Análise de Motor DC

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Problem 3.35PP
Assume that 
Jm = 0.01 kgm2, 
b = 0.001 N-msec,
Ke = 0.02 Vsec,
/C/=0.02Nm/A,
Ra = 10Q.
The equations of motion for the DC motor shown in Fig. were given in Eqs. (1-2) as
(a) Find the transfer function between the applied voltage va and the motor speed 9m-
(a) Find the transfer function between the applied voltage va and the motor speed Om-
(b) What is the steady-state speed of the motor after a voitage va = 10 V has been applied?
(c) Find the transfer function between the applied voltage va and the shaft angle dm.
(d) Suppose feedback is added to the system in part (c) so that it becomes a position servo 
device such that the applied voltage is given by 
va = K( dr - dm)
where K is the feedback gain. Find the transfer function between 0rand dm.
(e) What is the maximum value of K that can be used if an overshoot M1
Compare Equation (4) and (8).
(8)
5,(5) 5’+0.1045+«^
Step 18 of 20
Write the MATLAB program, 
close all
K1=[0.5 1.0 2.0 6.5e-2];
t=0:0.01:150;
for i=1:1:length(K1)
K = K1(i);
titleText = sprintf('K= % 1 .4 fK );
wn = sqrt(0.2*K);
num=wn''2;
den=[1 0.104 wn''2]:
zeta=0.104/2/wn;
sys = tf(num, den);
y= step(sys, t);
% Finding maximum overshoot 
if zeta 0.9));
t_r = t(idx_09)-t(idx_01);
risetimeText = sprintf('Rise time = %3.2f sec', t_r);
% Plotting 
subplot(3,2,i): 
piot(t,y); 
grid on; 
title(titleText);
text( 0.5, 0.3, overshootText); 
text( 0.5, 0.1, risetimeText); 
end
% Function for computing rise time 
fun tr = risetime(t,y)
% normalize y to 1:
y = y/y{length{y));
idxl = min(find(y>=0.1));
idx2 = min(find(y>=0.9));
if -isempty(idx1) & ~isempty(idx2);
tr = t(idx2)-t(idx1);
else
tr= 0
end
Step 19 of 20
The output for the MATLAB code is given in figure 1.
■ / V M ”-
IM IM
Figure 1
Step 20 of 20
Thus for the value of 8: