Transcript 非時變移動機械系統的

Ch.2
Modeling in the Frequency Domain
學習成果 Learning Outcomes
•
學習Transfer function技術：含Laplace transform、由
「微分方程」求「transfer function」、由「transfer
function」解 「微分方程」(2.1-2.3)
• 求線性、非時變電路系統的「轉移函數」 (2.4)
• 求線性、非時變移動機械系統的「轉移函數」
(2.5)
• 求線性、非時變轉動機械系統的「轉移函數」
(2.6)
• 求齒輪系統的「轉移函數」 (2.7)
• 求線性、非時變機電系統的「轉移函數」 (2.8)
• 機械系統的電路模擬(2.9)
• 非線性系統的線性化技術，為求「轉移函數」
(2.10-2.11)
求「轉移函數」的目的：
天線模型 → 天線運動的微分方程式
→ 天線的位移方程式
以牛頓第二定律為例: F(t) = M a(t)
Taking Laplace:
F(t) = M a(t) → F(s) = M a(s) 所有初始條件= 0
Output/input = 1/M;
a/F = 1/M ; F(s) = M a(s)
Transfer function 定義
Output = Transfer function × Input
Taking inverse Laplace:
X(s) → X(t)
Table 1.1 (測試系統性能的輸入指令)
Test waveforms used in control systems
物理意義
拍一下
定力拉
變力拉 (固定增幅)
變力拉 (加速增幅)
Sin 變力
Table 2.1
Laplace transform table
Table 2.2
Laplace
transform theorems
學習成果 Learning Outcomes
• 學習Transfer function技術：含Laplace
transform、由「微分方程」求「transfer function」、
由「transfer function」解 「微分方程」(2.1-2.3)
• 求線性、非時變電路的「轉移函數」 (2.4)
• 求線性、非時變移動機械系統的「轉移函數」
(2.5)
• 求線性、非時變轉動機械系統的「轉移函數」
(2.6)
• 求齒輪系統的「轉移函數」 (2.7)
• 求線性、非時變機電系統的「轉移函數」 (2.8)
• 機械系統的電路模擬(2.9)
• 非線性系統的線性化技術，為求「轉移函數」
(2.10-2.11)
§ 2.5 Translational Mechanical System
Transfer Functions 阻抗的觀念(in S)
阻抗觀念
阻尼器 Damper
一維系統
阻抗觀念 求系統方程式
Figure 2.17
a. Two-degrees-of-freedom translational mechanical system;
b. block diagram
需2方程式描述系統之運動
二維系統
A X1(s) + B X2(s) = F(s)
C X1(s) + D X2(s) = 0
Figure 2.18
Equation of Motion of M1
M1’s Free Body Diagram
a. Forces on M1 due only to motion of M1;
即 M1移X1、 M2固定時，M1之受力
b. forces on M1 due only to motion of M2;
即 M2移X2、M1固定時，M1之受力
c. all forces on M1
A X1(s) + B X2(s) = F(s)
Eq. 2.118a P.66
M1受力之自由體圖
Σ外力＝0
牛頓第二定律
Figure 2.19
Equation of Motion of M2
M2’s Free Body Diagram
a. Forces on M2 due only to motion of M2;
即 M2移X2、 M1固定時，M2之受力
b. forces on M2 due only to motion of M1;
即 M1移X1、 M2固定時，M2之受力
c. all forces on M2
M2受力之自由體圖
C X1(s) + D X2(s) = 0
Eq. 2.118b P.66
Σ外力＝0
牛頓第二定律
• System equation P.66
A X1(s) + B X2(s) = F(s)
C X1(s) + D X2(s) = 0
2.118a
2.118b
解2元1次聯立方程式 (By Cremer’s Rule)
X2(s)/ F(s) = G(s) (2.119)
P.67
由公式(2.120a； 2.120b) 直接求系統方程式
A X1(s) + B X2(s) = F(s)
C X1(s) + D X2(s) = 0
2.118a
2.118b
Figure 2.21
Translational
mechanical system
for Skill-Assessment
Exercise 2.8
P.67
由牛頓第二定律求系統方程式
Figure 2.20
Three-degrees-of-freedom translational mechanical system
P.68
由公式(2.121- 2.123) 直接求系統方程式
A1 X1(s) + B1 X2(s) + C1 X3(s) = 0
A2 X1(s) + B2X2(s) + C2 X3(s) = F(s)
A3 X1(s) + B3X2(s) + C3 X3(s) = 0
(2.124)
(2.125)
(2.126)
解3元1次聯立方程式 (By Cremer’s Rule)
X3(s)/ F(s) = G(s)
A1 X1(s) + B1 X2(s) + C1 X3(s) = 0
A2 X1(s) + B2X2(s) + C2 X3(s) = F(s)
A3 X1(s) + B3X2(s) + C3 X3(s) = 0
(2.124)
(2.125)
(2.126)
解3元1次聯立方程式 (By Cremer’s Rule)
X3(s)/ F(s) = G(s)
§2.6 Rotational Mechanical System
Week 3
學習成果 Learning Outcomes
• 學習Transfer function技術：含Laplace
transform、由「微分方程」求「transfer function」、
由「transfer function」解 「微分方程」(2.1-2.3)
• 求線性、非時變電路的「轉移函數」 (2.4)
• 求線性、非時變移動機械系統的「轉移函數」
(2.5)
• 求線性、非時變轉動機械系統的「轉移函數」
(2.6)
• 求齒輪系統的「轉移函數」 (2.7)
• 求線性、非時變機電系統的「轉移函數」 (2.8)
• 機械系統的電路模擬(2.9)
• 非線性系統的線性化技術，為求「轉移函數」
(2.10-2.11)
1D- Rotational System
1D- Rotational System
轉動慣量 J 的求法?
查 「應用力學」
View point
Equation of motion for J1
(J1S2 + D1S + K) θ1(s) － K θ2(s) ＝ T(s)
(2.127a)
Figure 2.23 a. Torques on J1 due only to the motion of J1 J1動
b. torques on J1 due only to the- motion of J2
J1不動 J2動
c. final free-body diagram for J1
J2不動
View point
Equation of motion for J1
(J1S2 + D1S + K) θ1(s) － K θ2(s) ＝ T(s)
(2.127a)
Figure 2.23 a. Torques on J1 due only to the motion of J1 J1動
b. torques on J1 due only to the- motion of J2
J1不動 J2動
c. final free-body diagram for J1
J2不動
View point
Equation of motion for J2
外力和=o: － K θ1(s) ＋ (J2S2 + D2S + K) θ2(s) ＝ 0
Figure 2.24 a. Torques on J2 due only to the motion of J2
J1不動
b. torques on J2 due only to the motion of J1
J1動
J2不動
c. final free-body diagram for J2
(2.127b)
J2動
Find T.F.
(J1S2 + D1S + K) θ1(s)
－ K θ2(s) ＝ T(s)
－ K θ1(s)
＋ (J2S2 + D2S + K) θ2(s) ＝ 0
By Cremer’s Rule → θ2(s) ＝ (K/△) T(s)
T.F. G(s) ＝ (eq. 2.128)
(2.127a)
(2.127b)
Ex. 2.20 3-D rotational system
Figure 2.25
Three-degrees-of-freedom rotational system
Equation of Motion: 公式 2.130a,b,c 直接寫出 如2.131a,b,c
Find T.F.
G(s) ＝ θ2(s) / T(s)
Figure 2.26 Homework
Rotational mechanical system for Skill-Assessment Exercise 2.9
Week 4
學習成果 Learning Outcomes
• 學習Transfer function技術：含Laplace
transform、由「微分方程」求「transfer function」、
由「transfer function」解 「微分方程」(2.1-2.3)
• 求線性、非時變電路的「轉移函數」 (2.4)
• 求線性、非時變移動機械系統的「轉移函數」
(2.5)
• 求線性、非時變轉動機械系統的「轉移函數」
(2.6)
• 求齒輪系統的「轉移函數」 (2.7)
• 求線性、非時變機電系統的「轉移函數」 (2.8)
• 機械系統的電路模擬(2.9)
• 非線性系統的線性化技術，為求「轉移函數」
(2.10-2.11)
1D Network
Figure 2.3
RLC network
By Kirchhoff’s Voltage Law
△VL＋△VR ＋ △VC ＋ V(t) = 0
Figure 2.3
RLC network
Figure 2.3
RLC network
G(s)
2D Network
Figure 2.6 Two-loop electrical network
需2方程式描述系統之狀態
方程式1: Loop i1
方程式2 : Loop i2
Kirchhoff’s Voltage Law
Input: V
Responses: i1, i2
Find T.F. : Vc / V
2D Network
2D Network
Kirchhoff’s Voltage Law
方程式1: Loop I1
-R1 I1 – Ls(I1-I2) + V(s) = 0 →
(R1+Ls) I1 – Ls I2 = V(s)
方程式2 : Loop I2
-R2 I2 – I2/Cs - Ls(I2-I1) = 0 →
– Ls I1 +【Ls + R2 + (1/Cs)】I2 = 0
2D Network
(R1+Ls) I1
– Ls I1
– Ls I2 = V(s)
+【Ls + R2 + (1/Cs)】I2 = 0
Figure 2.7
Block diagram of the network of Figure 2.6
Vc(s) = (1/Cs) I2
T.F. : Vc / V
Figure 2.9
Three-loop electrical network
3D Network
範例：外力在Mesh 2
A1I1(s) + B1I2(s) + C1I3(s) = 0
A2I1(s) + B2I2(s) + C2I3(s) = F(s)
A3I1(s) + B3I2(s) + C3I3(s) = 0
(2.91)
(2.92)
(2.93)
解3元1次聯立方程式 (By Cremer’s Rule)
I3(s)/ F(s) = G(s)
Figure 2.14
Electric circuit for kill-Assessment Exercise 2.6
Find G(s) = VL(s) / V(s) Homework
Figure 2.11 Homework
Inverting operational amplifier circuit for Example 2.14
Figure 2.12
General noninverting operational amplifier circuit
V0 = A (Vi – V1)
V1 = 【 V0/(Z1 + Z2) 】Z1
∵ A ﹥﹥1
V0/Vi = (Z1+Z2)/Z1
Find G(s) = V0/Vi
Figure 2.13 Homework
Noninverting operational amplifier
circuit for Example 2.15
Quiz on Oct. 8th 2010
Find G(s) = VL(s) / V(s)
學習成果 Learning Outcomes
• 學習Transfer function技術：含Laplace
transform、由「微分方程」求「transfer function」、
由「transfer function」解 「微分方程」(2.1-2.3)
• 求線性、非時變電路的「轉移函數」 (2.4)
• 求線性、非時變移動機械系統的「轉移函數」
(2.5)
• 求線性、非時變轉動機械系統的「轉移函數」
(2.6)
• 求齒輪系統的「轉移函數」 (2.7)
• 求線性、非時變機電系統的「轉移函數」 (2.8)
• 機械系統的電路模擬(2.9)
• 非線性系統的線性化技術，為求「轉移函數」
(2.10-2.11)
§ 2.7
T.F. for Gear System 1/4
Figure 2.27 A gear system
§ 2.7
T.F. for Gear System 2/4
r1θ1 ＝ r2θ2
→ N1θ1 ＝ N2θ2
Figure 2.28 Transfer functions for a. angular displacement in lossless gears
§ 2.7
T.F. for Gear System 3/4
T1θ1 ＝ T2θ2
→ T1N2 ＝ T2N1
Figure 2.28 Transfer functions for b. torque in lossless gears
§ 2.7
T.F. for Gear System
4/4
Figure 2.27 A gear system
Figure 2.28
T. F. for angular displacement
in lossless gears
T. F. for torque
in lossless gears
註1： T.F. for Gear System
Θ1 ＞ θ2
?
Θ1 ＜ θ2
?
註2： T.F. for Gear System
T1 ＞ T2
?
T1 ＜ T2
?
省略齒輪組後的等效阻抗推導 (a) → (c)
Figure 2.29
a. Rotational system driven by gears;
b. equivalent system at the output after reflection of input torque;
c. equivalent system at the input after reflection of impedances
1/5
2/5
＋
→
Case I
3/5
→
T.F. G(s) ＝ θ2(s) / T1(s)
(JS2 + DS + K) θ2(s) ＝ T1(s)
＋
T.F. G(s) ＝ θ1(s) / T1(s)
(JS2 + DS + K) θ1(s)
＝ T1(s)
→
Case
II
省略齒輪組的等效阻抗 (結論)
等效系統
→
4/5
5/5
Figure 2.31
T.F. of Gear train
Figure 2.32
Example 2.22 HOMEWORK 證明 Je and De; 求 G(s)=θ1/T1
a. System using a gear train;
b. equivalent system at the input;
c. block diagram
原系統
等效系統
Je = J1∪ J2 ∪ J3 ∪ J4 ∪ J5 對系統的綜合影響相同
De = D1∪ D2
對系統的綜合影響相同
綜合影響相同 = 施相同外力T1 獲相同反應θ1
原系統
等效系統
在θ1軸感到的慣量
在θ1軸感到的阻尼
等效系統
Figure 2.32
Example 2.22 HOMEWORK 證明 Je and De; 求 G(s)=θ1/T1
a. System using a gear train;
b. equivalent system at the input;
c. block diagram
Figure 2.33
HOMEWORK
Rotational mechanical system with gears for Skillassessment Exercise 2.10 求 G(s)=θ2/T
齒輪可省力 (證明)
等效系統
→
If N1 / N2 ＝ 1/ 2
小齒輪驅動大齒輪
於等效系統中
J, D, K 均降為原負載的 1/ 4
學習成果 Learning Outcomes
• 學習Transfer function技術：含Laplace
transform、由「微分方程」求「transfer function」、
由「transfer function」解 「微分方程」(2.1-2.3)
• 求線性、非時變電路的「轉移函數」 (2.4)
• 求線性、非時變移動機械系統的「轉移函數」
(2.5)
• 求線性、非時變轉動機械系統的「轉移函數」
(2.6)
• 求齒輪系統的「轉移函數」 (2.7)
• 求線性、非時變機電系統的「轉移函數」 (2.8)
• 機械系統的電路模擬(2.9)
• 非線性系統的線性化技術，為求「轉移函數」
(2.10-2.11)
§2.8 T.F. of Electromechanical System
Figure 2.34 NASA flight simulator robot arm with electromechanical
control system components (an example)
Typical Example 1/10
Figure 2.35 DC motor:
a. schematic12;
b. block diagram (T.F.)
Figure 2.36
Typical equivalent mechanical
loading on a motor
Figure 2.37 System
DC motor driving a
rotational mechanical
load
求機電系統的T.F. G(s)
2/10
νb (t) = κb (dθm／dt)
νb(t) : back emf 反電動勢 ;
Κb : back emf constant
dθm／dt : angular velocity of motor
求機電系統的T.F. G(s)
3/10
Tm(t) = κt ia(t)
Tm(t) : torque developed by motor ;
Κt : motor torque constant ;
ia(t) : armature current
求機電系統的T.F. G(s)
νb(t) = κb (dθm／dt)
4/10
νb(t) : back emf 反電動勢 ; Κb : back emf constant
dθm／dt : angular velocity of motor
Tm(t) = κt ia(t)
Tm(t) : torque developed by motor ;
Κt : motor torque constant ;
ia(t) : armature current
5/10
Laplace ↓
求機電系統的T.F. G(s)
By Kirchhoff’s current law
外加電壓與Tm ,θm 關係
6/10
-RaIa(s) – LasIa(s) – Vb(s) + Ea(s) = 0
(Ra+Las) Tm(s) / Κt + Κbsθm(s) = Ea(s) (2.149)
求機電系統的T.F. G(s)
6/10
Motor 轉子
機械系統
機電系統 機械負載部
外力 : Tm ,θm (機與電共有的項目)
θm 受機械負載影響
目標 找Ea 與θm 關係
7/10
Jm = Ja + JL(N1/N2)2
Dm = Da + DL(N1/N2)2
↓Laplace
外力 = Σ阻抗 * 反應
(JmS2 + DmS)θm(s) = Tm(s)
By Newton’s 2nd Law
JmS2θm(s) + DmSθm(s) = Tm(s) (2.150)
8/10
JmS2θm(s) + DmSθm(s) = Tm(s) (2.150)
By Newton’s 2nd Law
(Ra+LaS) Tm(s) / Κt + ΚbSθm(s) = Ea(s) (2.149)
(2.150) 代入 (2.149)
(Ra+LaS) (JmS2 + DmS)θm(s) / Κt + ΚbSθm(s) = Ea(s)
Sice Ra＞＞La
(Ra+LaS) → Ra usually true
(2.151)
i.e. La= 0
(Ra/ Κt ) (JmS2 + DmS)θm(s) + ΚbSθm(s) = Ea(s)
(2.152)
9/10
(Ra / Κt ) (JmS2 + DmS)θm(s) + ΚbSθm(s) = Ea(s) (2.152)
G(s) = θm(s) / Ea(s) = κ / S(S+α)
κ= Κt / (Ra Jm)
α= (Dm + Κt Κb / Ra) / Jm
if κ,α已知 G(s) 完全定義
κ,α為已知 if Κt, Kb已知 求Κt , Kb?
(2.154)
10/10
求Κt , Kb?
(Ra+LaS) Tm(s) / Κt + ΚbSθm(s) = Ea(s)
RaTm(s) / Κt + ΚbSθm(s) = Ea(s)
RaTm(t) / Κt + Κbωm(t) = ea(t)
(2.149)
(La= 0)
(after inverse Laplace)
RaTm / Κt + Κbωm = ea
(at steady state; ea :DC voltage)
(aTm + bωm = 常數) (ax + by = 常數)
Figure 2.38
Torque-speed curves
with an armature
voltage, ea, as a
parameter
At Tstall where ωm= 0 → Κt
At ωno-load where Tm = 0 → Κb
Dynamometer: 於定電壓下量
torque and speed of a motor
求Κt , Kb?
10/10
RaTm / Κt + Κbωm = ea (at steady state; ea :DC voltage)
(aTm + bωm = 常數) (ax + by = 常數)
At Tstall where ωm= 0 → Κt
At ωno-load where Tm = 0 → Κb
Dynamometer: 於定電壓下量torque
and speed of a motor
Figure 2.38
Torque-speed curves with an armature
voltage, ea, as a parameter
Example 2.23 Homework
求Κt , Kb
Figure 2.39
a. DC motor and load;
b. torque-speed curve;
c. block diagram
Figure 2.40 Homework
Electromechanical system for Skill-Assessment Exercise 2.11
Find G(s) = θL(s)/Ea(s)
Tm = －8ωm + 200
學習成果 Learning Outcomes
• 學習Transfer function技術：含Laplace
transform、由「微分方程」求「transfer function」、
由「transfer function」解 「微分方程」(2.1-2.3)
• 求線性、非時變電路的「轉移函數」 (2.4)
• 求線性、非時變移動機械系統的「轉移函數」
(2.5)
• 求線性、非時變轉動機械系統的「轉移函數」
(2.6)
• 求齒輪系統的「轉移函數」 (2.7)
• 求線性、非時變機電系統的「轉移函數」 (2.8)
• 機械系統的電路模擬(2.9)
• 非線性系統的線性化技術，為求「轉移函數」
(2.10-2.11)
§2.9 Electric Circuit Analogs 1/3
§2.9 Electric Circuit Analogs 2/3
Laplace
→
§2.9 Electric Circuit Analogs
3/3
外力 = Σ阻抗 * 反應
F(s) = (MS2+fvS+K) X(s)
-MSv(s)-fvv(s)-Kv(s)/S+F(s) = 0
v(s) = SX(s)
-MS2X(s)-fvSX(s)-K X(s)+F(s) = 0
MS2X(s)+fvSX(s)+K X(s)= F(s)
§ 2.10 Nonlinearities
Figure 2.45
a. Linear system;
b. nonlinear system
Figure 2.46
Some physical
nonlinearities
§ 2.11 Linearization
Example 2.26
Linearization of f(x) about x = /2 where f(x)= 5 cos x.
Taylor series (非線性系統線性化的工具)
f(x)= f(x0) + (df/dx) x=xo(x-x0)/1! + (d2f/dx2) x=xo(x-x0)2/2!
+ --- (2.181)
f(x0) = f(/2 ) = 5 cos /2 = 0
(df/dx) x=xo(x-x0)/1! = (-5 sin /2) x = -5 x
f(x)= -5 x = -5 (x-/2)
直線公式 當 x around x = /2 時為真
Figure 2.48 Example 2.26
Linearization of 5 cos x about x = /2
Skill assessment exercise
Skill assessment exercise
Skill assessment exercise
Skill assessment exercise
Skill assessment exercise