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Problem 7.55PP
The lateral motions of a ship that is 100 m long, moving at a constant velocity of 10 m/sec, are 
described by
[
^ 1 r -0.0895 -0.286 0 "I T ^ 1 [" 0.0145 "I
/; I = I -0.0439 -0.272 0 I I I +1 -0 ®I22 14.
where
jS = side slip angle(deg),
(fj = heading angle(deg),
6 = rudder angle{deg), 
r = yaw rate (see Fig. 1).
Figure 1 View of ship from above
r = yaw rate (see Fig. 1).
Figure 1 View of ship from above
(a) Determine the transfer function from 5to tp and the characteristic roots of the uncontrolled 
ship.
(b) Using complete state feedback of the form 
6 = -K - \p - K2r-K 3{iiJ - ijjd).
where ijjd is the desired heading, determine vaiues of K^, K2, and K3 that wiii place the closed- 
loop roots at s = -0.2,-0.2 ± 0.2).
(c) Design a state estimator based on the measurement of ijj (obtained from a gyrocompass, for 
example). Place the roots of the estimator error equation at s = -0.8 and -0.8 ± 0.8).
(d) Give the state equations and transfer function for the compensator Dc(s) in Fig. 2, and plot its 
frequency response.
(e) Draw the Bode plot for the closed-loop system, and compute the corresponding gain and 
phase margins.
(f) Compute the feed-fonvard gains for a reference input, and plot the step response of the 
system to a change in heading of 5°.
Figure 2 Ship control block diagram for Problem
S tep -b y -s te p so lu tion
step 1 of 3
The state space representation of the lateral motions of a ship is.
' f i -0.0895 -0.286 O ir^ ■ 0.0145
r = -0.0439 -0.272 0 r + -0.0122
0 1 oJ[»r 0
S
The matrix A and B ^re,
-0.0895 -0.286 O' 
-0.0439 -0.272 0 
0
0.0145 
- 0.0122 
0
B s
I
Step 2 of 3
(a)
As it is required to find the transfer function from g to the output equation is,
Y = [0 0 1]
B
V
V f ']
The matrix C is,
C » [ 0 0 1]
The transfer function of the system is,
G ( j ) = C [ i I - A ] 'B
> + 0.0895 0.286 O'
0.0439 J + 0.272 0 
0 -1 s
Determine the inverse matrix, [ j j —A ]" ' ■
_ 1
=[0 0 1]
0.0145
- 0.0122
0
« [(« + 0.0895)(i + 0.272) -0 .0439 x 0.286J
■ j(j+0.272) -0.0439J
-0.286J s ( f + 0.0895) 
0 0
-0.0439 
» +0.0895
( i + 0.0895)(j + 0.272) - (0.0439)(0.286)
i ( i ’ +0.36l5s + 0.0118) 
■ j(i+0.272) -0 .0439i
-0.286* *(* +0.0895)
0 0
-0.0439 
* + 0.0895
*’ +0.36l5* + 0 .0 ll8
Step 3 of 3
Determine the matrix, G (*)
“ *(*’ + 0.3615*+0.01l8)^®
0 1]
■*(* + 0.272) -0.0439* -0.0439 ■ 0.0145'
-0.286* *(* + 0.0895) * + 0.0895 -0.0122
0 0 *’ + 0.3615* + 0 .0 ll8 0
*(*’ + 0.36I5* + 0.01l8)
[0 0 1]
0.0145* (*+0.272)+0.000536 
-0.0122*(*+0.0895)-0.004147* 
0
As the transfer function is 0, it is not possible to determine the roots of the system.
Also, it is not possible to determine the state feedback, or to design a state estimator, or 
determine the state equations and to draw the Bode plot and step response of the system. 
Hence, subparts (b), (c), (d), (e) and (f) cannot be solved.