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Appendix 1 Control system design using MATLAB 383
(a) a text editor
(b) the MATLAB editor/debugger
(c) a word processor that can save as pure ASCII text files.
A script file
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Appendix 1 Control system design using MATLAB 407
chareqn
1 10 100
desired_eigenvalues
5:00008:6603i
5:00008:6603i
Ke
7:0000 77:0000
% Agrees with equation (8:150)
Example 8.
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Appendix 1 Control system design using MATLAB 395
Filename: examp64b.m
%Example 6.4(b)
%G(s)K/s(s 22s4)
%Creates a Bode Gain and Phase Diagrams for K4
%Determines Gain and Ph
12
13
Chapter 4 MATHEMATICAL MODELING OF PHYSICAL
SYSTEMS
4 -1 (a) F orce equations:
2
f ( t) = M 1
d y1
dt
dy1 dy 2 + K ( y y )
1
2
dt dt
dy1
+ B1
2
+ B3
dt
2
dy1 dy2 + K ( y y ) + M d y2 + B dy2
B3
1
2
2
2
2
dt
dt
dt dt
Rearrange the equations a
(b) T ransfer functions:
L ( s)
Tm ( s)
=
K
m ( s)
( s)
Tm ( s)
=
J L s2 + BL s + K
( s) = s J mJ L s3 + ( Bm J L + BL J m ) s 2 + ( KJ m + KJ L + Bm BL ) s + Bm K
(s )
(c) C haracteristic equation:
( s ) = 0
(d) Steady -state performance:
T (t )
m
Chapter 6
6 -1 (a)
STABILITY OF LINEAR CONTROL SYSTEMS
= 0 , 1. 5 +
Poles are at s
j 1. 6583 ,
1. 5
One poles at s = 0. M arginally stable .
j 1. 6583
(b) Poles are at s = 5, j 2 , j 2
(c) Poles are at s = 0 .8688 , 0 .4344 + j 2 . 3593 , 0 .4344 j 2 .
e t + et
( t) = 0.5 t
t
e + e
5 -14 (a)
e = K s ( r y
ia =
Solve for i
a
)
ea = e es
eu eb
eb = K b
Ra + Rs
in terms of
y
d
and
y
e +e
t
e + e
t
t
es = Rs ia
d y
t
eu = Kea
Tm = K i ia = ( J m + J L )
dt
d y
2
dt
2
, we have
dt
ia =
KKs (r y ) Kb
d y
d
Tm = K i ia = ( J m + JL )
T
D
=
d m
dt
d
(sec)
V
+ B m m
e
ea = Ra ia + La
= r b
b
dia
= Ksy
dt
+ K b m
E ( s)
a
=
y = n m
KG ( s ) E ( s )
c
B lock diagram:
(b) Forward-path transfer function:
Y (s )
=
E (s )
K Kin G c ( s ) e
TD s
s cfw_( Ra + La s )