ME 3600 Control Systems
Design Problem DP 5.1
DP 5.1: The block diagram for roll control of a jet fighter is shown below. The goal is to
select a suitable K value so the response of the system to a unit step command
d (t ) will provide a response  (t ) that is a fast response with less than 16%
overshoot.
aircraft dynamics
aileron actuator
+
roll
angle
-
a) The closed-loop transfer function is: T  s  

11.4 K
.
s 
d
s  s  1.4  s  10   11.4 K
b) The characteristic equation is: s3  11.4s 2  14s  11.4K  0 .
K
0.7
3
6
Poles
 10.09,  0.6545  0.602 j
 10.368,  0.5161  1.7414 j
 10.689,  0.3555  2.505 j
c) The real pole is insignificant in each case, so
T s 
11.4 K
11.4 K / p

 s  p  (s 2  2n s  n2 ) s 2  2n s  n2
Poles are:   j   n  jn 1   2 , so n   2   2 and    / n .
K
0.7
3
6
n
0.8893
1.816
2.53

0.7360
0.284
0.14
%OS
4-5
40
62
Tp (s)
5.17
1.79
1.26
Kamman – ME 3600 – page: 1/2
d) The plot shows a sample result for K = 0.7. Note that the response is accurately predicted
by the second order system approximation.
e) Using trial and error, found that for K = 1.275,   0.52 for approximate second order
system. This yields approximately a 16% overshoot.
f) What value of K yields an ITAE optimal response?
For a third order system, the optimal form of characteristic equation for a step input is
s3  11.4s 2  14s  11.4K  s3  1.75n s 2  2.15n2 s  n3
Comparing the coefficients of s 2 and s provides two different values for n , so a single K
value cannot be chosen using this approach. However, we note that the system is
approximately second order, so we need   0.7 to get ITAE optimal step response. Again,
using trial and error, we find that K =0.75. The step response of this system is very similar
to that shown in part (d).
Kamman – ME 3600 – page: 2/2