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Chapter 2 mathematical models of
control systems
1.what is mathematical models of control systems?
Used to describe the relation among input,output and internal
variables.
Two kinds:static model and dynamic model
2. mathematical models:
time domain:differential equation
complex domain:transfer function different forms
frequency domain: frequency transfer function
3.how to construct mathematical models:
analysis:analyze wok principles
experiment:system identification.
in fact,only small part system mathematical models can be gotten by
analysis。
2.1 construct time第一节
domain differential
equation
线性系统的微分方程
Assume loop current to be i (t)
（internal variable）,every component
is linear,and load effects is not
considered. According to Kichoof
rule,we get
from（2-2）,we can solve the relation between i(t) and uo(t) :
i(t) =Cdu0/dt
substitute into（2-1），and input ui(t), output uo(t), the differential
equation is following:
d 2u0 (t )
du0 (t )
L
( )  ( RC )

RC
 u0 (t )  ui (t )
2
R
dt
dt
Assuming T1=L/R,T2=RC which are the time constant in this
circuit.thus the differential equation is
Summary: a differential equation is a description among various
variables.the reference command is the input,the variables to
controlled is the output。
1、列写线性系统微分方程的主要过程和举例：
（1）define input and output;
（2）analyze work principle,write differential equation .
（3）eliminate internal variables.
（ 4 ） write input in the right side,output in the left in
decreasing power.
Assume the body mass to be m，the displacement 、velocity and
acceleration relative to the initial position x、dx/dt and d2x/dt2。Spring
force to m is k·x，and in the reverse direction to the displacement； the
force to m acceleration is m · d2x/dt2 ， and in the reverse direction to the
displacement；damped force is f · dx/dt ， and in the reverse direction to
the displacement 。According to Newton second rule,we get：
Fig 2-3 shows armature
control DC motor, input–
armature voltageUa(t)
（v），output—motor
revolution ωm（t）
（rad/s），write its
differential equ. 。In fig.
Ra(Ω)、La(H) are
armuture loop resistance
and inductance，
Mc(N·M) is total load
torque。If=const。
£«
if
La
£«
Ra
ia
Ua
Wm
Ea
SM
¸º
Jm, f m
ÔØ
ͼ2- 3 µçÊà¿Ø
ÖÆ
Ö±Á÷µç¶¯»úÔ-Àíͼ
armature control DC motor is to transform electrical energy
into mechanical energy.
armature Ua(t) → armature ia(t) →electrical-magnetic
torque Mm(t) →force load move。Its motion equ.
Includes:
• armature loop voltage equilibrium equ.
• electrical-magnetic torque equ.
• torque equilibrium equ in motor axis.
•armature loop voltage equilibrium equ. ：
dia (t )
U a (t )  La
 Ra ia (t )  Ea
dt
①
Ea---armature electromotive force whose
direction is opposite to Ua(t) ，that is
Ea=Ceωm（t） ② Ce－coefficient（v/rad/s）
•electrical-magnetic torque equ. ：
M m (t )  Cmia (t )
Cm
③
-coefficient （N·m/A）
• torque equilibrium equ in motor axis.
d (t )
M m (t ) J m m  f m m (t )  M m (t )  M c (t )
dt
Jm－revolving initia kg·m·
fm-friction efficient（N·m/rad/s）
④
③、④ solve ia(t)，substitute ：
d 2 m (t )
d m (t )
La J m
 ( La f m  Ra J m )
 ( Ra f m  C m C e ) m (t )
dt
dt
dM c (t )
 C mU a (t )  La
 Ra M c (t )
⑤
dt
If La is small，
Equ. ⑤ can be reduced as
dm (t )
Tm
 m (t )  K1U a (t )  K 2 M c (t )
dt
Tm 
Ra J m
Ra f m  C m Ce
Cm
K1 
Ra f m  Cm Ce
⑥
Motor time constant（s）
Ra
K2 
Raf m  Cm Ce
If Ra and Jm are all small,equ. ⑥ can reduced to
Ce m (t )  U a (t )
⑦
m (t ) is proportional to U a (t ) ,the motor can be used
to a generator to measure revolution.
The system is the most basic differential equation.
chief procedures :
① define system's input and output .
② divide system into some parts,from the input,
according to signal sequence,list each part
differential equ.。
③write outs differential equ.by cancelling
middle variables 。
2.2 some other examples
Exam.2-1
R1
• fig.2-1 is a passive
U1
network of RC。
assume input---U1（t），
output---U2(t),give it’s
differential equ.
C1
ͼ2- 1
R2
C2
RC×é³ÉµÄËÄ
¶ËÍøÂç
U2
Assume loop current i1、i2，from Kichihuff rules：
U 1  R1i1  U c1
U c1
1

(i1  i2 )dt

C1
U c1  R2 i2  U c 2
U c2 
1
i2 dt

C2
U 2  Uc2
由④、⑤得
①
②
③
i1
U1
R1
C1
i2
R2
C2
④
⑤
ͼ2- 1
RC×é³ÉµÄËÄ
¶ËÍøÂç
U2
dU c 2
dU 2
i2  C 2
 C2
dt
dt
From ② we get
dU c1
dU c1
dU 2
i1  C1
 i2  C1
 C2
dt
dt
dt
Substitute i1、i2 to ①、③，we get
U 1  R1  R2 i2  U c 2
dU c1
dU 2
dU 2
 R1 (C1
 C2
)  R2 C2
U2
dt
dt
dt
 R1[C1
dU 2
dU 2
d
( R2 i2  U 2 )  C2
]  R2 C2
U2
dt
dt
dt
d 2U 2
dU 2
dU 2
dU 2
 R1C1 R2 C 2
 R1C1
 R1C 2
 R2 C 2
U2
2
dt
dt
dt
dt
d 2U 2
dU 2
R1 R2 C1C 2
 ( R1C1  R1C 2  R2 C 2 )
 U 2  U1
2
dt
dt
It is a second-order differential equ. 。
Exam. 2-2
have a try to illustrate two system in fig.2-2(a)、
(b) to be similar system (same math model)。
ͼ2- 2 »úµçÏàËÆ
ϵͳ
R2
B1
B2
K1
Xr
C2
R1
Ur
C1
K2
Xc
( a) »úеϵͳ
( b) µçÆø
ϵͳ
Uc
Mechanical system (a)：
input--Xr，output --Xc，
from force equilibrium:
• Electrical system (b)，
from current and voltrage
equilibrium ：
R2 i 



K 1 (X r - X c )  B1 (X r - X c )  K 2 X c  B 2 X c


( B1  B2 ) X c  ( K1  K 2 ) X c  B1 X r  K1 X r
1
1
idt

R
i

idt  U r
1


C2
C1
C1U c1  C 2U c2
②
U c  R1i  U c1
③
(R 1  R2 )i  U c1  U c2  U r
④
From ②、③、④ we get
C1
)Uc
C2
i
C1
R1  R 2  (1 
) R1
C2
Ur  (1 
Substitute into ①,differentiate ① we get


1
1
1
( R1  R2 ) U c  ( 
)U c  R1 U r  U r
C1 C 2
C1
force-voltage similarity
Electrical
system
resistance
R1
mechanical ramp
system
B1
resistance
R2
ramp
B2
elastic
elastic
coefficient coefficient
K1
K2
1/C1
1/C2
• mechanical system （a）Electrical system（b）have same
math model，so called similar system。（ Electrical system
is equivalent to mechanical system ）
• similar system shows similar relation among different physical
value。
• We may utilize simple system to study complex system
In following system,there are two inputs（ua and ML）,The output is
ωc.give its mathematical model.
R2
ug
R3
R1
R1
R4
-k1
u1
ω
–
u2 功放 ua
-k2
kf
k3
cf
SM
n
ωc
Ce,
cm,
J,f,
ML
R2
amplifier1 : u1  k1 (u g  u f )   (u g  u f )
R1
u1
du1
amplifier 2 : u2   R4 (  C
)
R3
dt
gears :    c / n
techogenerator : u f  k f  c
Finally,we can get the relation between ug and ωc.
dug
d 2
d
T1 2  T2
   k1 '
 k2 ' ug  kc ' M c
dt
dt
dt
Solve differential equation:
Exam.
in exam.2-1，assume
when
U 1(t )  (t )
R1  20K, R2  3K, C1  0.1F , C2  20F
and the initial value is 0,solve
From exam. 2-1
R1
U1
C1
ͼ2- 1
R2
C2
RC×é³ÉµÄËÄ
¶ËÍøÂç
U2
U 2 (t )
d 2U 2
dU 2
R1 R2 C1C 2
 ( R1C1  R1C 2  R2 C 2 )
 U 2  U1
2
dt
dt
R1 R2 C1C 2 S 2U 2 ( s )  ( R1C1  R1C 2  R2 C 2 ) SU 2 ( s)  U 2 ( s)  U 1 ( s ) 
U 2 ( s) 
1
S[ R1 R2C1C2 S 2  ( R1C1  R1C2  R2C2 ) S  1]

1
S (1.2  10  4 S 2  0.462S  1)
10 4
a
b
c

 

1.2S ( S  2.166)( S  3847.85)
S S  2.166 S  3847.85
a  U 2 ( s) S
s 0
1
10 4
c
1.2S ( S  2.166)
10 4
b
1.2S ( S  3847.85)
s  3847.85
s  2.166
1
S
 1.000043
 5.63  10 4
U 2 (t )  1  1.00043e 2.166t  5.63  10 4 e 3847.85t
Exam.
in exam.2-1，assume L  1H , C  1F , R  1
when U1(t)=I(t), and the initial value is
uc(0)=0.1V,i(0)=0.1A, solve u0
From exam. 2-1
LC[ s 2U 2 ( s )  su2 (0)  u2 (0)]  RC [ sU 2 ( s )  u2 (0)]  U 2 ( s )  U1 ( s )
'
du2 du2
i ( 0)
'
i
i
iC
,

u2 (0) 

C
,
C
,
|
t 0
dt dt
C
U ( s)
0.1s  0.2
1
0.1s  0.2
U 2 ( s)  2 1
 2


s  s  1 s  s  1 s( s 2  s  1) s 2  s  1
1 1
1 0 .3
s


0
.
1
(
s

)
1
s 1
0.1s  0.2 1
2 2
2
2
[  2
] 2
[ 
]
s s  s 1
s  s 1
s
1
3
1
3
( s  )2  ( )2 ( s  )2  ( )2
2
2
2
2
1
3
0 .3 2
3
1
1



s
0.1( s  )
1
2
2
3
3 2
2
2
 



s
1 2
3 2
1 2
3 2
1 2
3 2
1 2
3 2
(s  )  ( )
(s  )  ( )
(s  )  ( )
(s  )  ( )
2
2
2
2
2
2
2
2
A
s
1 1
1
1
L ( )  I, L ( 2
)  sin At, L ( 2
)  cos At,
2
2
s
s A
s A
A
s 
t
1
t
L1[
]

e
sin
At
,
L
[
]

e
cos At
2
2
2
2
(s   )  A
(s   )  A
u2 (t )  1  e
 1  1.15e
1
 t
2
1
 t
2
1
1
 t
 t
3
1  12 t
3
3
0
.
3
3
cos
t
e sin
t  0.1e 2 cos
t
e 2 sin
t
2
2
2
2
3
3
1
 t
3
3
sin(
t  120 )  0.2e 2 sin(
t  30 )
2
2
Zero state response
Zero input response
chief procedures :
① consider initial value,transfer
differential equ.into algebraic equ. in sdomain .
②solve output in s-domain.
③ transfer inversely the equ. in s-domain
into the form in the time-domain .