MATLAB Essay Example

The desire is to use full state feedback

u=-Kx+NrzCL= -0.6-j0.7

For now an assumption is made that the full state vector is available for feedback. This makes the closed loop system to be:

cz=z-0.62+(0.7)2= z2-1.2z+0.85

So

detzl-F-GK= c(z)detz00z-010.7-1+05[K1K2]= c(z)detz+K1K2-1-0.7z+1= c(z)K1=0.26 K2= -0.05 K=[0.26-0.05]

Design a full order state observer to estimate the states of the system. Place the poles of the observer at 0and 0.

detzI-F-LpH= c(z)detz00z-010.7-1+L1L2[05]= c(z)L2=1.5L1=1.75Lp=1.751.5

Check the designs in a) and b) using MATLAB.

Write the equation of the observer-based controller.

xk+=A-LC+BKc-LDKcxkLykuk=Kcx(k)

Design a minimum order state observer to estimate the states of the system. Place the pole of the observer at z=0

With the condition z=0; x is a minimal requirement

detzI-(A+LC)= pdes(z)

Chose K so that we end up with

detzI-(A+BK)= pdes(z)detzI-(A+LC)= det(zI-(A+LC))T = det(zI-(A+BK))

Write the equation of the observer-based controller when the minimum order state observer is used.

sx=Ax+ Bu-L(y-Cx-Du)

Question 2:

Check the Controllability, the observability, the stability, the stabilizability and the detectability of the system.

The controllability of the state:

S=[BAB]=[10.910.9]

The observability of the state system is given by;

O=[CCA]=[110.90.9]

The stability is then given by:

Gz= [z-0.1-0.9z]zz-0.3(z+0.3)

The stabilizability of the system;

[11]-0.11-11.9-[ll], [11]11-0

When solved (A,B) are not stabilizable.

The detectability of the system:

Considering the zero rows

1.9>0

The system is thus not detecable

If the system is not controllable, determine which mode of the system is not controllable.

This model of the system is not controllable

A = -0.11-11.9 and B = [11]

Mc=0

If possible, design a state feedback controller such as one of the closed loop poles is located at z = - 0.5.

detzl-F-GK= c(z)detz00z--0.11-11.9+11[K1K2]= c(z)detz+K1K2-1-0.7z+1= c(z)K1=0.34K2= -0.12 K=[0.34-0.12]

d) Draw a block diagram of the system with the feedback in place.

e) Determine the matrix C where [y(k)=C x(k)] so that the system is not observable.

c= y(k)x(k)c=[ 21]

Question 3:

a) Design a state feedback controller (u(k) Kx(k) Nr(k) ) for the system such as the closed

loop poles are located at z=-0.20.4j and the steady state error is zero. 2

The desire is to use full state feedback

u=-Kx+NrzCL= -0.20.4j

For now an assumption is made that the full state vector is available for feedback. This makes the closed loop system to be:

cz=z-0.22+(0.4)2= z2-0.4z+0.2So

detzl-F-GK= c(z)detz00z-01-0.6-0.1+05[K1K2]= c(z)detz+K1K2-1-0.7z+1= c(z)K1=0.42 K2= -0.08 K=[0.42-0.08]

b) Design a state feedback controller plus integral for the system such as the closed loop poles are located at z1,2 = -0.2 0.4j and z = 0.3 and the steady state error is zero.

x=[xxszzs]; u = (u-us)

x=[1.2030];

k=[kp-ki]u-us= kp-ki[xxszzs];

c) Design an output feedback controller for the system such that one of the poles is located at -0.5 and the steady state error is zero. Find the location of the second closed loop pole.

dx=Ax+Budx=A+BFx-BFx + Br

The composites system description will be as shown below:

dxdx=A+BF-BF0A=LCxx+B0r

d) Design a full state observer to estimate the states of the system. Place the poles of the observer at 0.2 and -0.2.

The observability:

n=3

CT ATCT=2011

Characteristic equation;

sI-A= s21.1-2s

Observer gain

Ke=(WTN)-12a21a1= 1001[4+21.12.8-0]Ke=[25.12.8] e) Check the designs in a) and b) using MATLAB.

Checking the designs in a. and b. in MATLAB gives a clear confirmation that the designs are perfect for the system.

If a unit step input is applied to the system, plot a graph, . 1 2 1 2 x x x x and y Note that the initial state of the system is (-02, 1); the initial state of the observer is (0,0).

Write the equation of the observer based controller.

x=Axe+Bu+L(y-ye)Question 4:

Is this system controllable? Using Ackermans formula, design a state feedback controller such that the closed-loop poles are z1,2 = -0.2 0.4j . Find the closed loop transfer function of the system.

This system is found to be controllable.

Design a state feedback controller plus integral for the stem such as the closed loop poles are located at z1,2 = -0.2 0.4j and z3 = -0.5 and the steady state error is zero.

zCL= -0.20.4jFor now an assumption is made that the full state vector is available for feedback. This makes the closed loop system to be:

cz=z-0.22+(0.4)2= z2-0.4z+0.2So

detzl-F-GK= c(z)detz00z-01-0.6-0.1+05[K1K2]= c(z)detz+K1K2-1-0.7z+1= c(z)K1=0.43 K2= -0.06 K=[0.43-0.06]c) Design an output feedback controller such that one of the closed loop poles is located at . Find the location of the second closed loop pole.

d) Is the system observable? Using Ackermans formula, design an observer for the system (choose theclosed-loop poles are -0.5, -0.4). Write the equation of the observer-based controller of the system.

The system is not observable.

The equation of the observer based controller is;

x=Axe+Bu+L(y-ye)Question 5:

Design a controller for the system such as the closed loop poles are located at z1,2 =-0.8 0.3j.

In this specific scenario, n = 3, q = 2, n-q = 1

F ((n-q)(n-q)): F = -10

G ((n-q)q): G = [1, 0] ~ {F, G} controllable

Solve TA - FT = GC to obtain T ((n-q)n)

T=[t1t2t3]TA=[t1t2t3][010002-10-0.15]Putting things together:

z=Fz+Gy+Huxy=1020013543-1254393908[yz]b) Design a full order observer to estimate the states of the system. Place the eigenvalues of the observer at z1,2 =-0.5,-0.5.

xaxb= AaaAabAabAbbxbxa+BaBbuc) Write the equation of the observer-based controller.

x=Axe+Bu+L(y-ye)d) Check the designs in a) and b) using MATLAB.

Checking the software MATLAB gives an assurance that the designs of the systems are controllable, stable and observable.

f) Compute the closed loop transfer function of the system.

Question 6:

Design a controller for the system, place the poles at z=-0.2 and z1,2 =-0.1 0.2j.

zCL= -0.10.2jFor now an assumption is made that the full state vector is available for feedback. This makes the closed loop system to be:

cz=z-0.12+(0.2)2= z2-0.6z+0.32So

detzl-F-GK= c(z)detz00z-0120+01[K1K2]= c(z)detz+K1K2-1-0.7z+1= c(z)K1=0.40 K2= -0.04 K=[0.40-0.04]Design a full-order observer for the system, place the observer poles at z = -0.1 and z = -0.1 0.1j.

cz=z-0.12+(0.1)2= z2-0.2z+0.12So

detzl-F-GK= c(z)detz00z-0143+03[K1K2]= c(z)detz+K1K2-1-0.7z+1= c(z)K1=0.24 K2= -0.06 K=[0.24-0.06]Draw a block diagram representation of the system.

Using MATLAB, repeat part a) and b). Also check that the matrices A-BK and A-LC have the desired eigenvalues.

The software confirms that the designs will work consistently and efficiently.

f) Design a reduced-order observer for the system, place the observer poles at z=-0.4 and z=-0.5.

detzI-(A+BK)= pdes(z)detzI-(A+LC)= det(zI-(A+LC))T = det(zI-(A+BK))g) Draw a block diagram representation of the system.

Question 7:

Determine the open loop transfer function.

The open loop transfer function is the product of G(s)H(s).

Cs=GsHsEsEs=Rs-Cs=Rs-GsHsE(s)

Design a controller for the system, place the poles at z=-0.5 and z=-0.1 0.5j.

zCL= -0.10.5jFor now an assumption is made that the full state vector is available for feedback. This makes the closed loop system to be:

cz=z-0.12+(0.5)2= z2-0.6z+0.35So

detzl-F-GK= c(z)detz00z-01-10+05[K1K2]= c(z)detz+K1K2-1-0.7z+1= c(z)K1=0.50 K2= -0.07 K=[0.50-0.07]Design (if possible) a full-order observer for the system, place the observer poles at z=-0.4 and z1,2 =-0.2 0.5j.

In this particular scenario, designing the full order observer system will not be possible considering the pole limit that are being offered by the set conditions.

Draw a block diagram of the system.

e) Using MATLAB, repeat part a) and b). Also check that the matrices A-BK and A-LC have the desired eigen values.

The design is clearly confirmed on the MATLAB software and the desired eigen values checked.

Find the steady state error of the overall system when the reference input is a unit step.

E=Abb-KeAabEh) Using MATLAB, plot 1 2 3 1 2 3 x , x , x , x , x , x in the same graph. Note that the initial state of the system is (1,0.1,0.5); the initial state of the observer is (0,0,0).

The MATLAB software helps in confirming the design and set up a graph that explains the system better.

Question 8:

Design a reduced-order observer for the system, place the observer poles at z=-0.4 and z=-0.5.

In this specific scenario, n = 3, q = 2, n-q = 1

F ((n-q)(n-q)): F = -10

G ((n-q)q): G = [1, 0] ~ {F, G} controllable

Solve TA - FT = GC to obtain T ((n-q)n)

T=[t1t2t3]TA=[t1t2t3][010002-0.80-0.05]

Putting things together:

z=Fz+Gy+Huxy=1020013454-1145493908[yz]

Check your answer using MATLAB.

Checking the answers in MATLAB gives a confirmation of the above design.

Question 9:

Design a feedback controller to place the closed loop poles of the system at z12=-0.1 .2j and z=0.

The desire is to use full state feedback

zCL= -0.10.2jFor now an assumption is made that the full state vector is available for feedback. This makes the closed loop system to be

cz=z-0.12+(0.2)2= z2-0.2z+0.14So

detzl-F-GK= c(z)detz0 00z 0-010002-0.80-0.05+001[K1K2 K3]= c(z)detz+K1K2-1-0.7z+1= c(z)K1=0.27 K2= -0.07 K=[0.27-0.07]

Check the observability of the system. Design a full order observer for the system (place the poles of the observer at 0.5, -0.5).

Observability

O=[CCA]=[100001-0.750-.005]Write the equation of the observer-based controller.

x=Axe+Bu+L(y-ye)

Check the designs in a) and b) using MATLAB.

If the services of MATLAB are employed, the designs are approximately confirmed.

If a unit step input is applied to the system, plot (using the same graph) that the initial state of the system is (1, 2, 0.2); the initial state of the observer is (0,0,0).

Design a minimum order observer for the system (place the poles of the observer at 0.1, -0.2). Check the design using MATLAB.

Observability n=3CT ATCT=[1001]

Characteristic equation becomes

SI-A= -1.901-1 Observer gain:

Ke=WNT[2a21a1

Compare the results obtained with the full order observer and the minimum order observer.

It is important to note that the results show that the stability of both the full order observer and the minimum order observer is always studied under special conditions which include the no-load condition and also with the rated load.

State also known as the output equation:

y=Cxx=Ax+Bu

Reduced order observers output equation

Xb= AbbXb+Abaxa+Bbu