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Dynamic Modeling and Analysis of Inverted Pendulum Using Lagrangian-Differential Transform Method

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Dynamic Modeling and Analysis of Inverted Pendulum Using Lagrangian-Differential Transform Method Proceedings of the World Congress on Engineering 2017 Vol II WCE 2017, July 5-7, 2017, London, U.K. Dynamic Modeling and Analysis of Inverted Pendulum using Lagrangian-Differential Transform Method Michael C. Agarana, Member, IAENG, and Oluseyi O. Ajayi one's finger. The inverted pendulum is a classic Abstract— There has been an increase in application of problem in dynamics and control theory and is pendulum in robotics which is applicable in Medicine, Agriculture, Military, Industries Explorations and used as a benchmark for testing control Entertainment. Many Researchers have shown interest in strategies[3,4]. Differential transform method is a inverted pendulum stability and control in recent years. This semi analytical method which requires no study is concerned with the development of equation of motion of inverted pendulum and its analysis by adopting the discretization [4,5]. One of the advantages of Lagrangian and Differential Transform Method (DTM) DTM over others is that it is simple to apply respectively. The resulting equations were solved analytically. [5,6,7]. The solution of the governing equation is Results obtained are consistent with the ones in the literature. Index Terms—Applied Mechanics, Dynamics Differential very close to the analytical solutions [8]. The transform method. Inverted pendulum, Lagrangian. Lagrangian is a mathematical function of the generalized coordinates, their time derivatives, I. INTRODUCTION and time. It contains information about the dynamics of the system. Lagrangian mechanics is A N ordinary pendulum is one with the pivot at ideal for systems with conservative forces and for the top and the mass at the bottom. An bypassing constraint forces in any coordinate inverted pendulum is the opposite way round. The system. Dissipative and driven forces can be pivot is at the bottom and the mass is on top. The accounted for by splitting the external forces into inverted pendulum is a mechanism for carrying an a sum of potential and non-potential forces [9]. object form one place to another and this is how it When Newton's formulation of classical functions during walking. The ―passenger unit‖ as mechanics is not convenient, Lagrangian Perry would call it is carried forward by the mechanics is widely used to solve mechanical outstretched leg as it pivots over the foot [1]. So problems in physics and engineering. Lagrangian an inverted pendulum is a pendulum that has its mechanics applies to the dynamics of particles, centre of mass above its pivot point. It is often and used in optimisation problems of dynamic implemented with the pivot point mounted on a systems. Lagrangian mechanics uses the energies moving body that can move horizontally. Unlike a (both kinetic and potential) in the system instead normal pendulum which is always stable when of forces. Thus a function that summarizes the hanging downwards, an inverted pendulum is dynamics of an entire system is simply known as inherently unstable, and must be actively balanced the Lagrangian. Moreover, the Kinetic energy in order to remain upright. This can be done in reflects the energy of the system's motion. The different ways including the pivot point potential energy of the system reflects the energy horizontally as part of a feedback system [2,3]. A of interaction between the particles, i.e. how much simple demonstration of this is achieved by energy any one particle will have due to all the balancing an upturned broomstick on the end of others and other external influences.[9] Manuscript received March 30, 2017; revised April 10, 2017.. The purpose of this study was to use differential This work was supported in full by Covenant University. Dynamic transformation algorithm to solve the ordinary Modeling and Analysis of Inverted Pendulum using Lagrangian- second order differential equation resulted from Differential Transform Method the deployment of Lagrangian to model the M.C. Agarana is with the Department of Mathematics, Covenant University,Nigeria,+2348023214236,michael.agarana@covenantu Kinematics of motion of inverted pendulum. DTM niversity.edu.ng was then applied to solve the model so formed.[9] O.O. Ajayi is with the Department of Mechanical Engineering, The dynamic effects of the pendulum through its Covenant University, Nigeria angular displacement, angular velocity and ISBN: 978-988-14048-3-1 WCE 2017 ISSN: 2078-0958 (Print); ISSN: 2078-0966 (Online) Proceedings of the World Congress on Engineering 2017 Vol II WCE 2017, July 5-7, 2017, London, U.K. angular acceleration were estimated over a period The kinetic energy Ek of a point object is defined of time. by its mass m and velocity v: 2 Ek = 1/2mv (3) Equation of motion can be directly derived by substitution using Euler Lagrange equation: d L L dt (4) where is the angle the pendulum makes with the upward vertical. Fig. 1 The angular displacement of the inverted pendulum Equation (2) can be written as Ep =mgy (5) Where h = y, is the height along vertical axis From the figures 2a and 2b above, x = Lsin (6) y = Lcos (7) This implies dx d Lcos ( ) (8) dt dt Fig. 2 The vertical and horizontal components of the length of the inverted pendulum and dy d Fig. 1 represents an inverted pendulum, while Fig. Lsin ( ) (9) 2 is the schematic of the inverted pendulum, dt dt showing the vertical and horizontal components of From equation (1), the Lagrangian can be written as the length of the inverted pendulum. represents the angle the inverted pendulum makes with the vertical. 1 L = mv2 - mgy (10) 2 II. EQUATION OF MOTION where h = y Using Lagrange's equations, which employ a Velocity as a vector quantity is given as single scalar function rather than the vector 22 components, to derive the equations modeling an 2 dx dy v = + (11) inverted pendulum, we take partial derivatives. In dt dt classical mechanics, the natural form of the Substituting equations (6) and (7) into Lagrangian is defined as equation (11) gives: 22 2 2dd 2 2 2 L = Ek – Ep (1) v = L cos L sin (12) dt dt Where EK and EP are kinetic and potential energies are respectively. Ep is defined by its mass m, and the gravitational constant g: Ep = mgh (2) ISBN: 978-988-14048-3-1 WCE 2017 ISSN: 2078-0958 (Print); ISSN: 2078-0966 (Online) Proceedings of the World Congress on Engineering 2017 Vol II WCE 2017, July 5-7, 2017, London, U.K. 2 unknown function by using the initial data on the 2 2d 2 2 v = L cos sin (13) problem. The concept of differential transform dt method was first proposed by Zhou [10,11]. 2 22d v = L (14) dt B. Definitions and Properties of DTM Now substituting equations (7) and (14) into With reference to article [4, 5, 6] the differential equation (10) gives: transformation of a function f(x) can be written as: 2 k 1 2 d 1d f ( x ) L = mL mgLcos (15) 2 dt F( t ) x 0 (23) K! dx which implies L Where f(x) is referred to as the original function mgL sin (16) and F(x), the transformed function. The differential inverse of F(t) is Ld mL2 (17) d dt () k dt f( x ) F ( t ) x (24) k 0 d L d 2 Equation (26) is the Tailor series expansion of f(x) mL2 (18) d 2 about x=0. It can also be written as dt() dt dt xkk d f() x Substituting equation (16) and (18) into equation fx( ) (25) k! dx (4) gives: k0 d 2 Some operational properties of DTM are as mL2 mgL sin (19) dt 2 follows: Therefore i If f(x) = p(x) g(x), then F(t) = p(t) F(t) (26) dg2 ii If f(x) = cp(x), then F(t) = cP(t), where c is a 2 sin (20) dt L constant. (27) n Without oscillator and taking g =10 and L =5 d p( x ) ( k n )! iii If f(x) = n , then F(t) = p( k n ) (28) gives: dx k akk iv If f(x) = sin(ax + b), then F(x) = sin b (29) (t ) 2sin (21) k!2 subject to k v f(x) = ex , then F( t ) , where is a constant (30) (0) = 0, ' (0) 2 (22) k! C. Differential Transform Method on Second III MODEL SOLUTION Order ODE) To solve equation (20), is solved for using Second order linear differential equations can be Differential transform method (DTM). However used as follows [4] equation (20) is a linear second order differential equation solvable by applying differential 1 2 transform method. f()[()()] x r x cy x by x a (31) A. Differential Transformation Method (DTM) Applying the above properties, equation (31) Differential transform method is a numerical becomes: [4] method based on Taylor expansion. This method tries to find coefficients of series expansion of ISBN: 978-988-14048-3-1 WCE 2017 ISSN: 2078-0958 (Print); ISSN: 2078-0966 (Online) Proceedings of the World Congress on Engineering 2017 Vol II WCE 2017, July 5-7, 2017, London, U.K. 1 Fk(2) [()(1)(1) RxbkFk cFk ()] For k=5 a( k 1)( k 2) 1 Y(7) [2sin( 0.00002)] 0.00000002 (43) (32) 42 which is subject to the initial condition that: And so on. 1 F(0) = F0 = a1 and F (0) = F(1) = a2 (33) Genera lly speaking, n = k+2, where k=0,1,2,3,… Equation (33) is called recursive formula. Therefore, Using equation (33) on equation (32) gives the Y(n) = 0, -0.01163, 0, -0.00002, 0, -0.00000002 appropriate solution.[4] … From equation (24): IV. APPLICATION OF DTM f( x ) F ( t ) xk (44) k 0 To apply the discussed DTM to equation (20), equations (32) and (33) are re-written as: y( x ) Y ( t ) xk (45) k 0 y y(0) y (1) x y (2) x2 y (3) x 3 y (4) x 4 DT[(t ) 2sin ] (34) y(5) x5 y (6) x 6 y (7) x 7 ..
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