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Problem 3.51 PP
The block diagram of an autopilot designed to maintain the pitch attitude d of an aircraft is shown 
in Fig. The transfer function relating the elevator angle 5e and the pitch attitude d is
e(5) 5 0 (i+ l)(5 + 2 )
W j) (sZ + 5s+40)(l2+0.031+0.06)’
where d is the pitch attitude in degrees and 6e is the elevator angle in degrees.
The autopilot controller uses the pitch attitude error e to adjust the elevator according to the
transfer function
E(s) j + 1 0
Using Matlab, find a value of K that will provide an overshoot of less than 10% and a rise time 
faster than 0.5 sec for a unit-step change in d r . After examining the step response of the system 
for various values of K. comment on the difficulty associated with making rise time and overshoot
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faster than 0.5 sec for a unit-step change in d r . After examining the step response of the system 
for various values of K. comment on the difficulty associated with making rise time and overshoot 
measurements for complicated systems.
Figure Block diagram of autopilot
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Step-by-step solution
step 1 of 7
Consider the following block diagram of autopilot designed to maintain the pitch attitude $ of an 
aircraft:
Figure 1; Block diagram of autopilot.
Step 2 of 7
Consider the transfer function relating the elevator angle, S, in degrees and the pitch attitude, 
B in degrees.
5D(^+l)(^+2)_____ ^
+ a 0 S s+ O € 6 j
The autopilot controller uses the pitch attitude error e to adjust the elevator according to the 
following transfer function.
= D{s)
4 ^ + 3 )
s+W
Step 3 of 7
The error signal £ ( i ) is.
The transfer function of the system is.
g ( i ) G ( i ) 0 ( i ) 
6 l , ( i ) ° l + G ( i ) D ( i )
J
S 0 ( i+ l ) ( i + 2 )
(i^+ 5s+ 40 )(j*+ 0 .03s+ 0 .06 )J l, i+ lO J
5 0 ( i+ l ) ( i+ 2 ) V A :(i3 -3n
^ s ’ + 5 i+ 4 0 )(* ’ + 0.03j+ 0.06)Jt. J + 10 )
50K{s*
(j *+ 5 j + 4 0 )(j *
( soa: ( j + i )(j + 2 )( j +3) j
[ ( j ^+5 j + 40)(j ^+0.03j +0.06)(j +10) J
1 +
5 0 g ( j+ l ) ( j + 2 ) ( j+ 3 )
+5 j + 4 0 )(j ’ +0.03j +0.06)(j + 10) ]
Step 4 of 7
Simplify this equation further and we get.
( 5 0 y ( j + l ) ( j + 2 ) ( j+ 3 ) J
[ ( j ' +5 j + 4 0 )( j’ +0.03j +0.06)(j +10)J
e,(s) ( j " + 5j + 4 0 )(j =+0.03j +0.06)(j + 10)+50^C(j + 1)(j + 2 )(j + 3) 
j ’ +5j + 4 0 (j ’ +0.03j +0.06)(j +10)
5 0 y ( j + l ) ( j + 2 )( j+ 3 )
” ( j ' + 5j + 4 0 )(j =+0.03j +0.06)(j + 10 )+50K (j + 1)(j + 2 ) ( j +3) 
SOA: ( j ’ + 3j ’ + 2j * + 6j + j ' + 3j + 2j + 6 )
7 ( j ’ + 15j '+ 9 0 j +400)(j ’ +0.03j +0.06) 
[+50A: ( j ’ + 3j ' + 2j ’ + Sj + j ’ + 3j + 2j + 6 ) 
50Ar(j’ +ds’ + lU + 6 )
' r j ’ +15.03J*+(50AT+90.51)j’ +(300/C +403.6)j’ 1 
[ +(550A:+17.4) j +(300A :+24) J
Step 5 of 7
The output must be normalized to the final value of for easy computation of the overshoot
and rise-time. The next step is to find the value of K that will provide an overshoot of less than 
10% and a rise time faster than 0.5 sec for a unit-step change in
The following is the MATLAB function for computing the rise time:
%Function for computing rise time 
function tr = risteime(t,y)
% normalization of 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
Enter the following commands to plot the pitch attitude $ and also to find the peak overshoot 
and rise time.
% Program to find peak overshoot and rise time and also to plot the pitch attitude 
clear all 
close all
K1 =[3.5 3.0 0.3 0.03];
t=0:0.01:150;
for i=1:1:length(K1)
K = K1(i);
titleText = sprintf('K=%1.4f ,K);
wn = sqrt(0.2 *K);
num= wn''2;
den=[10.104wn^2];
zeta=0.104/2/wn;
sys = tf(num, den);
y= step(sys, t);
% Finding maximum overshoot 
if zeta ,1
Mp = {max(y)-1)*100:
overshootText = sprintf('Max overshoot = %3.2f, Mp); 
else
overshootText = sprintf('No overshoot') 
end
% Finding rise time 
idx_01 = max{find{y0.9)); 
t_r=t(idx_09)-t(idx_01);
risetimeText = sprintf('Rise time = %3.2f sec', t_r);
% Plotting the pitch attitude 
subplot(3,2,i):
Plot(t,y); 
grid on; 
title(titleText);
text( 0.5, 0.3, overshootText); 
text( 0.5, 0.1, risetimeText); 
end
Step 6 of 7
The following are the plots of pitch attitude verses time for various values of X - 
For a: = 3.5:
K= 3.5000
1.5
0.5
Max overs 
Rise time
hoot= 17. 
= 0.11 sei
>1
0 50 100
Time (sec)
Figure 2: Step response of autopilot system for K — Z.S
150
Step 7 of 7
For K = 3
K= 3.0000
1.5
0.5
Max overs 
Rise time
hoot =12. 
= 0.12 se :
0 50 100
Time (sec)
Figure 3; Step response of autopilot system for K — i - 
For a: = 0.3:
150
K= 0.3000
0 50 100
Time (sec)
Figure 4; Step response of autopilot system for K — 03- 
For a: = 0.03:
K= 0.0300
150