Chapter 2
2.1
Mathematical Models of Control Systems
Introduction
Mathematical models of control systems are mathematical expressions which describe the
relationships among system inputs, outputs and other inner variables. Establishing the
mathematical model describing the control system is the foundation for analysis and design of
control systems. Systems can be described by differential equations including mechanical systems,
electrical systems, thermodynamic systems, hydraulic systems or chemical systems etc. The
response to the input (the output of the system) can be obtained by solving the differential
equations, and then the characteristic of the system can be analyzed. The mathematical model
should reflect the dynamics of a control system and be suitable for analysis of the system. Thus,
when we construct the model, we should simplify the problem to obtain the approximate model
which satisfies the requirements of accuracy.
Mathematical models of control systems can be established by theoretical analysis or practical
experiments. The theoretical analysis method is to analyze the system according to physics or
chemistry rules (such as Kirchhoff’s voltage laws for electrical systems, Newton’s laws for
mechanical systems and Law of Thermodynamics). The experimental method is to approximate the
system by the mathematical model according to the outputs of certain test input signals, which is
also called system identification. System identification has been developed into an independent
subject. In this chapter, the theoretical analysis method is mainly used to establish the mathematical
models of control system.
There are a number of forms for mathematical models, for example, the differential equations,
difference equations and state equations in time domain, the transfer functions and block diagram
models in the complex domain, and the frequency characteristics in the frequency domain. In this
chapter, we shall study the differential equation, transfer function and block diagram formulations.
2.2
Time-Domain Mathematical Models of Control Systems
2.2.1 Differential Equations of Linear Components and Linear Systems
The steps for establishing the differential equations of a system or a component by the analytical
method are:
（1）Determine the inputs and outputs of the system or component;
（2）According to the physical (or chemical) rules, write the functions describing every part of
the system in the sequence of the signal flow;
（3）Eliminate the intermediate variables to obtain the differential equations of the inputs and
outputs.
（4）Rewrite the differential equations in the standard form by putting the input related terms
on the right hand side and the output related terms on the left hand side in descending order.
The following are some examples for establishing differential equations.
EXAMPLE 2-1
Consider the R-C-L network shown in Figure 2-1. Write the differential equation between the input
voltage u , and the output voltage u c.
Figure 2-1
R-L-C network
Solution:
This is an electrical system. According to Kirchhoff’s voltage law, we have,
di (t )
 u c (t )
dt
du (t )
i (t )  c c
dt
u r (t )  Ri (t )  L
(2-1)
(2-2)
Eliminating the intermediate variable i in the two equations above leads to:
d 2 u c (t )
du (t )
LC
 RC c  u c (t )  u r (t )
2
dt
dt
(2-3)
Assume that R, L, C are constants. The equation above is a second-order linear time-invariant
differential equation.
EXAMPLE 2-2
Consider the spring-mass-damper system shown in
Figure 2-2, K is the spring constant, F is
damping ratio of the damper, m is the mass of the car.
The friction between the car and the floor is ignored.
Obtain the differential equations of the system,
where the force F (t ) is the input and the
displacement y (t ) of the car is the output.
Solution：
This is a mechanical system. The force analysis of the
car is shown in Figure 2-3. By Newton’s second law,
the horizontal motion can be described as
F (t )  f
By setting T 
Figure 2-2
Figure 2-3
dy (t )
d 2 y (t )
 Ky (t )  m
dt
dt 2
f
m
, 
，
K
2 mK
spring-mass-damper system
force diagram of the car
(2-4)
Equation (2-4) becomes
T2
d 2 y (t )
dy (t )
F (t )
 2T
 y (t ) 
2
dt
K
dt
(2-5)
The Equation (2-5) is a second-order linear time-invariant differential equation when K, f and m
are constants.
EXAMPLE 2-3
Try to write the differential equation of the
armature-controlled DC motor shown in Figure
2-4. The armature voltage u a (t ) is the input. The
rotate speed of the DC motor  (t ) is the output.
Ra , La are the zxresistance and inductance of
the armature circuit. M c (t ) is the loading
torque to the motor output shaft. Assume that the
excitation current i f is constant.
Figure 2-4 elementary diagram of DC motor
Solution：
It is an electrical-mechanical system. The armature-controlled DC motor converts the electrical
energy into the rotational mechanical energy. The input armature voltage u a (t ) provides the
armature current ia (t ) in the armature loop. The current ia (t ) and the excitation magnetic flux
work interactively to generate the magnetic torque M m (t ) on the rotor of the motor. Thus, the
load is driven. The differential equation of the motor consists of the following three parts.
① The voltage balance equation of the armature loop:
u a (t )  La
dia (t )
 Ra ia (t )  E a
dt
(2-6)
where: E a is the back electromotive-force (EMF) voltage proportional to the motor speed. That is
E a  C e (t )
(2-7)
where C e (V / rad / s ) is the coefficient of the back electromotive-force voltage.
② The motor torque is:
M m (t )  C m (t )ia (t )
(2-8)
where: C m ( N .m / A) is the torque coefficient of the motor. M m (t ) is the motor torque which is
generated by the armature current.
③ The torque balance equation on the motor shaft is:
Jm
d (t )
dt
 f m (t )  M m (t )  M c (t )
(2-9)
where: f m ( N .m / rad / s ) is the viscous friction coefficient. J m ( kg .m.s 2 ) is the moment of
inertia.
Considering the Equation (2-6)-(2.9) and eliminating the intermediate variables
ia (t ), E a , M m (t ) , the differential equation from the input to the output of the motor is
La J m
d 2 (t )
dt
2
 Cmu a (t )  La
 ( La f m  Ra J m )
d (t )
dt
 ( Ra f m  CmCe ) (t )
dM c (t )
 Ra M c (t )
dt
(2-10)
Equation (2-10) is a second-order linear differential equation. The inductance La in the
armature loop is small in value and can be neglected in practice. Thus Equation (2-10) can be
rewritten as:
Tm
where: Tm 
Kc 
d (t )
  (t )  K a u a (t )  K c M c (t )
dt
(2-11)
Cm
Ra J m
is the time constant of the motor. K a 
,
Ra f m  C m C e
Ra f m  C m C e
Ra
are the transmission coefficients of the motor. When Tm 、K1 、K 2 are all
Ra f m  C m C e
constants, equation (2-12) is a first-order linear time-invariant differential equation.
The angle position  (t ) of the motor is usually regarded as an output in servo systems. Therefore,
submitting
 (t ) 
d (t )
into Equation (2-11) leads to
dt
d 2 (t) d
Tm
  Kaua (t)  Kc Mc (t)
dt
dt2
(2-12)
EXAMPLE 2-4
Figure 2-5 shows the control system for the rotating speed of a motor. Obtain the differential
equation of the system.
Figure 2-5 The control system for the rotating speed of a motor.
Solution：
In this system, the given voltage ur (t ) is the input. The rotating speed  (t ) of the motor is the
output. The differential equations of each component in the signal flow sequence from the error
signal are:
① Measuring component. Tachometer is the measuring component, which transfers the system
output  (t ) to the voltage u f (t ) :
u f (t )  K t  (t )
(2-13)
where K t is the transfer function of the tachometer, which can be seen as a constant.
② Comparing component. Comparing element compares the feedback voltage u f (t ) and the
given voltage u r (t ) to generate the error voltage u e (t ) :
(2-14)
u e (t )  u r (t )  u f (t )
③ Amplifying component. In this system, amplifying components are voltage amplifier and power
amplifier. They amplify the voltage and power of the error voltage u e (t ) :
First amplifier：
u1 (t ) 
R2
ue (t )
R1
(2-15a)
Second amplifier：
u 2 (t ) 
R4
u1 (t )
R3
(2-15b)
Power amplifier：
ua (t )  K 3 u2 (t )
(2-15c)
Thus
u a (t )  K u e (t )
where K 
(2-15d)
R2 R4
K 3 is amplifying constant for the amplifiers.
R1R3
④ Actuator. DC motor is the actuator. It converts the armature voltage u a (t ) to the shaft angle
speed  1 (t ) . According to Equation (2-11) in Example 2-3, the differential equation of DC motor
is:
Tm
d1 (t )
 1 (t )  K a u a (t )  K c M c (t )
dt
(2-16)
⑤ Reducer. Reducer is to reduce the speed and increase the moment. The differential equation of
the reducer is :
 (t )
1

 1 (t )
i
(2-17)
where i is the transmission ratio constant of the reducer.
Eliminating the intermediate variables u f (t ) , u e (t ) ,  1 (t ) in Equations (2-13)~(2-17)
leads to the differential equation of the motor rotating speed control system:
KK a
K
d (t ) i  KK a K t
u r (t )  c M c (t )
(
) (t ) 
dt
iTm
iTm
iTM
(2-18)
It is obvious from the above mathematical models that different components or systems may
have the same mathematical model. For example, the mathematical models in Example 2-1 and 2-2
are second-order differential equations, while the mathematical models in example 2-3 and 2-4 are
first-order differential equations. Systems with similar mathematical models are similar systems.
Similar systems have the similar characteristics although they are different in physical formality.
The similar principle reveals the relationship between different physical phenomena. The computer
simulation of control systems are based on the similar system idea.
2.2.2 Linearization of Nonlinear Differential Equations
The above components and systems are supposed to be linear, so the mathematical models of them
are linear differential equations. Actually, all components or systems are nonlinear to some extent.
For example, rigidity of a spring is related to its formation. It is not always a constant.
Some parameters such as resistant R, inductance L and capacitance C are related to the
environment (temperature, humidity, pressure, etc) and the current going through. Thus, they are
not always constants. Friction, dead-zone or some other nonlinear factors will make the differential
equation complex and nonlinear. Strictly speaking, mathematical models for real systems are
always nonlinear.
Unfortunately, so far there is no universal solution for nonlinear differential equations. Thus,
nonlinear systems are usually linearized based on reasonable rules. Nonlinear systems usually can
be represented by linear differential equations in a small value range of variables so that they can
be analysis and design by linear system theories. Although it is an approximate solution, it is
convenient for analysis and calculation in practice.
EXAMPLE 2-5
Figure 2-6(a) shows a solenoid coil. Obtain the differential equation for the system, whose input is
the voltage u r and the output is the current i .
Solution：
By Kirchhoff’s law, we have:
u r  u1  Ri
(2-19)
u
where 1 is induced potential of the coil. It is proportional to the changing ratio of the magnetic
flux in the coil. Thus:
u1  K 1
d (i )
dt
(2-20)
where K1 is a constant. The magnetic flux of the coil is a nonlinear function of the current i shown
in Figure2-6(b). Substituting Equation (2-20) into Equation (2-19) leads to:
K1
d (i ) di
 Ri  u r
di dt
(2-22)
It is evident that the equation is nonlinear.
(a)
(b)
Figure 2-6 A solenoid coil and the curved shape of its magnetic flux  (i )
If the voltage and the current of the coil varies vary in a small range around the equilibrium point
( u 0 ,i 0 ) and
series:
 (i) is smooth over the range of i 0 , the magnetic flux  could be expressed in Taylor
1 d 2
d
i 
(i ) 2  
  0 
2
2! dt i
di i0
0
where i ＝ i  i0 . The higher order terms could be neglected if i is small enough in value.
Thus
  0 
where
d
di
i0
d
di
i0
i
is the derivative of Φ( i ) at i 0 . By representing it by C 1 , we have:
   0  C 1 i
   0     C 1 i
Thus, the relation between the current and magnetic flux can be approximated by linear equation.
Rewrite u r ， ，i in the equation (2-21) into the increment equation around the equilibrium point.
u r  u 0  u r
i  i0  i
   0  C1 i
By substituting the above three equations into equation (2-21) and eliminating the intermediate
variables, we obtain:
K 1C1
di
 Ri  u r
dt
(2-22)
Equation (2-22) is the linearized increment differential equation of the solenoid coil. It is
usually written in the following form
di
(2-23)
 Ri  u r
dt
where rather than the actual value, ur and i now represent the increments around the equilibrium
K 1C1
point.
Systems that can not be linearized will be studied in Chapter 7.
§2.2.3
The solution of the linear time-variant differential equation
The aim of the differential equation is to analyze the dynamical characteristic by the
mathematic method quantificationally. Thus the differential equation needs to be solved. The
differential equation can be transformed into algebraic equation in complex domain, and then the
analytical solution of the differential equation will be obtained by taking inverse Laplace transform
of the solution of the algebraic equation.
EXAMPLE 2-6
The R-C passive network is shown in Figure2-7, ur (t )  U r  1(t ) , uc (0)  u0 . Obtain the
uc (t ) when K is closed.
Figure 2-7 R-C circuit
Solution：The voltage equation can be obtained by the Kirchhoff's law. Noticing i  C
duc
, we
dt
have
RC
duc (t )
 uc (t )  ur (t )
dt
（2-24）
Taking the Laplace transform of the both sides:
RC[ sU c ( s )  u0 ]  U c ( s) 
Ur
s
Solving U c ( s ) , we have
U c (s) 
u0
Ur
RC
U
Ur
u0  r 


1
s( RCs  1) RCs  1
s s 1
s
RC
RC
（2-25）
Taking the inverse Laplace transform of the Equation (2-25), the analytical solution of the
differential equation is
uc (t )  U r (1  e

t
RC
)  u0 e

t
RC
（2-26）
The first term of the above equation is the particular solution of the ur (t ) which is called as
the zero state response; the second term is the homogeneous solution of the u0 which is called
zero input response.
§2.2.4
The dynamic mode
The solution of the differential equation is constituted by the general solution of the
homogeneous solution and special solution of the given signal. The general solution reflects the
law of the natural response. If the eigenroot is
1 ， 2 ， , n
and has no multiple root, the
e  t ， e  t ， , e  t are called as the dynamical mode of the differential equation, also called as
1
2
mode shape.
n
If there exits multiple root
,
the mode is the function with the form of
te t ，
t 2 e t ,  .
    j ,
e t sin  t 、 e t cos  t .
If the eigenroot has conjugate multiple root
e (  j ) t 、 e (  j ) t can be written as
its conjugate multiple mode
Every mode can be viewed as the basic dynamical mode of the natural response of the linear
system which is the linear combination of the corresponding mode.
2.3
Complex Domain Mathematical Models of Control Systems
The differential equation is the mathematical model of control systems in the time domain. If the
external excitation and the initial condition are given, all the information of the output with time
can be obtained by solving the differential equation. Although this method is accurate, it is tedious
to rewrite the equation when the structure or the parameters of the system are changed.
The transfer function is the mathematical model in the S-plane, which is based on Laplace
transform. By using transfer function, we can study the dynamic response of the system, as well as
the effect of the structure or the parameter variation. The root locus method and the frequency
response method are based on the transfer function. It is the most important and fundamental
mathematical model for control systems.
2.3.1
Transfer function
1．Definition of transfer function
The transfer function of a linear time-invariant system is defined as the ratio of the Laplace
transform of the output to the Laplace transform of the input under the assumption that all initial
conditions are zero.
Let us consider that the input-output relation of a linear time-invariant system is described by
the following nth-order differential equation with constant real coefficients:
dnc(t)
dn1c(t)
dc(t)

a
...a1
a0c(t)
n1
n
n1
dt
dt
dt
an
 bm
d m r (t )
d m 1r (t )
dr (t )

b
 ...  b1
 b0 r (t )
m 1
m
m 1
dt
dt
dt
(2-27)
where c(t) is the output, r(t) is the input, and a n , a n 1 ,..., a 0 and bm , bm 1 ,..., b0 are the
coefficients decided by the system structure and parameters.
To obtain the transfer function of the linear system, we simply take the Laplace transform on
both sides of the equation and assume zero initial conditions. The result is
a s
n

n

 an 1s n 1  ....  a1s  a0 C ( s )

 bm s m b m 1 s m 1  ...  b1s  b0 R( s)
(2-28)
The transfer function of the system is:
C ( s ) bm s m  bm 1 s m 1  ...  b1 s  b0

 G ( s)
R( s ) a n s n  a n 1 s n 1  ...  a1 s  a 0
(2-29)
The transfer function is obtained under the assumption that all initial conditions are zero.
The assumption indicates that i) the input is connected to the system after the point t=0 , that is,
the input and its higher order derivatives were zero when t  0 ; ii) the system was in equilibrium
before the input is connected to the system, that is, the output and its higher order derivatives were
zero when t  0 . Most systems in practice satisfy this assumption. Under this assumption, the
calculation is simplified. This assumption is also essential to compare system responses under the
same condition.
EXAMPLE 2-7
Obtain the transfer function of the R-L-C network in Example 2-1.
Solution:
From Equation (2-3) of Example 2-1, the differential equation of this R-L-C network is
d 2 u c (t )
du (t )
LC
 RC c  u c (t )  u r (t )
2
dt
dt
When all initial conditions are zero, take the Laplace transform on both sides of the equation. The
transfer function of the network is consequently written
G (s) 
U c (s)

U r (s)
LCs
2
1
 RCs  1
In practice, loading and interaction between interconnected components or systems may occur.
If the loading of interconnected devices does occur, the engineer must account for this change in
the transfer function and use the corrected transfer function in the subsequent calculations.
2．Properties of the transfer function
（1）The transfer function is a rational fraction of the complex variable s. It has the properties
of the complex function. In practice, the highest power (n) of s in the denominator is greater or
equal to that (m) of the numerator, that is n≥m, because of the inertia of the system and the limited
power of the power source.
（2）The transfer function is a property of a system itself, independent of the magnitude and
nature of the input or driving function.
（3）The transfer function is dependent of the differential equations.
（4）The inverse Laplace transform of the transfer function gives the impulse response of the
system. Therefore, the transfer function reflects the dynamic characteristics of the system. Since the
Laplace transform of the impulse function is 1 (R(s)= L (t )  1 ), we have
 C ( s) 
L1 G ( s)  L1 
 L1 C ( s)  k (t )

 R( s) 
（5）The transfer function is related to the poles and zeros in s-plane.
2.3.2 Transfer Functions for Common Components of Control Systems
（1）Regulation resistance
Regulation resistance could transfer the displacement of linear or angular to voltage. In a
control system, single regulation resistance is always used for the equipment for signal
transformation. As what’s shown in Figure 2-8(a). A pair of regulation resistance can be
composed to be deviation detector. As what’s shown in Figure 2-8(b).
Figure 2-8 Regulation resistance and its characteristic
When it is idle load, relation between angular displacement  (t ) of a single regulation resistance
brush and output voltage can be expressed as:
u (t )  K 1 (t )
(2-30)
Here: K1  E /  max is the output voltage corresponding to the unit angular displacement of the
brush. It is called regulation resistance transfer coefficient. E is electric power source voltage of the
regulation resistance.  max is the maximum working angle of regulation resistance(rad).
G ( s) 
U (s)
 K1
( s )
(2-31)
Equation (2-30) shows that the transfer function of the regulation resistance is a constant decided
by electric power source voltage E and maximum working angle  max . The transfer function of the
regulation resistance can be represented by the block diagram as what is shown in Figure 2-8 (d).
When angular deviation detector is composed by a pair of same regulation resistances, the
output voltage is:
u (t )  u1 (t )  u 2 (t )  K 1 [ 1 (t )   2 (t )]  K 1  (t )
Here: K1 is the transfer function of a single regulation resistance.  (t )   1 (t )   2 (t ) is the
difference between the angular displacement of the two regulation resistances. It is called deviation
angle. Therefore, the transfer functions of deviation detector and single regulation resistance has
the same form when deviation angle is the input. As
G ( s) 
U (s)
 K1
( s )
Pay attention to the load effect when using regulation
resistance. Load effect is the influence which is
generated when the output port of the element is
loaded. Figure 2-9 is the circuit diagram of the
regulation resistance when its output is connected
with the loaded resistance Rl . Assume the resistance
of the regulation resistance is R p , so the loop electric
current is:
E  u (t )
RP  RP
'

Figure 2-9
the load-effect of the
regulation resistance
u (t )
RP
(2-32)
'

u (t )
Rl
After re-organizing the equation, the output voltage of the regulation resistance can be written as
below:
u (t ) 
E
Rp
'
R
R
 p (1  p )
'
Rp
Rp Rl
E (t )

R  (t )
 (t )
 max [1  p
(1 
)]
 max
Rl  max
So, because of the influence of the loaded resistance Rl , the relation between the output voltage
u (t ) and the angular displacement  (t ) is not he linear one anymore. If the loaded resistance
Rl is very strong, for example, when Rl  10 R p , it can be obtained approximately:
u (t )  E (t ) /  max  K 1 (t )
（2）Differential synchro
Differential synchro consists of a transmitter and a receiver (called control transformer in another
way). Figure 2-10 shows the elementary diagram and the method of connection.
Figure 2-10
the principle of the differential synchro
Figure 2-11
the structure of the differential synchro
The principle of work is written as below:
Input an alternating excitation voltage e1 (t )  E1 sin t on the single-phase winding of the
transmitter’s rotor. After, an impulsive magnetic flux  r will be generated on the transmitter.
Therefore, there will have current in the three-phase winding of the stator. This current will
generate an impulsive magnetic flux  c in the receiver. When  c  90    r , rotor winding will
not be linkage with the magnetic flux  c , output e(t )  0 . When  c  90    r ,  c will generate
inductive kick e(t ) in the rotor winding of the receiver, and the amount is:
e(t )  K s cos( r   c ) sin t
(2-33)
Ks 
E
E

r c e
(2-34)
Here K s -- the sensitivity of the differential synchro ( V/c)
E -- deviation voltage
 r -- the rolling angle of the differential synchro transmitter
 c -- the rolling angle of the differential synchro receiver
 -- the pulsatance of the AC signal
When  c and  r satisfies  c  90    r , equation (2-33) could be written as below:
e(t )  K s sin( r   c ) sin t
When  e   r   c  15  , sin  e   e (rad)
e(t )  K s e sin t  E sin t
(2-35)
As to the detail about the working principle of the differential synchro and the derivation, please
refer to the books about electric machine control.
The differential synchro and the deviation angle detector consisted of the potentiometer have
the same block plan, as shown in Figure 2-11. The difference is that the voltage generated by
differential synchro is AC voltage.
（3）Tachometer
Figure 2-12 shows the mechanism of a tachometer. The rotor of the tachometer is connected with
the rotating shaft of the measured plant. The output voltage is proportional to the angle speed of the
rotor. Thus
u  K t  K t
d
dt
(2-36)
where,  is rotating angle of the rotor,  is rotating speed, u is output voltage, and K t is slope of
the output voltage. When the rotor changes its rotating direction, the polarity and phase position of
the output voltage will be changed.
Figure 2-12
Tachometers
Laplace transform of Equation (2-34) under zero initial condition leads to
U ( s )  K t ( s )  K t s( s )
(2-37)
Therefore, the transfer function of the tachometer is:
G ( s) 
U (s)
 Kt
( s )
or
G(s) 
U ( s)
 Kt s
( s )
(2-38)
The above two equations are the transfer functions of the tachometer. The output of the first one is
the rotating speed  (t ) , and the output of the second one is angle position  (t ) .
It is obvious that a system may have different transfer functions for different inputs and
outputs.
The structure of tachometer can be found in Figure2-13.
Figure 2-13
The structure of tachometer
（4）DC motor
The differential equation of the DC motor is given in Example 2-3:
Tm
d (t )
  (t )  K a u a (t )  K c M c (t )
dt
Suppose the initial condition is zero. By applying Laplace transform to the above equation, we
have the following transfer function.
Ga ( s ) 
Ka
( s )

U a ( s ) Tm s  1
(2-39)
where the loading torque is neglected.
If the output is the angle position
 (t ) , the transfer function becomes
Ka
( s )
Ga ( s ) 

U a ( s ) s (Tm s  1)
(2-40)
The structure of DC motor is shown in Figure 2-14.
Figure 2-14 The structure of DC motor
（5）Two–phase asynchronous motor
Two–phase asynchronous motors have merits of light weight and low inertia. They are applied in
control systems widely as low power actuators.
The principle of two-phase asynchronous motor can be found in Figure2-15 and it consists of
a high resistance rotor and two stator coils that are mutual perpendicular. One of the stator coils is
the excitation winding, and the other one is the control winding, which is connected to the output
terminals of the power amplifier to generate the AC control voltage with variable value and
polarity.
The torque-speed characteristic curve of the two–phase asynchronous motor is nonlinear and
the slope is negative. Figure 2-15(b) shows the measured curves for different control voltage u a .
Since the asynchronous motor usually works around the speed of zero, the linear part of the low
speed characteristic is extended to the nonlinear part of the high speed characteristic, which is
presented by the dotted line shown in Figure 2-15(b). In addition, we can also use the following
linear equation.
M m  f mm  M s
where M m is the output torque of the motor,
(2-41)
 m is the angle speed of the motor,
f m  dM m d m is the damping coefficient, the slope of the dotted line, and M s is the torque to
lock the rotor. From Figure 2-14(b), we have
M s  CM ua
(2-42)
where C M  M s E .
Figure 2-15 Second-phase asynchronous electric motor and its performance curve
By Newton Law, we have
d 2 m
d
 fm m
(2-43)
2
dt
dt
where  m is angular displacement of the motor rotor, J m is the moment of inertia.
Eliminating the intermediate variables M s and M m in Equation (2-40) and (2-41) followed
Mm  Jm
by
Laplace Transformation leads to the following transfer function,
G(s) 
where
m (s)
Km
CM


U a ( s ) s ( J m s  f m ) s (Tm s  1)
(2-44)
U a ( s )  L[u a (t )],  m ( s )  L[ m (t )] , K m  CM f m is the transfer coefficient and
Tm  J m f m is the time coefficient of the motor.
Since  m ( s )  s( s ) , Equation (2-41) can be rewritten as
G ( s) 
Km
 m ( s)

U a ( s ) Tm s  1
(2-45)
The transfer function and structure (Figure 2-16) of the two-phase asynchronous motors have
the same form with the DC motor.
Figure 2-16 The structure of two–phase asynchronous motors
Equations (2-44) and (2-45) are the transfer functions of the two-phase asynchronous electric
motor in different forms. It is obvious that they are same as the transfer functions of the DC motor.
（6）Gear trains
A gear train is a set or system of gears arranged to
transfer rotational torque from one part of a mechanical
system to another. Since motors with high rotating
speed and low torque are used often in control systems,
the gear train is often adopted to reduce the speed and
increase the torque.
Consider the gear train shown in Figure 2-17. The
rotating speed and the number of teeth of the driving
gear are 1 and Z 1 . Those for the driven gear are
2
and Z 2 .
The transmission ratio of the gear train is:
i
Figure 2-17 gear train
1 Z 2

2 Z1
(2-46)
In control system, the gear train is usually used to reduce the speed, thus i  1 . The transfer
function of the gear train is:
G ( s) 
2 ( s) 1

1 ( s ) i
(2-47)
The structure of gear train can be found in Figure 2-17.
Figure 2-18
The structure of gear train
The moment of inertia and the viscous friction coefficient to the rotor shaft are
1
 J2
i2
1
f  f1  2  f 2
i
J  J1 
For gear trains with idler gears, the moment of inertia and the viscous friction coefficient to the
rotor shaft are
1
1 2
J  J1  ( )2 J 2  (
) J3 
i1
i1  i2
(2-48)
1
1 2
f  f1  ( ) 2 f 2  (
) f3  
i1
i1  i2
(2-49)
It is evident from equation (2-48) and (2-49) that the moment of inertia and the viscous friction
coefficient of the gears which are close to the motor shaft (or the input shaft or the gear train) have
the dominant effect on the load of the motor. Therefore, reducing the moment of inertia and the
viscous friction coefficient of the gears which are close to the motor shaft will improve the
transient performance of the motor.
2.3.3 Basic Factors
Typical factors shown in Table 2-1 are transfer functions representing various components used in
control systems.
Table2-1
Typical link
Num
Name of the
Differential Equation
Transfer Function
Example
Regulation
resistance
amplifier
differential
synchro
1
Gain
c  K r
K
2
First-order
factor
T c  c  r
1
Ts  1
CR circuit
AC/DC motor
3
Quadratic
factor
T 2 c  2T c  c  r
0  1
1
2 2
T s  2Ts  1
R-L-C circuit
Spring-mass-da
mper system
4
Integral
factor
c  r
1
s
5
Derivative
factor
c  r
s
6
Reciprocal
first-order
factor
c   r  r
 s 1
7
Reciprocal
quadratic
factor
c   2 r  2 r  r
 2 s 2  2 s  1
Note that different components may have the same transfer function while a component may have
different transfer functions for different inputs and outputs.
Basic factors are elements of transfer functions. They frequently occur in arbitrary transfer
functions.
2.3.4 Standard Forms of Transfer Functions
Transfer functions can be written into the highest 1 type and the lowest 1 type.
（1）Highest 1 form (zero-pole form)
Rewrite Equation 2-27 into Equation 2-48.
m
G (s) 
K *  (s  z j )
j 1
(2-50)
n
 (s  p )
i
i 1
where, z1 , z 2 ,..., z m are the zeros of the transfer function, and p1 , p 2 ,..., p n are the poles.
The coefficients of the highest order terms of both the denominator and numerator are unit.
The transfer function is in the highest 1 form. The denominator and numerator in this form can be
factorized into the poles and zeros form.
（2）The lowest 1 form ( Basic factor form)
Rewrite Equation 2-26 into Equation 2-48.
G ( s)  K
m1
m2
k 1
n1
l 1
n2
 ( k s  1) ( l2 s 2  2 l s  1)
s
v
 (T s  1) (T
i 1
i
j 1
(2-51)
2
j
s  2T j s  1)
2
The coefficients of the lowest order terms of both the denominator and numerator are unit. The
transfer function is in the lowest 1 form. The denominator and numerator in this form can be
factorized into basic factors.
*
The K is called as gain. The relationship between the K and K is
m
K
K * z j
j 1
（2-52）
n
p
i 1
i
EXAMPLE 2-8
The transfer function of closed-loop system is  ( s ) 
30( s  2)
s ( s  3)( s 2  2 s  2)
①Obtain the K ;
②Obtain the differential equation;
③Draw the zero-pole diagram.
Solution:
*
①We know that K =30
Figure 2-19
Zero-pole diagram
K
② ( s) 
30  2
 10
3 2
C (s)
30( s  2)
30( s  2)

 4
2
R ( s ) s ( s  3)( s  2s  2) s  5s 3  8s 2  6s
( s 4  5s 3  8s 2  6s )C ( s )  30( s  2) R( s )
Take the Laplace transform, we can obtain the differential equation:
d 4c(t )
d 3c ( t )
d 2 c(t )
dc(t )
dr (t )
5
8
6
 30
 60r (t )
dt
dt
dt
dt
dt
③The Zero-pole diagram is shown in Figure 2-19.
2.4
Block Diagrams of Control Systems
2.4.1 Block Diagram
A block diagram can be used simply to describe the composition and interconnection of a system.
Together with transfer functions, it can describe the cause-and-effect relationships throughout the
system.
There are two methods to construct the block diagram: If the differential equations are known, then
taking the Laplace transform of the sub-equation of the equations and drawing the corresponding
sub-block diagram. The block diagram of the system can be obtained by connecting the sub-block
diagram; if the structure diagram is known, then transfer the name of every component into its
corresponding transfer function and represent all the variables by the Laplace form, thus the block
diagram can be obtained.
EXAMPLE 2-9
Equation (2-23) ~ (2-29) show the differential equations of the DC motor. Obtain the block
diagram.
dia (t )

ua (t )  La dt  Ra ia (t )  Ea

 Ea  Cem (t )

 M m (t )  Cm (t )ia (t )
 dm (t )
 f mm (t )  M m (t )  M c (t )
Jm
dt

Solution: Write the sub-block diagram for each sub-equation which is shown in Figure 2-20(a).
Connect the sub-block diagram to construct the block diagram of the system which is shown in
Figure 2-20(b).
解. 列出各子方程对应的子结构图如图 2-20(a)所示，联接子结构图成为系统结构图如图
2-20(b)所示。
I a ( s)
1

U ( s )  E ( s )  L s  R
a
a
a
 a


 Ea ( s)  Ce
 m ( s)


 M ( s)
m

 Cm
I

a (s)



m (s)
1
 M ( s)  M ( s)  J s  f
c
m
m
 m
Figure 2-20(b)
Figure 2-20(a)
sub-block diagram
The block diagram of the DC motor
EXAMPLE 2-10
Obtain the corresponding block diagram for the structure diagram of function recorder control
system shown in Figure 2-21.
Figure 2-21
The structure diagram of function recorder control system
Solution: From § 2.3.2, replacing the name of each component by its transfer function,
representing all the variables by the Laplace form. Thus the block diagram can be found in Figure
2-22.
Figure 2-22 The block diagram of function recorder control system
2.4.2 Block Diagram Transformations
The block diagram representations of a given system often can be reduced to a simplified block
diagram with fewer blocks than the original diagram.
（1）Combining blocks in cascade
When two blocks are connected in cascade, as shown in Figure 2-23 (a), we assume that
C ( s )  G2 ( s )U ( s )  G2 ( s )G1 ( s ) R ( s )
holds true.
Figure 2-23 equivalent transformation of two series connection link
Thus,
G ( s) 
C (s)
 G2 ( s )G1 ( s )
U (s)
(2-53)
The equivalent diagram is shown in Figure 2-23(b).
The results can be extended to the case of multiple factors in cascade. The transfer function of
a system with factors in cascade is the product of the transfer functions of the factors.
（2）Combining blocks in parallel
When two blocks are connected in parallel, as shown in Figure 2-17 (a), we have
C ( s)  G1 ( s) R( s)  G2 ( s) R( s )  [G1 ( s)  G2 ( s)]R( s)
Thus the transfer function of the paralleled system is:
G ( s )  G1 ( s )  G2 ( s )
(2-54)
The equivalent diagram is shown in Figure 2-24 (b).
The above results can be extended to multiple factors connected in parallel. The transfer function of
a system with factors in parallel is the sum of the transfer functions of the factors.
Figure 2-24 equivalent transformation of two parallel connection link
Table 2-2
Transformatio
n
Original Diagram
Block Diagram Transformations
Equivalent Diagram
Algebra
Combining
blocks in
cascade
C ( s)  G1 ( s )G2 ( s) R( s)
Combining
blocks in
Prallel
C ( s)  [G1 ( s)  G2 ( s)]R( s)
G ( s) R( s)
Eliminating a
feedback loop
C ( s) 
Moving a
summing
point ahead of
a block
C ( s)  R( s)G ( s)  Q( s)
1  G ( s) H ( s)
 [ R( s) 
Q( s)
]G ( s )
G ( s)
Moving a
summing
point behind a
block
C ( s )  [ R ( s )  Q ( s )]G ( s )
 R ( s )G ( s )  Q( s )G ( s )
Moving a
pickoff point
ahead of a
block
C ( s )  G ( s ) R( s )
Moving a
pickoff point
behind a block
Changing the
position of a
summing
point and a
pickoff point
R ( s )  R( s )G ( s )
1
G (s)
C ( s )  G ( s) R( s)
C ( s)  R1 ( s)  R2 ( s)
（3）Eliminating a feedback loop
Consider the block diagram shown in Figure 2-25 (a). The following equation can be obtained as:
C ( s )  G ( s ) E ( s )  G ( s )[ R( s )  B( s )]  G ( s )[ R ( s )  H ( s )C ( s )]
Thus
G ( s)
R( s )
1  G ( s) H ( s)
Therefore, the transfer function relating the output C ( s ) to the input R( s ) is
G ( s)
(s) 
1  G ( s) H ( s)
C ( s) 
(2-55)
The equivalent diagram is shown in Figure 2-25(b).
Figure 2-25 equivalent transformation of feedback connection
When the transfer function of the feedback path is H ( s )  1 , the system is a unity feedback system.
Consequently, the closed-loop transfer function is:
( s) 
G( s)
1  G( s)
(2-56)
Table 2-2 summarizes the basic rules for moving the summing point and the pickoff point.
EXAMPLE 2-11
Reduce the block diagram shown in Figure 2-26. Obtain the closed-loop transfer function
( s) 
C (s)
.
R( s)
Figure 2-26 block diagram of system
Solution:
The block diagram reduction procedure is based on the utilization of Table 2-2, item 3, which
eliminates feedback loops. Therefore the other transformations are used to transform the diagram to
a form ready for eliminating feedback loops.
First, Moving the pickoff point a behind and the summing point b behind the block G2 ( s ) ,
we obtain Figure 2-27 (a).
Then, combining the blocks H 3 ( s ) and
G2 ( s)
in cascade and H 2 ( s ) in parallel,
G4 ( s)
followed by eliminating the inner loop, we obtain Figure 2-27(b).
Finally, by reducing the loop containing H1 ( s ) , we obtain the closed-loop system transfer
function as shown in Figure 2-27 (c).
Thus, the closed-loop transfer function of the system is:
( s) 

C (s)
R( s )
G1 ( s )G2 ( s)G3 ( s)G4 ( s )
1  G2 ( s )G3 ( s ) H 3 ( s )  G3 ( s )G4 ( s ) H 2 ( s )  G1 ( s)G2 ( s)G3 ( s )G4 ( s) H 1 ( s )
Figure 2-27 equivalent transformation of the block diagram in example 2-11
It is interesting to consider if the pickoff point a can be moved before the summing point d directly.
We leave this problem to the readers.
2.5
Signal Flow Graphs of Control Systems
A signal-flow graph (SFG) is a diagram that represents a set of simultaneous linear algebraic
equations. A signal flow graph contains essentially the same information as a block diagram. The
advantage of the signal flow graph is the availability of a flow graph gain formula, which provides
the relation between system variables without requiring any reduction procedure or manipulation of
the flow graph.
2.5.1 Signal Flow Graphs
There are 3 basic graph symbols in the signal-flow graph: branch, node and branch gain.
Node – Represented by “  ”. A node is a point representing a variable or signal.
Branch–Represented by “  ”. A branch is a directed line segment joining two nodes. The arrow
head represents the direction of the signal transmission.
Gain–Represented by the transfer function “ G ”. The gain is equivalent to the transfer function in
the block diagram. It represents the relation between the signals in the arrow head direction.
Figure 2-28(b) shows the corresponding signal flow graph of the block diagram shown in
Figure 2-28(a).
Before we discuss signal-flow graphs, we must define certain terms.
Input node or source. An input node or source is a node that has only outgoing branches. This
corresponds to an independent variable. Nodes R and N in Figure 2-28(b) are input nodes. They are
actually input signals.
Output node or sink. An output node or sink is a node that has only incoming branches. This
corresponds to a dependent variable. Node C in Figure 2-28 (b) is a output node. It is actually the
output signal.
Figure 2-28 block diagram and signal-flow graph of the system
Mixed node. A mixed node is a node that has both incoming and outgoing branches. Nodes E, P
and Q in Figure 2-28(b) are mixed nodes. The mixed mode is the equivalent of the pickoff point in
block diagrams.
Forward path. A forward path is a path from an input node (source) to an output node (sink) that
does not cross any nodes more than once. The paths REPQC and NPQC in Figure 2-28(b) are
forward paths.
Loop. A loop is a closed path that starts and ends at the same node and does not cross any other
nodes more than once. Path EPQE in Figure 2-28(b) is a loop.
Loop gain. The loop gain is the product of the branch transmittances of a loop.
Forward path gain. A forward path gain is the product of the branch transmittances of a forward
path.
Non-touching loops. Loops are non-touching if they do not possess any common nodes.
2.5.2 Mason’s Gain Formula
Given the signal flow graph or block diagram of a complex system, the task of solving for the
input-output relations by algebraic manipulation could be quite tedious. Fortunately, there is a
general gain formula available that allows the determination of the input-output relations of an SFG
by inspection.
Mason’s gain formula, which is applicable to the overall gain, is given by
G ( s) 
1 n
 Pk  k
 k 1
(2-57)
where,  = determinant of graph.
  1   La   Lb Lc   Ld Le L f  
(2-58)
n = total number of forward paths between the output node and the input node.
Pk — path gain or transmittance of kth forward path.
 L — sum of all individual loop gains.
 L L — sum of gain products of all possible combinations of two nontouching loops.
 L L L — sum of gain products of all possible combinations of three non-touching loops.
a
b
d
c
e
f
 k — cofactor of the kth forward path determinant of the graph with the loops touching the Kth
forward path removed, that is, the cofactor  k is obtained from  by removing the loops that
touch path Pk .
EXAMPLE 2-12
Consider the system shown in Figure 2-29. Obtain the closed-loop transfer function
Figure 2-29
C (s)
.
R(s)
signal-flow graph of the system
Solution:
There are three forward paths and the forward path gains are P1  G1G2 G3 G4 , P2  G5G3G4
and P3  G1G6 . There are three loops and the loop gains are L1  G1G2G3G4 H 2 ,
L2  G1G6 H 2 and L3  G3 H1 . From Figure 2-27, we can see that there is only one pair of
nontouching loops, that is, the two loops are L2 and L3 .
Thus, the determinant  is given by:
  1  ( L1  L2  L3 )  ( L1 L3 )
 1  G1G2G3G4 H 2  G1G6 H  G3 H1  G1G3G6 H1 H 2
All loops are in touch with the forward path P1 and P2 . Thus  1   2  1 . Since the forward
path P3 does not touch the loop L3 , we obtain the cofactor  3  1  ( L3 )  1  G3 H1 .
Therefore the closed-loop transfer function is given by:
C (s) 1
 ( P11  P2  2  P3 3 )
R( s) 
G1G2G3G4  G3G4G5  G1G6 (1  G3 H1 )

1  G1G2G3G4 H 2  G1G6 H 2  G3 H1  G1G3G6 H1H 2
EXAMPLE 2-13
Consider the system shown in Figure 2-30. Obtain the closed-loop transfer function
Figure 2-30
C ( s)
.
R( s)
block diagram of the control system
Solution:
The following results are obtained by inspection of the block diagram.
There are two forward paths. The forward path gains and their cofactors are
p1  G1G2G3 ,
p2  H 4 ,
1  1,
 2  1  G2G3 2
There are four individual loops. The gains of these loops are
L1   H 3 H 4 , L2  G1G2 G3 H 3 , L3  G2 G3 H 2 , L4  G1 H 1 .
There are two pairs of nontouching loops, that is, Loop L1 does not touch and loop L3 does not
touch L4 .
Using Equation 2-58, the closed-loop transfer function is written
C(s) p11  p22


R(s)

2.6
G1G2G3  H4(1G2G3H2）
1＋H3H4 G1G2G3H3 G2G3H2 G1H1 G2G3H2H3H4 G1G2G3H1H2
Transfer Functions of control systems
The transfer functions we have studied are subjected to
the input signal. Besides the input signal, a system in
practice is also subjected disturbances. Figure 2-31
shows a closed-loop system that is subjected to
disturbance, where R(s) is the input signal, N(s) is the
disturbance signal, C(s) is the output of the system and
Figure 2-31 block diagram of the
E(s) is the error signal. R(s) and N(s) are external
close-loop system
excitations, while C(s) and E(s) are the outputs of the
system. The closed-loop system shown in Figure 2-31 is a double-input-double-output system. The
responses to the two inputs can be considered independently. Using the superposition principle, the
response of the closed-loop can be obtained.
2.6.1 Transfer Functions of Open-Loop Systems
The opened-loop transfer function is the product of the forward path transfer function and the
feedback path transfer function, that is
G ( s ) H ( s ) . It is the ratio of the feedback signal B(s)
and the error signal E (s ) .
G ( s) H ( s) 
B( s )
 G1 ( s )G2 ( s ) H ( s)
E ( s)
(2-59)
Note that the open loop transfer function is obtained by opening the closed-loop at the output of
H (s ) , or the summing point. Differing from the transfer function of open loop systems, it is for
closed-loop systems.
2.6.2 Transfer function of the closed-loop system
（1）The closed-loop transfer function subjected to the input signal
By setting N ( s )  0 , the closed-loop transfer function  (s ) for the output C (s ) to the
input R(s ) is written
( s) 
C ( s)
G1 ( s)G2 ( s)

N ( s ) 1  G1 ( s )G2 ( s) H ( s )
(2-60)
（2）The closed-loop transfer function subjected to the disturbance
input
By setting R( s )  0 , the transfer function subjected to the disturbance is written
 n ( s) 
C (s)
G2 ( s )

N ( s ) 1  G1 ( s )G2 ( s ) H ( s )
(2-61)
（3）The output subjected to both the input and the disturbance
signals
By the superposition principle, the output subjected to both the input and the disturbance signals is
the sum of the outputs to the input signal and the disturbance signal. For the system shown in
Figure 2-29, the output to both the input and the disturbance signals are
C ( s) 
G1 ( s)G2 ( s) R( s)
G2 ( s) N ( s )

1  G1 ( s)G2 ( s ) H ( s) 1  G1 ( s)G2 ( s) H ( s )
2.6.3 Transfer Functions between The Error Signal and The Input Signal
（1）The error transfer function subjected to the input signal
By setting the disturbance signal N ( s )  0 , we obtain the error transfer function subjected to the
input.
 e ( s) 
E ( s)
1

R( s ) 1  G1 ( s )G2 ( s ) H ( s )
(2-62)
（2）The error transfer function subjected to the disturbance
By setting the disturbance signal R ( s )  0 , we obtain the error transfer function subjected to the
input disturbance.
 ne ( s ) 
E (s)
 G2 ( s ) H ( s )

N ( s ) 1  G1 ( s )G2 ( s ) H ( s )
（3）The error subjected to both the input and the output signals
(2-62)
From Equation (2-63) and (2-64), the error subjected to both the input and the output signals is
E ( s) 
 G2 ( s ) H ( s) N ( s)
R( s )

1  G1 ( s)G2 ( s ) H ( s) 1  G1 ( s)G2 ( s) H ( s )
It is evident that the denominator of the four closed-loop transfer functions  (s ) ,
 n ( s) ,
 e ( s ) and  en ( s ) is the same, that is, the denominator is
1  G1 ( s )G2 ( s ) H ( s )  1  G ( s ) H ( s )
This is an essential characteristic of closed-loop control systems. The denominator polynomial is
called the characteristic polynomial of the closed-loop system. 1  G K ( s )  0 is the
characteristic equation of the closed-loop system. The roots of this characteristic equation are
called the roots or the poles of the closed-loop system.
EXAMPLE 2-14
The block diagram is shown in Figure 2-32. Obtain the c(t ) and e(t ) when r (t )  1(t ) ，
n (t )   (t )
Solution: The open-loop transfer function is
2
G (s) 
s ( s  3)
The closed-loop transfer function is
Figure 2-32 block diagram of the system
2
C(S )
2
s( s  3)
 ( s) 


2
R( s) 1 
( s  1)( s  2)
s( s  3)
1
C (S )
s
s3
 n (s) 


( s  1)( s  2)
N (s) 1  2
s ( s  3)
The output is
C ( s )   ( s ) R ( s )   n ( s ) N ( s )) 

2
1
s
 
( s  1)( s  2) s ( s  1)( s  2)
s2  2
1
3
3
 

s ( s  1)( s  2)
s s 1 s  2
c(t )  1  3e t  3e2t
We should notice that the system is unit feedback when we solve e(t ) .
e(t )  r (t )  c(t )  1  (1  3e t  3e2t )  3e  t  3e 2t
Summary
Mathematical modeling is the foundation of the control theory and the precondition for analyzing
and designing the control system. The contents of this chapter are mainly: four types of
mathematical models for control systems, three methods for obtaining transfer functions, and five
forms of the transfer functions of control systems.
This chapter focuses on the construction of the mathematical model of the system by the analysis
method. The process is shown in Figure 2-33
Figure 2-33 The process of constructing the mathematical model for the system
（1）Mathematical model is a mathematical expression to show the relationship among the
system input, output and each inner variable. It is the basis for system analysis. A simplified but not
distorted mathematical model is usually obtained by analyzing the working principle of the system,
neglecting the minor factors, writing the mathematical expression and linearizing.
The differential equation is the mathematical model in time-domain. To write the proper
differential equation of a system, we need to understand the working process first.
The transfer function is the ratio of the Laplace transformations of the input and the output at
the zero initial condition. It is an important mathematical model in the classical control theory to
help us to analyze complicated systems.
The block diagram and the signal-flow graph are mathematical models in graphics, by which
we can apply the Mason’s gain formula to obtain the transfer function of complicated systems.
（2）There are three methods to obtain the transfer function: Laplace transformation of
differential equations, transformation of block diagrams and Mason’s gain formula.
（3）Three kinds of transfer functions of closed-loop systems are used often. They are the
open-loop transfer function G ( s ) H ( s ) , the closed-loop transfer function ( s ) ,  n (s) and the
error transfer function  e (s ) ,  en (s ) . They play important roles in the system analysis and
design.
Exercises
E2-1
F (t ) , displacement x(t ) and voltage u r (t ) ,
while the displacement y(t ) and voltage u c (t ) are outputs. k ( elastic coefficient ), f ( damping
coefficient ), R ( resistance ), C ( capacitance ) and m ( mass ) are constants. Obtain the differential
Some systems are shown in Figure 2-34 with inputs force
equations describing the systems.
Figure 2-34 principle diagram of the system
E2-2 Obtain the transfer functions of the passive networks shown in Fig.2-35 by the complex
impedance method.
（a）
（b）
（c）
Figure 2-35 passive networks
E2-3
The mechanical system is shown in Figure 2-36(a), and the circuit system in Figure 2-36(b). Prove they are
similar systems, namely, they have the same form of mathematical model.
Figure 2-36 principle diagram of the system
E2-4
The diode shown in Fig.2-37 is a nonlinear component. The relationship between the current
voltage ud is id  10
14
(e
ud
0.026
 1) . Suppose the system has
id and
slightly changed at the
3
working point u (0)  2.39V , i (0)  2.19  10 A , try to find the linearization equation
for the id  f (ud ) .
Figure 2-37
E2-5
The relationship between the liquid level
container is given by
dh 

dt S
h
The diode circuit
h and input Qr of a
1
Qr
S
where S is cross-sectional area of the container,  is a constant.
One may assume that Qr and h
is slightly changed at the
working point. Obtain the linearized equation of
Qr .
E-2-6
Figure 2-38 shows a pendulum system. The

h in terms of
l
is the length of
m.
rod,
is the angle, the mass is
Obtain the differential
equation of the system and linearized it.
E2-7 Obtain the image functions
Figure 2-38
Pendulum system
X (s) of the signals x(t ) shown
in Figure 2-39.
E2-8
Figure 2-39 signal diagam
Obtain the primitive functions for the following Laplace transformations.
(1)
X (s) 
(2)
X ( s) 
(3)
(4)
E2-9
e s
s 1
2
s 9
2
1
s ( s  2) 3 ( s  3)
s 1
X ( s) 
s ( s 2  2 s  2)
X ( s) 
Under the zero initial condition, a unit step response of the system can be represented as
c(t )  1  2e 2t  et
Determine the transfer function and impulsive response of the system.
The transfer function of a system is
E2-10
C (s)
2
 2
R ( s ) s  3s  2
and the initial condition is c(0)  1 , c(0)
 0.
Determine the output c(t ) of the system with the input
r (t )  1(t ) .
E2.11 Determine the transfer function
U c ( s ) for the network which is shown in Firgure2-40.
U r (s)
Figure 2-40 network
E2-12 The principle block of a position follow-up system is shown in Figure 2-41. Let the maxima working angle
of the regulation resistance be Qm  330 , and the amplify coefficient of the amplifier is k 3 .
（1）Obtain the transfer function k0 of regulation resistance, the first-order amplifier coefficients k1 , and
the second-order amplifier k 2 respectively.
（2）Draw the block diagram of the system.
（3）Determine the closed-loop transfer function
Qc ( s) .
Qr ( s)
Figure 2-41 working principle block diagram
E2-13 The block diagram of the longitudinal attitude control system is shown in Figure 2-42. Obtain the
closed-loop transfer function
Qc ( s) .
Qr ( s)
Figure 2-42 block diagram of the longitudinal attitude control system
E2-14 The equation set of the system is shown as follows. Draw the block diagram and determine the closed-loop
transfer function
C (s)
.
R( s)
 X 1 ( s )  G1 ( s ) R( s )  G1 ( s )[G7 ( s )  G8 ( s )]C ( s )
 X ( s )  G ( s )[ X ( s )  G ( s ) X ( s )]
 2
2
1
6
3

 X 3 ( s )  [ X 2 ( s )  C ( s )G5 ( s )]G3 ( s )
 C ( s )  G4 ( s ) X 3 ( s )
E2-15 Sketch the structure and signal diagram for the RC passive network shown in Figure2-43 and obtain the
transfer function
U c ( s)
.
U r ( s)
Figure 2-43
E2-16 Obtain the transfer functions
RC passive network
C (s)
for the systems which are shown in Figure 2-44 by the method of
R( s)
block diagram equivalent predigesting.
Figure 2-44
E2-17 A control system is shown in Figure 2-45. Draw the signal-flow diagram
Figure 2-45
The structure of system
E2-18 Draw the structure of system for the signal-flow which is shown in Figure 2-46.
Figure 2-46 The signal-flow graph
E2-19 Obtain the corresponding closed-loop transfer functions for the E2-16 by Mason’s equation.
E2-20 Using Mason’s signal-flow gain formula, obtain the closed-loop transfer functions of the systems which are
shown in Figure 2-47
E2-21 A block diagram of system is shown in Figure 2-48, determine the transfer functions
Figure 2-47
C (s) C ( s)
,
.
R( s) E ( s )
The structure of system
H1
G0
G1
G2
G3
H2
Figure 2-48
The structure of system
E2-22 A block diagram of system is shown in Figure 2-49, where
signal. Determine the transfer functions
R(s) is input signal and N (s) is disturbance
C (s) C ( s)
,
.
R( s) N ( s )
Figure 2-49
The structure of system
E2-22 A double pendulum system hung on the swivel without friction is shown in Figure 2-50. Suppose the
the length of the rod, M is the mass; the angular displacement
linearized; when

is small,
l is
sin  ,cos can be
1   2 , the spring has no deformation, elasticity coefficient is K ; the force f (t )
acts on the left rod, let a  g / l  K / 4 M , b  K / 4 M .
（1）Obtain the equation of the double pendulum;
（2）Obtain the transfer function and draw the pole-zero diagram;
（3）Draw the structure of the double pendulum.
Figure 2-50
double pendulums