CONTROL SYSTEM LAB REPORTS

Today is : Wednesday, 10 Mar 2010

>> Home

Control System Lab Reports

Thursday, 17 September 2009 09:59

Submitted by : Reza Ashraf(Rabbi), B.Sc. in EEE, 30 | B | Day, SUB.

EXPERIMENT NO - 01

NAME OF THE EXPERIMENT :

Modeling of Physical systems and study of their open loop response

OBJECTIVE :

The objective of this experiment is the modeling of physical systems and study of their open loop response.

Simulation of a cruise control system

Physical Interpretation & system equations:

Let us assume a car that travels only in one direction. Control to the car was applied in such a way that it has a smooth start up, along with a constant-speed ride. The force applied is ‘u’, velocity at any time is ‘v’ and frictional constant=b. The frictional force is linearly proportional to velocity. So the applied force, ‘u’ accelerates the mass while overcoming the frictional force.

If it is assumed that friction is opposing the motion of the car, then the modeling equations become
Mv + bv = u
Y = v

Mathematical analysis:

Taking the Laplace transform of the equations, we find
                                    msV(s)+bV(s) = U(s)
                                    Y(s)=V(s)

Substituting V(s) in terms of Y(s)
                                  msY(s)+bY(s) = U(s)

The transfer function of the system becomes
                                  Y(s)/U(s) = 1/(ms+b)

MATLab Code:

clear;
clc;
b=50;
u=602;

figure;
for i=1:5
m=200*i;
num=[1];
den=[m b];
sys=tf(num,den);
step(u*sys);
hold on;
title('Step response of the Cruise Control when Friction is fixed');
xlabel('time')
ylabel('velocity(m/s)');
end
figure;
m=1000;
for j=1:5
b=j*10;
num=[1];
den=[m b];
sys=tf(num,den);
step(u*sys);
hold on;
title('Step response of the Cruise Control when Mass is fixed');
xlabel('time')
ylabel('velocity(m/s)');
end

Output Curves:

Figure 1:

Figure 2:

MATLab Data:

For Figure 1:

For Figure 2:

Q-01

Define rise time, settling time, percentage overshoot and steady state error of a system for step input.

Answer:

Rise time (Tr):
          Time required for the response to rise from 0% to 100% of its final value.

Settling time(Ts):
          Time required for the response to reach and stay within the range about the final value of size specified by absolute percentage of final value.

Overshoot percentage :
The amount the system output response proceeds beyond the desired response.
         Percent Overshoot(P.O)={(Mpt-fv)/fv}X100%

Where,
Mpt= The peak value of the time response.
fv= The final value of the response.

Q-02

How steady state speed, rise time and overshoot of the output response vary with the variation of m and b?

Answer:

In Figure 1, Friction (b) is fixed Mass (m) has been varied. We take 5 different values for Mass (m) respectively 200, 400, 600, 800, and 1000. If we observe the graph, we can find that Rise time is increasing as Mass (m) increases. If we calculate overshoot percentage for given value we find it Zero.

In Figure 2, Mass (m) is kept fixed and Friction (b) varied. We measure the rise time for five different values of Friction (b) and the values are 10, 20, 30, 40, and 50. From the graph, we can see that Rise time and steady state speed is decreasing as Friction (b) increases. Overshoot percentage value is Zero for measured value.

Q-03

Justify that the steady state speed of the vehicle depends only on the friction coefficient b.

Answer:

We Know that, u = (dv/dt)+bv
Here v is fixed, so dv/dt = 0

Then u=bv
or v = u/b

Here u is fixed for figure 1 and Figure 2.
So, v is only depends on the friction Co-efficient b

All the 6 Lab Reports on Control System