Question:
Discuss about the Matlab Program.
Answer:
1
Given data:
The performance index J is given by
Where r > 0 scalar parameter
- Optimal gain vector K
The state feedback control law is given by u = -kx
To find the value of C (sI – A)-1 B
(SI –A)-1 =
C (sI – A)-1 B = [1 0] -1
= [1 0]
= [1 0] 1/s2+s
= 1/s2+s [1+0]
C (sI – A)-1 B = 1/s2+s
The above equation is representing the value of  = , B = , =[1 0]
A = , B =
Mcx = [B, AB]
AB =
Mcx =
 = , B =
Mcz = [B, ÂB]
Mcz =
To find the similarity transform relating the two. It is given by
Mcz[Mcx]-1 = T
[Mcx]-1 = 1/-1
=
T =
To find the optimal gian vector K
K = ǨT
 = , B =
Â-BǨ =
The desired pole polynomial is given by S2+2s+1
Now the above value is changed by
Where Ǩ = [3 1]
Now substitute the value for Ǩ in Â-BǨ equation
=
=
Therefore the original feedback system is given by
K = ǨT
K = [3 1]
Finally the optimal gain vector K = [-5 -3]
- K is critically dampled
Given
The closed loop system is given by
When K=-5
G(s) = c(s)/R(s)
= -5/s2+s-5
=-5/s2+2 ᶓ ωns+ωn2
The critically damped system ᶓ =1
The equation is given by
= -5/ s2+2 ωns+ωn2
= -5/s(s+ωn)2
The closed system with K is critically damped the given equation is
C(s) =-5/s(s+ωn)2
C(s) =-3/s(s+ωn)2
- The closed loop system poles
The given system is C(s) =-5/s(s+ωn)2
=A/s +B/(s+ωn)2 +C/(s+ ωn)
The pole values are determined by the above equation
= -1+4.582/2 = 1.791
P1 = 1.791
P2 = -2.791
- Part (a) and (c) using LQR command
- Kalman filter gain L
given
Q = 6xI2×2
R =1
G =I2×2
The given kalman filter gain L is given by
Q = 6x
Q =
G =
L = ½ +1/2 (1)
= +[0.5]
=
The gain value of the kalman filter is givevn by
L =
- Using LQE
2
(a)
Double pendulum system
Given
Solution
Concept of control system in state feedback
State feed back
Impact of state feedback on behaviour
Closed loop poles
It is clear that state feedback changes behavioura and allows us to move the closed –loop poles
The closed –loop poles are eigenvalues of A-BK
Find the impact of the state feedback
closed loop
apply these above values are given below
A =
14.7 -4.9 0 0
B = (0 0 0 -1/2(4))
= (0 0 0 -0.125)T
B =
C =
Find the closed-loop poles
A =
14.7 -4.9 0 0
B=
C = Now K = k1 k2 k3 k4
A-BK= -29.4-K1 19.6-K2
14.7- K3 -4.9-K3
Closed – loop poles
K2=19.6 k3=-29.4
(b)
= -29.4-K1 19.6
14.7- K3 -4.9-K3
K2=19.6 k3=-29.4
(c)
Change the transpose
X(0)=0.1
0
0
0
Open loop and closed loop different values an
step (A,B,C,D)
step (open loop,clos loop)
A = [0.1, 29.4]; B= [0, 19.6]; C = [0, 29.4]; D = [0, 0]; step (A,B,C,D) (d) LQR
U=-KX
The open loop and closed loop both are same in the LQR system
(e)
The behaviour of Q is unit step response (time domain) in the given following graph
3
- White noise
Ideal of white noise
Unconditional mean and variance are constant
Serially uncorrelated men/variance are same
Conditional and unconditional mean /varience are same
Mean and varience
Mean= 70,102.89 -8389.626 -3120
-8770 1026.0084 384
-3120 384 144
Varience = 73984 0 196
Program
output
References
Anon, (2017). [online] Available at: https://www.researchgate.net/post/How_can_we_obtain_the_Gain_Vector_or_Matrix_in_this_form_for_the_LQR_control [Accessed 28 Nov. 2017].
Filter, K. (2017). Kalmanfilter design, Kalman estimator – MATLAB kalman – MathWorks United Kingdom. [online] In.mathworks.com. Available at: http://in.mathworks.com/help/control/ref/kalman.html [Accessed 28 Nov. 2017].
Program
%% Plots
% outputs
figure; %create a new plot window
subplot(211); %create plots in 4 rows and 4 column and activate the 1st one
plot(t0,Y0(:,1),’Color’,’k’, ‘LineStyle’, ‘-.’); %plot in black color and
dotted line
hold on; %hold existing plot, don’t erase it and wait for the next plot
plot(t1,Y1(29.4 ,1),’Color’,’b’); % plot in blue color
plot(t2,Y2(19.6,1),’Color’,’r’); % plot in red color
plot(t3,Y3(:,1),’Color’,’b’); % plot in blue color
title(‘System Output’) % add title to the plot
ylabel(‘y_{1}(t)’); % add label to y-axis
xlabel(‘time (sec)’); % add label to x-axis
xlim([0,10]); % determine the range of x-axis
grid; % add grid
legend(‘closed-Loop Response’, ‘Q = 10*I_{4\times4}, R = 1’, …
‘Q = 100*I_{4\times4}, R = 1’, ‘Location’, ‘southeast’); % add legends
subplot(212); % in the current plot window with 4 rows and 4 column, active
the 2 nd one
plot(t0,Y0(:,2),’Color’,’k’, ‘LineStyle’, ‘-.’);
hold on;
plot(t1,Y1(:,2),’Color’,’b’);
plot(t2,Y2(:,2),’Color’,’r’);
plot(t3,Y3(:,2),’Color’,’b’);
subplot(2,1,2);
ylabel(‘y_{2}(t)’);
xlabel(‘time (sec)’);
xlim([0,10]);
grid;
legend(‘Open-Loop Response’, ‘Q = 10*I_{4\times4}, R = 1’, …
‘Q = 100_{4\times4}*I, R = 1’, ‘Location’, ‘southeast’);
%control inputs
figure;
plot(t1, u1(1,:)’);
hold on;
plot(t2, u2(1,:)’);
title(‘Control Input’);
xlim([0 10]);
ylabel(‘u(t)’);
xlabel(‘time (sec)’);
grid;
legend(‘Q = 10*I_{4\times4}, R = 1’, ‘Q = 100*I_{4\times4}, R = 1’,…
‘Location’, ‘southeast’);
