Ch.2 Modeling in the Frequency Domain

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」( ) 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」 ( )

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
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)

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Transfer function技術：含Laplace transform、由「微分方程」求「transfer function」、由「transfer function」解「微分方程」( ) 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」 ( )

§ 2.5 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 A X1(s) + B X2(s) = F(s) C X1(s) + D X2(s) = 0 二維系統

A X1(s) + B X2(s) = F(s) Equation of Motion of M1 Σ外力＝0 牛頓第二定律
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 M1受力之自由體圖 Σ外力＝0 牛頓第二定律 A X1(s) + B X2(s) = F(s) Eq a P.66

C X1(s) + D X2(s) = 0 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 Σ外力＝0 牛頓第二定律 C X1(s) + D X2(s) = 0 Eq b P.66

System equation P.66 A X1(s) + B X2(s) = F(s) a C X1(s) + D X2(s) = b 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) 2.118a
C X1(s) + D X2(s) = b

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( ) 直接求系統方程式

A1 X1(s) + B1 X2(s) + C1 X3(s) = 0 (2.124)
A2 X1(s) + B2X2(s) + C2 X3(s) = F(s) (2.125) A3 X1(s) + B3X2(s) + C3 X3(s) = (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

1D- Rotational System

轉動慣量 J 的求法? 查「應用力學」

(J1S2 + D1S + K) θ1(s) － K θ2(s) ＝ T(s) (2.127a)
View point Equation of motion for J1 (J1S2 + D1S + K) θ1(s) － K θ2(s) ＝ T(s) (2.127a) Figure a. Torques on J1 due only to the motion of J1 J1動 J2不動 b. torques on J1 due only to the- motion of J J1不動 J2動 c. final free-body diagram for J1

外力和=o: － K θ1(s) ＋ (J2S2 + D2S + K) θ2(s) ＝ 0 (2.127b)
View point Equation of motion for J2 外力和=o: － K θ1(s) ＋ (J2S2 + D2S + K) θ2(s) ＝ (2.127b) Figure a. Torques on J2 due only to the motion of J J1不動 J2動 b. torques on J2 due only to the motion of J J1動 J2不動 c. final free-body diagram for J2

Find T.F. (J1S2 + D1S + K) θ1(s) － K θ2(s) ＝ T(s) (2.127a) － K θ1(s) ＋ (J2S2 + D2S + K) θ2(s) ＝ (2.127b) By Cremer’s Rule → θ2(s) ＝ (K/△) T(s) T.F. G(s) ＝ (eq )

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 Homework Rotational mechanical system for Skill-Assessment Exercise 2.9

Week 4

By Kirchhoff’s Voltage Law
Figure 2.3 1D Network 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 Find T.F. : Vc / V Figure 2.6 Two-loop electrical network
Kirchhoff’s Voltage Law Input: V Responses: i1, i2 Find T.F. : Vc / V 需2方程式描述系統之狀態方程式1: Loop i1 方程式2 : Loop i2

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) I – Ls I2 = V(s) – Ls I 【Ls + R2 + (1/Cs)】I2 = 0

Figure 2.7 Block diagram of the network of Figure 2.6 T.F. : Vc / V
Vc(s) = (1/Cs) I2

Figure 2.9 Three-loop electrical network
3D Network

範例：外力在Mesh 2 A1I1(s) + B1I2(s) + C1I3(s) = (2.91) A2I1(s) + B2I2(s) + C2I3(s) = F(s) (2.92) A3I1(s) + B3I2(s) + C3I3(s) = (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
Figure 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

V0 = A (Vi – V1) V1 = 【 V0/(Z1 + Z2) 】Z1 ∵ A ﹥﹥1 V0/Vi = (Z1+Z2)/Z1
Figure 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 Homework Noninverting operational amplifier circuit for Example 2.15

Quiz on Oct. 8th 2010 Find G(s) = VL(s) / V(s)

§ T.F. for Gear System 1/4 Figure A gear system

§ 2.7 T.F. for Gear System 2/4 r1θ1 ＝ r2θ2 → N1θ1 ＝ N2θ2
Figure Transfer functions for a. angular displacement in lossless gears

§ 2.7 T.F. for Gear System 3/4 → T1N2 ＝ T2N1 T1θ1 ＝ T2θ2
Figure Transfer functions for b. torque in lossless gears

T. F. for angular displacement
§ T.F. for Gear System 4/4 Figure 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 ＞ θ ? Θ1 ＜ θ ?

註2： T.F. for Gear System T1 ＞ T ? T1 ＜ T ?

省略齒輪組後的等效阻抗推導 (a) → (c) 1/5
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

2/5 ＋ → Case I

→ ＋ → 3/5 T.F. G(s) ＝ θ2(s) / T1(s) (JS2 + DS + K) θ2(s) ＝ T1(s)
Case II

4/5 省略齒輪組的等效阻抗 (結論) 等效系統 →

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
Figure Example 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∪ D 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
Figure Example 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 HOMEWORK Rotational mechanical system with gears for Skill-assessment Exercise 求 G(s)=θ2/T

→ 齒輪可省力 (證明) 等效系統於等效系統中 J, D, K 均降為原負載的 1/ 4 If N1 / N2 ＝ 1/ 2
齒輪可省力 (證明) 等效系統 → 於等效系統中 J, D, K 均降為原負載的 1/ 4 If N1 / N2 ＝ 1/ 2 小齒輪驅動大齒輪

§2.8 T.F. of Electromechanical System
Figure NASA flight simulator robot arm with electromechanical control system components (an example)

Typical Example 1/10 Figure DC motor: a. schematic12; b. block diagram (T.F.) Figure 2.36 Typical equivalent mechanical loading on a motor Figure System DC motor driving a rotational mechanical load

求機電系統的T.F. G(s) /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) /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) /10 νb(t) = κb (dθm／dt) 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) /10 By Kirchhoff’s current law -RaIa(s) – LasIa(s) – Vb(s) + Ea(s) = 0 外加電壓與Tm ,θm 關係 (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

↓Laplace Jm = Ja + JL(N1/N2)2 Dm = Da + DL(N1/N2)2 外力 = Σ阻抗 * 反應
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) (2.151) Sice Ra＞＞La (Ra+LaS) → Ra usually true 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) (2.154) 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?

求Κt , Kb. (Ra+LaS) Tm(s) / Κt + ΚbSθm(s) = Ea(s) (2
求Κt , Kb? (Ra+LaS) Tm(s) / Κt + ΚbSθm(s) = Ea(s) (2.149) RaTm(s) / Κt + ΚbSθm(s) = Ea(s) (La= 0) RaTm(t) / Κt + Κbωm(t) = ea(t) (after inverse Laplace) RaTm / Κt + Κbωm = ea (at steady state; ea :DC voltage) (aTm + bωm = 常數) (ax + by = 常數) 10/10 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? RaTm / Κt + Κbωm = ea (at steady state; ea :DC voltage) (aTm + bωm = 常數) (ax + by = 常數)
10/10 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 Homework Electromechanical system for Skill-Assessment Exercise 2.11 Find G(s) = θL(s)/Ea(s) Tm = －8ωm + 200

§2.9 Electric Circuit Analogs 1/3

Laplace →

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 Taylor series (非線性系統線性化的工具)
Example 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 Example Linearization of 5 cos x about x = /2

Skill assessment exercise

Ch.2 Modeling in the Frequency Domain

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Ch.2 Modeling in the Frequency Domain

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