LIST OF DYNAMIC SYSTEMS 2008
Content
LIST OF DYNAMIC SYSTEMS 2008
The following are suggested systems for assignment 1 enclosed recommended references. You can collect more information from different sources and select appropriate numerical values of system parameters for your system. 1. Torsional Mechanical Feedback Control System (Dorf and Bishop, 2005)
A torsional mechanical system is shown in Figure 1(a). The torque due to the twisting of the shaft is − k θ ; the damping torque due to the braking device is −b θ ; the disturbance torque is d(t); the input torque is r(t); and the moment of inertia of the mechanical system is J. The transfer function of the torsional mechanical system is: G (s ) =
1/ J s2 + ( b / J ) s + k / J
A closed-loop control system for the system is shown in Figure 1(b). Suppose the desired angle θd = 0o , k = 5, b = 0.9 and J = 1.
(a) Determine the open-loop response, θ ( t ) , of the system for a unit step disturbance, d(t), using MATLAB (set r(t) = 0) (b) With the controller gain K 0 = 50, determine the closed-loop response, θ ( t ) , to a unit step disturbance, d(t), using MATLAB. (c) Co-plot the open-loop versus the closed-loop response to the disturbance input. Discuss your results and make an argument for using closed-loop feedback control to improve the disturbance rejection properties of the system.
Elastic
d(t), Disturbance torque
θ
Braking device
r(t), input torque (a)
d(t) θd(t)
G (s) =
K 0 r(t)
1/ J
θ(t)
s2 + ( b / J ) s + k / J
(b)
Figure 1 A torsional mechanical system 2. DC Motor Level Control System (Kuo and Golnaraghi, 2003) (pp. 208-209)
The following schematic diagram represents a control system whose purpose is to hold the level of the liquid in the tank at a desired level. The liquid level is controlled by a float whose position h(t) is monitored. The input signal of the open-loop system is e(t). The system parameters and equations are as follows: Motor resistance: R a = 10 Ohms Torque constant K i = 10 oz-in/A Back-emf constant K b = 0.0706 V/rad/sec Load inertia JL = 10 oz.-in-sec 2 Amplifier gain K a = 50 Motor inductance La = 0 H Rotor inertia J m = 0.005 oz-in.-sec
2
Gear ratio n = N 1/N2 = 1/100 Load and motor friction = negligible Area of tank A = 50 ft
2
Figure 2 Liquid level system with a DC motor
e a ( t ) = R a i a ( t ) + K b ωm ( t )
ωm ( t ) =
dθm ( t ) dt
Tm ( t ) = K i ia ( t ) = ( J m + n J L ) 2
dωn ( t ) dt
θ y ( t ) = nθm ( t ) The number of valves connected to the tank from the reservoir is N = 10. All the valves have the same characters are and controlled simultaneously by
θy . The equations that govern the volume of flow are
as follows:
q i ( t ) = K I Nθ y ( t )
K I = 10 ft3/sec-rad
qo ( t ) = K oh ( t )
K o = 50 ft 3/sec
h (t) =
volume of tank are of tank
=
1
∫
⎡q o ( t ) − q i ( t ) ⎤⎦ dt A ⎣
N inlet
valves
Figure 3 Liquid level control system
3. Permanent-magnet DC-motor control system (Kuo and Golnaraghi, 2003)
Figure 4 Permanent-magnet DC motor control system (See attached) 4. Process Control System (Michael L Luyben and Willim L. Luyben, Essentials of Process Control)
Figure 5 Process control system
5. DC motor speed control system (Kuo and Golnaraghi, 2003)
The following are parameters of an armature-controlled DC motor system: Jm = 0.0004 kg-m^2 Bm = 0.001 Nm/rad/sec Ra = 2 Ohms La = 0.008 H Km = 0.1 Nm/A Kb = 0.1 V/rad/sec Kt = speed sensor sensitivity (tachometer or encoder) Simulate a DC motor speed control system with PID control law using the above motor. 6. DC motor feedback control system (Kuo and Golnaraghi, 2003)
Simulate a DC motor feedback control system with PID control law using a DC motor as in Exercise 8 (pp. 122-123 Lecture Notes and also see Kuo and Golnaraghi, 2003). 7. Roll stabilization system
Extracted from Fossen (1994), Guidance and Control of Ocean Vehicles, Wiley and Sons Inc. x ′R
–0.5
c pr
0.0
cRrrr
–0.275
x p′
–0.526
τ
1.09
cRrrv
1.96
3.2 Nonlinear Course-keeping Equations of Motion (Sway, Roll and Yaw)
Consider a ship sailing nearly straight with an auto matic course-keeping device in operation. Hence, we can assume constant forward speed ( u′ = 1) which implies that the above equations of motion can be approximated by: 0 ⎤ ⎡(m′ + m′y ) − m′y l′y ⎢ − m ′ l′ ⎥ I′x + J ′x 0 y y ⎢ ⎥ ⎢⎣ 0 0 I′z + J ′z ⎥⎦
⎡ v ′⎤ ⎢ p ′⎥ = ⎢ ⎥ ′ ⎥⎦ ⎢⎣ r
⎡ Y′⎤ ⎢ K ′⎥ ⎢ ⎥ ⎢⎣ N′⎥⎦
(A3.5.9)
where Y = Yv v
m
+ Yrrφ r
2
φ
mx
Yr r
Yr φφ r φ 2
Yp p
Yφ φ
N
Yvφφ v φ
2
Yδδ
K = K p′ p′ − (W ′GM ′ − K ′φ )φ′ + K ′v v′ + (m′x l′x + K ′vφφ v′φ′ 2 + K rrφ r
Yvvφ v 2 φ
+ K ′r )r ′ + K ′vvφ v′ 2 φ′
(A3.5.10)
φ K r φφ r φ 2 K δ δ = N ′r r ′ + N ′v v′ + N p′ p′ + N ′φ φ′ + N′vvφ v′ 2 φ′ + N′vφφ v′φ′ 2 + N rr φ r 2 φ + N r φφr φ 2 N δ δ 2
with p′ = φ ′ = φ (L / U ) . The non-dimensional hydrodynamic derivatives for the course-keeping model with KG = 10.99 m and GM = 0.3 m are given below:
(m′ + m′ )
0.01497
N p′
0.000213
(I′z + J ′z ) (I′x + J ′x ) m′y α′y m′y l′y Yy′ (m′ + m′x − Yr ′ )
0.000875
N′φ
–0.0001468
0.000021
N′vvφ
–0.018191
0.0003525
N ′vφφ
–0.005299
0.0002205
N ′rr φ
–0.003684
y
Y p′ Yφ′
–0.012035
N′r φφ
0.0023843
0.00522
N′δ
0.00126
0.0
κφ K ′φ K ′v (m′x l′x + K ′r ) K ′vvφ
0.2
–0.000074
′φ Yvv
0.046364
Yv′φφ
0.003005
′φ Yrr
0.0093887
–0.000021 0.000314 –0.0000692 –0.0012094
Extracted from Fossen (1994), Guidance and Control of Ocean Vehicles, Wiley and Sons Inc. Yr ′φφ
–0.0013523
K ′vφφ
–0.0000784
Yδ′
–0.002578
K ′rr φ
–0.0002449
N ′r
–0.00243
K ′r φφ
0.00003528
N ′v
–0.0038436
K ′δ
0.0000855
3.3 Linearised Course-Keeping Equations of Motion (Sway, Roll and Yaw)
The linearised course-keeping equations of motion are:
′ m12 ′ 0 ⎡ m11 ⎢m′ m′ 0 22 ⎢ 21 ⎢ 0 0 m′33 ⎢ 0 0 ⎣ 0
⎡Δv ′⎤ ⎥⎢ ⎥ 0 Δ p ′ ⎥ ⎢ ⎥+ ′ ⎥ 0⎥ ⎢ Δr ⎥ ⎢ ′⎥ 1 ⎦ ⎣Δφ ⎦ 0⎤
d11
d12
d13
d14
d 21
d 22
d 23
d24
d 31
d32
d 33
d34
0
0
0
U L
⎡ Δv′⎤ ⎢ Δ p′⎥ ⎢ ⎥ = ⎢ Δr ′ ⎥ ⎢ ′⎥ ⎣Δφ ⎦
where m′11 = m′ + m′y m′12
= −m′y l′y m′21 = m12 m ′22 = I′x + J ′x m′33 = I′z + J ′z d11
Yv
′ d 12
= −Y p′
d13
m
2Yvvφ v0 φ0 mx
Yvφφφ0 2 2Yrrφ r0φ0
Yr
Yrφφ φ02
d ′14
= −Yφ′ − Yvv′ φ v′02 − 2Yv′φφ v′0 φ′0 − Yrr ′ φ r ′ 2 − 2Yr ′φφ r 0′φ′0 d 21 K v 2K vvφ v0 φ0 K vφφ φ 02 d ′22 = −K p′ d 23
m x lx
d 24
W GM
d 31
Nv
d ′32
= − N p′
d 33
Nr
d 34 b1′
Nφ
= Yδ′ b′2 = K ′δ b′3 = N′δ
Kr
2K rrφ r0 φ0
K vvφ v02
Kφ
2N vvφ v 0φ 0 2N rrφ r0 φ0
N vvφ v 02
K rφφφ 02
2K vφφ v0φ 0
K rrφ r0 2
N rrφ r0 2
2N rφφ r0φ 0
N vφφφ 02
N rφφφ02
2N vφφ v0φ 0
2K rφφ r0φ 0
⎡ b1′ ⎤ ⎢ b′ ⎥ ⎢ 2 ⎥ Δδ ⎢ b′3 ⎥ ⎢ ⎥ ⎣0⎦
The following are suggested systems for assignment 1 enclosed recommended references. You can collect more information from different sources and select appropriate numerical values of system parameters for your system. 1. Torsional Mechanical Feedback Control System (Dorf and Bishop, 2005)
A torsional mechanical system is shown in Figure 1(a). The torque due to the twisting of the shaft is − k θ ; the damping torque due to the braking device is −b θ ; the disturbance torque is d(t); the input torque is r(t); and the moment of inertia of the mechanical system is J. The transfer function of the torsional mechanical system is: G (s ) =
1/ J s2 + ( b / J ) s + k / J
A closed-loop control system for the system is shown in Figure 1(b). Suppose the desired angle θd = 0o , k = 5, b = 0.9 and J = 1.
(a) Determine the open-loop response, θ ( t ) , of the system for a unit step disturbance, d(t), using MATLAB (set r(t) = 0) (b) With the controller gain K 0 = 50, determine the closed-loop response, θ ( t ) , to a unit step disturbance, d(t), using MATLAB. (c) Co-plot the open-loop versus the closed-loop response to the disturbance input. Discuss your results and make an argument for using closed-loop feedback control to improve the disturbance rejection properties of the system.
Elastic
d(t), Disturbance torque
θ
Braking device
r(t), input torque (a)
d(t) θd(t)
G (s) =
K 0 r(t)
1/ J
θ(t)
s2 + ( b / J ) s + k / J
(b)
Figure 1 A torsional mechanical system 2. DC Motor Level Control System (Kuo and Golnaraghi, 2003) (pp. 208-209)
The following schematic diagram represents a control system whose purpose is to hold the level of the liquid in the tank at a desired level. The liquid level is controlled by a float whose position h(t) is monitored. The input signal of the open-loop system is e(t). The system parameters and equations are as follows: Motor resistance: R a = 10 Ohms Torque constant K i = 10 oz-in/A Back-emf constant K b = 0.0706 V/rad/sec Load inertia JL = 10 oz.-in-sec 2 Amplifier gain K a = 50 Motor inductance La = 0 H Rotor inertia J m = 0.005 oz-in.-sec
2
Gear ratio n = N 1/N2 = 1/100 Load and motor friction = negligible Area of tank A = 50 ft
2
Figure 2 Liquid level system with a DC motor
e a ( t ) = R a i a ( t ) + K b ωm ( t )
ωm ( t ) =
dθm ( t ) dt
Tm ( t ) = K i ia ( t ) = ( J m + n J L ) 2
dωn ( t ) dt
θ y ( t ) = nθm ( t ) The number of valves connected to the tank from the reservoir is N = 10. All the valves have the same characters are and controlled simultaneously by
θy . The equations that govern the volume of flow are
as follows:
q i ( t ) = K I Nθ y ( t )
K I = 10 ft3/sec-rad
qo ( t ) = K oh ( t )
K o = 50 ft 3/sec
h (t) =
volume of tank are of tank
=
1
∫
⎡q o ( t ) − q i ( t ) ⎤⎦ dt A ⎣
N inlet
valves
Figure 3 Liquid level control system
3. Permanent-magnet DC-motor control system (Kuo and Golnaraghi, 2003)
Figure 4 Permanent-magnet DC motor control system (See attached) 4. Process Control System (Michael L Luyben and Willim L. Luyben, Essentials of Process Control)
Figure 5 Process control system
5. DC motor speed control system (Kuo and Golnaraghi, 2003)
The following are parameters of an armature-controlled DC motor system: Jm = 0.0004 kg-m^2 Bm = 0.001 Nm/rad/sec Ra = 2 Ohms La = 0.008 H Km = 0.1 Nm/A Kb = 0.1 V/rad/sec Kt = speed sensor sensitivity (tachometer or encoder) Simulate a DC motor speed control system with PID control law using the above motor. 6. DC motor feedback control system (Kuo and Golnaraghi, 2003)
Simulate a DC motor feedback control system with PID control law using a DC motor as in Exercise 8 (pp. 122-123 Lecture Notes and also see Kuo and Golnaraghi, 2003). 7. Roll stabilization system
Extracted from Fossen (1994), Guidance and Control of Ocean Vehicles, Wiley and Sons Inc. x ′R
–0.5
c pr
0.0
cRrrr
–0.275
x p′
–0.526
τ
1.09
cRrrv
1.96
3.2 Nonlinear Course-keeping Equations of Motion (Sway, Roll and Yaw)
Consider a ship sailing nearly straight with an auto matic course-keeping device in operation. Hence, we can assume constant forward speed ( u′ = 1) which implies that the above equations of motion can be approximated by: 0 ⎤ ⎡(m′ + m′y ) − m′y l′y ⎢ − m ′ l′ ⎥ I′x + J ′x 0 y y ⎢ ⎥ ⎢⎣ 0 0 I′z + J ′z ⎥⎦
⎡ v ′⎤ ⎢ p ′⎥ = ⎢ ⎥ ′ ⎥⎦ ⎢⎣ r
⎡ Y′⎤ ⎢ K ′⎥ ⎢ ⎥ ⎢⎣ N′⎥⎦
(A3.5.9)
where Y = Yv v
m
+ Yrrφ r
2
φ
mx
Yr r
Yr φφ r φ 2
Yp p
Yφ φ
N
Yvφφ v φ
2
Yδδ
K = K p′ p′ − (W ′GM ′ − K ′φ )φ′ + K ′v v′ + (m′x l′x + K ′vφφ v′φ′ 2 + K rrφ r
Yvvφ v 2 φ
+ K ′r )r ′ + K ′vvφ v′ 2 φ′
(A3.5.10)
φ K r φφ r φ 2 K δ δ = N ′r r ′ + N ′v v′ + N p′ p′ + N ′φ φ′ + N′vvφ v′ 2 φ′ + N′vφφ v′φ′ 2 + N rr φ r 2 φ + N r φφr φ 2 N δ δ 2
with p′ = φ ′ = φ (L / U ) . The non-dimensional hydrodynamic derivatives for the course-keeping model with KG = 10.99 m and GM = 0.3 m are given below:
(m′ + m′ )
0.01497
N p′
0.000213
(I′z + J ′z ) (I′x + J ′x ) m′y α′y m′y l′y Yy′ (m′ + m′x − Yr ′ )
0.000875
N′φ
–0.0001468
0.000021
N′vvφ
–0.018191
0.0003525
N ′vφφ
–0.005299
0.0002205
N ′rr φ
–0.003684
y
Y p′ Yφ′
–0.012035
N′r φφ
0.0023843
0.00522
N′δ
0.00126
0.0
κφ K ′φ K ′v (m′x l′x + K ′r ) K ′vvφ
0.2
–0.000074
′φ Yvv
0.046364
Yv′φφ
0.003005
′φ Yrr
0.0093887
–0.000021 0.000314 –0.0000692 –0.0012094
Extracted from Fossen (1994), Guidance and Control of Ocean Vehicles, Wiley and Sons Inc. Yr ′φφ
–0.0013523
K ′vφφ
–0.0000784
Yδ′
–0.002578
K ′rr φ
–0.0002449
N ′r
–0.00243
K ′r φφ
0.00003528
N ′v
–0.0038436
K ′δ
0.0000855
3.3 Linearised Course-Keeping Equations of Motion (Sway, Roll and Yaw)
The linearised course-keeping equations of motion are:
′ m12 ′ 0 ⎡ m11 ⎢m′ m′ 0 22 ⎢ 21 ⎢ 0 0 m′33 ⎢ 0 0 ⎣ 0
⎡Δv ′⎤ ⎥⎢ ⎥ 0 Δ p ′ ⎥ ⎢ ⎥+ ′ ⎥ 0⎥ ⎢ Δr ⎥ ⎢ ′⎥ 1 ⎦ ⎣Δφ ⎦ 0⎤
d11
d12
d13
d14
d 21
d 22
d 23
d24
d 31
d32
d 33
d34
0
0
0
U L
⎡ Δv′⎤ ⎢ Δ p′⎥ ⎢ ⎥ = ⎢ Δr ′ ⎥ ⎢ ′⎥ ⎣Δφ ⎦
where m′11 = m′ + m′y m′12
= −m′y l′y m′21 = m12 m ′22 = I′x + J ′x m′33 = I′z + J ′z d11
Yv
′ d 12
= −Y p′
d13
m
2Yvvφ v0 φ0 mx
Yvφφφ0 2 2Yrrφ r0φ0
Yr
Yrφφ φ02
d ′14
= −Yφ′ − Yvv′ φ v′02 − 2Yv′φφ v′0 φ′0 − Yrr ′ φ r ′ 2 − 2Yr ′φφ r 0′φ′0 d 21 K v 2K vvφ v0 φ0 K vφφ φ 02 d ′22 = −K p′ d 23
m x lx
d 24
W GM
d 31
Nv
d ′32
= − N p′
d 33
Nr
d 34 b1′
Nφ
= Yδ′ b′2 = K ′δ b′3 = N′δ
Kr
2K rrφ r0 φ0
K vvφ v02
Kφ
2N vvφ v 0φ 0 2N rrφ r0 φ0
N vvφ v 02
K rφφφ 02
2K vφφ v0φ 0
K rrφ r0 2
N rrφ r0 2
2N rφφ r0φ 0
N vφφφ 02
N rφφφ02
2N vφφ v0φ 0
2K rφφ r0φ 0
⎡ b1′ ⎤ ⎢ b′ ⎥ ⎢ 2 ⎥ Δδ ⎢ b′3 ⎥ ⎢ ⎥ ⎣0⎦
