inverted pendulam

8/3/2019 Inverted Pendulam

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INVERTED PENDULUM FOR SELF BALANCING IN ROBOTS.

Aim:

The goal of this project was to build and implement an inverted pendulum balancer, in the vertical two

dimensional plane, using Proportional-Integral-Derivative (PID) feedback control.

Abstract:

An inverted pendulum is a pendulum which has its mass above its pivot point. It is often implemented

with the pivot point mounted on a cart that can move horizontally and may be called a cart and pole.

Whereas a normal pendulum is stable when hanging downwards, an inverted pendulum is inherently

unstable, and must be actively balanced in order to remain upright, either by applying a torque at the pivot

point or by moving the pivot point horizontally as part of a feedback system.

We return to the history of the seismographs themselves. In 1898, E. Wiechert of Gttingen introduced a

seismograph with a viscously-damped pendulum as a sensor (Wiechert, 1899). The damping was added to

lessen the effects of the pendulum eigen-oscillations. Wiechert's first seismograph was a horizontal-

pendulum instrument, which recorded photographically. After a trip to Italy to study seismometers usedin that country, he decided to build a mechanically-recording seismograph. For a sensor, he used an

inverted pendulum stabilized by springs and free to oscillate in any direction horizontally (Wiechert,

1904). The seismograph was completed in 1900.

The inverted pendulum is a classic problem in dynamics and control theory and is widely used as a

benchmark for testing control algorithms (PID controllers, neural networks, fuzzy control, geneticalgorithms, etc.). Variations on this problem include multiple links, allowing the motion of the cart to be

commanded while maintaining the pendulum, and balancing the cart-pendulum system on a see-saw. The

inverted pendulum is related to rocket or missile guidance, where the center of gravity is located behind

the center of drag causing aerodynamic instability. The understanding of a similar problem can be shown

by simple robotics in the form of a balancing cart. Aninertia wheel pendulum orgyroscopic


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pendulum is a pendulum with an inertia wheel attached. It can be used as a pedagogical problem

in control theory.

The Segway PT is a two-wheeled, self-balancing transportation machine invented by Dean Kamen.

Computers and motors in the base of the device keep the Segway PT upright when powered on with

balancing enabled. A user commands the Segway to go forward by shifting their weight forward on the

platform, and backward by shifting their weight backward. The Segway notices, as it balances, the change

in its center of mass, and first establishes and then maintains a corresponding speed, forward or

backward. Gyroscopic sensors and fluid-based leveling sensors are used to detect the shift of weight. To

turn, the user manipulates a control on the handlebar left or right.

Segway PTs are driven by electric motors and can go up to 12.5 miles per hour (20.1 km/h).

Problem setup and design requirements

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 otherwords, 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.

For this example, let's assume that

M mass of the cart 0.5 kg

m mass of the pendulum 0.2 kg

b friction of the cart 0.1 N/m/secl 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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For the PID, root locus, and frequency response sections of this problem we will be only interested in the

control of the pendulum's position. This is because the techniques used in these tutorials can only be

applied for a single-input-single-output (SISO) system. Therefore, none of the design criteria deal with

the cart's position. For these sections we will assume that the system starts at equilibrium, and experiences

an impulse force of 1N. The pendulum should return to its upright position within 5 seconds, and nevermove more than 0.05 radians away from the vertical.

The design requirements for this system are:

Settling time of less than 5 seconds.

Pendulum angle never more than 0.05 radians from the vertical.

However, with the state-space method we are more readily able to deal with a multi-output system.

Therefore, for this section of the Inverted Pendulum example we will attempt to control both the

pendulum's angle and the cart's position. To make the design more challenging we will be applying a step

input to the cart. The cart should achieve its desired position within 5 seconds and have a rise time under0.5 seconds. We will also limit the pendulum's overshoot to 20 degrees (0.35 radians), and it should also

settle in under 5 seconds.

The design requirements for the Inverted Pendulum state-space example are:

Settling time for x and theta of less than 5 seconds.

Rise time for x of less than 0.5 seconds.

Overshoot of theta less than 20 degrees (0.35 radians).

Force analysis and system equations

Below are the two Free Body Diagrams of the system.

Summing the forces in the Free Body Diagram of the cart in the horizontal direction, you get the

following equation of motion:


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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:

If you substitute this equation into the first equation, you get the first equation of motion for this system:

(1)

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:

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:

(2)

Since MATLAB can only work with linear functions, this set of equations should be linearized about

theta = Pi. Assume that theta = Pi + ( represents a small angle from the vertical upward direction).

Therefore, cos(theta) = -1, sin(theta) = - , and (d(theta)/dt)^2 = 0. After linearization the two equations

of motion become (where u represents the input):

1. Transfer Function

To obtain the transfer function of the linearized system equations analytically, we must first take theLaplace transform of the system equations. The Laplace transforms are:


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Since we will be looking at the angle Phi as the output of interest, solve the first equation for X(s),

then substitute into the second equation, and re-arrange. The transfer function is:

where,

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:

2. State-Space

After a little algebra, the linearized system equations equations can also be represented in state-space

form:


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The C matrix is 2 by 4, because both the cart's position and the pendulum's position are part of the output.

For the state-space design problem we will be controlling a multi-output system so we will be observing

the cart's position from the first row of output and the pendulum's with the second row.

MATLAB representation and the open-loop response

1. Transfer Function

The transfer function found from the Laplace transforms can be set up using MATLAB by inputting the

numerator and denominator as vectors. Create an m-file and copy the following text to model the transferfunction:

M = .5;

m = 0.2;

b = 0.1;

i = 0.006;

g = 9.8;

l = 0.3;

q = (M+m)*(i+m*l^2)-(m*l)^2; %simplifies input

num = [m*l/q 0];

den = [1 b*(i+m*l^2)/q -(M+m)*m*g*l/q -b*m*g*l/q];

pend=tf(num,den)

Your output should be:

Transfer function:4.545 s

----------------------------------s^3 + 0.1818 s^2 - 31.18 s - 4.455

To observe the system's velocity response to an impulse force applied to the cart add the following lines

at the end of your m-file:t=0:0.01:5;

impulse(pend,t)

axis([0 1 0 60])

You should get the following velocity response plot:


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As you can see from the plot, the response is entirely unsatisfactory. It is not stable in open loop. You can

change the axis to see more of the response if you need to convince yourself that the system is unstable.

Although the output amplitude increases past 60 radians (10 revolutions), the model is only valid for

small . In actuality, the pendulum will stop rotating when it hits the cart ( =90 degree).

2. State-Space

Below, we show how the problem would be set up using MATLAB for the state-space model. If you copythe following text into a m-file (or into a '.m' file located in the same directory as MATLAB) and run it,

MATLAB will give you the A, B, C, and D matrices for the state-space model and a plot of the response

of the cart's position and pendulum angle to a step input of 0.2 m applied to the cart.M = .5;

m = 0.2;

b = 0.1;

i = 0.006;

g = 9.8;

l = 0.3;

p = i*(M+m)+M*m*l^2; %denominator for the A and B matrices

A = [0 1 0 0;

0 -(i+m*l^2)*b/p (m^2*g*l^2)/p 0;

0 0 0 1;

0 -(m*l*b)/p m*g*l*(M+m)/p 0]B = [ 0;

(i+m*l^2)/p;

0;m*l/p]

C = [1 0 0 0;

0 0 1 0]

D = [0;

0]


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pend=ss(A,B,C,D);

T=0:0.05:10;

U=0.2*ones(size(T));

[Y,T,X]=lsim(pend,U,T);

plot(T,Y)axis([0 2 0 100])

You should see the following output after running the m-file:

The blue line represents the cart's position and the green line represents the pendulum's angle. It is

obvious from this plot and the one above that some sort of control will have to be designed to improve the

dynamics of the system. Several example controllers are included with these tutorials; select from belowthe one you would like to use.

Note: The solutions shown in the PID, root locus and frequency response examples may not yield a

workable controller for the inverted pendulum problem. As stated previously, when we put this problem

into the single-input, single-output framework, we ignored the x position of the cart. The pendulum canbe stabilized in an inverted position if the x position is constant or if the cart moves at a constant velocity

(no acceleration). Where possible in these examples, we will show what happens to the cart's position

when our controller is implemented on the system. We emphasize that the purpose of these examples is to

demonstrate design and analysis techniques using MATLAB; not to actually control an inverted

pendulum.

Example: Solution to the Inverted Pendulum Problem Using PID Control

Open-loop RepresentationClosed-loop transfer function

Adding the PID controller

What happens to the cart's position?

The transfer function of the plant for this problem is given below:


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where,

The design criteria (with the pendulum receiving a 1N impulse force from the cart) are:


Pendulum should not move more than 0.05 radians away from the vertical.

To see how this problem was originally set up, consult the inverted pendulum modeling page.

Open-loop Representation

The first thing to do when using PID control in MATLAB is to find the transfer function of the system

and to check to see if it makes sense. The transfer function found from the Laplace transforms for the

output Phi (the pendulum's angle) can be set up using MATLAB by entering the numerator and

denominator as vectors. Create anm-file (or a '.m' file located in the same directory as MATLAB) and

copy the following text to model the transfer function:

M = .5;

m = 0.2;

b = 0.1;

i = 0.006;

g = 9.8;

l = 0.3;


num = [m*l/q 0];

den = [1 b*(i+m*l^2)/q -(M+m)*m*g*l/q -b*m*g*l/q];pend=tf(num,den);

Closed-loop transfer function

The control of this problem is a little different than the standard control problems you may be used to.

Since we are trying to control the pendulum's position, which should return to the vertical after the initial

disturbance, the reference signal we are tracking should be zero. The force applied to the cart can be

added as an impulse disturbance. The schematic for this problem should look like the following.


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It will be easier to determine the appropriate transfer function to enter into MATLAB if we first rearrange

the schematic as follows:

Adding the PID controller

This closed-loop transfer function can be modeled in MATLAB by copying the following code to the end

of your m-file:

Kd = 1;

Kp = 1;

Ki = 1;

contr=tf([Kd Kp Ki],[1 0]);

sys_cl=feedback(pend,contr);

This transfer function assumes that both derivative and integral control will be needed along withproportional control. This does not have to be the case. If you wanted to start with a PI or PD controller,

just replace the contr line with your choice of controller as shown below.

contr=tf([Kp Ki],[1 0]); % PI control

contr=tf([Kd Kp],1); % PD control

Now we can begin the actual control of this system. First let's see what the impulse response looks like

with the numbers we already have. Enter the following code to the end of your m-file:

t=0:0.01:5;

impulse(sys_cl,t)

axis ([0 1.5 0 40])

You should get the following velocity response plot from the impulse disturbance:
http://www.library.cmu.edu/ctms/ctms/extras/plot.htm#axishttp://www.library.cmu.edu/ctms/ctms/extras/plot.htm#axis


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This response is still not stable. Let's start by increasing the proportional control to the system. Increase

the Kp variable to see what effect it has on the response. If you set Kp=100, and set the axis to axis([0,

2.5, -0.2, 0.2]), you should get the following velocity response plot:

The settling time is acceptable at about 2 seconds. Since the steady-state error has already been reduced to

zero, no more integral control is needed. You can remove the integral gain constant to see for yourself

that the small integral control is needed. The overshoot is too high, so that must be fixed. To alleviate this


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problem, increase the Kd variable. With Kd=20, you should get a satisfactory result. You should now see

the following velocity response plot:

As you can see, the overshoot has been reduced so that the pendulum does not move more than 0.05

radians away from the vertical. All of the design criteria have been met, so no further iteration is needed.

What happens to the cart's position?

At the beginning on this solution page, the block diagram for this problem was given. The diagram wasnot entirely complete. The block representing the the position was left out because that variable was not

being controlled. It is interesting though, to see what is happening to the cart's position when the

controller for the pendulum's angle is in place. To see this we need to consider the actual system block

diagram:

Rearranging a little bit, you get the following block diagram:


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The feedback loop represents the controller we have designed for the pendulum's. The transfer function

from the cart's position to the impulse force, with the PID feedback controller which we designed, is

given as follows:

Now that we have the transfer function for the entire system, let's take a look at the response. First we

need the transfer function for the cart's position. To get this we need to go back to the Laplace transforms

of the system equations and find the transfer function from X(s) to U(s). Below is this transfer function:

where,

For more about the Laplace transform please refer to the inverted pendulum modeling page.

The pole/zero at the origin cancelled out of the transfer function for Phi, has been put back in. So that

now den1 = den2, making calculations easier. Now, create a new m-file and run it in the command

window:

M = .5;

m = 0.2;b = 0.1;

i = 0.006;

g = 9.8;

l = 0.3;


num1 = [m*l/q 0 0];


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den1 = [1 b*(i+m*l^2)/q -(M+m)*m*g*l/q -b*m*g*l/q 0];

G1=tf(num1,den1);

num2 = [(i+m*l^2)/q 0 -m*g*l/q];

den2 = den1;G2=tf(num2,den2);

kd = 20;

kp = 100;

ki = 1;

contr=tf([kd kp ki],[1 0]);

xpos=feedback(1,G1*contr)*G2;

t=0:0.01:5;

impulse(xpos,t)

As you can see, the cart moves in the negative direction with a constant velocity. So although the PID

controller stabilizes the angle of the pendulum, this design would not be feasible to implement on an

actual physical system.

Modeling an Inverted Pendulum in Simulink


Force analysis and system equation setup

Building the model

Open-loop response

Extracting a linearized model

Implementing PID control

Closed-loop response
http://www.library.cmu.edu/ctms/ctms/simulink/examples/pend/pendsim.htm#Problemhttp://www.library.cmu.edu/ctms/ctms/simulink/examples/pend/pendsim.htm#equationshttp://www.library.cmu.edu/ctms/ctms/simulink/examples/pend/pendsim.htm#Buildhttp://www.library.cmu.edu/ctms/ctms/simulink/examples/pend/pendsim.htm#Openhttp://www.library.cmu.edu/ctms/ctms/simulink/examples/pend/pendsim.htm#Extracthttp://www.library.cmu.edu/ctms/ctms/simulink/examples/pend/pendsim.htm#Controlhttp://www.library.cmu.edu/ctms/ctms/simulink/examples/pend/pendsim.htm#Closedhttp://www.library.cmu.edu/ctms/ctms/simulink/examples/pend/pendsim.htm#Problemhttp://www.library.cmu.edu/ctms/ctms/simulink/examples/pend/pendsim.htm#equationshttp://www.library.cmu.edu/ctms/ctms/simulink/examples/pend/pendsim.htm#Buildhttp://www.library.cmu.edu/ctms/ctms/simulink/examples/pend/pendsim.htm#Openhttp://www.library.cmu.edu/ctms/ctms/simulink/examples/pend/pendsim.htm#Extracthttp://www.library.cmu.edu/ctms/ctms/simulink/examples/pend/pendsim.htm#Controlhttp://www.library.cmu.edu/ctms/ctms/simulink/examples/pend/pendsim.htm#Closed


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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 = 0 (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 0). Find a controller to satisfy all of the design requirements given below.

For this example, let's assume that

M mass of the cart 0.5 kg

m mass of the pendulum 0.2 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

In this example, we will implement a PID controller which can only be applied to a single-input-single-

output (SISO) system, so we will be only interested in the control of the pendulums angle. Therefore,

none of the design criteria deal with the cart's position. We will assume that the system starts at

equilibrium, and experiences an impulse force of 1N. The pendulum should return to its upright position

within 5 seconds, and never move more than 0.05 radians away from the vertical.

The design requirements for this system are:


Pendulum angle never more than 0.05 radians from the vertical.

Force analysis and system equation setup

Below are the two Free Body Diagrams of the system.


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This system is tricky to model in Simulink because of the physical constraint (the pin joint) between the

cart and pendulum which reduces the degrees of freedom in the system. Both the cart and the pendulum

have one degree of freedom (X and theta, respectively). We will then model Newton's equation for these

two degrees of freedom.

It is necessary, however, to include the interaction forces N and P between the cart and the pendulum in

order to model the dynamics. The inclusion of these forces requires modeling the x and y dynamics of the

pendulum in addition to its theta dynamics. In the MATLAB tutorial pendulum modeling example the

interaction forces were solved for algebraically. Generally, we would like to exploit the modeling power

of Simulink and let the simulation take care of the algebra. Therefore, we will model the additional x and

y equations for the pendulum.

However, xp and yp are exact functions of theta. Therefore, we can represent their derivatives in terms of

the derivatives of theta.


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These expressions can then be substituted into the expressions for N and P. Rather than continuing with

algebra here, we will simply represent these equations in Simulink.

Simulink can work directly with nonlinear equations, so it is unnecessary to linearize these equations as it

was in the MATLAB tutorials.

Building the Model in Simulink

First, we will model the states of the system in theta and x. We will represent Newton's equations for the

pendulum rotational inertia and the cart mass.

Open a new model window in Simulink, and resize it to give plenty of room (this is a large

model).

Insert two integrators (from the Linear block library) near the bottom of your model and connect

them in series.

Draw a line from the second integrator and label it "theta". (To insert a label, double-click where

you want the label to go.)

Label the line connecting the integrators "d/dt(theta)".

Draw a line leading to the first integrator and label it "d2/dt2(theta)".

Insert a Gain block (from the Linear block library) to the left of the first integrator and connect its

output to the d2/dt2(theta) line.

Edit the gain value of this block by double clicking it and change it to "1/I".

Change the label of this block to "Pendulum Inertia" by clicking on the word "Gain". (you can

insert a newline in the label by hitting return).

Insert a Sum block (from the Linear block library) to the left of the Pendulum Inertia block and

connect its output to the inertia's input.

Change the label of this block to Sum Torques on Pend.

Construct a similar set of elements near the top of your model with the signals labeled with "x"

rather than "theta". The gain block should have the value "1/M" with the label "Cart Mass", andthe Sum block should have the label "Sum Forces on Cart".

Edit the Sum Forces block and change its signs to "-+-". This represents the signs of the three

horizontal forces acting on the cart.


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Now, we will add in two of the forces acting on the cart.

Insert a Gain block above the Cart Mass block. Change its value to "b" and its label to "damping".

Flip this block left-to-right by single clicking on it (to select it) and selecting Flip Block from the

Format menu (or hit Ctrl-F).

Tap a line off the d/dt(x) line (hold Ctrl while drawing the line) and connect it to the input of the

damping block.

Connect the output of the damping block to the topmost input of the Sum Forces block. The

damping force then has a negative sign.

Insert an In block (from the Connections block library) to the left of the Sum Forces block and

change its label to "F".

Connect the output of the F in block to the middle (positive) input of the Sum Forces block.


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Now, we will apply the forces N and P to both the cart and the pendulum. These forces contribute torques

to the pendulum with components "N L cos(theta) and P L sin(theta)". Therefore, we need to construct

these components.

Insert two Elementary Math blocks (from the Non-linear block library) and place them one above

the other above the second theta integrator. These blocks can be used to generate simple functions

such as sin and cos.

Edit upper Math block's value to "cos" and leave the lower Math block's value "sin".

Label the upper (cos) block "Vertical" and the lower (sin) block "Horizontal" to identify the

components.

Flip each of these blocks left-to-right.

Tap a line off the theta line and connect it to the input of the cos block. Tap a line of the line you just drew and connect it to the input of the sin block.

Insert a Gain block to the left of the cos block and change its value to "l" (lowercase L) and its

label to "Pend. Len."

Flip this block left-to-right and connect it to the output of the cos block.

Copy this block to a position to the left of the sin block. To do this, select it (by single-clicking)

and select Copy from the Edit Menu and then Paste from the Edit menu (or hit Ctrl-C and Ctrl-

V). Then, drag it to the proper position.

Connect the new Pend. Len.1 block to the output of the sin block.


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Draw long horizontal lines leading from both these Pend. Len. blocks and label the upper one "l

cos(theta)" and the lower one "l sin(theta)".

Now that the pendulum components are available, we can apply the forces N and P. We will assume we

can generate these forces, and just draw them coming from nowhere for now.

Insert two Product blocks (from the Non-linear block library) next to each other to the left and

above the Sum Torques block. These will be used to multiply the forces N and P by their

appropriate components.

Rotate the left Product block 90 degrees. To do this, select it and select Rotate Block from the

Format menu (or hit Ctrl-R).

Flip the other product block left-to-right and also rotate it 90 degrees. Connect the left Product block's output to the lower input of the Sum Torques block.

Connect the right Product block's output to the upper input of the Sum Torques block.

Continue the l cos(theta) line and attach it to the right input of the left Product block.

Continue the l sin(theta) line and attach it to the right input of the right Product block.

Begin drawing a line from the open input of the right product block. Extend it up and the to the

right. Label the open end of this line "P".

Begin drawing a line from the open input of the left product block. Extend it up and the to the

right. Label the open end of this line "N".


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Tap a line off the N line and connect it to the open input of the Sum forces block.

Next, we will represent the force N and P in terms of the pendulum's horizontal and vertical accelerations

from Newton's laws.

Insert a Gain block to the right of the N open ended line and change its value to "m" and its label

to "Pend. Mass".

Flip this block left-to-right and connect it's to N line.

Copy this block to a position to the right of the open ended P line and attach it to the P line.

Draw a line leading to the upper Pend. Mass block and label it "d2/dt2(xp)".

Insert a Sum block to the right of the lower Pend. Mass block.

Flip this block left-to-right and connect its output to the input of the lower Pend. Mass block. Insert a Constant block (from the Sources block library) to the right of the new Sum block,

change its value to "g" and label it "Gravity".

Connect the Gravity block to the upper (positive) input of the newest Sum block.

Draw a line leading to the open input of the new Sum block and label it "d2/dt2(yp)".


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Now, we will begin to produce the signals which contribute to d2/dt2(xp) and d2/dt2(yp).

Insert a Sum block to the right of the d2/dt2(yp) open end.

Change the Sum block's signs to "--" to represent the two terms contributing to d2/dt2(yp).

Flip the Sum block left-to-right and connect it's output to the d2/dt2(yp) signal.

Insert a Sum block to the right of the d2/dt2(xp) open end.

Change the Sum block's signs to "++-" to represent the three terms contributing to d2/dt2(xp).

Flip the Sum block left-to-right and connect it's output to the d2/dt2(xp) signal.

The first term of d2/dx2(xp) is d2/dx2(x). Tap a line off the d2/dx2(x) signal and connect it to the

topmost (positive) input of the newest Sum block.


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Now, we will generate the terms d2/dt2(theta)*lsin(theta) and d2/dt2(theta)*lcos(theta).

Insert two Product blocks next to each other to the right and below the Sum block associated with

d2/dt2(yp).

Rotate the left Product block 90 degrees.

Flip the other product block left-to-right and also rotate it 90 degrees.

Tap a line off the l sin(theta) signal and connect it to the left input of the left Product block.

Tap a line off the l cos(theta) signal and connect it to the right input of the right Product block.

Tap a line off the d2/dt2(theta) signal and connect it to the right input of the left Product block.

Tap a line of this new line and connect it to the left input of the right Product block.


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Now, we will generate the terms (d/dt(theta))^2*lsin(theta) and (d/dt(theta))^2*lcos(theta).

Insert two Product blocks next to each other to the right and slightly above the previous pair of

Product blocks.

Rotate the left Product block 90 degrees.

Flip the other product block left-to-right and also rotate it 90 degrees.

Tap a line off the l cos(theta) signal and connect it to the left input of the left Product block.

Tap a line off the l sin(theta) signal and connect it to the right input of the right Product block.

Insert a third Product block and insert it slightly above the d/dt(theta) line. Label this block"d/dt(theta)^2".

Tap a line off the d/dt(theta) signal and connect it to the left input of the lower Product block.

Tap a line of this new line and connect it to the right input of the lower Product block. Connect the output of the lower Product block to the free input of the right upper Product block.

Tap a line of this new line and connect it to the free input of the left upper Product block.


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Finally, we will connect these signals to produce the pendulum acceleration signals. In addition, we will

create the system outputs x and theta.

Connect the d2/dt2(theta)*lsin(theta) Product block's output to the lower (negative) input of the

d2/dt2(yp) Sum block.

Connect the d2/dt2(theta)*lcos(theta) Product block's output to the lower (negative) input of the

d2/dt2(xp) Sum block.

Connect the d/dt(theta)^2*lcos(theta) Product block's output to the upper (negative) input of the

d2/dt2(yp) Sum block.

Connect the d/dt(theta)^2*lsin(theta) Product block's output to the middle (positive) input of the

d2/dt2(xp) Sum block.

Insert an Out block (from the Connections block library) attached to the theta signal. Label thisblock "Theta".

Insert an Out block attached to the x signal. Label this block "x". It should automatically be

numbered 2.


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Now, save this model as pend.mdl.

Open-loop response

To generate the open-loop response, it is necessary to contain this model in a subsystem block.

Create a new model window (select New from the File menu in Simulink or hit Ctrl-N).

Insert a Subsystem block from the Connections block library.

Open the Subsystem block by double clicking on it. You will see a new model window labeled

"Subsystem".

Open your previous model window named pend.mdl. Select all of the model components by

selecting Select All from the Edit menu (or hit Ctrl-A). Copy the model into the paste buffer by selecting Copy from the Edit menu (or hit Ctrl-C).

Paste the model into the Subsystem window by selecting Paste from the Edit menu (or hit Ctrl-V)

in the Subsystem window

Close the Subsystem window. You will see the Subsystem block in the untitled window with one

input terminal labeled F and two output terminals labeled Theta and x.

Resize the Subsystem block to make the labels visible by selecting it and dragging one of the

corners.


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Label the Subsystem block "Inverted Pendulum".

Now, we will apply a unit impulse force input, and view the pendulum angle and cart position. An

impulse cannot be exactly simulated, since it is an infinite signal for an infinitesimal time with timeintegral equal to 1. Instead, we will use a pulse generator to generate a large but finite pulse for a small

but finite time. The magnitude of the pulse times the length of the pulse will equal 1.

Insert a Pulse Generator block from the Sources block library and connect it to the F input of the

Inverted Pendulum block.

Insert a Scope block (from the Sinks block library) and connect it to the Theta output of the

Inverted Pendulum block.

Insert a Scope block and connect it to the x output of the Inverted Pendulum block.

Edit the Pulse Generator block by double clicking on it. You will see the following dialog box.

Change the Period value to "10" (a long time between a chain of impulses - we will be interested

in only the first pulse).

Change the Duty Cycle value to ".01" this corresponds to .01% of 10 seconds, or .001 seconds.

Change the Amplitude to 1000. 1000 times .001 equals 1, providing an approximate unit impulse.


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Close this dialog box. You system will appear as shown below.

We now need to set an appropriate simulation time to view the response.

Select Parameters from the Simulation menu.

Change the Stop Time value to 2 seconds.

Close this dialog box

You can download a version of the system here. Before running it, it is necessary to set the physical

constants. Enter the following commands at the MATLAB prompt.

M = .5;

m = 0.2;

b = 0.1;i = 0.006;

g = 9.8;

l = 0.3;

Now, start the simulation (select Start from the Simulation menu or hit Ctrl-t). If you look at the

MATLAB prompt, you will see some error messages concerning algebraic loops. Due to the algebraic

constraint in this system, there are closed loops in the model with no dynamics which must be resolved

completely at each time step before dynamics are considered. In general, this is not a problem, but often

algebraic loops slow down the simulation, and can cause real problems if discontinuities exist within the

loop (such as saturation, sign functions, etc.)

Open both Scopes and hit the auto scale buttons. You will see the following for theta (left) and x (right).


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Notice that the pendulum swings all the way around due to the impact, and the cart travels along with a

jerky motion due to the pendulum. These simulations differ greatly from the MATLAB open loop

simulations because Simulink allows for fully nonlinear systems.

Extracting the linearized model into MATLAB

Since MATLAB can't deal with nonlinear systems directly, we cannot extract the exact model from

Simulink into MATLAB. However, a linearizedmodel can be extracted. This is done through the use of

In and Out Connection blocks and the MATLAB function linmod. In the case of this example, will use

the equivalent command linmod2, which can better handle the numerical difficulties of this problem.

To extract a model, it is necessary to start with a model file with inputs and outputs defined as In and Out

blocks. Earlier in this tutorial this was done, and the file was saved as pend.mdl. In this model, one input,

F (the force on the cart) and two outputs, theta (pendulum angle from vertical) and x (position of the cart),

were defined. When linearizing a model, it is necessary to choose an operating point about which to

linearize. By default, the linmod2 command linearizes about a state vector of zero and zero input. Since

this is the point about which we would like to linearize, we do not need to specify any extra arguments in

the command. Since the system has two outputs, we will generate two transfer functions.


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At the MATLAB prompt, enter the following commands

[A,B,C,D]=linmod2('pend')

[nums,den]=ss2tf(A,B,C,D)

numtheta=nums(1,:)

numx=nums(2,:)

You will see the following output (along with algebraic loop error messages) providing a state-spacemodel, two transfer function numerators, and one transfer function denominator (both transfer functions

share the same denominator).

A =

0 0 1.0000 0

0 0 0 1.0000

31.1818 0.0000 0.0000 -0.45452.6727 0.0000 0.0000 -0.1818

B =

0

0

4.5455

1.8182

C =

1 0 0 00 1 0 0

D =

0

0

nums =

0 0.0000 4.5455 0.0000 0.0000

0 0.0000 1.8182 0.0000 -44.5455

den =

1.0000 0.1818 -31.1818 -4.4545 0.0000

numtheta =

0 0.0000 4.5455 0.0000 0.0000


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numx =

0 0.0000 1.8182 0.0000 -44.5455To verify the model, we will generate an open-loop response. At the MATLAB command line, enter the

following commands.t=0:0.01:5;

impulse(numtheta,den,t);

axis([0 1 0 60]);

You should get the following response for the angle of the pendulum.

Note that this is identical to the impulse response obtained in the MATLAB tutorial pendulum

modeling example. Since it is a linearized model, however, it is not the same as the fully-nonlinear

impulse response obtained in Simulink.

Implementing PID control

In the pendulum PID control example, a PID controller was designed with proportional, integral, and

derivative gains equal to 100, 1, and 20, respectively. To implement this, we will start with our open-loop

model of the inverted pendulum. And add in both a control input and the disturbance impulse input to the

plant.

Open your Simulink model window you used to obtain the nonlinear open-loop response.

(pendol.mdl)

Delete the line connecting the Pulse Generator block to the Inverted Pendulum block. (single-

click on the line and select Cut from the Edit menu or hit Ctrl-X).

Move the Pulse Generator block to the upper left of the window.

Insert a Sum block to the left of the Inverted Pendulum block.


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Connect the output of the second Sum block to the input of the PID block.

Connect the PID output to the first Sum block's free input.

Edit the PID block by double clicking on it.

Change the Proportional gain to 100, leave the Integral gain 1, and change the Derivative gain to

20.

Close this window.

You can download our version of the closed-loop system here

Closed-loop response

We can now simulate the closed-loop system. Be sure the physical parameters are set (if you just ran the

open-loop response, they should still be set.) Start the simulation, double-click on the Theta scope and hit

the autoscale button. You should see the following response:


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This is identical to the closed-loop response obtained in the MATLAB tutorials. Note that the PID

controller handles the nonlinear system very well because the angle is very small (.04 radians).

inverted pendulam

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