30. Figure P3.17 shows a free-body diagram of an an inverted pendulum, mounted o

Question

30. Figure P3.17 shows a free-body diagram of an an inverted pendulum, mounted on a cart with a mass, M. The pendulum has a point mass, m, concentrated at the upper end of a rod with zero mass, a length, l, and a frictionless hinge. A motor drives the cart, applying a horizontal force, u(t). A gravity force, mng, acts on m at all times. The pendulum angle relative to the y-axis, 0, its angular speed, 6, the horizontal position of the cart, its speed, x were to be the state variables. The state-space equations derived were heavily nonlinear 14 They were then linearized around the stationary point, xo 0 and uo 0, and manipulated to yield the following open-loop model written in perturbation form dt However, since xo 0 and uo 0, then let: x xo ox Ox and u uo Ou ou. Thus the state equation may be rewritten as (Prasad, 2012): x Ax Bu where 1 0 01 (MV m)g 0 0 0 Ml MI and B 0 0 1 mg 0 0 0 L M

Explanation / Answer

clear all
clc
M=2.4;% all the values in S.I unit in kg
m=0.23;% all the values in S.I unit in kg
l=0.36;% all the values in S.I unit in meter
g=9.81;% all the values in S.I unit in m/s^2

syms s % defining the variable s

A=[0 1 0 0;((M+m)*g/(M*l)) 0 0 0; 0 0 0 1;((-m*g)/M) 0 0 0];
B=[0;(-1/(M*l));0;(1/M)];
C=[l 0 1 0];
D=0;

Phi=inv(s*eye(4)-A);% finding the state transition matrix here eye(4)will produce identity
% matrix of order 4 by 4

G=C*Phi*B+D;% formula of transfer function from state variable

G1=(simplify(G));% simplified value of transfer function G;
display(G1)

copy paste the code and u will see the transfer function

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