1. 1
Controls II Final
Arna Friend
Giventhe transfermatrix of a two-input,two-outputsystem:
𝐺( 𝑠) = [
7
𝑠
𝑠 + 2
𝑠2 + 6𝑠 + 5
𝑠 + 3
𝑠2 + 6𝑠 + 5
10
𝑠 + 1
]
Designa discrete time decouplingcontrol systemwithtimestep ∆𝑡 = .005 seconds.Channel one must
have an outputfeedbackcontrollerandachieve settlingtimeof 2 secondsandpercentovershootof the
stepresponse of 10%. Channel 2must have a state variable feedbackcontrollerwithastate observer
and achieve settlingtime of 4secondsand0% overshootof the stepresponse.
FirstI convertedG(s) tothe z-domainforthe expresspurposeof designingthe decouplingfilterW(z):
𝐺( 𝑧) = [
.035
𝑧 − 1
.00495𝑧 − .004901
𝑧2 − 1.97𝑧 + .9704
. 004963𝑧 − .004889
𝑧2 − 1.97𝑧 + .9704
.04998
𝑧 − .995
]
I designedW(z) suchthat:
𝑄( 𝑧) = 𝐺( 𝑧) 𝑊( 𝑧) = [
𝑞11(𝑧) 0
0 𝑞22(𝑧)
]
Thisresultedin:
𝑊( 𝑧) = [
−.04998
. 004963𝑧 − .004889
𝑧 − 1
𝑧2 − 1.97𝑧 + .9704
𝑧 − .995
𝑧2 − 1.97𝑧 + .9704
−.035
.00495𝑧 − .004901
]
And:
𝑞11( 𝑧) =
−.3468𝑧4 + 1.366𝑧3 − 2.019𝑧2 + 1.326𝑧 − .3265
𝑧6 − 5.926𝑧5 + 14.63𝑧4 − 19.27𝑧3 + 14.27𝑧2 − 5.637𝑧 + .9277
𝑞22( 𝑧) =
−.3477𝑧4 + 1.37𝑧3 − 2.024𝑧2 + 1.329𝑧 − .3273
𝑧6 − 5.926𝑧5 + 14.63𝑧4 − 19.27𝑧3 + 14.27𝑧2 − 5.637𝑧 + .9277
To designthe outputfeedbackcontrollerforchannel one,Ibeganbydefiningansdomainmodel
throughpole placement.Ichose the real partof the dominantpolesaccordingtothe settlingtime:

2. 2
𝑝𝑑
𝑟𝑒𝑎𝑙
= −
4
𝑇𝑠𝑒𝑡
= −2
I chose the imaginarypartsof the dominantpolesaccordingtothe dampingratioof .6, correspondingto
overshootof 10%:
𝑝𝑑
𝑖𝑚
=
√1 − 𝜁2
𝜁2 = ±2.6667
I chose the remaining polesas -8, -17, -18 and -20. I thenconstructedthe model transferfunctioninthe
s domainas:
𝐺 𝑀( 𝑠) =
5.44 ∗ 105
𝑠6 + 67𝑠5 + 1709𝑠4 + 20652𝑠3 + 1.217 ∗ 105 𝑠2 + 5.44 ∗ 105
I thenconvertedthe model tothe z-domain:
1.125 ∗ 10−11 𝑧5 = 6.117 ∗ 1010 𝑧4 + 3.09 ∗ 10−9 𝑧3 + 2.946 ∗ 10−9 𝑧2 + 5.299 ∗ 10−10 + 8.86 ∗ 10−12
𝑧6 − 5.678𝑧5 + 13.43𝑧4 − 16.94𝑧3 + 12.01𝑧2 − 4.542𝑧 + .7153
To get the numeratorof the feedbackcontroller,Isubtractedthe coefficientsof the characteristic
polynomial of the model fromthose of the characteristicpolynomial of the plant:
𝐹 =.2478𝑧5 − 1.2012𝑧4 + 2.3296𝑧3 − 2.2589𝑧2 + 1.0952𝑧 − .2124
The denominatorof the feedbackcontrollerwasjustthe numeratorof the plant,delayedbyone sample
to make it the same orderas the numerator.Ithendesignedthe pre-filtertohave the numeratorof the
model andthe numeratorof the plant(delayedbytwosamples) asthe denominator. Simulation
yielded:

3. 3
For channel 2, I designedastate variable controllerwithastate observer. Asinthe case of the output
feedbackcontrollerI beganbydesigningamodel toachieve the designspecificationsof 4second
settlingtime and0%overshoot.Since there isno overshoot,there isjustone real dominantpolewhichI
chose as:
𝑝 𝑑 = −
4
𝑇𝑠𝑒𝑡
= −1
I chose the remainingpolesas -8, -13,-10,-12 and-4. This resultedinthe followingsdomainmodel:
𝐺 𝑀( 𝑠) =
49920
𝑠6 + 48𝑠5 + 905𝑠4 + 8410𝑠3 + 39264𝑠2 + 81632𝑠 + 49920
Convertingtothe z-domainIobtained:
1.047 ∗ 10−12 𝑧5 + 5.766 ∗ 10−11 𝑧4 + 2.952 ∗ 10−10 𝑧3 + 2.853 ∗ 10−10 𝑧2 + 5.203 ∗ 10−11 𝑧 + 8.819 ∗ 10−13
𝑧6 − 5.766𝑧5 + 13.85𝑧4 − 17.75𝑧3 + 12.79𝑧2 − 4.914𝑧 + .7866
I thenconvertedboththe plantandthe model toCCFform, resultinginthe followingfundamental
matrices:
𝐴 𝑃 =
[
0 1 0 0 0 0
0 0 1 0 0 0
0 0 0 1 0 0
0 0 0 0 1 0
0 0 0 0 0 1
−.9277 5.6367 −14.2694 19.2653 −14.6306 5.9257]

4. 4
𝐴 𝑀 =
[
0 1 0 0 0 0
0 0 1 0 0 0
0 0 0 1 0 0
0 0 0 0 1 0
0 0 0 0 0 1
−.7866 4.9135 −12.7868 17.7457 −13.8518 5.7661]
To obtainfeedbackcontrollerFI used:
𝐹 = [0 0 0 0 0 1] ∗ ( 𝐴 𝑃 − 𝐴 𝑀)
𝐹 = [−.1411 . 7232 −1.4825 1.5196 −.7788 .1596]
I thenbeganthe designof the state observer.Idesignedthe state observertohave 50% of the settling
time of the controlledplant,or2 seconds.Ichose the dominantpole as:
𝑝 𝑑 = −
4
𝑇𝑠𝑒𝑡
= −2
I chose the remainingpolesas -16,-17,-19,-20 and -22. Thisresultedinthe followingmodel of the
observer:
𝐺 𝑂( 𝑠) =
1
𝑠6 + 96𝑠5 + 3711𝑠4 + 72852𝑠3 + 744228𝑠2 + 3.499 ∗ 106 + 4.548 ∗ 106
I thenconvertedtothe z domain:
2.027 ∗ 10−17 𝑧5 + 1.079 ∗ 10−15 𝑧4 + 5.338 ∗ 10−15 𝑧3 + 4.984 ∗ 10−15 𝑧2 + 8.782 ∗ 10−16 + 1.438 ∗ 10−17
𝑧6 − 5.542𝑧5 + 12.79𝑧4 − 15.75𝑧3 + 10.9𝑧2 − 4.024𝑧 + .6188
Andthe to CCF, resultinginthe followingfundamentalmatrix:
𝐴 𝑂 =
[
0 1 0 0 0 0
0 0 1 0 0 0
0 0 0 1 0 0
0 0 0 0 1 0
0 0 0 0 0 1
−.6188 4.0240 −10.9015 15.7478 −12.7932 5.5417]
Usingthe fundamental matrix of the observerandthe fundamentalmatrix of the plantIsolvedfor
vectorK as follows:
𝑘1 = −𝐴 𝑂(6,6) + 𝐴 𝑃(6,6)
𝑘2 = −𝐴 𝑂(6,5) + 𝐴 𝑃(6,5) + 𝐴 𝑃(6,6) 𝑘1
𝑘3 = −𝐴 𝑂(6,4) + 𝐴 𝑃(6,4) + 𝐴 𝑃(6,5) 𝑘1 + 𝐴 𝑃(6,6) 𝑘2
𝑘4 = −𝐴 𝑂(6,3)+ 𝐴 𝑃(6,3)+𝐴 𝑃(6,4) 𝑘1 + 𝐴 𝑃(6,5) 𝑘2 + 𝐴 𝑃(6,6) 𝑘5
𝑘5 = −𝐴 𝑂(6,2)+𝐴 𝑃(6,2)+ 𝐴 𝑃(6,3) 𝑘1 + 𝐴 𝑃(6,4) 𝑘2 + 𝐴 𝑃(6,5) 𝑘3 + 𝐴 𝑃(6,6) 𝑘4

5. 5
𝑘6 = −𝐴 𝑂(6,1) + 𝐴 𝑃(6,1)+𝐴 𝑃(6,2) 𝑘1 + 𝐴 𝑃(6,3) 𝑘2 + 𝐴 𝑃(6,4) 𝑘3 + 𝐴 𝑃(6,5) 𝑘4 + 𝐴 𝑃(6,6) 𝑘5
Thisresultedin:
𝐾 =
[
.3840
.4380
.4952
.5554
.6189
.6856]
I thenintroducedafiltertocancel the zerosof the plantso that itsC matrix wouldbe:
𝐶 𝑃 = [1 0 0 0 0 0]
The filterwas:
𝐹𝑍𝐶 =
1
−.3477𝑧4 + 1.37𝑧3 − 2.024𝑧2 + 1.329𝑧 − .3273
I designedmatrix Mas:
𝑀 = 𝐴 𝑃 − 𝐾𝐶 𝑃 =
[
−.3840 1 0 0 0 0
−.4380 0 1 0 0 0
−.4952 0 0 1 0 0
−.5554 0 0 0 1 0
−.6189 0 0 0 0 1
−.1.6133 5.6367 −14.2694 19.2653 −14.6306 5.9257]
FinallyIdesignedafilterinthe reference channel equaltothe numeratorof the model inorderto make
the overall closedlooptransferfunctionequal tothe transferfunctionof the model.Simulationof
channel 2 resultedin:

6. 6
The control effortU1, withreference appliedtochannel one,and nodisturbance,was:
AndU2:
The control effortU1 withreference appliedtochannel twoandnodisturbance was:

7. 7
AndU2:
Whena disturbance wasappliedtochannel one, withnoreference, the systemresponse was:

8. 8
The control effortU1 was:
AndU2:

9. 9
Whena stepdisturbance wasappliedonchannel two,withnoreference signal,the system response
was:
Andcontrol effort U1:

10. 10
AndU2:
I thenintroducedfeasabilityconstraintsonthe control effort,boundingU1 between -1and1 and
boundingU2 between -.75and.75.
WhenI appliedastepreference of magnitude5to channel one,U1 reacheditsuppersaturationlimit,as
shownbelow:

11. 11
U2 was still well withinitslimits:
The systemwasstill able tofollowhigherreferencesbutthe control effortssimplybecame toolarge to
be feasible.Forinstance here isthe systemresponsetoa stepof magnitude 100:

12. 12
On channel two,the systemcouldtrackstepreferencesuptomagnitude 7.5before feasibilitylimitson
the control effortU2 were exceeded.Below are U1 andU2 fora stepreference of 7.5:
AndU2:

13. 13
As inthe previouscase the systemcouldstill trackveryhigh magnitude referencesbutthe control
effortswere beyondfeasibilitylimits.Here isthe responseof channel twotoa stepof magnitude 100:
So althoughthe systemremainsdecoupledandfollowsthe reference forlarge signals,inorderforthe
designconstraintsonbothcontrol effortstobe met,the reference mustbe 5 or below inmagnitude.
Simulationdiagrams:

14. 14
Diagram forstepresponsestoa unitstepinput andfor control efforts:
Diagram forresponsestoa unitstepdisturbance (withthe disturbance onchannel 2for thatcase):
[P1]
[Z1]
[R2]
netsum9
netsum10
NMz2(z)
1
z
1
Unit Delay6
z
1
z
1
z
1
z
1
z
1
U2
U1
G_12s
M(6,1)
-K-
M(5,1)
-K-
M(4,1)
M(3,4)
M(3,1)
M(2,3)
F(6)
F(5)
F(4)
F(3)
F(2)
F(1)
M(1,2)
M(1,1)
M(6,2)
M(2,1)
FZC
Pz
G_21s
G_22s
G_11s
W_22
W_21
Fz
W_12
W_11
[Z2]
[Z1]
[P2]
[P1]
[A]
[Y2]
[Y1]
[Z6]
Goto13
[Z5][Z4]
[Z3]
[U2]
[U1]
M(6,5)
M(6,4)
M(6,3)
K(5)K(4)
K(2)
K(1)
Bp(6)
K(6)
-K-
K(3)
[Z1]
[U2]
[U2]
[P2]
[Z5]
[Z6]
[P2]
[Z4]
[Z3]
[Z2]
[Z1]
[Y2]
From35
[Z6]
[Z5]
[Z4]
[Z3]
[Z2]
[P1]
[Y2]
[Z6]
[Z1]
[Y2]
[Z5]
[Z1]
[Y2]
[Y2]
[Z4]
[U1]
[Z3]
[Z1]
[Y2]
[Y1]
[A]
[Z2]
[Z1]
[Y1]
[Y2]
[U1]
[R2]
[R1]
[P1]
[Z1]
[R1]
netsum9
netsum10
NMz2(z)
1
z
1
Unit Delay6
z
1
z
1
z
1
z
1
z
1
U2
U1
G_12s
M(6,1)
-K-
M(5,1)
-K-
M(4,1)
M(3,4)
M(3,1)
M(2,3)
F(6)
F(5)
F(4)
F(3)
F(2)
F(1)
M(1,2)
M(1,1)
M(6,2)
M(2,1)
FZC
Pz
G_21s
G_22s
G_11s
W_22
W_21
Fz
W_12
W_11
[Z2]
[Z1]
[P2]
[P1]
[A]
[Y2]
[Y1]
[Z6]
Goto13
[Z5][Z4]
[Z3]
[U2]
[U1]
M(6,5)
M(6,4)
M(6,3)
K(5)K(4)
K(2)
K(1)
Bp(6)
K(6)
-K-K(3)
[Z1]
[U2]
[U2]
[P2]
[Z5]
[Z6]
[P2]
[Z4]
[Z3]
[Z2]
[Z1]
[Y2]
From35
[Z6]
[Z5]
[Z4]
[Z3]
[Z2]
[P1]
[Y2]
[Z6]
[Z1]
[Y2]
[Z5]
[Z1]
[Y2]
[Y2]
[Z4]
[U1]
[Z3]
[Z1]
[Y2]
[Y1]
[A]
[Z2]
[Z1]
[Y1]
[Y2]
[U1]
[R2]
[R1]
0
Constant

15. 15
Matlab code:
%% Arna Friend Controls II Final Take Home Exam
% declare s domain transfer matrix to be used for simulation
G_11s = tf(7,[1 0]);
G_12s = tf([1 2],[1 6 5]);
G_21s = tf([1 3],[1 6 5]);
G_22s = tf(10,[1 1]);
% declare time step
dt = .005;
% convert transfer matrix into z domain
G_11z = c2d(G_11s,dt);
G_12z = c2d(G_12s,dt);
G_21z = c2d(G_21s,dt);
G_22z = c2d(G_22s,dt);
% define numerators and denominators in code to avoid rounding errors
[G_11zN G_11zD] = tfdata(G_11z,'v');
[G_12zN G_12zD] = tfdata(G_12z,'v');
[G_21zN G_21zD] = tfdata(G_21z,'v');
[G_22zN G_22zD] = tfdata(G_22z,'v');
% define decoupling filter W(z)
W_12 = tf(G_11zD,G_12zD,dt);
W_22 = tf(-1*G_11zN,G_12zN,dt);
% check that Q_12 = 0
Q_12 = minreal(W_12*G_11z) + minreal(W_22*G_12z);
% this is non-zero but I am certain it is due to rounding error
W_11 = tf(-G_22zN,G_21zN,dt);
W_21 = tf(G_22zD,G_21zD,dt);
% check that Q_21 = 0
Q_21 = minreal(W_11*G_21z)+minreal(W_21*G_22z);
% similarly to Q_12 this is non-zero but the numerator is on the order of
% 10^-18
% construct Q_11 and Q_22
Q_11 = minreal(W_11*G_11z+W_21*G_12z);
[z1 p1 ki] = zpkdata(Q_11,'v'); % assure n0 poles or zeros outside the unit
circle and no uncancelled pole zero pairs.
Q_22 = minreal(W_12*G_21z + W_22*G_22z);
[z2 p2 k2] = zpkdata(Q_22,'v');% assure no poles or zeros outside the unit
circle and no uncancelled pole zero pairs.
% begin design of output feedback controller for channel one
% channel one settling time 2 seconds. Overshoot 10% zeta=.6
% define settling time, damping ratio and proceed with pole placement
T_set = 2;
pd_real = 4/T_set;

16. 16
zeta = .6;
pd_im = pd_real*sqrt(1-zeta^2)/zeta;
pd1 = pd_real+pd_im*1i;
pd2 = pd_real-pd_im*1i;
pnd1 = 8;
pnd2 = 17;
pnd3 = 18;
pnd4 = 20;
% contruct characteristic polynomial
d1 = conv([1 pd1],[1 pd2]);
d2 = conv(d1,[1 pnd1]);
d3 = conv(d2,[1 pnd2]);
d4 = conv(d3,[1 pnd3]);
DMs = conv(d4,[1 pnd4]);
% insure gain of 1
NMs = DMs(end);
GMs = tf(NMs,DMs);
% step(GMs);
% convert to discrete time
GMz = c2d(GMs,dt);
[NMz,DMz]=tfdata(GMz,'v');
[NPz,DPz] = tfdata(Q_11,'v');
F = DMz(2:end)-DPz(2:end);
% feedback controller
Fz = tf(F,[NPz 0],dt);
% pre filter
Pz = tf(NMz,[NPz 0 0],dt);
%%
% channel 2 state variable feedback controller with state observer
% 4 second settling time 0% overshoot
T_set = 4;
p1 = 4/T_set;
p2 = 13;
pnd1 = 4;
pnd2 = 8;
pnd3 = 10;
pnd4 = 12;
d1 = conv([1 p1],[1 p2]);
d2 = conv(d1,[1 pnd1]);
d3 = conv(d2,[1 pnd2]);
d4 = conv(d3,[1 pnd3]);
DMs2 = conv(d4,[1 pnd4]);
NMs2 = DMs2(end);
GMs2 = tf(NMs2,DMs2);
GMz2 = c2d(GMs2,dt);
[NMz2,DMz2] = tfdata(GMz2,'v');
[NPz2,DPz2] = tfdata(Q_22,'v');
% plant and model in CCF form
Ap = [0 1 0 0 0 0;
0 0 1 0 0 0;
0 0 0 1 0 0;
0 0 0 0 1 0;
0 0 0 0 0 1;
-1.*fliplr(DPz2(2:end))];
Bp = [0;0;0;0;0;1];
Cp = fliplr(NMz2(2:end));
Am =[0 1 0 0 0 0;

17. 17
0 0 1 0 0 0;
0 0 0 1 0 0;
0 0 0 0 1 0;
0 0 0 0 0 1;
-1.*fliplr(DMz2(2:end))];
Bm =[0;0;0;0;0;1];
Cm = fliplr(NMz2(2:end));
F = [0 0 0 0 0 1]*(Ap-Am); % state variable controller
% design state observer to have settling time 2 seconds
T_set_est = 2;
pd = 4/T_set_est;
pnd1 = 16;
pnd2 = 17;
pnd3 = 19;
pnd4 = 20;
pnd5 = 22;
d1 = conv([1 pd],[1 pnd1]);
d2 = conv(d1,[1 pnd2]);
d3 = conv(d2,[1 pnd3]);
d4 = conv(d3,[1 pnd4]);
DMEs = conv(d4,[1 pnd5]);
GEs = tf(1,DMEs);
GEz = c2d(GEs,dt);
[NEz DEz] = tfdata(GEz,'v');
AE = [0 1 0 0 0 0;
0 0 1 0 0 0;
0 0 0 1 0 0;
0 0 0 0 1 0;
0 0 0 0 0 1;
-1.*fliplr(DEz(2:end))];
% recursive formula to solve for K
k1 = -AE(6,6)+Ap(6,6);
k2 = -AE(6,5)+Ap(6,5)+Ap(6,6)*k1;
k3 = -AE(6,4)+Ap(6,4)+Ap(6,5)*k1+Ap(6,6)*k2;
k4 = -AE(6,3)+Ap(6,3)+Ap(6,4)*k1+Ap(6,5)*k2+Ap(6,6)*k3;
k5 = -AE(6,2)+Ap(6,2)+Ap(6,3)*k1+Ap(6,4)*k2+Ap(6,5)*k3+Ap(6,6)*k4;
k6 = -AE(6,1)+Ap(6,1)+Ap(6,2)*k1+Ap(6,3)*k2+Ap(6,4)*k3+Ap(6,5)*k4+Ap(6,6)*k5;
K = [k1;k2;k3;k4;k5;k6];
FZC = tf(1,NPz2,dt); %zero cancelling filter
M = Ap-K*[1 0 0 0 0 0]; %because of zero cancelling filter Cp is now [1 0 0 0
0 0]
% filter in the reference channel
FR = tf(NMz2,1,dt);