Class 34 - 35 Modeling Of A Inverted Pendulum

System Modeling Coursework

Class 34-35: Modeling of Inverted Pendulum

P.R. VENKATESWARAN Faculty, Instrumentation and Control Engineering, Manipal Institute of Technology, Manipal Karnataka 576 104 INDIA Ph: 0820 2925154, 2925152 Fax: 0820 2571071 Email: [email protected], [email protected] Web address: http://www.esnips.com/web/SystemModelingClassNotes

WARNING! • I claim no originality in all these notes. These are the compilation from various sources for the purpose of delivering lectures. I humbly acknowledge the wonderful help provided by the original sources in this compilation. • For best results, it is always suggested you read the source material. July – December 2008

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Contents • Dynamics of Inverted Pendulum • Transfer function of Inverted Pendulum

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Problem statement • The cart with an inverted pendulum, shown below, is "bumped" with an impulse force, F. • Determine the dynamic equations of motion for the system, and linearize 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 Pi). • Find a controller to satisfy all of the design requirements given below. July – December 2008

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System Diagram

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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

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Force analysis

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System equations • Summing the forces in the Free Body Diagram of the cart in the horizontal direction, you get the following equation of motion: • Note that you could also sum the forces in the vertical direction, but no useful information would be gained. • Summing the forces in the Free Body Diagram of the pendulum in the horizontal direction, you can get an equation for N:

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System equations • If you substitute this equation into the first equation, you get the first equation of motion for this system: • To get the second equation of motion, sum the forces perpendicular to the pendulum. Solving the system along this axis ends up saving you a lot of algebra. You should get the following equation:

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System equations • To get rid of the P and N terms in the equation above, sum the moments around the centroid of the pendulum to get the following equation: • Combining these last two equations, you get the second dynamic equation:

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System equations • The equations should be linearized about theta = Pi. Assume that theta = Pi + ø (ø represents a small angle from the vertical upward direction). Therefore, cos(theta) = -1, sin(theta) = -ø, and (d(theta)/dt)^2 = 0. • After linearization the two equations of motion become (where u represents the input):

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Transfer function • To obtain the transfer function of the linearized system equations analytically, we must first take the Laplace transform of the system equations. The Laplace transforms are: • Since we will be looking at the angle Phi as the output of interest, solve the first equation for X(s),

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Transfer function • then substituting into the second equation:

• Re-arranging, the transfer function is:

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Transfer function • From the transfer function above it can be seen that there is both a pole and a zero at the origin. These can be canceled and the transfer function becomes:

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And, before we break… A monk was imprisoned. Within a week, he was to be killed. In the prison, he heard a beautiful verse from a scripture sung by his co-prisoner. He requested him to teach that verse. The co-prisoner asked, “ what is the purpose of learning if you are going to die within one week?” The monk answered: “Exactly for the same reason you learn something if you are going to die within forty five years”. Thanks for listening… July – December 2008

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