This document describes the modeling and control of an inverted pendulum system mounted on a cart. It begins with defining the control objectives of stabilizing the pendulum angle and cart position. Mathematical models are derived using equations of motion and linearized. Both transfer function and state space models are obtained. Open loop responses show the system is unstable. PID and state feedback controllers are designed in closed loop, but do not fully meet requirements. An optimal LQR controller is then designed using state weighting matrices to satisfy all requirements, stabilizing both pendulum angle and cart position with minimal control effort.

1. Optimum controller design
Dr.R.SUBASRI
Associate. Professor
Dept .of EIE
17.03.2015
soamisuba@yahoo.comDr.R.Subasri,KEC,INDIA

2. Problem Definition
To control the deflection of pendulum () and
cart’s position(x).
Dr.R.Subasri,KEC,INDIA

3. Control Requirements
-SISO System
• The pendulum should return to its upright
position within 5 seconds (Settling time)
• The position of pendulum should never move
more than 0.05 radians away from the
vertical.
Transfer function model with impulse
input
Dr.R.Subasri,KEC,INDIA

4. Parameters
• mass of the cart (M) :0.5 kg
• mass of the pendulum (m) :0.5 kg
• friction of the cart (b) :0.1 N/m/sec
• length to pendulum center
of mass(l) : 0.3 m
• Inertia of the pendulum (I) : 0.006 kg*m^2
• Force applied to the cart :F
• Cart position coordinate : x
• pendulum angle from vertical :
Dr.R.Subasri,KEC,INDIA

5. Mathematical modeling
• Summing the forces in the Free Body Diagram
of the cart in the horizontal direction
• Summing the forces in the Free Body Diagram
of the pendulum in the horizontal direction
Mx bx N F+ + =
2
N mx mlθcosθ mlθ sinθ= + −
2
(M m)x bx mlθcosθ mlθ sinθ F+ + + − =
First equation of motion:
Dr.R.Subasri,KEC,INDIA

6. • sum the forces perpendicular to the pendulum
• Second equation of motion:
2
(I ml )θ mglsinθ mlxcosθ+ + = −
Linearise the equations about vertical equilibrium angle
( = )
Let  represent the deviation of the pendulum's
position from equilibrium, that is  =  + 
Approximations of small angle
sin ( + ) =- 
cos ( + ) = -1
Dr.R.Subasri,KEC,INDIA

7. Mathematical modeling
• The two linear equations of motions are:
( )2
I ml φ mglφ mlx
(M m) x bx mlφ u
+ − =
+ + − = Control Force F
replaced by u
Dr.R.Subasri,KEC,INDIA

8. Transfer function model
• Use Laplace transform to derive the transfer
function model
• Where,
2
2
4 3 2
ml
s
φ(s) q
U(s) b(I ml ) (M m)mgl bmgl
s s s s
q q q
=
+ +
+ − −
( )( )2 2
q M m I ml (ml) = + + − 
Dr.R.Subasri,KEC,INDIA

9. open-loop response
-transfer function model
• M = .5; m = 0.2; b = 0.1;
• i = 0.006; g = 9.8;
• l = 0.3;
• q = (M+m)*(i+m*l^2)-(m*l)^2; %simplifies input
• num = [m*l/q 0]
• den = [1 b*(i+m*l^2)/q -(M+m)*m*g*l/q -b*m*g*l/q]
output should be:
• num = 4.5455 0
• den = 1.0000 0.1818 -31.1818 -4.4545
Dr.R.Subasri,KEC,INDIA

10. open-loop response
-transfer function model
• To observe the system's velocity response to an
impulse force applied to the cart add the following
lines at the end m-file:
• t=0:0.01:5;
• impulse(num,den,t)
• axis([0 1 0 60])
The response is entirely
unsatisfactory.
It is not stable in open loop.
Dr.R.Subasri,KEC,INDIA

11. Closed loop response-transfer function model
• As the pendulum's position should return to
the vertical after the initial disturbance, the
reference signal should be zero.
• The force applied to the cart can be added as
an impulse disturbance.
Dr.R.Subasri,KEC,INDIA

12. Effect of PID controller
• kd = 1;k = 1;ki = 1;
• numPID = [kd k ki];
• denPID = [1 0];
• numc = conv(num,denPID)
• denc = polyadd(conv(denPID,den),
conv(numPID,num))
numc = 4.5455 0 0
denc = 1.0000 4.7273 -26.6363 0.0910 0
Dr.R.Subasri,KEC,INDIA

13. Closed loop Impulse response with PID
• t=0:0.01:5;
• impulse(numc,denc,t)
• axis([0 1.5 0 40])
This response is still
not stable.
Dr.R.Subasri,KEC,INDIA

14. Closed loop Impulse response with PID
• increasing the proportional control (k) ,
k=100;
axis([0, 2.5, -0.2, 0.2])
The settling time is
acceptable at about
2 sec.
The overshoot is high
Dr.R.Subasri,KEC,INDIA

15. Closed loop Impulse response with PID
• increase kd as 20
The overshoot has been
reduced so that the
pendulum does not
move more than 0.05
radians away from the
vertical.
Dr.R.Subasri,KEC,INDIA

16. Control Requirements
-MIMO System
• To control both the pendulum's angle and the
cart's position.
• The cart should achieve it's desired position
within 5 seconds (settling time)
• Rise time under 0.5 seconds.
• Pendulum's overshoot to 20 degrees (0.35
radians), and it should also settle in under 5
seconds.
-State space model with step inputDr.R.Subasri,KEC,INDIA

17. State space model
• After a little algebra, the linearized system
equations can also be represented in state-
space form:
Dr.R.Subasri,KEC,INDIA

18. • R - commanded step input to the cart (step
input of 0.2 m)
• 4 states X– the position and velocity of the
cart and the angle and angular velocity of the
pendulum.
• output y - both the position of the cart and
the angle of the pendulum.
Dr.R.Subasri,KEC,INDIA

19. Open loop response
• M = 0.5; m = 0.2; b = 0.1; i = 0.006; g
= 9.8; l = 0.3;
• p = i*(M+m)+M*m*l^2; %denominator
• A = [0 1 0 0;
0 -(i+m*l^2)*b/p (m^2*g*l^2)/p 0;
0 0 0 1;
0 -(m*l*b)/p m*g*l*(M+m)/p 0];
B = [0; (i+m*l^2)/p; 0; m*l/p];
C = [1 0 0 0; 0 0 1 0];
D = [0;0];
p = eig(A)
Dr.R.Subasri,KEC,INDIA

20. • Eigen Values are
p = 0 -0.1428 5.5651 -5.6041
one right-half-plane pole at 5.5651,the system
is unstable in open loop.
• blue line - the cart's position
• green line - the pendulum's angle.Dr.R.Subasri,KEC,INDIA

21. State feedback controller
Open Loop:
Closed Loop:
Dr.R.Subasri,KEC,INDIA

22. State feedback controller
• t = 0:0.01:3;
• u = 0.2*ones(size(t));
• p1=-0.8+1.351i;
• p2=-0.8-1.351i;p3=-4;p4=-5
• K = place(A,B,[p1 p2,p3,p4])%%% controller
gain
• %input
• [y,x]=lsim(A-B*K,B,C,D,u,t);
• plot(t,y(:,1),'r',t,y(:,2))
Dr.R.Subasri,KEC,INDIA

23. Closed Loop response
Dr.R.Subasri,KEC,INDIA

24. Boost converter
• A=[0 -0.2 ; 40 -20]
• B=[ 0.5;0];
• C=[ 1 0];
• Pole: -3 , -3
Dr.R.Subasri,KEC,INDIA

25. Linear Quadratic Regulator (LQR) Controller
• The lqr function allows to choose two
parameters, R and Q, which will balance the
relative importance of the input and state in
the cost function to be optimized.
• The simplest case is to assume R=1, and
Q=C'*C.
• Essentially, the lqr method allows for the
control of both outputs. The controller can be
tuned by changing the nonzero elements in
the Q matrix to get a desirable response.
Dr.R.Subasri,KEC,INDIA

26. • The structure of Q, is C'*C
• 1 0 0 0
• 0 0 0 0
• 0 0 1 0
• 0 0 0 0
• The element in the (1,1) position will be used
to weight the cart's position and the element
in the (3,3) position will be used to weight the
pendulum's angle.
Find the K matrix that will give a good controller.
Dr.R.Subasri,KEC,INDIA

27. • x=1;y=1;
• Q=[x 0 0 0; 0 0 0 0; 0 0 y 0; 0 0 0 0];
• R = 1;
• K = lqr(A,B,Q,R)
• Ac = [(A-B*K)];
• Bc = [B];
• Cc = [C];
• Dc = [D]; T=0:0.01:5;
• U=0.2*ones(size(T));
• [Y,X]=lsim(Ac,Bc,Cc,Dc,U,T);plot(T,Y)
Dr.R.Subasri,KEC,INDIA

28. Closed loop response-LQR
• K = -1.0000 -1.6567 18.6854 3.4594
Dr.R.Subasri,KEC,INDIA

29. • Using x=5000 and y=100, the following value
of K and step response were found:
• K = -70.7107 -37.8345 105.5298 20.9238
All of the design requirements have been met
with the minimum amount of control effortDr.R.Subasri,KEC,INDIA

30. Thank you
Dr.R.Subasri,KEC,INDIA