# Chapter 1 Introduction

```Chapter 1 Introduction
1. Control system definition
2. Control system configurations
＊Open Loop
3. Design Process
★ Laplace Transformation
＊Closed Loop (Feedback Loop)
★ Solving the differential equation
y  3 y  2 y  1
Exp 1.
Initial condition y (0)  y (0)  0
Exp 2.
y  4 y  5 y  5
Initial condition y (0)  1 ; y (0)  2
Chapter 2 Models of Physical System
1. RLC Circuit (electronic circuit)
V(s)=I(s)*Z(s) ;
V : voltage ; I: current ; Z : impedance
resistance
inductance
v
v  i*R
Take Laplace
di
L
dt
v
Take Laplace
V ( s)  I ( s) * R
Zt  Z1  Z 2  Z3  
※ Parallel
1
1
1
1




Z t Z1 Z 2 Z 3
R
Vi
R
V0
C
V0 ( s )
？
Vi ( s )
Sol：

1 

Vi  I   R  Cs 



V  I  1
 0
Cs
V (s) 
1
○
2
○
Vi
1 I ( s)

C
s
Z c (s) 
Z l ( s)  Ls
※ Series
1
* i  dt
C 
Take Laplace
V ( s )  sI ( s ) * L
Z (s)  R
Exp1.
capacitance
R
1
Cs
V0
1
Cs
1
V0
cs

1
Vi

I R  
cs 

1
V0
1
 cs 
Rcs  1 Rcs  1
Vi
cs
I
Exp1.
i
Ls
Vi
1
Cs
V0
Exp2.
Z2
Z1
Va
I
Z31
node a：
V1  Va V2  Va V3  Va


0
Z1
Z2
Z3
2. Operational Amplifier Circuit
※Three assumptions for an ideal operational amplifier :
(a) It has infinite impedance at both inputs, hence no current is drawn
from the inpu circuits, that means i1  i2  0
(b) It has infinite gain A. As a consequence, the difference between the
input V1 and V2 must be zero, that means V1  V2 .
(c) It has zero output impedance. Therefore, the output voltage does not
depend on the current, that means Vo  Vout .
Application of OPA
2-1 Inverting amplifier
Ro
Ri
Va i1
a
Vi
node a
b
Vi  Va V0  Va

0
Ri
R0
i2
Vout
V0
Vout
R
  out
V in
Rin
; Z out  Rout ; Zin  Rin ;
Vout
Z
  out
V in
Z in
2-2 Integrating amplifier
Z out 
1
; Z in  R
Cs
;
Vout
1

V in
CRs
;
Vout
  RLs
V in
2-3 Differential amplifier
Z out  R ; Z in  Ls
2-4 Difference amplifier
Rf
Va
R
a
V1
V2
b
V
b
R
V0
Rf
V0
node a：
V1  Va V0  Va

0
R
Rf
node b：
V2  Vb 0  Vb

0
R
Rf
Va  Vb
V2  Vb
 Vb

R
Rf
V2  Rf  Vb  Rf  RVb
Rf
 V2  Va
R  Rf
Rf
Rf
V1 
 V2 V0 
V2
R  Rf
R  Rf

0
R
Rf
Vb 
……
V0 
Rf
V2  V1 
R

R
3. Mechanical translation system (Mass –Spring –Damping System)
Exp1 :
Exp2 .
4. Mechanical rotational system
5. Electromechanical systems
(a) DC generator
Assumption : The angular velocity of the armature has been assumed to be
constant.
d
 0
dt
(b) Field current controlled DC motor
Assumption : The armature current I a is constant
(c) Armature current controlled DC motor (Servo DC motor)
Assumption : The field current I f is constant;
that means : flux  k  I f  cons tan t
Chapter 3 Feedback Control System
1. Block diagram models
Example 1
Example 2
Example 3
Mason’s formula
C ( s)
P 
 T ( s)  i i
R( s )

Pi = path gain or transmittance of ith forward path
 = determinant of graph
= 1-(sum of all individual loop gains)+(sum of gain
products of all possible combinations of two
non-touching loops)-( sum of gain products of all
possible combinations of three non-touching loops)+…
=1- Li + Li  L j - Li  L j  Lk +…
Li =sum of all loop gains
 i = cofactor of the ith forward path determinant of graph
with the loops non-touching the ith path
＊Signal flow chart
Exp1.
Exp2.
Exp3.
Exp4.
Chapter 4 System Response
※ Input signal type
4-1 First order system
C (s)
K
Standard form
 T (s) 
R( s )
s  1
Parameters :
K : gain ；
a. Step input
r(t)=A ;
R( s) 
τ : time constant
A=constant ；A=1
1
;
s
: unit step input
1 K
C ( s)  
s s  1
1
Take Laplace inverse c(t )   {C (s)}  K  Ke

t

Where c(t) : system’s response or performance
cs s  lim c(t ) : steady-state response
t 
ct
: transient response
Exp 1. The first order plant has the unit step response given in following
figure. Please find the parameters of the transfer function. And if
this plant is connected into the closed-loop system in following
figure. Please sketch the unit step response of the closed-loop
system.
Exp2.
Please plot the displacement x(t) vs time when f(t)=100N.
b. Ramp input
r(t)=At ;
A=constant
R( s) 
; A=1 unit ramp input
1
K
1
; C ( s)  2 
2
s
s s  1
Take Laplace inverse c(t )   {C (s)}  Kt  K  K  e
1

t

4-2 Second order system
n
C ( s)
Standard form
 T ( s)  2
2
R( s )
s  2n s  n
2
Parameters : ωn natural frequency
ζ damping ratio
1. Step input
r(t)=A ; A=constant
R( s) 
1
;
s
; A=1 unit step input
n
1
C ( s)   2
s s  2n s  n 2
2
Take Laplace inverse c(t )   1{C (s)}
Characteristic equation Q(s)=0 ;
Q(s)  s 2  2n s  n  0
2
Case I    2  1  0 ；   1 over-damped system
then characteristic roots are different negative real
roots λ1 and λ2
c(t )   1{C ( s)}  1  C1e 1t  C1e 2t
Case II    2  1  0 ；   1 critical-damped system
then characteristic roots are equal negative real roots λ1
c(t )   1{C ( s)}  1  C1e1t  C1t  e1t
Case III    2  1  0 ；   1 under-damped system
then characteristic roots are complex conjugates
λ1,2 =    n  n 1   2  i
c(t )   1{C ( s)}  1 
e  nt
1 2
sin( d t   )
ωd :damped natural frequency d  n 1   2
  tan 1
1  2

Response specifications in design
1
1. Rise time Tr    cos 
2
n 1  
2. Peak time
Tp 
3. Settling time

n 1   2
Ts 
4
n
4. Maximum overshoot M pt  100  e

 
1 2
Exp1. A unity negative feedback control system has the loop transfer
function
G(s) 
25
s ( s  6)
(a) Find the characteristic equation.
(b) Draw the unit input response.
Exp2.
Find the range of K for which the system is (a) under damped (b) over
damped (c) critical damped (d) Find the value of K that will result in
the system having minimum settling time.
Exp3 .
b. Ramp input
r(t)=At ;
A=constant ; A=1 unit ramp input
1
n
1
R ( s )  2 ; C ( s)  2  2
s
s s  2n s  n 2
2
Take Laplace inverse c(t )   1{C ( s)}  t 
2 
n

e  n
d
sin( d t   )
4-3 High order system
Method I : To expand the high order system into first or second order
partial fractions.
Method II : As a rule of thumb, the ratio b/a is 5 to 10 or greater the
pole at –b can be ignored.
Exp1 . T ( s) 
8
( s  10)( s 2  2s  4)
function response for unit step input.
The s-plane root location and the transient response
characteristic roots are complex conjugates
λ1,2 =    n  n 1   2  i
Chapter 5 Control system Characteristics
5-1 Stability
1. Absolutely stability ( Bounded input bounded output)
The requirement for a linear time-invariant system to be
stable is that all roots of characteristic equation Q(s)
must lie in the left half of the s-plane.
The closed loop Transfer function
T ( s) 
P( s )
Q( s )
▲ Routh-Hurwitz criterion
Characteristic equation can be expressed :
Q(s)  an s n  an1s n1  an2 s n2  .......  a2 s 2  a1s  a0  0
First step : All coefficients an an1 an2 a1 a0 must be the positive
real parts (same sign) and no loss.
Second step : To express the Routh-hurwitz array
First colun elements must be the same sign, the system is stable.
Exp1 . Q( s)  s 4  3s 3  s 2  5s  6  0
Exp1 . Q( s)  s 4  2s 3  s 2  2s  2  0
Exp3 . Q(s)  s 5  3s 4  5s 3  15s 2  4s  12  0
The system error is the difference between the desired output and practical
output.
e(t )  cd (t )  c(t )
cd (t )  r (t )
e(t )  r (t )  c(t )
E ( s )  Cd ( s )  C ( s )
Cd ( s )  R ( s )
E ( s )  R( s )  C ( s )  R( s ) 
•
Case 1 : r(t)=1
Gc ( s)  G p ( s)
1  Gc ( s)  G p ( s)
 R( s ) 
Es  s  lim e(t )  lim sE ( s )
； R( s) 
t 
1
s
s 0
1
 R( s )
1  Gc ( s)  G p ( s)
Final value theory
unit step input
R( s) 
1
s
Es s  lim s
s 0
1
1
1
1
 lim

1  Gc ( s )  G p ( s ) s s0 1  Gc ( s )  G p ( s ) 1  lim Gc ( s )  G p ( s )
s 0
K p  lim Gc ( s )  G p ( s )
s 0
Ess 
1
1 K p
Kp is c a l l e tdh p
e o s i t i eo rnr ocro n s t a n t
； R( s) 
Case 2 : r(t)=t
R(s) 
1
s2
unit ramp input
1
s2
E s s  lim s
s 0
1
1
1
1
1
1
 lim
 lim

2
s 0 1  G ( s )  G ( s ) s
s 0 s  sG ( s )  G ( s )
1  Gc ( s )  G p ( s ) s
lim sGc ( s )  G p ( s )
c
p
c
p
s 0
Kv  lim sGc ( s )  G p ( s )
s 0
Ess 
1
Kv
Kv is called the velocity error constant
Case 3 : r(t)=0.5t^2
R( s) 
； R( s) 
1
parabolic input
s3
1
s3
E s s  lim s
s 0
1
1
1
1
 lim 2

3
2
2
s

0
1  Gc ( s )  G p ( s ) s
s  s Gc ( s )  G p ( s ) lim s Gc ( s )  G p ( s )
s 0
Ka  lim s Gc ( s )  G p ( s )
2
s 0
Ess 
1
Ka
Ka is called the acceleration error constant
ΔSystem Type j :
G c ( s)  G p ( s) 
P( s)
s j Q ' ( s)
Since the transfer function of an integrator
is 1/s, then j is the number of free integrators in the transfer function
Gc ( s )  G p ( s ) . The number of free integrators, j, is called the system
type.
Case a Type
j=0
K p  lim Gc ( s )  G p ( s )  K p
Ess 
s 0
K v  lim sGc ( s )  G p ( s )  0
s 0
K a  lim s 2Gc ( s )  G p ( s )  0
s 0
Case b
Type
Ess
K v  lim sGc ( s )  G p ( s )  K v E s  s
s 0
K a  lim s 2Gc ( s )  G p ( s )  0
s 0
1
0
1 
1

Kv
Ess 
s 0
Type
1

0
1
 
0
Ess 
j=1
K p  lim Gc ( s )  G p ( s )  
Case b
1
1 K p
Ess 
1

0
j=2
K p  lim Gc ( s )  G p ( s )  
s 0
K v  lim sGc ( s )  G p ( s )  
s 0
K a  lim s 2Gc ( s )  G p ( s )  K a
s 0
1
0
1 
1
 0

1

Ka
Ess 
Ess
Ess
Exp 1: Supposed the open loop transfer function G (s ) has a unity negative
feedback.
Find the steady-state error when r (t )  5  2t .
(a) G ( s) 
10
( s  1)( s  3)
(b) G ( s) 
10
s( s  1)( s  6)
Exp 2: Supposed the open loop transfer function of controller and plant are
100
.This control system has a unity negative feedback. Find
Gc ( s); G p ( s) 
( s  1)( s  3)
the transient-response terms and steady-state for cases when r (t )  10 for following
controller type.
(a) P controller Gc( s )  K . K  0.25
0 .1
s
(c) PD controller Gc ( s)  1  0.3s .
(b) PI controller Gc ( s )  1 
Exp 3.
5-3 Disturbance rejection
To reduce the effects of disturbance, gain Td(s) should be small enough.
(1) Reduce the gain Gd(s)=0 between the disturbance input and output. It is hard to do
that.
(2) Increase the loop gain Gc(s). This is usually accomplished by the choice of the
compensator.
(3) Reduce the magnitude of the disturbance D(s). This should always be attempted, if
reasonable. ;
(4) Use feedforward compensation, if the disturbance can be measured.
Exp1 : Consider the model of the telescope-pointing system shown in Figure 6. Please
find the value of Ka let the system’s final output reaches 0.9 to unit step input and
disturbance.
5-3 Sensitivity
System sensitivity is the ratio of the change in the system transfer function to
the change of a process transfer function (or parameter) for a small incremental
change.
ST 
T / T T 


 /   T
S GT p 
T / T
T G p
1



G p / G p G p T
1  Gc G p H
S HT 
 Gc G p H
T / T
T H

 
H / H H T 1  Gc G p H
Exp1.
Exp2.
Exp3 .
Chapter 6 Root Locus Method
A close-loop control system is described by the transfer function
T ( s) 
Gc G p
1  Gc G p H
;
Q(s)  s 2  2n s  n  0
2
characteristic roots are complex conjugates
λ1,2 =    n  n 1   2  i
The s-plane root location and the transient response
Exp1. Gc ( s)  K ; G p ( s) 
1
( s  1)( s  3)
Q( s )  s 2  4 s  3  K  0
λ1,2 =  2  4  (3  K )  i
＊ The root locus is the path of the roots of the characteristic equation
traced out in the s-plane as a system parameter is changed.
When the controller is a variable parameter K that means
Gc ( S )  K . Then T ( s ) 
KG p
1  KG p H

P( s)
Q( s)
Characteristic equation Q(s)=0 ; Q( s)  1  KG p H  0 ;
Gp H  
1
1
   180 &deg;
K K
Root locus principle
Gp H 
z s  z s  z s 
z s  z s  z s 
(s  z1)( s  z 2)( s  z3)
 1 2 3
 1 2 3
 zi   pi
(s  p1)( s  p2)( s  p3)    p1s  p2 s  p3 s    
p1s  p2 s  p3 s    
ZEROS : z1 . z2 . z3 ……
POLES : p1 . p2 . p3 ……
&lt;1&gt; Magnitude criterion
1
Gp H 
K
1
; Gp H  =
K
z1s  z 2 s  z3 s     
p1s  p2 s  p3 s    
&lt;2&gt; Angle criterion
G p H   zi   pi  180
＊ Root locus technique
1. Root locus is symmetrical with respect to the real axis.
2. The root locus originates on the poles of GH and terminates
on the zeros of GH, including those zeros at infinity.
3. The root locus on the real axis to the left of an odd sum of
Np &amp; Nz.
Exp1.
Exp2.
G p ( s) H ( s) 
s
(s  2)( s  4)
G p ( s) H ( s) 
s
(s  2)( s  4)( s  6)
4. The loci proceed to the zeros at infinity along asymptotes
centered at  a and with angles a .When the number of finite
zeros of P(s), M, is less than the number of poles n by the number
N=n-M, then N sections of loci must end at zeros at infinity. These
sections of loci proceed to the zeros at infinity along asympotes as K
approaches infinity. These linear asympotes are centered at a point on
the real axis given by
a 
 Pi   Z j
nM
The angle of the asympotes with respect to the real axis is
a 
Exp3.
2  1
 180;   0,1,2,3,4....
nM
G p ( s) H ( s) 
1
(s  2)( s  4)( s  6)
5. The breakaway points on a root locus will appear among the
roots of the polynomial obtain from
GH
0
s
Exp4.
G p ( s) H ( s) 
1
(s  8s  32)( s  4)
Exp5.
G p ( s) H ( s) 
(s  1)
s 2 (s  10)
2
6. Determine the angle of departure of the locus from a pole
and the angle of arrival of the locus at a zero, using the phase
angle criterion..
7. Complete the root locus sketch.
Exp5.
G p ( s) H ( s) 
1
s(s  8s  32)( s  4)
2
Exp6. page 409 figure 7.2
page 410 figure 7.5
Exp 7 . G p ( s) H ( s) 
1
(s  2s  5)( s 2  6s  10)
2
(s 2  6s  10)
Exp 8 . G p ( s) H ( s)  2
(s  2s  5)
＊The
Design of Feedback Control Systems
1. P-D controller
Gc ( s)  K p  K d  s
Exp 9 .Open loop transfer function G p ( s) H ( s) 
1
.
( s  2s)
2
Requirements of the
unit step response specifications are settling time is equal to 0.5 seconds and
damping ratio is less than 0.707.
2. P-I controller Gc ( s )  K p  K i 
1
s
Exp10. Open loop transfer function G p ( s) H ( s) 
0.25 .
( s  0.1)
Requirements of the
step response specifications are steady-state error is 0.4 for unit step input and
damping ratio is equal to 1.
Gc (s) 
K (s  z)
and z  p ( Place the zero of the
( s  p)
compensator directly below the desired root )
Exp11. Open loop transfer function G p ( s) H ( s) 
1
.
s ( s  1)( s  5)
Requirements of
the unit step response specifications are settling time is equal to 3 seconds and
damping ratio is equal to 0.707.
4. Phase Lag compensator
Gc (s) 
K (s  z)
and p  z ( Place the zero of the
( s  p)
compensator near the origin of the s-plane in comparison to desired root )
Exp11. Open loop transfer function G p ( s ) H ( s ) 
1
.
s ( s  10) 2
Requirements of the
response specifications are velocity error constant K v  20 and damping ratio is
equal to 0.707.
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