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EE380-1 Design Project Control of an Inverted Pendulum on a Cart Dr. Ahmad A. Masoud, Semester 122, 2013 The project is due a week before the last lecture in the semester. The cart with an inverted...

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EE380-1 Design Project
Control of an Inverted Pendulum on a Cart
Dr. Ahmad A. Masoud, Semester 122, 2013
The project is due a week before the last lecture in the semester.
The cart with an inverted pendulum, shown below, is "bumped" with an impulse force, F. The dynamic equations
of motion for the system are:
( )???? ?? ???? cos( ) ?? sin( )
( )???? sin( ) ???? cos( )
M mX bX m l m l F
I m l m g l m l X
+ + + · · · - · · =
+ · + · · · = - · · ·
? ? ? ?
? ? ?
2
2
M mass of the cart 0.5 kg
m mass of the pendulum 0.5 kg
b friction of the cart 0.1 N/m/sec
l length to pendulum center of mass 0.3 m
I inertia of the pendulum 0.006 kg*m^2
F force applied to the cart
x cart position coordinate
theta pendulum angle from vertical
1- Linearize the above dynamic equations about the pendulum's angle, theta = Pi (in other words, assume that
pendulum does not move more than a few degrees away from the vertical, chosen to be at an angle of B) and
obtain the transfer function of the system,
2- Assume that the system starts at equilibrium, and experiences an impulse force of 1N. Design a PID controller
that returns the pendulum to its upright position within 5 seconds, and never move more than 0.05 radians away
from the vertical position.
3- Suggest a control scheme that would keep the pendulum in an upright position and x around zero after the
impulse force is applied. Demonstrate your controller with one working example.
You are expected to hand in an organized structured report
with all the necessary drawings, derivations and simulations.
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EE380-1 Design Project Control of an Inverted Pendulum on a Cart Dr. Ahmad A. Masoud, Semester 122, 2013 The project is due a week before the last lecture in the semester. The cart with an inverted pendulum, shown below, is "bumped" with an impulse force, F. The dynamic equations of motion for the system are: 2  () Mm++XbX+m·l·?? · cos( )-m·l·? sin(? )=F 2 () Im +·l?? +·mg·l·sin( )=-m·l·X· cos(? ) M mass of the cart 0.5 kg m mass of the pendulum 0.5 kg b friction of the cart 0.1 N/m/sec l length to pendulum center of mass 0.3 m I inertia of the pendulum 0.006 kg*m^2 F force applied to the cart x cart position coordinate theta pendulum angle from vertical 1- Linearize the above dynamic equations about the pendulum's angle, theta = Pi (in other words, assume that pendulum does not move more than a few degrees away from the vertical, chosen to be at an angle of B) and obtain the transfer function of the system, 2- Assume that the system starts at equilibrium, and experiences an impulse force of 1N. Design a PID controller that returns the pendulum to its upright position within 5 seconds, and never move more than 0.05 radians away from the vertical position. 3- Suggest a control scheme that would keep the pendulum in an upright position and x around zero after the impulse force is applied. Demonstrate your controller with one working example. You are expected to hand in an organized structured report with all the necessary drawings, derivations and simulations.    

Answered Same Day Dec 23, 2021

Solution

David answered on Dec 23 2021
130 Votes
Sol: System Dynamics
The system consists of an inverted pole on a cart which is free to move in
the x direction as shown in figure (1).
Fig (1) Cart and Inverted Pendulum System
Table.1 Parameters of the inverted pendulum
For the analysis, Newton’s second law of motion is applied. Fig (2)
epresents the free body diagram (FBD) of the mechanism.
The force distribution of the mechanism is shown in Fig (2). While the
pendulum rod tilts with some angle, it resolves two force components along
horizontal and vertical direction. ‘P’ denotes the force in vertical direction,
and ‘N’ in horizontal direction, when
090 , N 0, and P maximum.   
M Mass of the cart 0.5 kg
m Mass of the pendulum 0.5 kg
b Friction of the cart 0.1 N/m/sec
l Length to pendulum center of mass 0.3 m
I Inertia of the pendulum 0.006 kg*m^2
F Force applied to the cart 2kg m.s
X Cart position coordinate 9.8
2m s
 Pendulum angle from vertical In degree
Free body diagram of the inverted pendulum
From the Free body diagram, summing all the forces of the cart in horizontal
direction
M x b x N F
 
   …………………….. (1),
Following equation for N is obtained:
2
cos sinN m x ml ml   
  
   …………………. (2),
After substituting equation (2) into equation (1), the equation of motion for
 
2
cos sinM m x b x ml ml F   
   
     ……………. (3),
The forces along in the...
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