Controller design of inverted pendulum using pole placement and lqr

IJRET: International Journal of Research in Engineering and Technology ISSN: 2319-1163
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Volume: 01 Issue: 04 | Dec-2012, Available @ http://www.ijret.org 532
CONTROLLER DESIGN OF INVERTED PENDULUM USING POLE
PLACEMENT AND LQR
P. Kumar1
, O.N. Mehrotra
2
, J. Mahto3
Abstract
In this paper modeling of an inverted pendulum is done using Euler – Lagrange energy equation for stabilization of the pendulum. The
controller gain is evaluated through state feedback and Linear Quadratic optimal regulator controller techniques and also the results
for both the controller are compared. The SFB controller is designed by Pole-Placement technique. An advantage of Quadratic
Control method over the pole-placement techniques is that the former provides a systematic way of computing the state feedback
control gain matrix.LQR controller is designed by the selection on choosing. The proposed system extends classical inverted
pendulum by incorporating two moving masses. The motion of two masses that slide along the horizontal plane is controllable .The
results of computer simulation for the system with Linear Quardatic Regulator (LQR) & State Feedback Controllers are shown in
section 6.
Keyword-Inverted pendulum, Mathematical modeling Linear-quadratic regulator, Response, State Feedback controller,
gain formulae.
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1. INTRODUCTION
One of the most celebrated and well – publicized problems in
control system is Inverted Pendulum[3,4] or broom balancer
problem. This is an unstable system[1,17]] that may model a
rocket before launch. Almost all known and novel control
techniques have been tested on IP problem.This is a classical
problem in dynamics and control theory[2,5] and is widely
used as a benchmark[16] for testing control algorithms(PID
controllers, Linear Quadratic Regulator (LQR), neural
networks, fuzzy logic control, genetic algorithms, etc)[7,8].The
inverted pendulum is unstable[11] in the sense that it may fall
over any time in any direction unless a suitable control force is
applied. The control objective of the inverted pendulum is to
swing up[4] the pendulum hinged on the moving cart by a
linear motor[12] from stable position (vertically down state) to
the zero state(vertically upward state)[6,9] and to keep the
pendulum in vertically upward state in spite of the
disturbance[10,13].It is highly nonlinear[12,15], but it can be
easily be controlled by using linear controllers in an almost
vertical position[18]. If the system is controllable or at leasr
stabilizable, this method gives excellent stability margins. The
guaranteed margins in LQR design are 60 degree phase margin,
infinite gain margin ,and -6dB gain reduction margin.
2. MATHEMATICAL MODEL OF PHYSICAL
SYSTEM
The inverted pendulum is a classical problem in dynamics and
control theory and is widely used as a benchmark for testing
control algorithms.
𝑥1 = 𝑥 + 𝑙 sin 𝜃
𝑦1 = 𝑙 cos 𝜃 𝜃 𝑚𝑔 𝑙
𝐹 𝑀 𝑥
Fig 1 : The Inverted Pendulum System
Let the new ordinate of the centre of gravity of the pendulum
be (x1, y1).
Define the angle of the rod from the vertical (reference) line
as θ and displacement of the cart as x. Also assume the force
applied to the system is F , g be the acceleration due to gravity
and l be the half length of the pendulum rod, v , and w be the
translational and angular velocity of the cart and pendulum.
The physical model of the system is shown in fig (1).
Therefore,
𝑥1 = 𝑥 + 𝑙 sin 𝜃
𝑦1 = 𝑙 cos 𝜃
𝑥1 = 𝑥 + 𝑙𝜃 cos 𝜃
𝑦1 = 𝑙𝜃 sin 𝜃

__________________________________________________________________________________________
Let V1 be the resultant velocity of pendulum & cart,
V1
2
= x1
2
+ y1
2
=x2 +2lxθ cos θ + l2θ2
Therefore kinetic energy of the pendulum,
k1 =
1
2
mv2
+
1
2
Iω2
=
1
2
m(x2
+2lxθ cos θ + mI2 θ2
) +
1
2
Iθ2
Kinetic energy of the of the cart,
k2 =
1
2
Mx2
Now the Potential energy of the pendulum,
P1=mgy1 = mgl cos θ
Potential energy of the cart P2 = 0
The Lagrangian of the entire system is given as,
L =kinectic energy-potential energy
L=
1
2
(mx2 +2mlx cos�+ml2
θ2+Mx2)+
1
2
Iθ2 ]-mglcos�
The Euler- Lagrangian equations are given by,
d
dt
δL
δθ
−
δL
δθ
+ dθ = 0 and
d
dt
δL
δx
−
δL
δx
+ bx = F
Simplifying the above equations we get
I + ml2
θ + ml cos θ x − mgl sin θ + dθ = 0 ………..(1)
M + m x + ml cos θ θ − ml sin θ θ2
+ bx = F ………(2)
The above equation shows the dynamics of the entire system.
In order to derive the linear differential equation modelling, we
need to linearaize the non linear differtional equation obtained
as above so far. For small angle deviation around the upright
equilibrium (fig.2) point assume
sin θ = θ, cos θ = 1, θ2
= 0
Using above relation we can write as,
rθ+qx-kθ+d =0………..(3)
px + qθ + bx = F …….(4)
Where, (M + m)= p, mgl=k, ml=q, I + ml2
= r
Eq (3&4) is the linear differential equation modeling of the
entire system.
3. STATE SPACE MODELLING
Let,θ=x1, θ=x2=x1, θ=x2=x1 and, x = x3, x=x4=x3, x=x4=x3
From state space modeling, the system matrices are found in
matrix from given below.
A=
0
4.0088
1
−0.047753952
0 0.0000000
0 0.0139155
0 0 0 1.0000000
0.116806166 0.0939155 0 0.0344200
And
B=
0
−0.27831
0
0.68842
Y=
1 0 0 0
0 0 1 0
𝑥1
𝑥2
𝑥3
𝑥4
, Y be the output equation
4. STATE FEEDBACK CONTROLLER DESIGN
CONDITIONS:
Our problem is to have a closed loop system having an
overshoot of 10% and settling time of 1 sec. Since the
overshoot
MP=e−π ξ 1− ξ2
=0.1.
Therefore, ξ = 0.591328 and wn=6.7644 rad/sec. The dominant
poles are at=-4 ± j5.45531, the third and fourth pole are placed
5 & 10 times deeper into the s-plane than the dominant poles.
Hence the desired characteristics equation:-
s4 + 68s3 +845.7604s2 +9625.6s + 36608.32 =0
Let gain, k= [k1 k2 k3 k4]
A – Bk=
A – Bk= 0 0 1 0
0 0 0 1
0 0 0 1
- k1 0.2346-k2 6.8963-k3 -0.0765- k4
Closed loop characteristics equation:
S4 +(0.0765-k4)s3 +(-6.8963 + k3)s2 +(-0.2346 +k2)s +k1 = 0
Comparing all the coefficient of above equation we found
K 1 = [36608.32 9625.8346 852.6567 -67.9235]

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Similarly if the poles are are placed 2 & 3 times and also 12 &
14 times deeper into the s-plane than the dominant poles then
we got the value of k 2 & k3 as,
K 2= [123002.88 26263.2746 2823.8263 -111.9235]
K 3= [4392.96 859.7346 308.6563 -27.9235]
Where, K1, K2, K3 and K4 are gain vectors for different sets
of desire poles.
Table 1: Parameters of the system from feedback instrument
.U.K.
Parameter Value Unit
Cart mass(𝑀) 1.206 Kilo gram
Mass of the pendulum(𝑚) 0.2693 Kilo gram
Half Length of
pendulum(𝑙)
0.1623 Meter
Coefficient of frictional
force(𝑏)
0.05 Ns/m
Pendulum damping
coefficient(q)
0.005 𝑀𝑚/rad
Moment of inertia of
pendulum(𝐼)
0.099 𝑘𝑔/𝑚2
Gravitation force(𝑔) 9.8 𝑚/𝑠2
5. LQR DESIGN
A system can be expressed in state variable form as
x = Ax + Bu
with x(t)∈R N
, u(t)∈ R N
. The initial condition is x(0) . We
assume here that all the states are measurable and seek to find a
state-variable feedback (SVFB) control
u = −Kx + v
that gives desired closed-loop properties.
The closed-loop system using this control becomes
x = (A- BK) x+ Bv= Ac x+ Bv
with Ac the closed-loop plant matrix and v the new command
input.
Ackermann's formula gives a SVFB K that places the poles of
the closed-loop system at desired location . To design a SVFB
that is optimal, we may define the performance index J as
J =
1
2
XT∞
0
Q + KT
RK Xdt
We assume that input v(t) is equal to zero since our only
concern here are the internal stability properties of the closed
loop system.
The objective in optimal design is to select the SVFB K that
minimizes the performance index J.
The two matrices Q (an n× n matrix) and R (an m × n matrix)
are of appropriate dimension.
One should select Q to be positive semi-definite and R to be
positive definite. Since the plant is linear and the performance
index is quadratic, the problem of determining the SVFB K to
minimize J is called the Linear Quadratic Regulator (LQR).To
find the optimal feedback gain matrix K we proceed as
follows. Suppose there exists a constant matrix P such that
d
dt
xT
Px = −xT
Q + KT
RK X
After some mathematical manipulation, the equation becomes,
J = −
1
2
d
dt
xT
Px dt =
1
2
∞
0
xT
0 Px(0)
Where, we assumed that the closed-loop system is stable so
that XT
goes to zero as time t goes to infinity. Substituting the
values we get,
xTPx + xT
Px + xT
Qx + xT
KT
RKx = 0
xT
AT
cPx + xT
PAcx + xT
Qx + xT
KT
RKx = 0
xT
(Ac
T
P + PAc + Q+KT
RKx)x = 0
It has been assumed that the external control v(t) is equal to
zero. Now note that the last equation has to hold for every XT
.
Therefore, the term in brackets must be identically equal to
zero. Thus, proceeding one sees that
(A − BK)T
P + P A − BK + Q + KT
RK = 0
AT
P + PA + Q + KT
RK − KT
BT
P − PBK = 0
This is a matrix quadratic equation. Exactly as for the scalar
case, one may complete the squares. Though this procedure is a
bit complicated for matrices, suppose we select
K = R−1
BT
P
Then, there results
AT
P + PA + Q − PBR−
BT
P = 0

__________________________________________________________________________________________
This result is of extreme importance in modern control theory.
The above Equation is known as the algebraic Riccati equation
(ARE). It is a matrix quadratic equation that can be solved for
the auxiliary matrix P given (A,B,Q,R).
The design procedure for finding the LQR feedback K is:
1. Select design parameter matrices Q and R
2. Solve the algebraic Riccati equation for P
3. Find the SVFB using K = 𝑅−1
𝐵 𝑇
𝑃
6 SIMULATION & RESULTS
6.1.a Response due to LQR
Fig2:- LQR response
Q=[1 0 0 0;0 1 0 0;0 0 1 0;0 0 0 10]
R=0.1000
Fig3:- LQR response
Q=[1 0 0 0;0 1 0 0;0 0 1 0;0 0 0 10]
R=0.033
Fig4:- LQR response
Q=[1 0 0 0;0 1 0 0;0 0 1 0;0 0 0 10]
R=0.044
Fig5:- LQR response
Q=[1 0 0 0;0 1 0 0;0 0 1 0;0 0 0 10]
R=0.088

__________________________________________________________________________________________
Fig6:- LQR response
Q=[1 0 0 0;0 1 0 0;0 0 1 0;0 0 0 10]
R=0.090
6.2.a-Stabilisation of Angle of the Pendulum by State
Feedback Controller with Initial Condition
K 1 = [36608.32 9625.8346 852.6567 -67.9235]
Fig7:- Response of state feedback controller considering initial
condition when poles are placed 5 & 10 times deeper into the
s-plane.
K 2= [123002.88 26263.2746 2823.8263 -111.9235]
condition when poles are placed 12 & 14 times deeper into the
s-plane.
K 3= [4392.96 859.7346 308.6563 -27.9235]
condition when poles are placed 2 & 3 times deeper into the s-
plane.
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
-0.2
0
0.2
response due to initial conditions
statevariableX1
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
-0.5
0
0.5
X2
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
-5
0
5
X3
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
-100
0
100
X4

__________________________________________________________________________________________
6.2.b-Step Response of the System by State Feedback
Controller
K 1 = [36608.32 9625.8346 852.6567 -67.9235]
Fig10:- Response of Angle as output using state feedback
controller
K 2= [123002.88 26263.2746 2823.8263 -111.9235]
controller
K 3= [4392.96 859.7346 308.6563 -27.923]
CONCLUSIONS
Modeling of inverted pendulum shows that system is unstable
with non-minimum phase zero.Results of applying state
feedback controllers show that the system can be stabilized.
while LQR controller method is cumbersome because of
selection of constants of controller. Constant of the controllers
can be tuned by some soft computing techniques for better
result.Fuzzy logic controller can be use in equation (1&2)
would help finding out the solution of non-linear differential
equations thus helping towards the design of non-linear
controller.
REFERENCES
[1]. DONGIL CHOI and Jun-Ho Oh “Human-friendly Motion
Control of a Wheeled Inverted Pendulum by Reduced-order
Disturbance Observer” 2008 IEEE International Conference on
Robotics and Automation Pasadena, CA, USA, May 19-23,
2008.
[2] Elmer P. Dadias, Patrick S. Fererandez, and David
J,”Genetic Algorithm on Line Controller For The Flexible
Inverted Pendulum Problem”, Journal Of Advanced
Computational Intelligence and Intelligent Informatics
[3] R. Murillo Garcia1, F. Wornle1, B. G. Stewart1 and D. K.
Harrison1, “Real-Time Remote Network Control of an Inverted
Pendulum using ST-RTL”, 32nd ASEE/IEEE Frontiers in
Education Conference November 6 - 9, 2002, Boston, MA.
[4] W. Wang, “Adaptive fuzzy sliding mode control for
inverted pendulum,” in Proceedings of the Second Symposium
International Computer Science and Computational
Technology(ISCSCT ’09) uangshan, P. R. China, 26-28, Dec.
pp. 231-234, 2009.
[5] Berenji HR. A reinforcement learning-based architecture
for fuzzy logic control. International Journal of Approximate
Reasoning 1992;6(1):267–92.
[6] I. H. Zadeh and S. Mobayen, “ PSO-based controller for
balancing rotary inverted pendulum, ” J. AppliedSci., vol. 16,
pp. 2907-2912 2008.
[7] I. H. Zadeh and S. Mobayen, “ PSO-based controller for
balancing rotary inverted pendulum, ” J. AppliedSci., vol. 16,
pp. 2907-2912 2008.
[8] Mohd Rahairi Rani,Hazlina Selamat,Hairi Zamzuri,”Multi
Objective Optimization For PID Controller Tuning Using The
Global Ranking Genetic Algorithm”,International Journal Of
Innovative Computing, Information and Control, VOL-8,
Number 1(A),January-2012
[9] Ohsumi A, Izumikawa T. Nonlinear control of swing-up
and stabilization
of an inverted pendulum. Proceedings of the 34th Conference
on Decision and Control, 1995. p. 3873–80.
[10] Eiben, A.E., Hinterding, R. and Michalewicz, Z.
Parameter Control in Evolutionary Algorithms. IEEE
Transactions on Evolutionary Computation, 3, 2 (1999), 124-
141.[]Stefani,Shahian,Savant,Hostetter
[11] Kumar,P, Mehro gain margin tra, O.N, Mahto, J,
Mukherjee, Rabi Ranjan,”Modelling and Controller Design of

__________________________________________________________________________________________
Inverted Pendulum”, National Conference on Communication,
Measurement and Control, Vol-I, 14th August, 2012,in press.
[12]Kumar,P, Mehrotra, O.N, Mahto, J, Mukherjee, Rabi
Ranjan,”Stabilization of Inverted Pendulum using LQR”,
National Conference on Communication, Measurement and
Control, Vol-I, 14th August, 2012 in press.
[13]Stefani,Shahian,Savant,Hostetter :Design Of Feedback
Control Systems,4th edition,New York, Oxford University
Press 2002 ,Page (675 – 732)
[14] Behra Laxmidhar & Kar Indrani; Intelligent Systems and
Control Principals and Applications; Oxford University Press
[15]Ogata, K.; System Dynamics, 4th Edition Englewood
Cliffs, NJ: Prentice-Hall, 2003.
[16]Goldberg, D.E. Genetic Algorithms in search,
optimization, and machine learning. Reading, Mass. : Addison-
Wesley, 1989.
[17] feedback instrument,U.K
BIOGRAPHIES
Mehrotra, a Gold Medalist at B.Sc.
Engineering (B.U), M.E.(Hons)(U.O.R)
and Ph.D. (R.U) all in Electrical
Engineering, has the industrial exposure
at SAIL as Testing & Commissioning
Engineer. Served Department of Science
& Technology, Govt. of Bihar & Govt. of Jharkhand for 35
years and retired as Professor in Electrical Engineering. Served
as coordinator of various projects sanctioned through MHRD
and AICTE, including TEQIP, a World Bank Project. His
research interests include control and utilization of renewable
energies, power quality and power system.
Jagdeo Mahto was born in Madhubani,
Bihar, India, in 1943. He obtained the
B.Sc (Engg) degree in Electrical
Engineering from Bhagalpur University in
1964, M.Tech. in Control System from IIT
Kharagpur, India in 1970 and Ph.D in
Control System in 1984 from IIT Delhi, India. He served MIT
Muzaffarpur from 1964 to 1971 in the capacity of Lecturer and
Assistant Professor. From 1971 to 1980 he served as Asst.
Professor, from 1980 to 1985 as Associate Professor and from
1985 to 1988 as Professor in the Department of Electrical
Engineering at BIT Sindri, India. He taught at Bright Star
University, Brega (Libya) from 1988 to 1989. From 1989 to
2003 he was again at BIT Sindri. From 2004 till date he is
Professor at Asansol Engineering College, India.
Pankaj Kumar was born in Muzaffarpur,
India, in 1970 and received the B.Sc and
M.Sc. degree in Electronics Honours and
Electronic Science respectively from
Magadh University and Gauhati University
Assam. He received M.Sc Engineering in
Control System Engineering from Patna University in 2004. He
began his career as Lecturer in Bihar University Muzaffarpur.
Currently he is an Assistant Professor in the Department of
Electrical Engineering, Asansol Engineering College, Asansol,
India.

Controller design of inverted pendulum using pole placement and lqr

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Controller design of inverted pendulum using pole placement and lqr