Question

Use matlab to solve the following:

A field-controlled DC motor can be described by the following differential equation:

ay³(t)+by²(t)+cy¹(t) = dx(t)

Where y(t) is the angle displacement of the motor’s load and x(t) is the applied voltage to the motor. The applied voltage is DC that turns on at t = 0, which is a step function. The values for a,b,c and d are derived from the model of the field controlled DC motor. This is a concept that is slightly advanced from this class. For now, go with this:

% Motor Parameters

J = .01; % Gain (positive)

f = .10; % Friction (0

Rf = 10;

Lf = .01;

kt = 10;

% System coefficients

a = J;

b = f+ J*Rf/Lf;

c = f*Rf/Lf;

d = kt/Lf;

(1) Plot the impulse response and the step response of the system.

(2) Analyze the plots by answering these questions:

(a)If an impulse is applied, how many degrees does the motor turn? And how long does it take to turn those degrees?

(b)After 1 minute of applying a step voltage (1 volt DC), how many revolutions of the motor ? How many revolutions if the DC voltage applied is 12 volts ?

(c)The only parameters we can control in the motor are J, f, and the input voltage x(t). Play with these to evaluate the motor. In other words try a few different sets of values for J, f, and the input voltage x(t) and explain what happens.

Answer 1

Hello,
          Please find the answer attached as under. Please give a thumbs up rating if you find the answer useful! Have a rocking day ahead!

******* Matlab Code *******

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% dc motor characteristics

% Motor Parameters
J = .01; % Gain (positive)
f = .10; % Friction
Rf = 10;
Lf = .01;
kt = 10;
% System coefficients
a = J;
b = f+ J*Rf/Lf;
c = f*Rf/Lf;
d = kt/Lf;

G = tf(d,[a b c 0]);
subplot(2,1,1);
impulse(G);
grid;
subplot(2,1,2);
step(G,60);
grid;

%%%%%%%%%% Study of effect of parameters %%%%%%%%%
J1 = 0.03; J2 =0.07;
a1 = J1;
b1 = f+ J1*Rf/Lf;
G = tf(d,[a b c]);
G1 = tf(d,[a1 b1 c]);
a2 = J2;
b2 = f+ J2*Rf/Lf;
G2 = tf(d,[a2 b2 c]);
fprintf('=========== System response characteristics for J = 0.01======\n')
stepinfo(G)
fprintf('=========== System response characteristics for J = 0.03======\n')
stepinfo(G1)
fprintf('=========== System response characteristics for J = 0.07======\n')
stepinfo(G2)

f1 = 0.40; f2 = 0.8;
b1 = f1+ J*Rf/Lf;
c1 = f1*Rf/Lf;
G1 = tf(d,[a b1 c1]);
b2 = f2+ J*Rf/Lf;
c2 = f2*Rf/Lf;
G2 = tf(d,[a b2 c2]);
fprintf('=========== System response characteristics for f = 0.1======\n')
stepinfo(G)
fprintf('=========== System response characteristics for f = 0.4======\n')
stepinfo(G1)
fprintf('=========== System response characteristics for f = 0.6======\n')
stepinfo(G2)

****** Output *****

The output as seen above is in radians.

(a) For an impulse, the motor will turn 10 radians (572 degrees), in approximately 0.5 seconds.

(b) After 1 minute = 60 secs, the motor turns 600 radians = 95 rotations in 1 minute.

For 12 volts, rpm = 12 * 95 = 1140

(c) When you run the above matlab code, you will get the impulse response characteristics for various values of J and f. The input voltage has not been simulated because all it does is amplify the magnitude of the response, the dynamic characteristics remain the same:

=========== System response characteristics for J = 0.01======

ans =

struct with fields:

        RiseTime: 0.2197
    SettlingTime: 0.3922
     SettlingMin: 9.0354
     SettlingMax: 9.9997
       Overshoot: 0
      Undershoot: 0
            Peak: 9.9997
        PeakTime: 1.0546

=========== System response characteristics for J = 0.03======

ans =

struct with fields:

        RiseTime: 0.6591
    SettlingTime: 1.1746
     SettlingMin: 9.0418
     SettlingMax: 9.9997
       Overshoot: 0
      Undershoot: 0
            Peak: 9.9997
        PeakTime: 3.1638

=========== System response characteristics for J = 0.07======

ans =

struct with fields:

        RiseTime: 1.5379
    SettlingTime: 2.7394
     SettlingMin: 9.0436
     SettlingMax: 9.9997
       Overshoot: 0
      Undershoot: 0
            Peak: 9.9997
        PeakTime: 7.3821

=========== System response characteristics for f = 0.1======

ans =

struct with fields:

        RiseTime: 0.2197
    SettlingTime: 0.3922
     SettlingMin: 9.0354
     SettlingMax: 9.9997
       Overshoot: 0
      Undershoot: 0
            Peak: 9.9997
        PeakTime: 1.0546

=========== System response characteristics for f = 0.4======

ans =

struct with fields:

        RiseTime: 0.0550
    SettlingTime: 0.0988
     SettlingMin: 2.2513
     SettlingMax: 2.4983
       Overshoot: 0
      Undershoot: 0
            Peak: 2.4983
        PeakTime: 0.1831

=========== System response characteristics for f = 0.6======

ans =

struct with fields:

        RiseTime: 0.0276
    SettlingTime: 0.0499
     SettlingMin: 1.1261
     SettlingMax: 1.2491
       Overshoot: 0
      Undershoot: 0
            Peak: 1.2491
        PeakTime: 0.0915

Thus, if you increase J you can see that the response becomes slow. The reverse happens if we increase f.