Electromechanical Systems and Devices (Solutions, Instructor Solution Manual) [1 ed.] 9781420072952, 9781420069723, 1420069721

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SOLUTIONS MANUAL FOR Electromechanical Systems and Devices

by Sergey Edward Lyshevski

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SOLUTIONS MANUAL FOR Electromechanical Systems and Devices

by Sergey Edward Lyshevski

Boca Raton London New York

CRC Press is an imprint of the Taylor & Francis Group, an informa business

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CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2009 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed in the United States of America on acid-free paper 10 9 8 7 6 5 4 3 2 1 International Standard Book Number-13: 978-1-4200-7295-2 (Softcover) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

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Possible Solutions of Homework Problems The instructor should be advised that for many problems, there are no unique solutions. E.g., students may obtain other solutions which are correct. Different problem formulation, methods, procedures, assumptions, simplifications and hardware (devices, electronics, ICs, etc.) solutions result in various solutions. There fore, the majority of solutions provided must be treated as possible solutions.

Chapter 1 Problem 1. 1. Provide examples of electromechanical systems and electromechanical motion devices. Solution The electromechanical systems are various electric drives and servos (with corresponding electronics and controllers), e.g., propulsion systems (automotive, locomotive, marine and other applications), speaker systems (comprised form speakers, amplifiers, circuits, power supplier, etc.), etc. The electromechanical motion devices are stand-alone rotational and translational (linear) electric machines, actuators, sensors, etc. For example, for the aforementioned examples, traction motor and speaker. Those stand-alone electromechanical motion devices do not possess electronics, ICs and other components. Problem 1. 2. What is the difference between electromechanical systems and electromechanical motion devices? Solution The electromechanical system comprises of various components which are integrated in a unified functional system, e.g., electromechanical motion devices – sensors – power electronics – controlling/processing and driving/sensing circuitry (ICs and DSPs). In electromechanical systems, • Energy conversion, actuation and sensing; • Conversion of physical stimuli and events to electrical and mechanical quantities and vice versa; • Control, diagnostics data acquisition and other features are performed. Correspondingly, electromechanical systems integrate various components and modules, including electromechanical motion devices. There could be many electromechanical motion devices (electric machines, actuators, sensors, etc.) in an electromechanical system. The electromechanical motion devices are one of the major components of systems. The electromechanical motion devices convert energy and stimuli, performing actuation, conversion and sensing. Problem 1. 3. Choose a specific electromechanical system or device (propulsion system, traction electric drive, control surface actuator, speaker, microphone, etc.) and explicitly identify problems needed to be addressed, studied and solved to design and analyze the examined system or device. Formulate and report the specifications and requirements imposed on the system of your interest. Develop the high-level functional diagram with the major components. Formulate and report the structural and behavioral designs tasks. Solution There are various electromechanical systems which can be examined. For example, consider a temperature controlled fan. The problems under the consideration in order to perform the design are: • Size estimates (blades, housing, etc.) and what is maximal heat (thermal energy) to be removed;

2

• • •

Estimate the power (torque, angular velocity, voltage, current, etc.) required to operate fan (as a system) and electric machine (as an electromechanical motion device); Define the specified (desired) and achievable performance characteristics (losses, vibaration, heat, etc.), capabilities, and the system dynamics; Choose motor, sensors and electronics to be utilized. Cooling Controller

Power Electronics

Voltage

Electric Motor

Fan Kinematics

Environment

Angular Velocity Desired / Specified Temperature

+

Actual Temperature / Velocity

Sensors

Temperature

Figure. High-level functional diagram Problem 1. 4. Explain why systems and devices must be examined in the functional, structural and behavioral domains? Solution To guarantee the electromechanical systems - Functionality and operationability; - Sound organization; - Optimal achievable performance, the corresponding design with sequential tasks must be performed in the functional, structural and behavioral domains, respectively. The following problems are usually emphasized: • Design and optimization of electromechanical systems components and units (actuators, generators, sensors, electronics, etc.) according to their applications and overall systems requirements; • Design of high-performance electromechanical motion devices and electronics for specific applications; • Integration of all components and modules (actuators, generators, sensors, power electronics and ICs) emphasizing compliance, integrity, regularity, modularity, compliance, matching and completeness; • Control of actuators, generators, sensors and electronics.

3

Chapter 2 Problem 2. 1. A force F=3i+2j+4k acts through the point with position vector r=2i+j+3k. Derive a torque about a perpendicular axis. That is, find T=r×F. Solution From the formula a×b=(a2b3-a3b2)i+(a3b1-a1b3)j+(a1b2-a2b1)k, we have the resulting expression for the torque as T=r×F=(1×4-3×2)i+(3×3-2×4)j+(2×2-1×3)k=-2i+j+k. Problem 2. 2. A spherical electrostatic actuator, as documented in Figure 2.38, is designed using spherical conducting shells separated by the flexible material (for example parylene, teflon and polyethylene have relative permittivity ~3). The inner shell has total charge +qi and diameter ri. The charge of the outer shell q0(t) is semi-negative and time-varying. The diameter of the outer shell is denoted as r0. u(t) ! !q0(t)

"r

!

+

+ +qi

ri

!

r0

Figure 2.38. Electrostatic actuator 2.2.1. 2.2.2.

For a spherical actuator, derive the expression for the capacitance C(r). Calculate the numerical value for capacitance if ri=1 cm, r0 =1.5 cm, qi=1 C and q0(t)=[sin(t)+1] C. Derive the expression for the electrostatic force using the coenergy Wc[u,C(r)]=C(r)u2/2. Recall that the force is Fe (u , r ) =

!Wc [u , C (r )] . Calculate the electrostatic force between the inner and !r

outer shells assigning the various values for the applied voltage u which could be up to ~1000 V. For a flexible materials (parylene, teflon or polyethylene), find the resulting displacements. Use the expression for the restoring force of the flexible media chosen. The approximation Fs=ksr can be applied, where ks=1 N/m. 2.2.4. Develop the differential equations which describe the spherical actuator dynamics. Examine the actuator performance and capabilities. The simulation in MATLAB should be performed. 2.2.3.

2.2.1.

First, we derive the capacitance for the studied electrostatic actuator.

Solution From the Gauss law

# E " ds = r0

Q(t ) Q , we have E = . ! 4"! 0! r r 2

Q(t ) This yields )V = * E ( dr = 4,+ 0+ r ri

r0

r

1 Q(t ) 1 0 Q(t ) dr = = *r r 2 4,+ 0+ r r ri 4,+ 0+ r i

&1 1# $$ ' !! % ri r0 "

4

One obtains the expression for the capacitance as C =

Q(t ) 4$# 0# r . = 1 1 "V ! ri r0

Recalling that ε0=8.85×10–12 F/m, we have

C=

4#$ 0 3 Q(t ) =1×10–11 F. = 1 1 "V ! 0.01 0.015

Assume that ri is constant, we find C as a function of r, where r is the varying r0. We have

C (r ) = 2.2.2.

4#" 0" r 4#" 0" r ri r = 1 1 r ! ri ! ri r

The electrostatic force Fe is the negative (or positive, depending on the assigned direction)

derivative of co-energy Wc. That is, Fe (u , x) = " the expression C (r ) =

Fe = !2#" 0" r

!Wc (u , r ) . We recall that Wc (u , r ) = 12 C (r )u 2 . Using !r

4#" 0" r ri r , one finds the expression for electrostatic force as r ! ri ri 2 u2 (r ! ri ) 2

Using the MATLAB file, as documented below, the electrostatic force is calculated for various applied voltages with a fixed r=1.5 cm and ri=1 cm. The resulting plot is illustrated. echo on; clear all % Electrostatic actuator parameters e0=8.85e-12; er=3; r=0.015; ri=0.01; u=0:1:1000; Fe=-2*pi*e0*er*ri*ri*u.*u./(r-ri)^2; % Plot of the electrostatic force plot(u,Fe); xlabel('u, [V]','FontSize',14); ylabel('F_e, [N]','FontSize',14); title('Electrostatic Force, F_e','FontSize',14); Electrostatic Force, F

-4

3

x 10

e

2 1

] N [

0 -1 -2

, e F

-3 -4 -5 -6 -7

0

100

200

300

400

500

600

700

800

900

1000

u, [V]

Figure. Electrostatic force 2.2.3.

In equilibrium, the restoring elastic force Fs is equal and opposite to the electrostatic force Fe. An expression for the device outer radius r is found approximating the elastic force as Fs=ks(r0–r). From Fe=Fs we have

5

r=r0–Fs/ks. For the specified voltage applied, the resulting r can be obtained. 2.2.4.

A set of two first-order differential equations which describe the actuator dynamics are

# ri 2 dv 1 1& = ( Fe ' Fs ) = $' 2)( 0( r u 2 + k s (r0 ' r )! 2 dt m m% (r ' ri ) " dx =v dt The SIMULINK model to simulate the dynamics is reported in the Figure, and the resulting transient dynamics is documented (m=0.001 kg). In order to enhance the model accuracy, the elastic damping is integrated. The differential equations become

# ri 2 dv 1 1& = ( Fe ' Fs ) = $' 2)( 0( r u 2 ' kv v + k s (r0 ' r )! 2 dt m m% (r ' ri ) " dx =v. dt Scope feout

Step: Applied Voltage

rout

-2*pi*e0*er*u(1)^2*ri^2/(u(2)-ri)^2

d^2r/dt^2

dr/dt

1 s

1/m

1 s

r

kv

ks fsout

ro

Figure. Electrostatic actuator SIMULINK diagram and transient behavior for the actuator outer shell displacement (u=1000 V and kv=0.01) Problem 2. 3. The power generation device is illustrated in Figure 2.39. Assume that the magnetic field is uniform, and the top and/or low magnets or current loop vibrate such as the magnetic field relative the current loop vary as B(t)=sin(t)+sin(2t)+cos(5t)+sin(t)cos2(t). That is, permanent magnets are stationary, while a current loop is on the movable (suspended) structure. The variations of the magnetic field are shown in Figure 2.40. The area of the conducting current loop in the magnetic field is 1 mm2. The resistance is 1 ohm.

6

sin(t)+sin(2 t)+cos(5 t)+sin(t) cos(t)2 3

Permanent Magnet S

B(t) i(t) N, r

Damper Bv

y(t)

Suspended Structure

2

Spring k s

A

Permanent Magnet N

1

0

emf u(t)

-1

-2

-3 -6

-4

-2

0 t

2

4

6

Figure 2.40. Power generation device and assumed variations of the magnetic field 2.3.1.

2.3.2. 2.3.3. 2.3.4.

Find the induced emf (generated voltage) and the current in the current loop. Assume that the number of turns N can be 1 or 100. How the emf changes if N=1 and N=100. The MATLAB Partial Differential Toolbox can be used to plot the resulting variables, perform differentiation, calculations, plotting, etc. In particular, the MATLAB statement is t=sym('t','positive'); B=sin(t)+sin(2*t)+cos(5*t)+sin(t).*cos(t)^2; ezplot(B); dBdt=diff(B) The resulting plot for the magnetic field and derivative for the magnetic field density are found. In the Command Window we have the result dBdt = cos(t)+2*cos(2*t)-5*sin(5*t)+cos(t)^3-2*sin(t)^2*cos(t) Assign the dimensions of the device to derive the parameters of the power generation system (mass of the suspended structure with the current loop m, spring constant ks, etc.) Derive differential equations for the device. Simulate and examine the power generation system performance.

Solution 2.3.1. The magnetic field density B(t) is given as B(t)=sin(t)+sin(2t)+cos(5t)+sin(t)cos2(t). Then, the emf induced in the loop is found by using emf=–NdΦ/dt.

#

From " = B ! ds = BA , one obtains the expression for the emf as emf(t)= –NA[cos(t)+2cos(2t) –5sin(5t)+cos3(t) –2sin2(t)cos(t)]. The emf induced is proportional to N and A. The plot of B(t) and dB(t)/dt are calculated using MATLAB. The calculations and plotting are performed using the MATLAB statement as t=0:0.0001*pi:20; B=sin(t)+sin(2*t)+cos(5*t)+sin(t).*cos(t).^2; dBdt=cos(t)+2*cos(2*t)-5*sin(5*t)+cos(t).^3-2*(sin(t).^2).*cos(t); N=100; A=0.01; emf=-N*A*dBdt; subplot(2,1,1); plot(t,B); title('Field B(t), [T]'); xlabel('time, [sec]'); subplot(2,1,2); plot(t, emf); title('Emf Induced emf(t), [V]'); xlabel('time, [sec]');

Assign N=100 and A=0.01 m2, the plots for B(t) and emf(t) are given in Figure 1.

7

Field B(t), [T]

4 2 0 -2 -4

0

2

4

6

0

2

4

6

10

8 10 12 time, [sec] Emf Induced emf(t), [V]

14

16

18

20

14

16

18

20

5 0 -5 -10

8

10 12 time, [sec]

Figure 1. Plots of B(t) and dB(t)/dt as functions of time 2.3.2.

The parameters are assigned to the device under consideration. Let the device dimensionality is ~10×10×2 cm. It can be estimated that Bmax=1 T, ks=0.1 N/m, Bv=0 N-sec/m, m=0.01 kg, N=100 and A=0.01 m2.

2.3.3. We assume that: (i) The generator is unloaded, therefore no current; (ii) Vibrational acceleration is restricted to the y-direction; (iii) Uniform magnetic field established by the permanent magnet. Using the expression for the y-displacement, which is due to multi-frequency vibrations, we let y(t)=0.5sin(t)+0.001sin(10t)+0.0001cos(100t). The vibration magnitude and frequency of the object (and suspended mass) depends on the specific application (aerospace, automotive, marine, etc), device placement, suspension, etc. For the chosen y(t) (which can be measured in the specific application), we have dy(t)/dt=0.5cos(t)+0.01cos(10t) –0.01sin(100t) and d2y(t)/d2t=–0.5sin(t) –0.1sin(10t) –1cos(100t). The magnetic flux density, as a function of y(t), can be approximated as B(y)=Bmagnet/y(t). Hence, dB(y)/dy=–Bmagnet/y2(t) and dB(y)/dt=(dB(y)/dy)(dy/dt)=–vBmagnet/y2(t). Recalling that emf=–NdΦ/dt, we have emf=–NAdB/dt. Thus, emf=NAvBmagnet/y2(t). The differential equations for the suspended mass are

dv 1 1 = ( Fa ! Ffriction ! Fs ) = (maa ! Bv v ! k s y ) dt m m dy =v. dt 2.3.4. The SIMULINK diagram to simulate the vibrational generator is shown in Figure 2. The dynamics for y(t) and the waveforms for the emf induced are reported in Figure 3. As expected, the voltage is induced in the output terminal.

8

Figure 2. Vibrational generator SIMULINK model

) s / s / m (

Applied Acceleration vs Time 2

0

a -2

0

5

10

15

20

25 T (s) Y vs Time

30

35

40

45

50

0

5

10

15

20

25 Time (s) EMF vs Time

30

35

40

45

50

0

5

10

15

20

25 Time (s)

30

35

40

45

50

0.1 ) m (

0

Y -0.1

) V (

5

F M E

0

-5

Figure 3. Applied (vibration-induced) force, displacement and induced emf Problem 2. 4. One-, two- and three-dimensional magnetic levitation systems have been foreseen to be used in control of underwater and flight vehicles (so-called moving mass concept). Possible magnetic levitation systems with a ball are illustrated in Figure 2.41. u1 (t )

N

!1

u (t )

u (t )

Magnetic Force Fe y (t )

N

N

!

!

Spring ks Magnetic Force Fe

y (t )

y (t )

!2

N

Spring ks

Magnetic Force Fe

Permanent Magnet

u2 (t )

Figure 2.41. Magnetic levitation systems 2.4.1.

For the chosen device, derive a mathematical model by applying Newton’s and Kirchhoff’s laws. That is, derive the differential equations which describe the system dynamics. Find the expression for

9

2.4.2.

2.4.3. 2.4.4.

the electromagnetic force developed by using the coenergy. Note that the reluctance in the magnetic system varies. Develop the equivalent magnetic circuit. Assign magnetic levitation system dimensions and derive the parameters. For example, let the total length of the magnetic path is ~0.1 m. Assuming that the diameter of copper wire is 1 mm, one layer winding can include ~10 turns, but you may have multi-layered winding. The geometry (shape) and diameter of the moving mass (ball) and its media density will result in the value for m, A, µr, etc. In MATLAB, develop the files and perform numerical simulations of the studied magnetic levitation system. Analyze the dynamics and assess the performance. Examine what happens if in the suspended mass (coated with an insulator) will be surrounded by N turns conducting current loop as illustrated in Figure 2.41. Assume that the current loop has a resistance r and area A. u (t )

Magnetic Force Fe

y (t )

A

uemf(t) B(t)

Permanent Magnet

Figure 2.41. Magnetic levitation systems with the conducting current loop Solution 2.4.1. The equivalent magnetic circuit for the device is given in Figure 1.

Ni

Rfs

Ry

Rfm

Ry

Figure 1. Equivalent magnetic circuit The magnetic reluctances Rfs, Rfm and Ry are

R fs = R fm = Ry =

l fs

µ r µ0 A l fm µ r µ0 A x02 + y 2 µ0 A

where x0 is the horizontal constant airgap; y is the vertical varying airgap (vertical displacement). Using these reluctances, the magnetizing inductance L is derived as

10

N2 L= R fm + R fs + 2 Ry L=

N 2 µ r µ0 A l fm + l fs + 2 µr x02 + y 2

Hence, we have

!2 N 2 µr 2 µ0 A dL = dy (l fm + l fs + 2 µr x0 2 + y 2 ) 2

y x0 2 + y 2

.

Using the magnetizing inductance, applying Kirchhoff’s voltage law, one finds

u = ir +

d! dt

! = Li di dL dx +i dt dx dt di 1 dL = ["ir + u + iv ] dt L dx

u = ir + L

2 2 2 N 2 µ r 2 µ0 A di l fm + l fs + 2 µr x0 + y = [ " ri + dt N 2 µ r µ0 A (l fm + l fs + 2 µr x0 2 + y 2 ) 2

y x0 2 + y 2

iv + u ]

Using Newton’s second law, we have the following equations 2 2 2 N 2 µ r 2 µ0 A di l fm + l fs + 2 µr x0 + y = [ ! ri + dt N 2 µ r µ0 A (l fm + l fs + 2 µr x0 2 + y 2 ) 2

N 2 µ r 2 µ0 A dv 1 = [ dt m (l fm + l fs + 2 µr x0 2 + y 2 ) 2

y x0 + y 2

2

y x0 2 + y 2

i 2 ! mg ! Bm v ! K s y ]

dy =v dt 2.4.2.

The parameters assigned for simulation of device are: u Applied = 5 l fs = 0.1 r = 21 g = 9.8 m = 0.01 N = 1000 u 0 = 4 ! " ! 10 - 7 u r = 5000

2.4.3.

l fm = 0.025 A = 0.01 Ks = 1 Bv = 0.1 u f= 0 x0 =0.01

The SIMULINK diagram for the levitation device is shown in Figure 2.

iv + u ]

11

Figure 2. SIMULINK diagram for a levitation device Applying the voltage, as shown in Figure 3, the ball (suspended mass) is displaced. The transient dynamics for y(t) is illustrated. We conclude that the suspended ball can be displaced by applying the armature voltage to the winding. The settling time is ~0.5 sec. Applied Voltage vs Time

10 ) m (

5

Y 0 0

2

4

6

8

0

2

4

6

8

0.05

10 12 U (s) Y vs Time

14

16

18

20

14

16

18

20

0 ) m ( -0.05 Y

-0.1 -0.15

10 Time (s)

12

Figure 3. Vertical displacement of the suspended mass (ball) 2.4.4.

There are several possible ways to generate (induce) voltage. The most obvious that the voltage can be generated as an induced emf (as in Problem 2.3). If the voltage is applied to the coil, the force produced will result in the motion of the the suspended ball with coil on it. The voltage in the coin on the ball will be induced. Problem 2. 5. Let for a solenoid the inductance is L(x)=4sin(3x). Derive the expression for the electromagnetic force and calculate the force for i=10 A at x=0. Solution The electromagnetic torque is found to be

Fe =

!Wc 1 !L( x) 2 1 d (4 sin(3 x)) 2 1 =2 i =2 i = 2 12 cos(3 x)i 2 = 6 cos(3 x)i 2 !x !x dx

Thus, for 10A current at x=0, we have Fe=600 N.

12

Problem 2. 6. Consider a solenoid as documented in Figure 2.18 in Example 2.17. This solenoid integrates movable (plunger) and stationary (fixed) members made from high-permeability ferromagnetic materials. The windings wound with a helical pattern. The performance of solenoids depends on the electromagnetic system, mechanical geometry, magnetic permeability, winging resistivity, inductance, friction, etc. 2.6.1. Assign the dimensions of solenoid (say ~5 cm length and ~2 cm outer diameter) and make the estimates for all other dimensions. Chose the materials that can be used in the solenoid emphasizing high performance and capabilities. 2.6.2.

Derive the magnetizing inductance, winding resistance and other parameters of interest. In particular, friction coefficient, relative permeabilities of movable and stationary members, resistivity, spring constant, etc.

2.6.3.

Derive a mathematical model by applying Newton’s and Kirchhoff’s laws. Find the expression for the electromagnetic force developed using coenergy. 2.6.4. In MATLAB, develop the files and perform numerical simulations of the studied solenoid. Analyze the results examining steady-state and dynamic performance. Solution: Relay shematics +

uu

-

Length 1 to 2 = 10 cm Length 3 to 4 = 2.5 cm Cross Sectional Area = 1 cm2 Relative Magnetic Permittivity = 5000 Number of Coils = 1000 Coil Resistance = 21.9 ohm Mass of Moveable Member = 0.01 kg

2 1

3

4 x

d~0

ks Bv

lfm Figure 1. Solenoid (relay) schematic 2.6.1.

The equivalent magnetic circuit is given in Figure 2.

Rfs Ni Rfm

Rx

Figure 2. Equivalent magnetic circuit The magnetic reluctances are found as

13

R fs =

l fs

µ r µ0 A x R fm = µ r µ0 A l !x Rx = fm µ0 A 2.6.2.

The magnetizing inductance is

L=

N2 R fm + R fs + Rx

L=

N 2 µ r µ0 A x + l fs + µr (l fm ! x)

Hence, we have

( µr ! 1)( N 2 µr µ0 A) dL . = dx ( x + l fs + µr (l fm ! x)) 2

The resistance of the windings can be calculated as ρl/A, where ρ is the resistivity of copper, l is the length of the winding, and A is the cross sectional area of the wire. Assuming 1000 turns of 38 gauge "9 wire around a 4 cm core, the resistance is 21.9 ohms. In fact, # = 17.2 E (#m) $1000 $ .04(m) = 21.9 ohm. 2 2

! 0.0001 (m )

2.6.3.

Using Kirchhoff’s voltage law, we have

u = ir +

d! dt

! = Li di dL dx +i dt dx dt di 1 dL = ["ir + u + iv ] dt L dx x + l + µ ( l ( µr " 1)( N 2 µr µ0 A) di fs r fm " x ) = ["ri + iv + u ] dt N 2 µ r µ0 A ( x + l fs + µr (l fm " x)) 2

u = ir + L

The derived equation, integrated with the translational dynamics as found using the Newtonian mechanics, results in

( µr ! 1)( N 2 µr µ0 A) di x + l fs + µr (l fm ! x) = [ ! ri + iv + u ] dt N 2 µ r µ0 A ( x + l fs + µr (l fm ! x)) 2 dv 1 ( µr ! 1)( N 2 µr µ0 A) = [ ! mg µ f ! Bm v ! K s x] dt m ( x + l fs + µr (l fm ! x)) 2 dx =v dt

2.6.4.

The SIMULINK diagram to perform simulations is shown in Figure 3.

14

Figure 3. SIMULINK diagram The transient dynamics for the displacement x(t) is provided in Figure 4. Applied Voltage vs Time

3 2

) v (

1

U

0 -1

0

2

4

6

8

0

2

4

6

8

0.03

10 12 T (s) X vs Time

14

16

18

20

14

16

18

20

) m 0.02 ( X

0.01 0

10 Time (s)

12

Figure 4. The applied voltage waveform and displacement dynamics Solution: Consider a solenoid as documented in Figure 2.18 in Example 2.17. The solenoid schematics and equivalent magnetic circuit are illustrated in Figure 5. The reluctances of the stationary member ℜfs, the stationary member which faces the plunger ℜfsp, the air gap

ℜx and of the plunger ℜfp are given as ! fs =

! fp =

l fp

µ 0 µ r A2

.

l fs

µ 0 µ r A1

, ! fsp =

l fsp

µ 0 µ r A2

, !x =

x µ 0 A2

and

15

u(t) Stationary Member

"1

XXXXXXXXXXXXXXXXXXXXXXX XXXXXXXXXXXXXXXXXXXXXXX XXXXXXXXXXXXXXXXXXXXXXX

lfsp x(t) ! lfp A2 Magnetic Flux µ

r

························ Windings ························ ························

Spring k s

Force Fe Plunger A1

! fs

_ Ni

µr lfs

! fsp

!x

! fs "3

"2

_ Ni

! fp

Figure 5. Schematic and magnetic circuit of a solenoid Our goal is to derive the expression for Fe and obtain the resulting differential equations. From the magnetic circuit, illustrated in Figure 3.8, one has 1 1 2 Ni = " fs !1 + (" fsp + " x + " fp )! 3 and 2 Ni = " fs ! 2 + (" fsp + " x + " fp )! 3 .

Ni = " fs (!1 + ! 2 ) + 2(" fsp + " x + " fp )! 3 ,

From

using

Φ1+Φ2 =Φ3,

we

obtain

Ni = (" fs + 2" fsp + 2" x + 2" fp )! 3 . Hence, the total magnetic flux and flux linkages ψ=NΦ3 are

"3 =

Ni N 2i and " = . ! fs + 2! fsp + 2! x + 2! fp ! fs + 2! fsp + 2! x + 2! fp

The magnetizing inductance L(x) is given as

L( x) =

N 2 µ 0 µ r A1 A2 N2 . = ! fs + 2! fsp + 2! x + 2! fp l fs A2 + 2l fsp A1 + 2 A1µ r x + 2l fp A1

Using the coenergy concept, the electromagnetic force Fe is found to be

Fe =

(

)

" 12 L( x)i 2 N 2 µ 0 µ r2 A12 A2 =! i2 . "x (l fs A2 + 2l fsp A1 + 2 A1µr x + 2l fp A1 )2

Form the Kirchoff voltage law and the second Newton law, we have

di 1 = dt L( x) + Ll

& # 2 N 2 µ 0 µ r2 A12 A2 ' ri ' iv + u ! , $ 2 (l fs A2 + 2l fsp A1 + 2 A1µr x + 2l fp A1 ) $% !"

# N 2 µ 0 µ r2 A12 A2 dv 1 & = $' i 2 + k s1 x + Bv v + FL ! , 2 dt m $% (l fs A2 + 2l fsp A1 + 2 A1µ r x + 2l fp A1 ) !"

dx =v. dt The SIMULINK diagram is virtually identical as was reported in Figure 3. The transient dynamics is similar as documented in Figure 4. The solenoid parameters are available form the manufacturer or can be measured or estimated. Problem 2. 7. Figure 2.43 illustrates a motion devise with windings on rotor. The electromagnetic torque is developed due to the interaction of the 20-turn rectangular coil (winding) in the yz plane and stationary magnetic field (established by magnets on stator windings).

16

2.7.1.

2.7.2.

Determine the magnetic moment and derive the electromagnetic torque acting on the coil. Let the current in the coil (l=15 cm and w=5 cm) is i=10 A, and the magnetic field density as B=2×102(ax+2ay) T. Derive at what angle φ Te=0. At what angle φ, Te is maximum? Determine the value of Temax. Derive the solution from: (a) physical (electromagnetic) viewpoint; (b) mathematical (mini-max problem) viewpoint.

z N turns

w

i(t)

l y !

n

x Figure 2.43. Rotor winding in the magnetic field Solution 2.7.1. The direction of the magnetic flux density B is φ=tan-1(y/x)=tan-1(2/1)=63.43° From N=20, i=10 A and A=15×10-2×5×10-2 m2, the magnetic dipole moment of the loop is found to be m=NIAan=20×10×(15×5) ×10-4an=1.5an, where a n is the surface normal unit vector in accordance with the right hand rule. When the current loop is in the negative –y of the yz plane, an=ax. When the plane of the current loop is moved to an angle φ, an becomes an=axcosφ+aysin φ, Using B=2×10-2(ax+2ay), one has T=m×B=an1.5×2×10-2(ax+2ay)=(axcosφ+aysin φ)1.5×2×10-2(ax+2ay)=3×10 –2(2cos φ–sinφ)az N-m 2.7.2 The electromegnetic torque is zero when 2cosφ–sinφ=0, or tanφ=2. That is, φ=63.43° or –116.57° Physical interpretation: The torque is zero when an is parallel to B, and T=0. The torque is a maximum when an is perpendicular to B which occurs at φ=63.43°±90°=–26.57° or +153.43°. Thus, Tmax=±0.067az N-m. One can find the values of φ at which |T| is a maximum by differentiating T and equating the derived expression to 0. In particular, we have "T " ( = 3 x10 ! 2 (2 cos # ! sin # ))= 0 "# "# Hence, –2sinφ–cos φ=0 or –2sinφ=cos φ One has, tan φ=–1/2 or φ=–26.57° or +153.43°. This is the angle φ at which the electromagnetic torque T is maximum. We obtain Tmax=3×10–2(2cosφ–sinφ)az N-m. The derived result yields that Tmax occurs at 2 values of angle φ, e.g., at φ=–26.57° or +153.43°. We substitute these two values of angle φ in Tmax, yielding Tmax=3×10–2[2cos(–26.57°)-sin(–26.57°)]az N-m and Tmax=3×10–2(2cos153.43°–sin153.43°)az N-m Hence, one finds Tmax=±0.067az N-m.

17

18

Chapter 3 Problem 3. 1. Using the operational amplifiers, develop the schematics to implement the proportional-integral controller. Report at least two possible schematics. Derive the transfer functions G(s) and express the feedback gains kp and ki using the circuitry parameters (resistors and capacitors). Solution The proportional-integral-derivative (PID) controller can be implemented using the configuration as depicted in Figure 3.9. The transfer function realized by an inverting operational amplifier is found to be R C + R2 C2 1 R2 C1s 2 + 1 1 s+ R1C1s + 1)(R2 C2 s + 1) ( U 0 ( s) R1C2 R1C2 G ( s) = =! =! . One recall that the transfer U 1 ( s) R1C2 s s function of the PID controller is GPID ( s ) = k p +

k d s 2 + k p s + ki ki + kd s = . The feedback gains kp, ki s s

and kd are defined by the values of resistors and capacitors of the input and feedback paths. Ee have R C + R2 C2 1 kp = ! 1 1 , ki = ! and k d = ! R2 C1 . The proportional, integral and derivative feedback R1C2 R1C2 gains (kp, ki and kd) must be positive. Due to the minus signs, these proportional, integral and derivative feedback should be inverted. Hence, an additional inverting amplifier is needed. The commonly applied implementation of the PID controller is illustrated in Figure 3.10. The U ( s) R2 p 1 transfer function is G ( s) = 0 = + + R2 d C1d s . The proportional, integral, and derivative U 1 ( s) R1 p R1i C2i s feedback coefficients are k p =

R2 p R1 p

, ki =

1 and k d = R2 d C1d . R1i C2i

Correspondingly, the PI controller schematics is reported in Figure 1 R2p R1p

R

!

+

C2i R1i

R R

!

!

+ u1(t) –

R

+ +

R2d

u0+(t) –

Figure 1. Implementation of an analog PI control law Two possible PI control circuit schematics, as given in Figures 2 and 3, are examined.

19

Proportional Stage

Rp2

umax -

R

Rp1

Summing Stage R

u1 + umin

umax Ci

Integral Stage u3 umax

+ umin

-

uo

Ri ui

R

u2

+ umin

Figure 2. PI Controller schematics A R2

Ci

umax

R umax

R1 u1

+

ui

umin

R

u2 + umin

uo

Figure 3. PI Control schematics B The transfer function for the PI controller reported in Figure 2 is derived by finding the transfer functions of the proportional and integral stages, and, then, using the summing stage. We have - Proportional part !

ui u ! 1 =0 R p1 R p 2

u u1 =! i Rp 2 R p1 u1 = !

Rp 2 R p1

ui

- Integral part

20

!

ui u2 ! =0 1 Ri Ci s

u u2 =! i 1 Ri Ci s 1 ui Ri Ci s The summation gives u3 ! u2 u3 ! u1 u3 ! uo + + =0 R R R uo = !u2 ! u1 R 1 uo = ( + p 2 )ui Ri Ci s R p1 u2 = !

R uo 1 = + p2 ui Ri Ci s R p1

The expressions for kp and ki are kp=Rp2/Rp1 and ki=1/(RiCi). The transfer function of the circuit shown in Figure 2 can be found, and the feedback coefficients results. One obtains 1 R2 + Cs u u1 = ! i R1

R u1 R uo R2 1 = + ui R1 R1Cs Hence, kp =R2/R1 and ki=1/(R1Ci). uo = !

Problem 3. 2. Using the operational amplifiers, develop the schematics to implement the PID controller. Report at least two possible schematics. Derive the transfer functions G(s) and express the feedback gains kp, ki and kd using the circuitry parameters. Note that there are many solutions which approximate an ideal GPID(s) to make the implementable circuit and relax some deficiencies of an ideal PID controller, including cases when the controller input is step-like signals or contain noise. Perform the trade-off studies for ideal and practical PID controllers. You may utilize the Laplace transform, Bode plots, frequency-domain and other analyses. Solution A PID controller can be implemented using operational amplifiers in various configurations. Two possible analog PID-control-implanting circuit schematics are given in Figures 1 and 2.

21

Proportional Stage

Rp2

umax -

R

Rp1 u1 + umin

Derivative Stage Rd

umax

Cd -

R

Summing Stage R

u2

+

u4

umin

umax Ci

Integral Stage -

umax

+ uo

umin

Ri ui

u3

+

R

umin

Figure 1. PID control law: Schematics A C1

C2 R2 umax

R u2

umax

R1

+

ui

u1 umin

R + umin

uo

Figure 2. PID control law: Schematics B The transfer function for the PID controller in Figure 1 is derived by finding the transfer functions for the proportional, derivative and integral stages, and, then performing the summation. Using the derivations reported in the book, we have the expressions for the feedback gains

kp =

Rp2 R p1

, ki =

1 and k d = Rd Cd . Ri Ci

The transfer function of the circuit shown in Figure 2 is found as

22

Z1 =

R1 1 = 1 + C1s R1C1s + 1 R1

Z 2 = R2 + u1 = !

R C s +1 1 = 2 2 C2 s C2 s

Z2 ui Z1

R u1 R uo R2C2 s + 1 R1C1s + 1 = ui C2 s R1

uo = !

uo R2C2 R1C1s 2 + R1C1s + R2C2 s + 1 = ui R1C2 s uo C R 1 = R2C1s + 1 + 2 + ui C2 R1 R1C2 s

Hence, we have k p =

C1 R2 1 , ki = and k d = R2C1 . + C2 R1 R1C2

The derived transfer functions can be used to calculate the Bode plots, obtain the frequency response, as well as dynamic behavior. The control law gains must be assigned. We let kp=2, ki=1 and kd =1. For the reported PID controller schematics, the resulting Both plots are calculated and plotted using MATLAB. We apply the bode command, and the statements are % PID Controller: Schematics A rp1=1000;rp2=2000; ri=1000; ci=0.001; rd=1000; cd=0.001; PIDA=tf([rd*cd*rp1*ri*ci (rp2*ri*ci) rp1], [rp1*ri*ci 0]); figure(1); bode(PIDA); figure(2); bode(PIDA,{1,10000}); title('Practical PID Controller Frequency Response, Schematics A'); % PID Controller: Schematics B r1=1000; r2=1000; c1=0.001; c2=0.001; PIDB=tf([r2*c2*r1*c1 (r1*c1+r2*c2) 1], [r1*c2 0]); figure(3); bode(PIDB); figure(4); bode(PIDB,{1,10000}); title('PID Controller Frequency Response, Schematics B');

The Bode plots of the analog PID controllers are illustrated in Figures.

23

) B d ( e d u t i n g a M ) g e d ( e s a h P

) B d (

Bode Diagram 25

20

15

10

5 90 45

e d u t i n g a M

80

) g e d (

0 90

e s a h P

0 -45

Practical PID Controller Frequency Response, Schematics A 100

-90

60 40 20

45

0 -1

10

0

10

1

0

10

10

1

10

Frequency (rad/sec) ) B d ( e d u t i n g a M ) g e d ( e s a h P

2

10

3

10

4

10

Frequency (rad/sec)

) B d (

Bode Diagram 25

20

15

10

5 90 45

e d u t i n g a M

80

) g e d (

0 90

e s a h P

0 -45

PID Controller Frequency Response, Schematics B 100

60 40 20

45

0

-90 -1

10

0

10

1

10

0

10

Frequency (rad/sec)

1

10

2

10

3

10

4

10

Frequency (rad/sec)

Figure 3. Analog PID controller frequency responses The frequency responses for both PID circuits are identical. However, it is virtually impossible to refine the feedback coefficients if the schematics B is applied. One may not be able to tune (adjust) the feedback gains kp, ki and kd. Therefore, schematics A is a preferable choice. Hence, though the PID controller, implemented as schematics B, results in lower number of components, simplicity, low cost, low power consumption, this implementation is restricted to very specific possible applications. Problem 3. 3. Why converters and power amplifiers should be utilized in electromechanical systems? Solution Because high voltage and current are required to be supplied to the armature, field and phase windings of electric machines. Some motors (automotive, marine and other applications) require ~1,000 A and higher. Furthermore, the voltage must be controlled.

Problem 3. 4. Study the PWM switching concept applying the sinusoidal-like (not triangular ut) signal to the comparator. E.g., the switching signal us, which drives the switch S, is generated by comparing a signallevel control voltage uc with a repetitive sinusoidal-like signal usin. Propose the schematics to generate usin (transistors or operational amplifiers can be used). Report how to define and vary (if needed) the frequency of usin. Report the voltage waveforms.

24

Solution The typical switching technique utilizes a triangular or sinusoidal wave and a threshold voltage to determine the duty cycle of a PWM signal. When the amplitude of the periodic signal is higher than the control voltage the PWM signal is driven low, otherwise it is driven high. This means that as the control voltage approaches its maximum, the duty cycle of the PWM will become 100%. Using a sinusoidal signal-level voltage is a simple and very effective way to generate a PWM signal since these are typically easier to produce then a true triangular wave. This method, however, will result in a nonlinear mapping of threshold voltage to duty cycle. Two possible approaches for producing a sinusoid like waveform are the ring oscillator shown in Figure 1, and the Wien Bridge oscillator shown in Figure 2.

Figure 1. CMOS Ring Oscillator

}

Rf

Z1

{

Z2

umax Rg +

X umin

R1

} }

Z3

uo

C1

R2

C2

Z4

Figure 2. Wien Bridge Oscillator The ring oscillator operates by indefinitely looping a high/low signal around the circuit. Any signal will take two trips around the circuit to complete a full period. If the propagation delay tP for a single gate is known, the oscillating period can be calculated as T=2tPNNg , where N is the number of gates in series; Ng is the number of gates in parallel at each junction. For the ring oscillator, reported in Figure 1, there are 5 series gates each with 1 gate in parallel at each junction. A method to calculate propagation delay is using the effective resistance of the CMOS devices during transition, and calculating the time to reach half VDD via an “RC” time constant. Assuming 180 nm technology node and load capacitance of 47 pF, we have

25

L Req CL W 10um = .69 (30 K !)(47 pF ) = 19.46ns 500um 10um = .69 (12.5 K !)(47 pF ) = 20.27 ns 200um

t P = .69 t PLH t PHL

Therefore, the oscillation period is calculated to be T=2tPNNg=2(19.4 nsec)x5x1=198 nsec The simulated output of this ring oscillator is shown in Figure 3.

Figure 3. Ring oscillator: Simulation result To change the period of the oscillator, additional inverters can be added in series or additional load capacitance can be placed at each junction. This results in a longer or shorter per stage propagation delay. Variance of the source voltage can also affect the delay. Increasing the source voltage will drop the delay time, and the inverse for decreasing the source voltage. The Wien Bridge Oscillator, as reported in Figure 2, is one of the simplest versions of the circuit to establish the sinusoidal output signal. The tuning network attached to the positive terminal of the operational amplifier is used to set the oscillation frequency. The negative feedback provides amplification of the oscillation signal chosen to ensure stability. By breaking at point X of the positive feedback and applying a test source, an expression for the positive terminal voltage can be found. In particular, we have

26

U + = UT

Z3 Z3 + Z 4

Z 3 = R1 + C1 Z 4 = R2 || C2 U + = UT

R2 R2C2 s + 1

R2 1 + R1 + R2C2 s + 1 C1s

U+ R2 = U T R + R R C s + R + R2C2 s + 1 2 1 2 2 1 C1s U+ 1 = U T R C s + 1 + R1 + C2 + 1 1 2 R2 C1 R2C1s

Since the positive and negative terminal voltages are equal, an expression for the output voltage and transfer function are found to be. In particular, U o = !U +

Z1 + Z 2 Z1

Z1 = Rg Z2 = R f Rg + R f Uo 1 =! R C 1 UT Rg R1C2 s + 1 + 1 + 2 + R2 C1 R2C1s

Oscillation occurs when HolHf=–1, where Hol is the open loop system transfer function, and Hf is the feedback transfer function. To satisfy this requirement, the passive components must be chosen to ensure a gain of 1 and a phase shift of –180o. The phase shift is accomplished by the negative feedback. Hence, we need to ensure a gain of 1 and no change in phase. We choose a desired oscillation frequency of ω0 s = j!0

Rg + R f Uo 1 =" R1 C2 1 UT 1+ + + j ( R1C2!0 " ) Rg R2 C1 R2C1!0

!0 =

1 1 = R2C1 R1C2

To assure no phase shift, we must set the passive components with R = R = R 0 2 1 C0 = C2 = C1 in the following expression for the transfer function as Rg + R f Uo 1 =! R C UT Rg 1+ 0 + 0 R0 C0 Uo 1 Rg + R f =! UT 3 Rg

. This results

27

To satisfy, the magnitude condition, Rg and Rf are found to result in a negative feedback gain of 3. Let Rg=1000 ohm and Rf =1000 ohm, resulting in the oscillation conditions being satisfied. Furthermore, Uo/UT=–1. To change the frequency of oscillation !0 , we vary R0 and C0 . We have

ω0=1/(R0C0) and f0=1/(2πR0C0). The output waveform of the Wien Bridge is a near-ideal sine wave at the circuit natural frequency with an amplitude of the operational amplifiers supply voltages. A possible schematic of the sine-wave generator with a comparator is reported in Figure 4. -12Vdc V1

0

C3 1u

C2 1u

2

4 -

AD648A VOUT

3

+ 8 U3A

R2

1

6

4 -

1k V

V+

AD648A VOUT

5

+ 8 U3B

2

7

3 V3 89.2718uV

4

AD648A V-

V

OUT

-

+ 8 U4A

V+

1

V

V+ R3 1k

R4 1

C1 1u R1 1k

V2 12Vdc

0

Figure 4. Schematic for the sine-wave generator The equation for the frequency of the sine-wave generator is f=1/(2πRC), where R is the resistor value; C is the capacitor value. The simulations were performed, and the resulting waveforms are documented in Figures 5 and 6.

28

-89.26uV

(13.291m,-89.272u)

-89.27uV

-89.28uV -15.468270mV

(19.760m,-89.272u)

V(R2:2) (8.4345m,-15.468m)

-15.468271mV

SEL>> 5.4824m,-15.468m) -15.468272mV 0s 10ms V(U4A:OUT)

(11.929m,-15.468m) 20ms

30ms

40ms

50ms

Time

Figure 5. Voltage waveform, R=1000 ohm and C=1 µF (duty cycle is 0.4579) -89.2633uV -89.2688uV

-89.2750uV

-89.2813uV -15.439820mV

V(R2:2) (11.467m,-15.440m)

-15.439821mV

SEL>> (6.4724m,-15.440m) (9.925m,-15.440m) -15.439822mV 0ms 10ms 20ms V(U4A:OUT)

30ms

40ms

50ms

Time

Figure 6. Voltage waveform, R=800 ohm and C=1 µF (duty cycle is 0.691) As seen from Figures 5 and 6, the duty cycle when the resistor value is changed also varies. The sine-wave is compared with voltage uc. The second (rectangular) waveforms in Figures 5 and 6 indicate the switching takes place to control the switch (transistor) of the PWM converter. Problem 3. 5. Report the four-quadrant H-configured power stage schematics. Explain how it operates. Solution A four-quadrant H-bridge permits bidirectional control of motors and actuators. Figure 1 provides a schematic of a possible configuration of a four-quadrant H-configured power stage with a permanentmagnet DC motor.

29

5 Vdc

V Motor

5 Vdc

RL

RL P MOS

Pin 1

P MOS

NMOS

NMOS

Pin 2

CL M

CL RS

Pin 3

NMOS

NMOS

Pin 4

Figure 1. Four-quadrant H-bridge schematic The bridge is operated with PWM (pulse width modulated) signals that are applied to the gates of the MOSFETS (pins 1-4). To drive the motor in a particular direction, for example clockwise, a PWM signal would be applied to pins 1 and 4 while pins 2 and 3 remained grounded. To drive the motor in the opposite direction, for example reverse (counterclockwise), a PWM signal would be applied to pins 2 and 3 leaving 1 and 4 grounded. This creates an average DC voltage applied to the motor in either direction based on the duty cycle of the PWM signal. The reason for using an nMOS–pMOS configuration on the high side of the bridge is to ensure a low voltage switching. Various schematics are used, as reported in the textbook. For example, the following schematics (http://www.cadvision.com/blanchas/hexfet/np-s.htm) can be utilized.

Four switches (transistors) control the armature voltage supplied. The motor can be rotated clockwise and counter clockwise, as well as ensure the braking. When Q1 and Q4 are closed (and Q2 and Q3 are open), the voltage applied to the motor is positive, resulting in, say, clockwise rotation. When Q1 and Q4 are open (and Q2 and Q3 are closed), the voltage applied to the motor is negative, resulting in counterclockwise rotation. Switching Q1 and Q4 (one closed, one open) will result in disaceleration.

30

Chapter 4 Problem 4.1. The torque-speed characteristic of a permanent-magnet DC motor is given in Figure 4.36 as one applies the rated voltage ua=10 V. Calculate the values for the back emf constant ka, rated armature current ia, and the armature resistance ra. Assume that the viscous friction can be neglected.

!r

100 80

Te, TL 0

1

Figure 4.36. Torque-speed characteristic of a permanent-magnet DC motor Solution One recalls that the electromagnetic torque is Te=kaia, and in the steady-state operation Te=TL. steady-state torque-speed characteristics are give by (4.3), e.g., u "r i u r ! r = a a a = a " a2 Te . ka ka ka

The

At the no-load conditions, ωr=ua/ka. Thus, ka=ua/ωr=10/100=0.1 V-sec/rad. The electromagnetic torque is given as Te=kaia. Hence, the rated current is ia=Te/ka =1/0.1=10 A. To calculate the armature resistance one recalls that the slope of the torque-speed characteristic is ra/ka2. Furthermore, the angular velocity difference between no-load and rated load is 20 rad/sec. Thus, u "r i u r from ! r = a a a = a " a2 Te we have ka ka ka ra=∆ωrka2/Te=20Η0.01/1=0.2 ohm. Problem 4.2. Perform simulation (in MATLAB using SIMULINK) and analysis (study the steady-state and dynamic behavior, assess the losses for the unloaded and loaded motor, examine efficiency, etc.) of a 30 V, 10 A, 300 rad/sec permanent-magnet DC motor. The motor parameters are: ra=1 ohm, ka=0.1 V-sec/rad (Nm/A), La =0.005 H, Bm=0.0001 N-m-sec/rad and J=0.0001 kg-m2. Solution The differential equations describing the behavior of a permanent-magnet DC motor are derived in the textbook. In particular, we have equations (4.1) dia r k 1 = ! a ia ! a " r + ua , dt La La La d! r k B 1 = a i a " m ! r " TL . dt J J J Using the results reported in section 4.1.2, the SIMULINK diagram is developed as illustrated in Figure 1.

31

Figure 1. Permanent-magnet DC motor SIMULINK model At no load conditions, as reported in Figure 2, the motor accelerates to the rated angular velocity of 300 rad/sec with a the current spike ~19 A. While the current value is higher the rated value 10 A, the overloading is acceptable for a short period. The motor accelerates with a small overshoot which is acceptable in many applications. The electrical losses correspond to the small armature current (due to the friction) and mechanical losses which must be considered only at the steady-state operation. That is, during the transient (acceleration, dynamics, etc.), the losses should not be examined. Applied Voltage vs Time

Current vs Time 20

30 ) v ( U

) A (

20

i

10

15 10 5 0

0 ) d a r ( a t e h T

0

0.1

0.2

0.3

0.4

0.5

T (s) Theta vs Time 150

100

) s / d a r (

0

0.1

0.2 0.3 Time (s)

0.4

0

0.1

0.5

0.4

0.5

0.4

0.5

200 100 0

0.5

0

0.1

0.2 0.3 Time (s) Mech Losses vs Time

10 ) W (

300 200

m l P

100 0

0.4

300

400

e l P

0.3

400

Electric Losses vs Time ) W (

0.2

Time (s) Omega vs Time

a g e m O

50

0

-5

0

0.1

0.2 0.3 Time (s)

0.4

0.5

8 6 4 2 0

0

0.1

0.2 0.3 Time (s)

Figure 2. Permanent-magnet DC motor simulation results, TL=0 N-m We study the motor dynamics and its performance (acceleration capabilities, losses, efficiency, etc.) at the rated load. At TL=1 N-m, the steady state angular velocity is 200 rad/sec as illustrated in Figure 3. This is due to the downward slope of the torque speed characteristic with increasing load. Like in the

32

unloaded case, there is the current spike while the motor is accelerating to its steady state value. Since the motor is at the maximum rated load, the steady state current is ~10A. This large steady-state current results in the electrical losses being the main loss component, instead of the mechanical losses as seen in the unloaded case. Applied Voltage vs Time

Current vs Time 25

30

20 ) v (

20

) A (

U

10

i

15 10 5

0

0

0.1

0.2

0.3

0.4

) s / d a r (

T (s) ) d a r (

Theta vs Time 100

50

a t e h T

0

0.1

0.2 0.3 Time (s)

0.4

0.5

0

0.1

0.5

0.4

0.5

0.4

0.5

200 100 0 -100

0

0.1

0.2 0.3 Time (s) Mech Losses vs Time

600

e l P

0.4

300

Electric Losses vs Time ) W (

0.2 0.3 Time (s) Omega vs Time

a g e m O

0

-50

0

0.5

5 ) W (

400

m l P

200

4 3 2 1

0

0

0.1

0.2 0.3 Time (s)

0.4

0

0.5

0

0.1

0.2 0.3 Time (s)

Figure 3. Permanent-magnet DC motor simulation results, , TL=1 N-m The efficiency varies as a function of load and angular velocity. The machine efficiency must be examined at the steady-state when the motor variables (current, angular velocity, etc.) are at the equilibrium. The analysis of efficiency is performed using the simulation results. The steady-state efficiency is provided by Figure 4. One finds that depending on the operating conditions, the efficiency varies from 70 to 80 %. Electical Losses for Tl = 0.2,0.4,0.6,0.8 600 .2 ) W ( e l P

.4 .6

400

.8 200

0

0

0.01

0.02

0.03

0.04 0.05 0.06 Time Mech Losses for Tl = 0.2,0.4,0.6,0.8

0.07

0.08

0.09

0.1

10 ) W ( m l P

.2 .4 .6

8 6

.8 4 2 0

y c n e i c i f f E

0

0.01

0.02

0.03

0.04 0.05 0.06 Time Efficiency for Tl = 0.2,0.4,0.6,0.8

0.07

0.08

0.09

0.1

1

0.6

.2 .4 .6

0.4

.8

0.8

0.2 0 0

0.01

0.02

0.03

0.04

0.05 Time

0.06

0.07

0.08

0.09

0.1

Figure 4. Efficiency of a permanent-magnet DC motor at different loads: The efficiency is a steady-state characteristic, e.g., the efficiency is found when the system in equilibrium

33

Problem 4.3. Develop the mathematical model for an axial topology limited angle actuator (computer hard drive) for which −100≤θr≤100 or −0.175≤θr≤0.175 rad. Let B(θr)=Bmaxsin(aθr), a>0, Bmax=1 T. Perform simulation in SIMULINK and carry-out analysis assigning a=1, a=3 or a=6. Estimate and calculate the parameters of the actuator (or use the parameters reported in section 4.2.2). Propose other sound expressions for B(θr) not reported in the chapter. For the proposed B(θr), derive the expression for Te, develop a mathematical model, simulate the actuator, and analyze it performance. Solution The winding filament is schematically represented in Figure 1. Current Loop Resistance – r Number - N

R

l

A

_0

Figure 1. Axial topology actuator: Current loop schematics The actuator is actuated due to the Lorenz force developed by the left and right radial components of the current loop filaments. We have Fe=FeL+FeR. This force (sum of the left and right forces) at the median distance from the pivot point R produces the electromagnetic torque. We have Te = R " Fe

Te = (aˆ z ) N # l # i # R # Bmax [sin(a (! r $ ! 0 )) $ sin(a (! r + ! 0 ))] Integrating the friction and spring torques, expressions for angular acceleration and velocity are. J ! = (aˆ z )% T

d" 1 = (aˆ z ) [Te $ T f $ Ts ] dt J d" 1 = (aˆ z ) [ NliRBmax [sin(a (# r $ # 0 )) $ sin(a (# r + # 0 ))] $ Bm" $ K s# $ K sr# 3 ] dt J The Kirchhoff voltage law is used to derive an expression for the current though the coil of the axial topology actuator. We have d" u = ir + dt " = Li + AB

u = ir + L where

di dB d! +A dt d! dt

34

dB d! dB = " d! dt d! dB = Bmax # a[cos(a (! r $ ! 0 )) $ cos(a (! r + ! 0 ))] d!

Thus, one finds di 1 = [u # ir # A $ Bmax $ a $ ! $ [cos(a (" r # " 0 )) # cos(a (" r + " 0 ))]] dt L A set of differential equations describing the motion of the axial topology hard drive actuator is di 1 = [u # ir # A $ Bmax $ a $ ! $ [cos(a (" r # " 0 )) # cos(a (" r + " 0 ))]] dt L d! 1 = (aˆ z ) [ NliRBmax [sin(a (" r # " 0 )) # sin(a (" r + " 0 ))] # Bm! # K s" # K sr" 3 ] dt J d" =! dt The SIMULINK model is illustrated in Figure 2. N*Bmax*R*u*l sin(a*(u-theta0))-sin(a*(u+theta0))

alpha

1/J Inverse inertia

omega

1 s

1 s

theta

theta

Friction Bm

omega

Ks Spring Ksr

u^3

Non-Linear Spring

current uApplied

1 s

1/L

i

Reluctance

r Resistance Bmax*A*a*u cos(a*(u-theta0))-cos(a*(u+theta0))

t Clock

Figure 2. SIMULINK model of an axial topology actuator Figures 3, 4, and 5 provide plots of angular position, angular velocity, and current for a=1, 3 and 6, respectively. The axial topology actuator is driven by a square wave voltage with a frequency of 1 Hz and amplitude of 1 V.

) s d a r (

35

Theta vs Time

0.2

a 0 t ) es h / -0.2 Ts 0 d a r 5 ( ) s p m A (

a 0 g e -5 m 0 O

t n e r r u C

0.05

0.5

1

1.5

2

2.5 3 Time (s) Omega vs Time

3.5

4

4.5

5

0.5

1

1.5

2

2.5 3 Time (s) Current vs Time

3.5

4

4.5

5

0.5

1

1.5

2

3.5

4

4.5

5

0 -0.05

0

2.5 Time (s)

3

) Figure 3. Dynamics for θ(t), ω(t) and i(t) for a=1 s d a r (

Theta vs Time

0.2

a 0 t) es h/ -0.2 Ts 0 d a r 10 ( ) a 0 sg pe mm -10 0 AO ( t n e r r u C

0.5

1

1.5

2

2.5 3 Time (s) Omega vs Time

3.5

4

4.5

5

0.5

1

1.5

2

2.5 3 Time (s) Current vs Time

3.5

4

4.5

5

0.5

1

1.5

2

3.5

4

4.5

5

0.05 0 -0.05

0

2.5 Time (s)

3

Figure 4. Dynamics for θ(t), ω(t) and i(t) for a=3

) s d a r (

36

Theta vs Time

0.2

a 0 t) es h/ -0.2 Ts 0 d a r 20 ( ) a 0 sg pe mm -20 0 AO ( t n e r r u C

0.5

1

1.5

2

2.5 3 Time (s) Omega vs Time

3.5

4

4.5

5

0.5

1

1.5

2

2.5 3 Time (s) Current vs Time

3.5

4

4.5

5

0.5

1

1.5

2

3.5

4

4.5

5

0.05 0 -0.05

0

2.5 Time (s)

3

Figure 5. Dynamics for θ(t), ω(t) and i(t) for a=6 Figures 3, 4 and 5 provide plots of the actuators performance for a=1, 3 and 6. The variable a is the actuator “magnetization parameter”. Comparison of the dynamics reveals that increases in a yield an improved transient response of the actuator due to the higher value of B(θ) within the limited displacement angle. For a=1, the settling time is approximately 0.1 sec, while for a=6, the settling time is reduced to ~0.02 sec. Another case of interest to examine is when the magnets magnetized as B(θr)=Bmaxθr3. We have Fe = Fl + Fr

Fe = $&("aˆr ) Nli # (aˆ z ) Bl %' + $&(aˆr ) Nli # (aˆ z ) Br %' Fe = (aˆ! ) Nli[ B(! r " ! 0 ) " B(! r + ! 0 )] and

Te = R $ Fe

[

3 3 Te = (aˆ z )N # l # i # R # Bmax (! r " ! 0 ) " (! r + ! 0 )

]

The resulting dynamics is reported in Figure 6. The overall actuator performance is significantly poorer as compared with the previous studies. This is due to low magnetic flux density for the assigned displacement angle. For the last case the simulations were performed assigning the applied voltage to be 12 V.

) s d a r (

37

Theta vs Time

0.1

a 0 t ) es / h -0.1 Ts 0 d a r 5 ( ) a s g pe mm AO ( t n e r r u C

0.5

1

1.5

2

2.5 3 Time (s) Omega vs Time

3.5

4

4.5

5

0

0.5

1

1.5

2

2.5 3 Time (s) Current vs Time

3.5

4

4.5

5

0

0.5

1

1.5

2

3.5

4

4.5

5

0 -5

0.5 0 -0.5

2.5 Time (s)

3

Figure 6. Dynamics for θ(t), ω(t) and i(t) if B(θr)=Bmaxθr3, Bmax=1 T

38

Chapter 5 Problem 5. 1. Consider a four-phase, 60 Hz, two pole induction motor. 5.1.1. Derive the expression for ψar assuming that the stator-rotor mutual inductances obey a cos3θr distribution. Hence, the stator-rotor mutual inductances become to be functions of cos3(θr+phase). 5.1.2. Derive the expression for the emfar induced in the rotor winding ar. Justify why the studied ac motor is called the induction motor. 5.1.3. The induction motor accelerates, and the motor “instantaneous” angular velocity is denoted as ωrInst in Figure 5.29. What is the final angular velocity with which the motor will operate (report the numerical value of ωr). !r TL !e !r Inst

TL0 0 Te start Te Figure 5.28. Torque-speed and load characteristic

Solution 5.1.1. We have ψar=Larasias+Larbsibs+Larcsics+Lardsids+Larariar+Larbribr+Larcricr+Lardridr. Hence, ψar=Lsrcos3θrias+Lsrcos3(θr+π)ibs+Lsrcos3(θr+2π)ics+Lsrcos3(θr+3π)ids+(Llr+Lmr)iar+0ibr+Lmricr+0idr =Lsrcos3 θrias+Lsrcos3(θr+π)ibs+Lsrcos3(θr+2π)ics+Lsrcos3(θr+3π)ids+(Llr+Lmr)iar+Lmricr. 0idr 5.1.2. The emfas is emfar=dψar/dt=Lsrcos3θr(dias/dt)+Lsrcos3(θr+π)(dibs/dt)+Lsrcos3(θr+2π)(dics/dt) +Lsrcos3(θr+3π)(dids/dt)+3Lsrsinθr[cos2θrias+cos2(θr+π)ibs+cos2(θr+2π)ics+cos2(θr+3π)ids]ωr +(Llr+Lmr)(diar/dt)+Lmr(dicr/dt). The motional emf induced in the rotor winding is found to be emfωar=3Lsrsinθr[cos2θrias+cos2(θr+π)ibs+cos2(θr+2π)ics+cos2(θr+3π)ids]ωr As the phase voltages are supplied to the stator phase windings, the voltages are induced in the rotor windings. Therefore, these machines are called induction motors. rad , one concludes that 5.1.3. Motor accelerates passing the ωrInst. Having found "e = 4!f = 4!f = 377 sec P P ωr=365 rad/sec (the intersection of the motor torque-speed and load characteristics) Problem 5. 2. Consider an A-class two-phase, 115 V (rms), 60 Hz, four-pole (P=4) induction motor. Let Lasar=Laras=Lms1cos θr+Lms3cos5θr, Lasbr=Lbras=Lms1sinθr+Lms3sin5θr, Lbsar=Larbs=Lms1sinθr+Lms3sin5θr and Lbsbr=Lbrbs=Lms1cos θr+Lms3cos5θr. The motor parameters are: r s=1.2 ohm, rr' =1.5 ohm, Lms1=0.14 H, Lms2=0.004 H, Lls=0.02 H, Lss=Lls+Lms1 +5Lms3, L'lr =0.02 H, L'rr =L'lr + Lms1 + 5 Lms 3 , Bm=1×10−6 N-m-

39

sec/rad and J=0.005 kg-m2. The phase voltages supplied are

uas (t ) = 2115 cos(377t ) and

ubs (t ) = 2115 sin(377t ) . 5.2.1. In MATLAB, calculate, plot and compare Lasar=Lmscos θr (Lms=0.16 H) and Lasar=Lms1cos θr +Lms3cos5θr; 5.2.2. Derive the circuitry-electromagnetic equations of motion; 5.2.3. Report emfas and emfar; 5.2.4. Find the expression for the electromagnetic torque; 5.2.5. Using Newton’s second law, obtain the torsional-mechanical model; 5.2.6. Report the equations of motion in non-Cauchy’s form; 5.2.7. Develop the SIMULINK mdl model or MATLAB file using ode45 differential equations solver to simulate induction motor dynamics for A and D-class induction motors. 5.2.8. Plot the transient dynamics for all state variables, e.g., report ias(t), ibs(t), i‘ar(t), i’br(t), ωr(t) and θr(t); 5.2.9. Plot the motional emfarω and emfbrω induced in the rotor windings; 5.2.10. Plot the torque-speed characteristics ! r = " T (Te ) using the simulation results; 5.2.11. Analyze the induction motor performance (acceleration, settling time, load attenuation, etc.). Solution The solution of this problem is similar as given in the textbook, see section 5.1.5. Example 5.3 provides the results closely related to this homework problem.

Problem 5. 3. Consider the arbitrary reference frame for two-phase electric machines and electromechanical motion devices. Using the Park transformations, the machine (ab) variables can be transformed to the quadrature and direct (qd) quantities. The Figure 5.29, given in Table 5.2, illustrates the machine stator and rotor magnetic axes, as well as the quadrature and direct magnetic axes. Letting the angular velocity of the reference frame to be ω (not specified), ω=0, ω=ωr and ω=ωe, one deals with the arbitrary, stationary, rotor and synchronous reference frames. The direct and inverse transformations, as well as corresponding matrices for different reference frames, are reported in Table 5.2. Problems to be solved: 5.3.1. Find the quadrature- and direct-axis components of stator voltages in the arbitrary reference frame if the voltages uas (t ) = 100 cos(377t ) and ubs (t ) = 100 sin(377t ) are supplied to the as and 5.3.2.

bs windings. e e (t ) , uds Find uqs (t ) and uqss (t ) , udss (t ) in the synchronous and stationary reference frame assigning ! = 377

5.3.3.

rad sec

and ! = 0 .

Prove that uqs (t ) and uds (t ) are the dc or ac voltage components in the synchronous and stationary reference frames, respectively.

Solution. 5.3.1.

"cos( To solve the problem, one applies the Park transformation matrix K s = $ # sin ( !u $ !cos( Taking note of u qds = K s u abs , # qs & = # #"uds &% " sin ( we have

sin ( $ !uas $ , ' cos( &% #"ubs &%

sin ( % . ! cos( '&

40

uqs (t ) = cos!uas (t ) + sin !ubs (t ) ,

uds (t ) = sin "uas (t ) ! cos"ubs (t ) . Thus, in the arbitrary reference frame, the quadrature- and direct-axis components of voltages are found to be uqs (t ) = 100(cos! cos(377t ) + sin ! sin(377t )),

(

uds (t ) = 100 sin " cos(377t ) ! cos" sin(377t )).

5.3.2. Specifying the angular velocity of the reference frame, one obtains the quadrature- and directaxis components of voltages in the stationary, rotor, and synchronous reference frames. The synchronous reference frame rotates with the constant angular velocity ! = 377 rad and let !0 = 0 . Thus, we have sec

! = "t = 377t . Hence, e uqs (t ) = 100 cos2 (377t ) + sin 2 (377t ) = 100 ,

( (

)

e uds (t ) = 100 sin(377t )cos(377t ) ! cos(377t )sin(377t ))= 0 .

One concludes that in the synchronous reference frame the quadrature- and direct-axis e e (t ) and uds components uqs (t ) are the dc voltages (recall that it is assigned that the angular frequency of the supplied voltages and the reference frame angular velocity are equal). This result cannot be generalized to all references frames used. For example one finds that in the stationary reference frame, uqss (t ) and udss (t ) are the ac voltages. In the stationary reference frame, the frame angular velocity is assigned to be zero. Making use "cos( sin ( % ! = 0 and thus ! = 0 in K s = $ ' , we have the following transformation matrix needed to # sin ( ! cos( & be used in the stationary reference frame "1 0 % K ss = $ '. #0 ! 1& Therefore, the following expressions result !u s $ !1 0 $ !uas $ s u qds = K ss u abs , # qss & = # &# & . #"uds &% "0 ' 1% "ubs % Thus, in expanded form one obtains uqss (t ) = uas (t ) , udss (t ) = ! ubs (t ) . Hence, making use the expressions for the phase voltages we have uqss (t ) = 100 cos(377t ) and udss (t ) = !100 sin(377t ) . 5.3.3. e e (t ) and uds It was found that uqs (t ) are the dc voltages because e e uqs (t ) =100 V and uds (t ) =0 V.

Using the stationary reference frame, we have found that uqss (t ) and udss (t ) are the ac voltages because

uqss (t ) = 100 cos(377t ) and udss (t ) = !100 sin(377t ) .

41

Chapter 6 Problem 6. 1. Consider a two-phase permanent-magnet synchronous motor. Let ψasm=ψmcos7θr and ψbsm=ψmsin7θr. 6.1.1. Using Kirchhoff’s voltage law, derive the differential equation for the phase current ias and ibs. E.g., obtain and report the circuitry-electromagnetic differential equations. 6.1.2. Report the emfas induced (as found in 1.1). Plot the derived emfas as a function of θr at the steadystate operation if ωr=100 rad/sec and ψm=0.1. Report the MATLAB statement to calculate and plot emfas. 6.1.3. Derive an explicit expression for the electromagnetic torque Te. 6.1.4. For Te found, derive the balanced voltage and current sets with the goal to eliminate the torque ripple and current chattering. Report the problems one faces. Solution 6.1.1. From the Kirchhoff voltage law, by using ψas=Lssias+ψasm= Lssias+ψmcos7θr and ψbs=Lssibs+ψbsm=Lssibs+ψmsin7θr, we have dias r 7$ 1 = ! s ias + m #r sin " r cos 6 " r + uas dt Lss Lss Lss dibs r 7$ 1 = ! s ibs ! m #r cos " r sin 6 " r + ubs dt Lss Lss Lss

6.1.2.

The emfas is found to be emfas=7ψmωrsinθrcos6θr. The emfas is calculated and plotted by using the following MATLAB statement t=0:0.0001*pi:4*pi;psim=0.1;wr=100; emfas=-7*psim*wr*sin(t).*cos(t).^6;plot(t,emfas) 20 15 10 5 0 -5 -10 -15 -20

The plot is reported in the Figure 6.1.3.

2

4

6

8

10

12

14

The electromagnetic torque is Te = =

6.1.4.

0

P $Wc P $ø i = 2 $! r 2 $! r

T abs abs

=

P 2

$[ø asm

*i ' ø bsm ]( as % 7 7 )ibs & = P " $ (cos ! r ias + sin ! r ibs ) m $! r 2 $! r

P 7" m (# sin ! r cos 6 ! r ias + cos ! r sin 6 ! r ibs ). 2

The balanced current and voltage sets are ias=iMsinθr/cos6 θr or ias=iM/sin θrcos4θr, ibs=iMcosθr/sin6θr or ibs=iM/cosθrsin4θr,

and uas=uMsinθr/cos6θr or ias=uM/sinθrcos4θr,

42

ubs=uMcos θr/sin6θr or ibs=uM/cos θrsin4θr. Even mathematically, the “ideal” balanced voltage set cannot be implemented due to the phase voltage saturation. In particular, |uas|≤umax and |ubs|≤umax. The problems to implement this balanced voltage set is the saturation, singularity, hardware complexity (controllers should be implemented using DSPs), as well as (probably most impossible) inability to implement those voltages by six-, twelve- and other multioutput stages by PWM amplifiers. Problem 6. 2. Consider a two-phase axial topology permanent-magnet synchronous motor. The effective fluxe with respect to the as and bs windings are Bas=BMmaxsin7θr and Bbs=Bmaxcos7θr. The motor parameters are N=100, Aag=0.0001, Bmax=0.75 and Nm=10. Let the phase currents are ias=iMcos θr and ibs=iMsinθr. 6.2.1. Drive the expression for the electromagnetic torque. 6.2.2. Examine and document how the electromagnetic toque varies as a function of the rotor displacement. Plot torque versus displacement. 6.2.3. Make the conclusions how to improve the motor performance and enhance it capabilities. E.g., how to maximize the torque and ensure that Te does not have the ripple. 6.2.4. Document how to use MATLAB to solve the problems 6.2.1 through 6.2.3. Solution 6.2.1 and 6.2.4. The coenergy is given as Wc (ias , ! r ) = NAag Bmax sin 7 ! r ias + cos 7 ! r ibs , and

Te =

(

N m "Wc (ias , ! r ) . 2 "! r

)

The following MATLAB file is developed using the Symbolic Math Toolbox. % To use a symbolic variable, create an object of type SYM x=sym('x'); N=100; Aeq=0.0001; Bmax=0.75; Nm=8; iM=1; % Derive, calculate and plot the results w1=N*Aeq*Bmax*sin(x)^7; w2=N*Aeq*Bmax*cos(x)^7; % Differentiate w1 and w2 using the DIFF command d1=diff(w1); d2=diff(w2); % Phase currents ias=iM*cos(x); ibs=-iM*sin(x); % Derive and plot the electromagnetic torque Te=Nm*(d1*ias+d2*ibs)/2; Te=simplify(Te), ezplot(Te); pause % Derived phase currents set ias=iM*cos(x)/sin(x)^6; ibs=-iM*sin(x)/cos(x)^6; % Electromagnetic torque equation T=Nm*(d1*ias+d2*ibs)/2; Te=simplify(T), ezplot(Te);

The results of the calculations are given in the Command Window as reported below

Te = -21/100*cos(x)^2*(-4*cos(x)^4-1+3*cos(x)^2+2*cos(x)^6)

Having performed the differentiations and calculations, one finds the expression for the electromagnetic torque as 21 Te = ! 100 cos 2 " r ! 4 cos 4 " r + 3 cos 2 " r + 2 cos 6 " r ! 1 N-m.

(

)

6.2.2 and 6.2.3 21 The electromagnetic torque Te = ! 100 cos 2 " r ! 4 cos 4 " r + 3 cos 2 " r + 2 cos 6 " r ! 1 is plotted in

(

the Figure.

)

43

-21/100 cos(x)2 (-4 cos(x)4-1+3 cos(x)2+2 cos(x)6)

0.025

0.02

0.015

0.01

0.005

0 -6

-4

-2

0 x

2

4

6

Figure. Electromagnetic torque The electromagnetic torque varies as a function of the rotor angular displacement. Therefore, the torque ripple results. This is highly undesirable phenomena due to noise, vibration, degradations, etc. To minimize the torque ripple, one can perform structural machine redesign (attaining close to sinusoidal airgap flux distribution) and / or refine the balanced phase current set. The balanced current set (though cannot be implemented) is ias=iMcosθr/sin6θr and ibs=iMsin θr/cos6θr. For this current set Te=21/100 N-m. Problem 6. 3. Simulate and analyze an electromechanical system actuated by a NEMA 23 size, two-phase 1.8o full-step, 5.4 V (rms) and 1.4 N-m permanent-magnet stepper motor. The parameters are: RT=50, rs=1.68 ohm, Lss=0.0057 H, ψm=0.0064 V-sec/rad (N-m/A), Bm=0.000074 N-m-sec/rad and J= 0.000024 kg-m2. Study the motor performance for: (1) uas = ! u M sgn sin(RT"rm ) and ubs = u M sgn cos(RT!rm ) ; (2)

(

)

(

)

uas = ! 2u M sin(RT"rm ) and ubs = 2u M cos(RT!rm ) ; (3) Phase voltages uas and ubs with the magnitude uM are applied as sequences of pulses with different frequency to ensure a step-by-step operation. Explain why stepper motors can be a favorable solution in some direct drives and servos applications. Discuss the possibility to use stepper motors in the open-loop systems without Hall-effect sensors to measure θr. Report the challenges one faces using the stepper motors in open-loop configuration and not using the rotor displacement sensor. Solution The SIMULINK diagram is given in the Figure.

44

MATLAB Function

*

0.32

sin( RT! rm )

Angular velocity

*

+ +

0.595

*

0.0034s+1 Torque TL

-1 Voltage uas

Current ias

Voltage ubs

* Voltage

+ -

Current ibs

0.595 0.0034s+1

+

0.32

+

1 0.000024s+0.000074

1 s Rotor displacement

Torque Te

*

2u M 0.32

*

MATLAB Function

cos( RT! rm )

Figure. SIMULINK diagram to simulate a stepper motor if the phase voltages are

uas = ! 2u M sin(RT"rm ) and ubs = 2u M cos(RT!rm ) The motor transient dynamics is illustrated in Figure. The analysis of the transient responses indicates that the settling time is 0.02 sec. Due to the excellent acceleration capabilities, permanentmagnet stepper motors are used in high-performance electromechanical systems. The steady-state mechanical angular velocity of permanent-magnet stepper motors is relatively low. Using ! r = RT! rm , the steady-state value of the electrical angular velocity is found to be 975 rad/sec. Stepper motors develop high electromagnetic torque because the number of rotor tooth is 50 or higher. Therefore, stepper motors are commonly used in direct drive applications.

45

Rotor angular velocity , ! rm

Stator voltage , uas [V ]

[ ] rad sec

25

5

20 15

0

10 -5

5

0

0.005

0.01 0.015 Time (second)

0.02

0.025

0

0

Stator voltage , ubs [V ]

0.005

0.01 0.015 Time (second)

0.02

0.025

Electromagnetic torque , Te [ N " m]

0.4 0.3

5

0.2 0

0.1

-5

-0.1

0

-0.2 0

0.005

0.01 0.015 Time (second)

0.02

0.025

0

0.01 0.015 Time (second)

0.02

0.025

Rotor angular displacement , # rm [ rad ]

Stator current , ias [ A]

0.6

0.005

0.5 0.4

0.4

0.2

0.3 0.2

0

0.1 -0.2 0

0.005

0.01 0.015 Time (second)

0.02

0.025

0.02

0.025

0

0

0.005

0.01 0.015 Time (second)

0.02

0.025

Stator current , ibs [ A] 1.2 1 0.8 0.6 0.4 0.2 0 -0.2 0

Figure.

0.005

Dynamics

0.01 0.015 Time (second)

of

a

permanent-magnet

stepper

ubs = 25.4 cos(RT!rm )

motor,

uas = ! 25.4 sin(RT"rm )

and

For the permanent-magnet stepper motor studied, the phase voltages can be applied as uas = ! u M sgn sin(RT"rm ) and ubs = u M sgn cos(RT!rm ) . The motor dynamics is documented in Figure.

(

)

(

)

The torque production is degraded compared with the results obtained as the sinusoidal voltages to the stator

46

windings uas = ! 2u M sin(RT"rm ) and ubs = 2u M cos(RT!rm ) were supplied. One can assess and analyze the waveforms for the currents ias (t ), ibs (t ) as well as the electromagnetic torque developed. Due to the chattering of the phase currents and the electromagnetic torque ripple, motor is overheating and ripple of the angular velocity result. However, from software and hardware standpoints, the pulse train algorithms can be easily implemented. Furthermore, as the phase voltages applied as the sequence of pulses (not measuring r), stepper motors can be applied in the open-loop configuration. However, steps can be missed or passed depending on the frequency of applied voltages, TL, etc. Rotor angular velocity , ! rm

Stator voltage , uas [V ] 6

[ ] rad sec

35

4

30

2

25

0

20 15

-2

10 -4

5

-6 0

0.005

0.01 0.015 Time (second)

0.02

0.025

0

0

0.005

Stator voltage , ubs [V ]

0.01 0.015 Time (second)

0.02

0.025

Electromagnetic torque , Te [ N " m]

6 0.4

4 2

0.2

0 0

-2 -4

-0.2

-6 0

0.005

0.01 0.015 Time (second)

0.02

0.025

0

0.005

0.01 0.015 Time (second)

0.02

0.025

Rotor angular displacement , # rm [ rad ]

Stator current , ias [ A] 0.5 0

0.4 0.3

-0.5 0.2 -1

0.1

0

0.005

0.01 0.015 Time (second)

0.02

0.025

0.02

0.025

0

0

0.005

0.01 0.015 Time (second)

0.02

0.025

Stator current , ibs [ A]

1 0.8 0.6 0.4 0.2 0 -0.2 -0.4 0

Figure.

0.005

Dynamics

(

0.01 0.015 Time (second)

of

)

ubs = 5.4 sgn cos(RT!rm )

a

permanent-magnet

stepper

motor,

(

)

uas = !5.4 sgn sin(RT"rm ) ,

47

Chapter 7 Problem 7. 1. Why one should control electromechanical systems? Solution In order to improve the closed-loop system performance achieving the best achievable performance, e.g., stability, high efficiency, robustness, desired accuracy, minimum settling time, etc. Problem 7. 2. List the specifications imposed on systems in the behavioral domain. Solution Stability and robustness to parameter variations and disturbances, desired transient dynamics waveforms (overshoot, first and second peaks, etc.), accuracy, minimum settling time, etc. Problem 7. 3. Explain the differences between bounded and unbounded control laws. Provide the examples. Explain how control bounds influence the system performance and capabilities. Solution In general, all controllers are bounded due to the rated (maximum) control efforts (variables) imposed by the hardware. For example, the magnitude of the applied voltage is bounded. If one changes the voltage using ICs, the duty ratio is bounded. The voltage of the power amplifier is bounded by the bus voltage. The control bounds significantly influence the system performance. With the bounded control, different transient dynamics, accuracy and stability result. The linear methods of the control theory cannot be applied. Problem 7. 4. What are the challenges in the design of bounded control laws. How the designer may approach and solve control problems for electromechanical systems? Solution Laplace transforms, eigenvalues concept, superposition principles and other methods of linear control theory cannot be applied. One may find analytic solution for a limited class of low-order systems as well as apply the high-performance computational environment to solve the problems. For example, MATLAB can be effectively used. Problem 7. 5.

!

2 3 Let the control law is expressed as u = k p1e + k p 2 e + k p 3e + ki edt . Elaborate the feedback terms used.

Make the conclusion if this control law can be applied. Explain why one should study the bounded control

(

!

)

2 3 law u = sat k p1e + k p 2 e + k p 3e + ki edt to examine the system performance. Propose the terms to

improve the systems performance. Also provide the terms that degrade the system performance and lead to unstable closed-loop systems. Solution The first term is the conventional proportional term. The second term is the absolute value of the squired tracking error. The third term is the cubed tracking error. The fourth term is the conventional integral term. This control law can be used. However, it is difficult to implement the controller due to the nonlinear terms (look-up tables, specialized ICs or DSPs should be used that complicates microcontroller and may lead to

48

high execution time), In general, the saturated controllers are due to real physics of controlling, energy conversion and energy transfer. Correspondingly, one may utilize the realistic features and analyze systems with the bounded controllers. Among the additional terms to be possibly utilize in the control law to improve the performance is k p 4 e 5 , while the term k p 5e 2 will significantly degrade the performance and must not be applied (but the closed-loop systems will be stable due to the term k p 3e 3 ). The term k 6 e 6 will lead to unstable closed-loop system. Problem 7. 6. Let the performance functional is given as J = min t , x ,e

"

! (x

2

)

+ e 4 + t e dt . Explain what performance

0

characteristics are specified and how. Justify the results. If strict specifications should be imposed on the settling time and tracking error, propose the additional integrand in the performance functional to be used. Solution The first integrand imposes the specifications on the squired state variable deflections. The second integrand imposes the specific requirements on the deflections of the tracking error by making use of e4. The third integrand imposes the specifications on the settling time and the tracking error. If one would like to impose 4 4 strict integrated specification on the time and error, one may apply the following integrand t e

or

t 10 e10 . Problem 7. 7. Let the system is modeled as

d" = !" + 100u . The control is bounded as –10≤u≤10 V. Propose the dt

bounded PID control law and simulate the closed-loop system in SIMULINK. Let the desired speed is 200 rad/sec. Find the feedback coefficients by varying the feedback gains. Report the simulation results for the closed-loop system. Solution The bounded control law, as a function of the tracking error is given by (7.5) . That is, de(t ) $ ! u(t ) = sat uumax # k p e(t ) + ki e(t )dt + k d & , umin≤u≤umax. The feedback gains vary, and for various kp, ki min " dt % and kd the simulations are performed. For the SIMULINK diagram, as reported in the Figure, the output dynamics is reported for kp=1, ki=0 and kd=0. It is found that the tracking error e(t) tends to zero. The effect of the derivative term is examined. The system can become very sensitive for large kd.

'

Figure. SIMULINK diagram and simulation results

49

Problem 7. 8. Consider a pointing system actuated by a geared permanent-magnet DC motor. The angular displacement of the pointing stage is y(t)=kgearθr. The motor data is: umax=24 V (−24≤ua≤24 V), iamax=10 A, ωrmax=240 rad/sec, ra=1 ohm, La=0.005 H, ka =0.1 V-sec/rad (N-m/A), J=0.0005 kg-m2 and Bm =0.0005 N-m-sec/rad. The reduction gear ratio is 10:1. Design and analyze • not bounded linear and nonlinear PID control laws; • bounded PID control laws. For different control laws and feedback coefficients, study the transient dynamics for r(t)=0.1 rad and r(t)=1 rad. Simulations must be performed by developing the MATLAB (SIMULINK) files. Solution There are numerous possible solutions for this problem, as reported in various book’s examples.

50

Chapter 8 Problem 8. 1.

dx1 dx2 = ! x1 + x2 , = x1 + u , find a control law that minimizes the dt dt

For the second-order system quadratic functional J =

" 1 2

! (x

2 1

)

+ 2 x22 + 3u 2 dt applying MATLAB. Study the closed-loop system

0

stability. Solution Using the state-space notations, we have the system evolution matrix differential equation x! (t ) = Ax + Bu as

& x!1 # &' 1 1# & x1 # &0# &' 1 1# &0 # $ x! ! = $ 1 0! $ x ! + $1!u , A = $ 1 0!, B = $1! . " % 2" % " % " % " % 2" % Using the quadratic return function

V (x )= 12 k11 x12 + k12 x1 x2 + 12 k 22 x22 =

1 2

[x1

&k x2 ]$ 11 %k 21

k12 # & x1 # , k12 = k 21 , k 22 !" $% x2 !"

the control law is

u = !G !1 B T Kx = !

1 [0 3

'k 1]% 11 &k 21

k12 $ ' x1 $ 1 = ! (k 21 x1 + k 22 x2 ). " % " k 22 # & x2 # 3

&k In MATLAB, the unknown matrix K = $ 11 %k 21

k12 # is found solving the Riccati equation k 22 !"

&0 0 # ' Q ' AT K ' K T A + K T BG '1 B T K = $ ! %0 0 "

The stability is examid using the eigenvalues. Utilizing the MATLAB lqr function, one has the following script >> A=[-1 1;1 0],B=[0;1],[K_feedback,K,Eigenvalues] = lqr(A,B,[1 0;0 2],[3]) A = 0 0

1 0

B = 0 1 K_feedback = 0.3162 0.8558 K = 2.7064 3.1623 3.1623 8.5584 Eigenvalues = -0.4279 + 0.3648i -0.4279 - 0.3648i

Hence, we found the control law as given as u = !0.32 x1 ! 0.86 x2 . The eigenvalues have negative real parts, and the closed-loop system is stable.

Problem 8. 2.

51

Using the Lyapunov stability theory, study stability of the system that is described by the differential equations x!1 (t ) = ! x1 + 10 x2 ,

x!2 (t ) = !10 x1 ! x27 , t ! 0 . Solution The positive-definite scalar Lyapunov candidate is chosen in the following form V ( x1 , x2 ) = 12 x12 + x22 .

(

)

Thus, V ( x1 , x2 ) > 0 . The total derivative is

dV ( x1 , x2 ) = x1 x!1 + x2 x!2 = x1 (! x1 + 10 x2 ) + x2 (!10 x1 ! x27 ) = ! x12 ! x28 . dt

dV ( x1 , x2 ) < 0 . We conclude that the system is stable. dt Hence, the equilibrium state of the system is uniformly asymptotically stable, and the quadratic function V ( x1 , x2 ) = 12 x12 + x22 is the Lyapunov function. Therefore,

(

)

Problem 8. 3. Let the electromechanical system is described by the differential equations x!1 (t ) = ! x1 + 10 x2 ,

x!2 (t ) = x1 + u . Derive (synthesize) the control law that will stabilize this system. Using the Lyapunov stability theory, prove the stability of the closed-loop system. Solution Let the control law is given as u = !11x1 ! x2 . Thus, the closed-loop system dynamics is described as

x!1 (t ) = ! x1 + 10 x2 , x!2 (t ) = x1 + u = !10 x1 ! x2 . The positive-definite scalar Lyapunov candidate is chosen as V ( x1 , x2 ) =

1 2

(x

2 1

)

+ x22 , and

V ( x1 , x2 ) > 0 . The total derivative of the Lyapunov function is

dV ( x1 , x2 ) = x1 x!1 + x2 x!2 = x1 (! x1 + 10 x2 ) + x2 (!10 x1 ! x2 ) = ! x12 ! x22 . dt Therefore, V ( x1 , x2 ) > 0 and

dV ( x1 , x2 ) < 0 . We conclude that the closed-loop system is stable. dt

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