

0.0001 s + 1

>> k=0.01;

>> tfr=feedback(k*GS,1)

Transfer function:1

------------0.0001 s + 2>>rltool

In the Architecture Menu and the System Data submenu, importing G(s), and changing the compensatorvalue in the Compensator Editor C=0.01 in the SISO we have the following graph


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The System response is stable because its always the root ever stay in the left of the complex axis andincrementing (over damped), therefore, the Dr. Mallity can use this configuration.

Exercise 9

Dr. Abner Mallity needs your help. He has been working on a system, and he got his two grad students -Willy Nilly and Millie Farad involved. Here is the transfer function of the system, and a block diagram forthe system.


Mallity's problem is that due to various things changing within the system there is a zero that canmove and be anywhere on the negative real axis. His grad students say that can present problems. Hedoesn't see that. He claims the following is true.

No matter where the zero is, the real axis segments are always in the LHP. Since there are three poles and one zero there are only two poles that go to infinity.

o

Since there are only two branches that go to infinity, they go at +90o and -90o.

o Therefore - no problem.

Mallity is convinced that this system will never be unstable, and that it doesn't matter what gain value youhave for K, or where the zero is. The trouble is, he can't prove it, and can't convince Willy and Millie thathe is right. As a matter of fact, they claim that is not true and he is getting very angry about the wholedeal.

Help Mallity out. His personal pride is at stake. Show that the system is always stable for any gain orzero position.

Solution:

Using Matlab to solve this problem we have the following:

G(s) =5(s- s

z)

(s+1)(s+ 3)(s+ 5)


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For use several values for Sz we need to define a function without zeros and after we define a zero for the4 cases.

>> GS = zpk( [], [-1 -3 -5], 5 )

Zero/pole/gain:5

-----------------(s+1) (s+3) (s+5)

>>rltool(GS)In the Architecture Menu and the System Data submenu, import ing G(s), we have the root locus plot:

Adding a Zero between the pole that is in -1 and the imaginary axis, in the Root Locus Editor the system

continues being stable:

Moving the zero between the pole that is in -1 and the pole that is in -3 the system continues being stable:


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Moving the zero between the pole that is in -3 and the pole that is in -5 the system continues being stable:

Moving the zero between the pole that is in -5 and the negative infinite (- ) the system continues beingstable:


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We have proved 4 possible cases of Sz for the System, and we conclude that the Dr. Mallity is at stake,the System is stable for all the values of Sz and the gain K because the response is damped and in theroot locus graph the poles ever are left of the imaginary axis.

Exercise 11:

In this problem, the system has three poles:

The system is in a feedback loop as shown below.

It is desired that the closed loop system dominant poles (Every other pole is at least 5 or 10 times as farinto the LHP so that any response from those poles dies out so quickly that the system looks second order- the dominant pole(s)) have a damping ratio of 0.7.

Determine the gain that produces poles with the required damping ratio. Determine if those poles are - in fact - dominant, i.e. the third pole is at least 5x as far into the

LHP.

Solution:


>> GS = zpk( [ ], [0 -1 -2], 0.5 );>> GS

Zero/pole/gain:

0.5-----------------s (s+1) (s+2)

G(s) =0.5

s(s+1)(s+ 2)


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>> rltool

In the Architecture Menu and the System Data submenu, importing G(s),in the plot click right, on the pop-up menu selecting Design requirement>>New>> type search Damping ratio, and write on Designrequirement parameters < 0.7, after we have the following:

The plot below shows that the gain is 1.32 when the damping ratio is 0.7 and we move the poles :

The current location for the pinks points on the blacks lines is: -0.381 0.389i, this value multiplied by 5only in the real axis is; -1.905, and the current point for the pink point on the real axis is -2.24, this tell usthat at least this 5 times away, therefore, these parallel poles are Dominant because the lowest polesdetermine the velocity of the control system.

Exercise 13:

Here is the transfer function for a system to control.

This open loop system has three poles: -1 + j2.5, -1 - j2.5, -9.

The open loop system has one zero: -0.5.The system is being controlled in a feedback system as shown above at the right.

Here is the root locus for the system.

G(s) =4(s+ 0.5)

s3+11s

2+ 25.25s+ 65.25


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Determine the gain, K, which will produce the minimum overshoot in the system by finding thegain for the largest damping ratio in the complex roots.

Estimate where the third pole is for the gain found in the first part.


>> GS = tf( [4 2], [1 11 25.25 62.25] )>>Transfer function:

4 s + 2-----------------------------------------s^3 + 11 s^2 + 25.25 s + 62.25

>> rltool

In the Architecture Menu and the System Data submenu, importing G(s), in the plot click right, on the pop-up menu selecting Design requirement>>New>> type search Damping ratio>>OK, adjusting the black lineto the minimum value to the exponential function, then moving the poles to these place, we have the

following:


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The gain observed is k=5.37, the third pole is localized in 3.7 in the Real- axis, with a damping ratio of8.24, finally we find the minimum overshoot percentage on 194% as we can see in the below figures:


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Information Sources:

WEB:

E. J. Mastascusa. An Introduction To The Root Locus. Lewisburg, Pennsylvania, USA,URL[http://www.facstaff.bucknell.edu/mastascu/econtrolhtml/RootLocus/RLocus1A.html]consulted November 22nd, 2010.

MC. Manuel Amarante Rodrguez. Prctica N 7 Laboratorio de Ingeniera de Control Anlisisde Sistemas de Control por Lugar Geomtrico de las Races.Nuevo Len Mxico. URL[amarante.fime.uanl.mx/ago-Dic-2010/Labs-ago-2010/ic.../pract7.PDF]. Consulted November13, 2010

Source Code Matlab

Practice performed on October 28, 2010, directed by MC Guillermo Ramirez Prado.

Annex

Source code 10/28/2010%Control toolbox(Sistemas de control en Matlab)%tf transfer function tf([numerador],[denominador])%zpk zero/pole/gain([ceros],[polos],[ganacia o k])%impulse impulso%step escaln unitariofuncion1=tf([1 2],[1 1 1])funcion2=zpk([],[-2 -3],1)%grafica impulsoimpulse(funcion1)

impulse(funcion2)%grafica escalonstep(funcion1)step(funcion2)%poniendo en seriefuncion3= funcion1*funcion2%poniendo en paralelofuncion3= funcion1+funcion2%Con retro alimentacinfuncion3=feedback(funcion1,funcion2)%lugar de las racesrlocus(funcion1)rlocus(funcion2)

%poner en lazo cerrado con k=1funcion4=feedback(1*funcion1,1)rlocus(funcion4)%con k= 10funcion4=feedback(10*funcion1,1)rlocus(funcion4)rltoolfuncion7=zpk([-1 -3],[2 4],1)funcion8=zpk([[],[2 4],1)
http://en.wikipedia.org/wiki/Lewisburg,_Pennsylvaniahttp://en.wikipedia.org/wiki/Lewisburg,_Pennsylvania


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