Principles of Automatic Control
School of Aeronautics and Astronautics
SHEN Qiang
2
Chapter 2 Dynamic Models
2.1 Concepts of Dynamic Models
2.2 Model of Mechanical System
2.3 Model of Electric System
2.4 Model of Eletromechanical System
2.5 Nonlinear Model of Aircraft System & Linearization
2.6 Typical Elements
2.4 Models of Electromechanical Systems
A common actuator used in
electromechanical system (机电系
统) is the direct current (DC) motor
to provide rotary motion.
2.4 Models of Electromechanical Systems
Example 1: The armature (电枢) – Controlled DC Motor
𝑇𝑚
𝑇𝑑
𝐽 : moment of inertia
𝑏: damping coefficient
𝑇𝑑 : torque from load (disturbance)
𝑇𝑚: torque from DC motor
𝐸: back electromotive force (emf)
𝜃: rotating angle
𝜔: rotating angular velocity
𝑉𝑎: control voltage
𝐼𝑎: control current
2.4 Models of Electromechanical Systems
Example 1: The armature (电枢) – Controlled DC Motor
𝑇𝑚
The Motor Equations:
𝑇𝑑
𝑇m = 𝐾m 𝐼𝑎
𝐸 = 𝐾𝑒 𝜔 = 𝐾𝑒 𝜃
armature current
back emf
The generated electromotive force (emf) works against the
applied armature voltage, so it is called back emf.
2.4 Models of Electromechanical Systems
Example 1: The armature (电枢) – Controlled DC Motor
𝑇𝑚
𝑇𝑑
Output:
𝜽: rotating angle
Input:
𝑽𝒂: control voltage
armature current
back emf
Question:
Please Establish dynamic model of this system.
2.4 Models of Electromechanical Systems
Example 1: The armature (电枢) – Controlled DC Motor
𝑉𝑎 (𝑠) = 𝐸(𝑠) + (𝑅𝑎 + 𝑠𝐿𝑎 )𝐼𝑎 (𝑠)
𝑇𝑚
𝑇𝑑
𝑇𝑚 (𝑠) = 𝐾𝑚 𝐼𝑎 (𝑠)
𝐸(𝑠) = 𝐾e 𝜔(𝑠)
𝑇𝑚 (𝑠) − 𝑇𝑑 (𝑠) − 𝑏𝜔(𝑠) = 𝐽𝑠𝜔(𝑠)
𝜔(𝑠) = 𝑠𝜃(𝑠)
2.4 Models of Electromechanical Systems
Example 1: The armature (电枢) – Controlled DC Motor
𝑉𝑎 (𝑠) = 𝐸 + (𝑅𝑎 + 𝑠𝐿𝑎 )𝐼𝑎 (𝑠)
𝑉𝑎 (𝑠)
1
𝑅𝑎 + 𝑠𝐿𝑎
𝐸
𝑇𝑚 (𝑠) = 𝐾𝑚 𝐼𝑎 (𝑠)
𝐼𝑎 (𝑠)
𝐾𝑚
𝜔(𝑠)
𝐾e
𝐸 = 𝐾e 𝜔
𝜔(𝑠) = 𝑠𝜃(𝑠)
𝜔(𝑠)
1
𝑠
𝐼𝑎 (𝑠)
𝑇𝑚 (𝑠)
𝐸(𝑠)
𝜃(𝑠)
𝑇𝑑 (𝑠)
𝑇𝑚 (𝑠) − 𝑇𝑑 (𝑠) − 𝑏𝜔(𝑠) = 𝐽𝑠𝜔(𝑠)
𝑇𝑚 (𝑠)
1
𝐽𝑠 + 𝑏
𝜔(𝑠)
2.4 Models of Electromechanical Systems
Example 1: The armature (电枢) – Controlled DC Motor
𝑇𝑚 (𝑠) = 𝐾𝑚 𝐼𝑎 (𝑠)
𝑇𝑚 (𝑠) − 𝑇𝑑 (𝑠) − 𝑏𝜔(𝑠) = 𝐽𝑠𝜔(𝑠)
𝑉𝑎 (𝑠) = 𝐸 + (𝑅𝑎 + 𝑠𝐿𝑎 )𝐼𝑎 (𝑠)
𝐸 = 𝐾e 𝜔
𝜔(𝑠) = 𝑠𝜃(𝑠)
𝐾e
2.4 Models of Electromechanical Systems
Example 2: Position Tracking Systems
TG: Tachogenerator （测速发电机）
SM: Servomotor （伺服电机）
𝐾𝑝
𝑢𝜔 = 𝐾𝑇 𝜔
𝜃𝑖𝑛
𝑢𝑖𝑛 = 𝐾𝑝 𝜃𝑖𝑛
𝐾𝑇
𝐾𝑝
𝑢𝜃 = 𝐾𝑝 𝜃𝑜𝑢𝑡
𝜃𝑜𝑢𝑡
2.4 Models of Electromechanical Systems
Example 2: Position Tracking Systems
𝐾𝑝
𝑢𝜔 = 𝐾𝑇 𝜔
𝜃𝑖𝑛
𝑣𝑎
𝑢1
𝑢2
𝑢𝑖𝑛 = 𝐾𝑝 𝜃𝑖𝑛
𝐾𝑇
𝐾𝑝
𝑢𝜃 = 𝐾𝑝 𝜃𝑜𝑢𝑡
𝑢𝑖𝑛 −𝑢𝜃 𝑢1
+
+
=0
10
10
30
𝑢1 −𝑢𝜔 𝑢2
+
+
=0
10
10
20
𝑣𝑎 = 𝐾3 𝑢2
𝜃𝑜𝑢𝑡
2.4 Models of Electromechanical Systems
Example 2: Position Tracking Systems
𝑣𝑎
DC Servo Motor
𝜔
𝑇𝑑
𝑣𝑎
𝐾𝑚
𝑅𝑎
𝜔
1
𝐽𝑠 + 𝑏
𝐾𝑏
1
𝑠
𝜃𝑜𝑢𝑡
2.4 Models of Electromechanical Systems
Example 2: Position Tracking Systems
𝑇𝑑
𝜃𝑖𝑛
𝐾𝑝
3
2
𝑣𝑎
𝐾3
𝐾𝑚
𝑅𝑎
1
𝐽𝑠 + 𝑏
𝐾𝑏
𝐾𝑇
Speed feedback
𝐾𝑝
Position feedback
Block Diagram of Position Tracking System
𝜔
1
𝑠
𝜃𝑜𝑢𝑡
14
2.5 Nonlinear Model of Aircraft & Linearization

Many complex processes are nonlinear in physics,
for example the aircraft system.

However, the analysis and design of linear system
is much easier than that of nonlinear system.

So, we need Linearization to find a linear model to
approximate a nonlinear one.
15
2.5 Nonlinear Model of Aircraft & Linearization
6 DOF nonlinear aircraft dynamic model
16
2.5 Nonlinear Model of Aircraft & Linearization
Force equations
𝑋𝑎𝑒𝑟𝑜 + 𝑇𝑥
− 𝑔𝑠𝑖𝑛𝜃
𝑚
𝑌𝑎𝑒𝑟𝑜 + 𝑇𝑦
𝑣 + 𝑟𝑢 − 𝑝𝑤 =
+ 𝑔𝑐𝑜𝑠𝜃𝑠𝑖𝑛𝛷
𝑚
𝑍𝑎𝑒𝑟𝑜 + 𝑇𝑧
𝑤 + 𝑝𝑣 − 𝑞𝑢 =
+ 𝑔𝑐𝑜𝑠𝜃𝑐𝑜𝑠𝛷
𝑚
𝑢 + 𝑞𝑤 − 𝑟𝑣 =
Moment equations
𝑢, 𝑣, 𝑤 – velocity of X,Y,Z
𝑝, 𝑞, 𝑟 – angular velocity of
roll, pitch, yaw
𝜃, 𝛷, 𝜓 – angles of
roll, pitch, yaw
𝑔 – gravitational force
per unit mass
𝑙, 𝑚, 𝑛 – aerodynamic torques
𝐼𝑖𝑖 – the inertias in body axes
𝐼𝑥𝑥 𝑝 + 𝐼𝑧𝑧 − 𝐼𝑦𝑦 𝑞𝑟 − 𝐼𝑥𝑧 𝑟 + 𝑝𝑞 = 𝑙
𝐼𝑦𝑦 𝑞 + 𝐼𝑥𝑥 − 𝐼𝑧𝑧 𝑝𝑟 − 𝐼𝑥𝑧 𝑟 2 − 𝑝2 = 𝑚
𝐼𝑧𝑧 𝑟 + 𝐼𝑦𝑦 − 𝐼𝑥𝑥 𝑝𝑞 − 𝐼𝑥𝑧 𝑝 − 𝑞𝑟 = 𝑛
𝑋𝑎𝑒𝑟𝑜, 𝑌𝑎𝑒𝑟𝑜, 𝑍𝑎𝑒𝑟𝑜 – aerodynamic
force in X,Y, Z
𝑇𝑥 , 𝑇𝑦, 𝑇𝑧 – thrust in X, Y, Z
17
2.5 Nonlinear Model of Aircraft & Linearization
Kinematic equations
𝛷 = 𝑝 + 𝑞𝑠𝑖𝑛𝛷 + 𝑟𝑐𝑜𝑠𝛷 𝑡𝑎𝑛𝜃
𝜃 = 𝑞𝑐𝑜𝑠𝛷 − 𝑟𝑠𝑖𝑛𝛷
𝛹 = 𝑞𝑠𝑖𝑛𝛷 + 𝑟𝑐𝑜𝑠𝛷 𝑠𝑒𝑐𝜃
The related force and moment components
𝑇𝑥, 𝑇𝑦, 𝑇𝑧 are thrust related force
𝑋𝑎𝑒𝑟𝑜, 𝑌𝑎𝑒𝑟𝑜, 𝑍𝑎𝑒𝑟𝑜 are aerodynamic related force
𝑙, 𝑚, 𝑛 are aerodynamic related moment
They are function of elevator deflection 𝞭𝑒 (升降舵偏转)，aileron deflection
𝞭𝑎 (副翼偏转), rudder deflection 𝞭𝑟 (方向舵偏转), and throttle setting
𝞭𝜋 (油门调节).
2.5 Nonlinear Model of Aircraft & Linearization
19
2.5 Nonlinear Model of Aircraft & Linearization
So, 6 DOF flight dynamics is a set of 9 ODEs
𝑥 = 𝑓(𝑥, 𝑢)
states:
controls:
{𝑢, 𝑣, 𝑤, 𝑝, 𝑞, 𝑟, 𝛷, 𝜃, 𝛹}𝑇
𝑥=
or {𝛼, 𝛽, 𝑉𝑇 , 𝑝, 𝑞, 𝑟, 𝛷, 𝜃, 𝛹}𝑇
𝑢 = {𝛿𝑎, 𝛿𝑒, 𝛿𝑟, 𝛿π}𝑇
𝛼: angle of attack
𝛽: sideslip angle
20
2.5 Nonlinear Model of Aircraft & Linearization
Linear model of aircraft
Basic trim condition:
– steady, symmetric, level flight
旋转
侧滑
倾斜
21
2.5 Nonlinear Model of Aircraft & Linearization
For a moderate rate in states, the behavior of aircraft
dynamics can be decoupled into 2 sets
纵向
横向
22
2.5 Nonlinear Model of Aircraft & Linearization
Linearization of nonlinear aircraft model
23
2.5 Nonlinear Model of Aircraft & Linearization
Linearization of nonlinear aircraft model
24
2.5 Nonlinear Model of Aircraft & Linearization
With linearization, we have 2 sets of linear ODEs
𝑢
{𝛼,
, 𝑞, 𝜃}𝑇
𝑉𝑇0
𝑥= 𝐴 𝑥+ 𝐵 𝑢
State: 𝑥 =
∆ is omitted
Input: 𝑢 = {𝛿𝑒}
𝐴 =
𝑍𝛼
𝑋𝛼
𝑚𝛼
0
𝑍𝑢 1 + 𝑍𝑞
𝑋𝑢
𝑋𝑞
𝑚𝑢 𝑚𝑞
0
1
0
−𝑔/𝑉𝑇0
0
0
𝐵 =
𝑍𝛿
𝑋𝛿
𝑚𝛿
0
25
2.5 Nonlinear Model of Aircraft & Linearization
𝑥= 𝐴 𝑥+ 𝐵 𝑢
Input: 𝑢 = {𝛿𝑎, 𝛿𝑟}
∆ is omitted
𝐴 =
𝑌𝛽
𝑙𝛽
𝑛𝛽
0
State: 𝑥 = {𝛽, 𝑝, 𝑟, 𝛷}𝑇
𝑌𝑝 𝑌𝑟 − 1 𝑔/𝑉𝑇0
𝑙𝑝
𝑙𝑟
0
𝑛𝑝
𝑛𝑟
0
0
0
1
𝐵 =
𝑍𝛿
𝑋𝛿
𝑚𝛿
0
26
2.6 Typical Elements of Linear Dynamic Systems
1. Proportion Element（or Amplifying Element）
2. Differential Element
3. Integral Element
4. Inertial Element (First-order Integral )
5. Oscillation Element (Second-order Integral )
6. First-order Differential Element
7. Second-order Differential Element
27
1. Proportion Element（or Amplifying Element）
R (s )
K
C (s )
Features: Output depends on input according to a certain proportion.
Dynamic equation :
c(t)=Kr(t)
K——Amplification coefficient, are usually have dimension
Transfer function：
Frequency response：
H(s) 
C(s)
K
R(s)
H(j  ) 
C(j )
R(j  )
K
28
1. Proportion Element（or Amplifying Element）
电位器：将线位移或角位移变换为电压量的装置
Input：(t)——angle
E——constant voltage
Output：u(t)——voltage
+
E
-
u(t)
 (s )
 (t )
U (s )
K

+
Dynamic equation ：𝑢 𝑡 = 𝐾θ(𝑡)
Transfer function ：𝐻 𝑠 =
𝑈(𝑠)
=𝐾
𝜃(𝑠)
Frequency response: 𝐻 𝑗ω = 𝐾
29
1. Proportion Element（or Amplifying Element）
Gear 传动装置
Ex 2：Input：n1(t)——rotate speed, Z1—— No. of teeth of active wheel
Output：n2(t)——rotate speed, Z2—— No. of teeth of passive wheel
passive wheel
active wheel
Z1
n1 (t )
n2 ( t )
N1  s 
Z2
𝑧1
𝑛 (𝑡)
𝑧2 1
𝑁2 (𝑠)
𝑧1
=
= =𝐾
𝑁1 (𝑠)
𝑧2
𝑁2 (𝑗𝑤)
𝑧1
𝑗𝑤 =
=
𝑁1 (𝑗𝑤)
𝑧2
z1
z2
Dynamic equation ：𝑛2 𝑡 =
Transfer function：𝐻 𝑠
Frequency response: 𝐻
=𝐾
N2  s
30
1. Proportion Element（or Amplifying Element）
Other Proportion Elements
voltage divider（分压器）
proportional circuit
R(s)
-
r1
r2
r2
+
r (t )
c (t )
C s
r1  r2
𝐶(𝑠)
𝑟2
𝐻(𝑠) =
=
𝑅(𝑠) 𝑟1 + 𝑟2
+ Ec
R2
R1
r (t )
Triode(三极管)
K
c (t )
R3
R(s)

ic (t )
R
R2
C s
ib (t )
Ib (s)
R1
𝐶(𝑠)
𝑅2
𝐻(𝑠) =
=−
𝑅(𝑠)
𝑅1
𝐻(𝑠) =

Ic (s)
𝐶(𝑠)
=𝛽
𝑅(𝑠)
31
2. Differential Element
Features：Dynamic process, output is proportional to the rate
(differentiation) of input.
R (s )
S
Equation of motion：𝐶 𝑡 = 𝐾
C (s )
d𝑟 (𝑡)
𝑑𝑡
𝐶(𝑠)
𝑅(𝑠)
= 𝐾𝑠
Frequency response：𝐻 𝑗𝑤 =
𝐶(𝑗𝑤)
𝑅 𝑗𝑤
Transfer function：𝐻 𝑠 =
= 𝑗𝐾𝑤
32
2. Differential Element
Ex1. RC circuit
i (t )
Input——ur(t)
C
u r (t )
𝐶
R
𝑑𝑢(𝑡)
= 𝑖(𝑡)
𝑑𝑡
u c (t )
Output——uc(t)
1
𝑢 𝑡 =
𝐶
Try to write transfer function.
𝑖(𝑡) 𝑑𝑡
33
2. Differential Element
EX1 RC circuit
C
i (t )
Input——ur(t)
1
𝑢𝑟 𝑡 =
𝐶
u r (t )
R
u c (t )
𝑖 𝑡 =
Output——uc(t)
𝑢𝑐 (𝑡)
𝑅
1
Equation of motion：𝑢𝑟 𝑡 = 𝑅𝐶 𝑢𝑐 (𝑡)𝑑𝑡 + 𝑢𝑐 (𝑡)
Transfer function：𝐻 𝑠 =
𝑈𝑐 (𝑠)
𝑈𝑟 (𝑠)
=
𝑇𝑐 𝑠
𝑇𝑐 𝑠+1
(Tc=RC – product of R and C)
𝑈 (𝑠)
When Tc<<1，the up formula is change to：𝐻 𝑠 = 𝑈𝑐 (𝑠) = 𝑇𝑐 𝑠
𝑟
Frequency response：𝐺 𝑗𝑤 = 𝑗𝑇𝑐 𝑤
𝑖 𝑡 𝑑𝑡 + 𝑖 𝑡 R
34
2. Differential Element
Ex2：Mathematical Model of Generator
 (t )
u d (t )
F
D
u f (t )
Input： (𝑡)——angle of the rotor of motor D
Output： 𝑢𝑓(𝑡)——armature voltage of generator F
Dynamic equation：𝑢𝑓 𝑡 = 𝐾
𝑑𝜑(𝑡)
𝑑𝑡
Transfer function：𝐻 𝑠 = 𝐾𝑠
Frequency response ：
𝐻 𝑗𝑤 = 𝑗𝐾𝑤
35
2. Differential Element
Other Differential Elements
C
i (t )
C
i (t )
I (s)
Cs
U (s)
eL (t )
L
u (t )
uc (t )
U c (s)
i (t )
R
Cs
1
R
+
+
I (s)
I (s)
Ls
EL ( s )
Try to write transfer function
I ( s)
H ( s) 
 Cs H ( s)  I ( s)  Cs  1
U ( s)
U ( s)
R
E ( s)
H ( s) 
 Ls
I ( s)
36
3. Integral Element
R (s )
C (s )
1
s
Feature： The change rate of output is proportional
to the input. (or the output is proportional to the
integration of input).
Dynamic
𝑑𝑐(𝑡)
equation：
𝑑𝑡
= 𝐾𝑟(𝑡)
Transfer function：𝐻 𝑠 =
𝐾
𝑠
Frequency Response： 𝐻 𝑗𝑤 =
𝐾
𝑗𝑤
37
3. Integral Element
Ex1：Integral circuit
Input : r(t); Output: c(t)
Transfer function?
38
3. Integral Element
Ex1：Integral circuit
ic (t )
i1 (t ) R1
+
C
K
r (t )
c (t )
R3
Input : 𝑟(𝑡)，Output: 𝑐(𝑡)
𝑖𝑐 𝑡 = 𝑖1 𝑡 =
R (s )
𝑟(𝑡)
𝑅1
Dynamic equation：𝑐 𝑡 = −
1
𝐶
𝑖𝑐 𝑡 𝑑𝑡 = −
𝐶(𝑠)
𝑠
Transfer function:（𝑇 = 𝑅1𝐶）𝐺 𝑠 = 𝑅
Frequency function: 𝐺 𝑗𝑤 =
𝐶(𝑗𝑤)
𝑅(𝑗𝑤)
=−
1
𝑅1 𝐶
1

𝐾
−𝑗𝑇𝑤
R1Cs
𝑟 𝑡 𝑑𝑡 = −
𝐾
= − 𝑇𝑠 = − 𝑠
C (s )
1
1
𝑇
𝑟 𝑡 𝑑𝑡
39
3. Integral Element
Other Integral Elements
i (t )
u (t )
Current
I (s)
Voltage
1
Cs
U (s)
𝑈(𝑠)
1
𝐻(𝑠) =
=
𝐼(𝑠) 𝐶𝑠
40
4. Inertia Element
Features： This link has an independent energy storage components,
so that the input cannot be transferred to output immediately,
existing delay in time.
R (s )
𝑑𝑐(𝑡)
+𝑐
𝑑𝑡
𝐾
𝑠 =
𝑇𝑠+1
Dynamic equation: 𝑇
Transfer function:𝐺
1
Ts  1
𝑡 = 𝐾𝑟(𝑡)
Frequency response：𝐺 𝑗𝑤 =
𝐾
𝑗𝑇𝑤+1
C (s )
41
4. Inertia Element
id
Ex1：DC Motor
+
Input: 𝑢𝑑 ——armature voltage
Output: 𝑖𝑑 ——armature current
Dynamic equation： 𝐿𝑑
𝜏𝑑
ud
𝑑𝑖𝑑
+ 𝑅𝑑 𝑖𝑑 = 𝑢𝑑
𝑑𝑡
𝑑𝑖𝑑
𝑢𝑑
+ 𝑖𝑑 =
𝑑𝑡
𝑅𝑑
1
𝐼 (𝑠)
𝑅𝑑
Transfer function: 𝐻(s) = 𝑑
=
𝑈𝑑 (𝑠) 𝜏𝑑 𝑠 + 1
𝐿𝑑 ——armature (电枢) inductance
𝑅𝑑 ——armature resistance
𝜏𝑑 ——armature time constant
𝜏𝑑 =
𝐿𝑑
𝑅𝑑
D
42
4. Inertia Element
Other inertial elements
L
r(t)
R
c(t)
v(t )
f (t )
M
T (t )
 (t )
B
J
B
R( s)
1
L
s 1
R
C (s)
F (s)
1
𝐵
𝑀
𝑠+1
𝐵
V (s)
T ( s)
1
B
J
s 1
B
(s)
43
5. Oscillation Element
Features：Contains two independent energy storage components,
when the input of the change, two energy storage components
exchange the energy, to make the output with the nature of the
oscillation.
Dynamic equation：
𝑇2
d2 c(t)
d𝑐(t)
+ 2𝜍𝑇
+ 𝑐(t) = Kr(t)
2
dt
dt
Transfer function：
——Damping ratio， 𝑇——Time constant
Frequency response：𝐺(𝑗𝜔) = 𝐶(𝑗𝜔) =
𝑅(𝑗𝜔)
1
(1 − 𝑇 2 𝜔 2 ) + 𝑗2𝜍𝑇𝜔
44
5. Oscillation Element
L
R
+
EX1：RLC circuit
+
r(t)
i (t )
_
𝑟(𝑡) = 𝐿
𝑑𝑖(𝑡)
1
+ 𝑅𝑖(𝑡) +
𝑑𝑡
𝐶
𝑐(𝑡) =
1
𝐶
𝑖(𝑡)𝑑𝑡
Transfer
𝑖(𝑡)𝑑𝑡
Elimination
Variables
𝑖(𝑡)
c(t)
C
_
d2 c(t)
dc(t)
LC
+
RC
+ c(t) = r(t)
dt 2
dt
ODE model
1
function：G(s) = LCs2 + RCs + 1
Frequency
response：G(jω) =
1
1
=
LC(jω)2 + RC(jω) + 1 (1−LCω2 ) + jRCω
_
45
5. Oscillation Element
Ex2: Armature-controlled DC Motor
i (t )
+
R
L
 (t )
eb (t )
ea (t )
+
_
D
_
𝑒𝑎(𝑡) --- Input voltage applied to the armature
(𝑡) --- Output : angular displacement of the shaft
𝑅 --- Armature resistance；
𝐿 --- Armature (电枢绕组) inductance ；
𝑖(𝑡)--- Armature current；
𝑒𝑏(𝑡) --- Generator back-EMF(反电势)；
𝑇(𝑡) --- Generator torque ；
𝐽 --- moment of inertia ；
𝐵 --- Viscous friction coefficient
J
B
46
5. Oscillation Element
1）𝑇 𝑡 = 𝐾𝑖 𝑡 , 𝑇(𝑡)——Torque 𝐾——Torque coefficient
𝑑𝜃(𝑡)
2）𝑒𝑏 𝑡 = 𝐾𝑏 𝑑𝑡 , 𝑒𝑏(𝑡)——Back-EMF, 𝐾𝑏——Back-EMF coefficient
𝑒𝑎(𝑡)——Armature Voltage
3）𝐿
𝑑𝑖(𝑡)
𝑑𝑡
+ 𝑅𝑖 𝑡 + 𝑒𝑏 𝑡 = 𝑒𝑎 (𝑡)
i a (t )
𝑑 2 𝜃(𝑡)
4）𝐽
𝑑𝑡 2
+
𝑑𝜃(𝑡)
𝐵
𝑑𝑡
L
+
= 𝑇(𝑡)
Laplace transform:
1）
2）
3）
4）
R
𝑇(𝑠) = 𝐾𝐼(𝑠)
𝐸𝑏(𝑠) = 𝐾𝑏𝑠(𝑠)
𝐸𝑎(𝑠) = (𝐿𝑠 + 𝑅)𝐼(𝑠) + 𝐸𝑏(𝑠)
𝑇(𝑠) = (𝐽𝑠2 + 𝐵𝑠) (𝑠)
 (t )
eb (t )
ea ( t )
_
+
_
D
J
B
47
5. Oscillation Element
Elimination Variables 𝐸𝑏(𝑠), 𝑇(𝑠) and 𝐼(𝑠)
𝜃(𝑠)
𝐾
=
𝐸𝑎 (𝑠) 𝑠[𝐿𝐽𝑠 2 + 𝐿𝐵 + 𝑅𝐽 𝑠 + (𝑅𝐵 + 𝐾𝐾𝑏 )]
Input 𝐸𝑎(𝑠), output speed (𝑠), we can get：
Ω(𝑠)
𝐾
=
𝐸𝑎 (𝑠) 𝐿𝐽𝑠 2 + 𝐿𝐵 + 𝑅𝐽 𝑠 + (𝑅𝐵 + 𝐾𝐾𝑏 )
This is a typical oscillation element.
48
5. Oscillation Element
If we Ignore the effect of 𝐿, i.e., 𝐿=0:
𝜃(𝑠)
𝐾𝑚
=
𝐸𝑎 (𝑠) 𝑠(𝑇𝑚 𝑠 + 1)
Ω(𝑠)
𝐾𝑚
=
𝐸𝑎 (𝑠) 𝑇𝑚 𝑠 + 1
where:𝐾𝑚 =
𝐾
,
𝑅𝐵+𝐾𝐾𝑏
Generator gain constant
𝑇𝑚 =
𝑅𝐽
,
𝑅𝐵+𝐾𝐾𝑏
Generator time constant
When 𝑇𝑚 approximates to zero (𝑅 → 0), we can get：
𝜃(𝑠)
1/𝐾𝑏 𝐾 ′
1
′
=
= , (𝐾 = )
𝐸𝑎 (𝑠)
𝑠
𝑠
𝐾𝑏
Ω(𝑠)
= 𝐾′
𝐸𝑎 (𝑠)
49
5. Oscillation Element
EX3：Mechanical System
K
f (t )
Input---Force 𝑓(𝑡)，
Output---Displacement 𝑥(𝑡)
𝐾---Elastic coefficient
𝑀---Mass of the object
𝐵---Viscous friction coefficient
Transfer function?
M
x (t )
B
图2-16 机械振荡
Mechanic
Movement
50
5. Oscillation Element
K
f (t )
EX3：Mechanical System
M
x (t )
Input-----Force 𝑓(𝑡)
Output----Displacement 𝑥(𝑡)
B
Differential equation:
𝑑 2 𝑥(𝑡)
𝑑𝑥(𝑡)
𝑓 𝑡 =𝑀
+𝐵
+ 𝐾𝑥(𝑡)
2
𝑑𝑡
𝑑𝑡
where：𝐾 — Elastic coefficient
𝑀 — Mass of the object，
𝐵 — Viscous friction coefficient
Transfer function：
𝑋(𝑠)
1/𝐾
𝐻 𝑠 =
=
𝐹(𝑠) 𝑀 𝑠 2 + 𝐵 𝑠 + 1
𝐾
𝐾
图2-16 机械振荡
Mechanic
Movement
51
6. First-order Differential Element
Features：the output not only is related to input itself, but also
related to the change rate of input
Dynamic equation：𝑐 𝑡 =
𝑑𝑟(𝑡)
𝑇
𝑑𝑡
+ 𝑟(𝑡)
Transfer function： 𝐻 𝑠 = 𝑇𝑠 +1
Frequency response： 𝐻 𝑗𝑤 = 𝑗𝑇𝑤 + 1
52
6. First-order Differential Element
RC Circuit
Input：𝑢(𝑡)，Output：𝑖(𝑡) ，hence
𝑖 𝑡 = 𝑖1 𝑡 + 𝑖2 𝑡 = 𝐶
=
Transfer
1
𝑅
𝑅𝐶
𝑑𝑢 𝑡
𝑑𝑡
+𝑢 𝑡
𝐼(𝑠)
function：
𝑈(𝑠)
（assume 𝑅 = 1, 𝑅𝐶 =
𝑑𝑢(𝑡)
𝑑𝑡
=𝜏
+
𝑢(𝑡)
𝑅
𝑑𝑢(𝑡)
𝑑𝑡
+ 𝑢(𝑡)
= 𝜏𝑠 + 1
）
Frequency response ：𝐻 𝑗𝑤 = 1 + 𝑗𝜏𝑤
u (t )
i1 (t )
C
i2 (t )
R
i (t )
53
7. Second-order Differential Element
Dynamic equation：𝑐 𝑡 =
Transfer function：𝐺 𝑠 =
𝑑 2 𝑟(𝑡)
2
𝑇
𝑑𝑡 2
𝐶(𝑠)
𝑅(𝑠)
+ 2𝜉𝑇
𝑑𝑟(𝑡)
𝑑𝑡
+ 𝑟(𝑡)
= 𝑇 2 𝑠 2 + 2𝜉𝑇𝑠 + 1
Frequency function：𝐺 𝑗𝑤 = 𝑇 2 (𝑗𝑤) 2 +2𝜉𝑇 𝑗𝑤 + 1
= (1 + 𝑇 2 𝑤 2 ) + 𝑗2𝜉𝑇𝑤)
2.6 Typical Elements of Linear Dynamic Systems
Example
?
How many typical elements the transfer function have?
2.6 Typical Elements of Linear Dynamic Systems
Solution:
56
Summary of Chapter 2
2.1 Concepts of Dynamic Models
2.2 Model of Mechanical System
2.3 Model of Electric System
2.4 Model of Eletromechanical System
2.5 Nonlinear Model of Aircraft System & Linearization
2.6 Typical Elements
57
Homework
Page 99-100，Problem 2.20
Thanks！