Control System: Double Pendulum System, Kalman Filter, LQR Command

control system
Dell

Table of Contents
Question-1.................................................................................................................................1
Question -2................................................................................................................................6
Question-3...............................................................................................................................13
References...............................................................................................................................20

Question-1
Given data:
The performance index J is given by
Where r > 0 scalar parameter
(a) 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 = adj(SI-A)det(SI-A)
C (sI – A)-1 B = [1 0][s -1 0 s+1 ] -1 [0 1 ]
= [1 0] 1s(s+1) [s+1 1 0 s ][0 1 ]
= [1 0] 1/s2+s [1 s ]
= 1/s2+s [1+0]
C (sI – A)-1 B = 1/s2+s
The above equation is representing the value of  = [1 0 1 0 ] , B =[1 0 ] , Ĉ =[1 0]
A = [0 1 0 -1 ], B = [0 1 ]
Mcx = [B, AB]
AB = [1 -1 ]
Mcx = [1 1 0 -1 ]
 = [1 0 1 0 ] , B =[1 0 ]

Mcz = [B, ÂB]
Mcz = [0 1 1 1 ]
To find the similarity transform relating the two. It is given by
Mcz[Mcx]-1 = T
[Mcx]-1 = [1 1 0 -1 ] 1/-1 [-1 -1 0 1 ]
= [0 1 1 1 ][-1 -1 0 1 ]
T = [0 1 1 2 ]
To find the optimal gian vector K
K = ǨT
 = [1 0 1 0 ] , B =[1 0 ]
Â-BǨ =[1-k -k 1 0 ]
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
= [1-3 -1 1 0 ]
=[-2 -1 1 0 ]
Therefore the original feedback system is given by
K = ǨT
K = [3 1] [0 1 1 2 ]
Finally the optimal gain vector K = [-5 -3]
(b) 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
(c) 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
(d) Part (a) and (c) using LQR command
When comparing the closed loop system the below figure show in the matlab window.
This plot indicates the closed loop system poles using the LQR command
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
n (samples)
-0.7
-0.6
-0.5
-0.4
-0.3
-0.2
-0.1
0
Amplitude
Step Response

(e) Kalman filter gain L
given
Q = 6xI2x2
R =1
G =I2x2
The given kalman filter gain L is given by
Q = 6x[1 0 0 1 ]
Q =[6 0 0 6 ]
G = [1 0 0 1 ]
L = ½ [6 0 0 6 ] +1/2 (1)
= [3 0 0 3 ]+[0.5]
= [1.5 0 0 1.5 ]
The gain value of the kalman filter is givevn by
L = [1.5 0 0 1.5 ]