Modeling a Cruise Control System

Modeling a Cruise Control System

June 8

2010

Murad [Type the document subtitle]

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Abstract

The aim of our project is to find an Open loop for the Cruise Control System and find the performance.and Design a linear feedback proportional controller, and redesign it for the close loop and find the performance.


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Page No.Subjects

2Abstract

41. Introduction

41.2 State Space1

51.3 Transfer Function

71.3Design requirements

82. Open loop

103.digits of our student ID

124.Close Loop

124.1P controller

154.2 PI controller

185.Root Locus Method

205.1 Lag controller

236.References


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1. Introduction:

The model of the cruise control system is relatively simple. If the inertia of the wheels is neglected, and it is assumed that friction (which is proportional to the car's speed) is what is opposing the motion of the car, then the problem is reduced to the simple mass and damper system shown below.

Using Newton's law, the dynamic equation for this system is:

where u is the force from the engine. The state-space model of this equation now becomes:

For this example, let's assume that

m = 1000kgb = 50Nsec/m

u = 500N1.2 State Space

The state equations for this problem are:


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1.3 Transfer Function

>> m = 1000;

b = 50;

u = 500;

A = [0 1; 0 -b/m]

B = [0; 1/m]

C = [0 1]

D = 0

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

A =

0 1.0000

0 -0.0500

B =

1.0e-003 *


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0

1.0000

C =

0 1

D =

0

num =

0 0.0010 0

den =

1.0000 0.0500 0


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>> TF(num,den)

Transfer function:

0.001 s

------------

s^2 + 0.05 s

1.3 Design requirements

The next step in modeling this system is to come up with some design criteria. When the engine gives a 500 Newton force, the car will reach a maximum velocity of 10 m/s (22 mph). An automobile should be able to accelerate up to that speed in less than 5 seconds. Since this is only a cruise control system, a 10% overshoot on the velocity will not do much damage. A 2% steady-state error is also acceptable for the same reason.

Keeping the above in mind, we have proposed the following design criteria for this problem:

Rise time < 5 secOvershoot < 10%Steady state error < 2%

2. Open-loop Modeling a Cruise Control System

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Below we show how this problem would be solved using Matlab. If you copy the following text into a m-file (or create a '.m' file located in the same directory as Matlab) and run it, Matlab will give you the A, B, C, and D matrices of the state-space model and a plot of the response to a step response of the input force.

m = 1000; b = 50; u = 500; A = [0 1; 0 -b/m] B = [0; 1/m] C = [0 1] D = 0 step(A,u*B,C,D)

After running the m-file, Matlab should output to the command window the following state-space matrices:

A =

0 1.0000 0 -0.0500

B =

1.0e-003 *

0 1.0000

C =

0 1

D =

0

You should also get the following plot of the open-loop response of the velocity (initially at rest at time t=0) for a step change in the applied force (going from 0-500N):


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As you can see, this step response does not meet the design criteria placed on the problem. The system is overdamped, so the overshoot response is fine, but the rise time is too slow. Therefore a controller will have to be designed to improve the dynamics of the system. Listed at the bottom of the page are four methodologies for designing a controller (PID, root locus,frequency response, and state space). Click on whichever one you would like to see.


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3. Design a feedback proportional controller with gain equals to the last digits of our student ID.

3.1 Murad al-mutairi 426102080

The first thing to do in this problem is to transform the state-space equations into transfer function form. Next, close the loop and add some proportional control to see if the response can be improved. Create a new m-file with the following lines of code:

m = 1000;b = 50;u = 500;A = [0 1; 0 -b/m]B = [0; 1/m]C = [0 1]D = 0

[num,den]=ss2tf(A,B,C,D)k = 2080;[numc,denc] = cloop(k*num,den,-1);t = 0:0.1:20;step(10*numc,denc,t)axis([0 20 0 10])


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3.2 Mohammed al-wenais 427103009

m = 1000;b = 50;u = 500;A = [0 1; 0 -b/m]B = [0; 1/m]C = [0 1]D = 0


As you can see at the gain 2080 has steady state error 2% and settling time 2 sec, and at the gain 3009 has steady state error 1% and settling time 1.5 sec.


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4. Close Loop

The state equations for this problem are:

4.1 P controller

The first thing to do in this problem is to transform the state-space equations into transfer function form. Next, close the loop and add some proportional control to see if the response can be improved. Both of these steps can be done sequentially in Matlab. Create a new m-file with the following lines of code (or create a '.m' file located in the same directory as Matlab):

m = 1000;b = 50;u = 500;A = [0 1; 0 -b/m]B = [0; 1/m]C = [0 1]D = 0


The number 10 that is multiplied by numc in the step command is used to make the reference signal 10 the original step response had a steady state value of 10m/sec for an applied force of 500N. This m-file should now represent that same desired steady-state value. Running your m-file should give you the following plot of the velocity response to the step change in the reference.


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As you can see, the step response is not very good. The steady state error is more than 10%, and the rise time is still too slow. You can adjust the proportional gain to make the response better, but you will not be able to make the steady state value go to 10m/sec without getting rise times that are too fast. You must always keep in mind that you are designing a real system, and for a cruise control system to respond 0-10m/sec in less than half of a second is unrealistic. To illustrate this idea, adjust the k value to equal 10,000, and you should get the following velocity response:


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The steady state error has been dropped to near zero, but the rise time on this step response is way too fast (about 0.2 seconds). The solution to this problem is to add some integral control to eliminate the steady state error. Adjust k until you get a reasonable rise time. For example, with k = 600, the response should now look like:


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4.2 PI controller

Now, try PI control. Change your m-file to look like the following:

k = 600;ki = 1;num1 = [k ki];den1 = [1 0];num2 = conv(num,num1);den2 = conv(den,den1);[numc,denc] = cloop(num2,den2,-1);t = 0:0.1:20;step(10*numc,denc,t)axis([0 20 0 10])

You should get the following velocity response plot:


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We suggest starting with a ki value that is small, i. e. around 1, to get a feel for what the integral control is doing to the response. Remember, integral control makes the transient response worse, so adding too much initially can make the response unrecognizable. Now you can adjust the ki value to reduce the steady state error. With ki = 40 and k adjusted up a little more to 800, the response looks like the following:


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As you can see, this step response meets all of the design criteria, and therefore no more iteration is needed. It is also noteworthy that no derivative control was needed in this example.

The system was originally critically damped, making the overshoot equal to zero. Had more integral control been needed to eliminate the steady state error, the transient response may have worsen to the point of needing derivative control.

It is easy to implement the derivative control in Matlab. Just change the num1 = [k ki]; line to num1 = [kd k ki];. However, this will make it necessary to adjust three variables simultaneously. The best way to do this would be to adjust one variable at a time starting with k, then kd, and then ki. You may have to go back and readjust, or tweak, the variables (gains) after they have all been set.


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5. Root Locus Method

Proportional controller

Since this system is first order, the problem can be solved using proportional control, although a lag controller can be added if the steady state error becomes too high.

The design criteria of a rise time of less than 5 seconds implies that the natural frequency must be greater than 0.36 (remember: tr = 1.8/Wn). The design criteria also states that the overshoot must be less than 10% which implies that the damping ratio zeta must be greater than 0.6. We will use this fact to start the design of the controller (note: the sgrid command can show the regions of the root locus where these two criteria are satisfied).

m = 1000; b = 50; u = 500; A = [0 1; 0 -b/m] B = [0; 1/m] C = [0 1] D = 0

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

You should get the following output:

A = 0 1.0000 0 -0.0500 B = 1.0e-03 *

0 1.0000 C = 0 1 D = 0


http://www.ee.usyd.edu.au/tutorials_online/matlab/extras/function.html

http://www.ee.usyd.edu.au/tutorials_online/matlab/extras/conver.html

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num = 0 0.0010 0 den = 1.0000 0.0500 0

If you look at the transfer function closely, you will notice that there is a pole-zero cancellation at the origin (both num and den have a root at zero, as can be easily seen since they each have a zero constant term). Matlab does not automatically cancel the pole and zero. As a result, it can cause confusion in the root locus plot, so it is best to eliminate it now. This can be done by manually eliminating both the pole and zero. At the end of your m-file, add the following code:

num = [num(1), num(2)]den = [den(1), den(2)]

The transfer function is now:

num = 0.0010

den = 1.0000 0.0500

Now, we shall plot the root locus and step response to see the effects of proportional control on the system. Copy the following code to the end of you m-file:

figurehold;axis([-0.6 0 -0.6 0.6]);rlocus(num,den)sgrid(0.6,.36)[k,poles] = rlocfind(num,den);figuret = 0:0.1:20;[numc,denc] = cloop(k*num,den, -1);step(10*numc,denc,t)axis([0 20 0 10])

When prompted to select a point on the root locus, click on the real axis just to the left of the natural frequency requirement (about -0.4). The 10 in the step response is multiplied to the numerator so that the desired reference signal will be 10 instead of the default of 1. If you look


http://www.ee.usyd.edu.au/tutorials_online/matlab/extras/step.html

http://www.ee.usyd.edu.au/tutorials_online/matlab/extras/plot.html#axis

http://www.ee.usyd.edu.au/tutorials_online/matlab/extras/PZ.html

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back to the original cruise control problem, you will see that the steady state value should be 10. The first plot given by Matlab should be the root locus. The second plot should be the velocity response to the step change in the reference. The two plots should look similar to the following:

5.1 Lag controller

A lag controller can be added to improve the steady state error. For this lag controller, we will place the zero at -0.3 and the pole will at -0.03; this will result in an improvement of the steady-state error by a factor of 10 = 0.3/0.03.

Change your m-file to look like the following to implement this new controller:

m = 1000;b = 50;u = 500;A = [0 1; 0 -b/m];B = [0; 1/m];


http://www.ee.usyd.edu.au/tutorials_online/matlab/examples/cruise/cc.html

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C = [0 1];D = 0;

[num,den] = ss2tf(A,B,C,D);num=[num(1), num(2)];den=[den(1), den(2)];

numlag = [1 .3];denlag = [1 .03];num1 = conv(num,numlag);den1 = conv(den,denlag);

axis([-0.6 0 -.4 .4]);hold;rlocus(num1,den1)sgrid(.6,.36)[k,poles] = rlocfind(num1,den1);figuret = 0:0.1:20;[numc,denc] = cloop(k*num1,den1, -1);step(10*numc,denc,t)axis([0 20 0 12])

The root locus should look like the following:


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When prompted to select a point on the root locus, again click on the real axis just to the left of the natural frequency requirement to achieve the desired rise time. Now you should have the following velocity response:


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6. References

WWW.engin.umich.edu


Modeling a Cruise Control System

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