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##### ECE 321 - Electromechanical Motion Devices - Purdue Study Resources
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###### HW1 Solution

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Homework 1, Problem 1 Let the elements of the vector correspond to the x, y, and z components 1 H := 0 0 Now 1 Point1 := 1 1 2 Point2 := 5 1 Since the H-field is constant Point2 H dl = H ( Point2 Point1) Point1 Thus the MMF drop is given by T H (

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###### Hw9

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EE321 Spring 2008 / Homework 9 Problem 37 Discrete winding function The number of conductors in each slot of the a-phase of the stator of the machine are as follows: N as = [10 20 20 10 10 20 20 10 10 20 20 10 10 20 20 10]T Compute and graph the winding f

• 2 Pages
###### Hw3[1]

School: Purdue

Course: Electromechanical Motion Devices

Fall 2010 ECE 321 Homework Set 3 Due Wed. Sept. 29 Work on separate sheets of paper. Must be turned in at beginning of class. First page blank with only your name and should be stapled. Homework will be collected promptly at 2:30. If not submitted in time

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###### Hw1

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EE321 Spring 10 Homework 1 Problem 1 Review of line-integral and application to MMF drop. Consider a Cartesian co-ordinate system (x,y,z). Suppose a uniform H-field of 1 A/m exists in the direction of the x-axis. Calculate the MMF drop from the point (1,1

• 6 Pages
###### Hw 4 Solutions

School: Purdue

Course: Electromechanical Motion Devices

EE321, Spring 2013 Homework 4 Problem 1 1 2 Wc = 5 + 2 sin 4 rm i 2 ( ( ( ) Te = 4 cos 4 rm i ) 2 Problem 2 We may express the system as 1 2 7 5 2 5 + x i1 = 7 i 2 2 2 5+x which is of the form i 1 = L 1 2 i2 where L is independent of both c

• 2 Pages
###### Hw3

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EE321 Spring 2010 Homework 3 Problem 1 UI Inductor Analysis Consider the UI inductor design we did in class. Recall we had N = 260 Turns d = 8.4857 cm g = 13.069 mm w = 1.813 cm aw = 21.5181 mm2 ds = 8.94 cm ws = 8.94 cm In our design, we assumed that the

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###### HW5 Solution

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EE321 HW #5 Problem 1 Problem 2 Problem 3

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###### HW6

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ECE321/ECE595 Spring 2012 Homework 6 Problem 1 Permanent Magnet DC Machine A permanent magnet dc machine has ra = 8 and kv = 0.01 Vs/rad. The shaft load torque is approximated as TL = Kr, where K = 510-6 Nms. The applied voltage is 6 V and Bm = 0. Calcula

• 4 Pages
###### Abet

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EE321. ABET Exam Spring 2004 Name: Student ID: Instructions: Work ALL Problems. When you have completed exam, turn in to Professor Sudhoff, Brandon Cassimere, or Brant Cassimere, any of whom will check it on the spot, and let you know which ones are wrong

• 27 Pages
###### Lecture Set 0

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Lecture Set 0 ECE321/ECE595 S.D. Sudhoff Electromechanical Motion Devices Spring 2012 Courses Meeting Together Courses ECE321 Live (57) cfw_321L ECE321 Video (9) cfw_321V ECE595 On Campus (3) cfw_595C ECE595 Off Campus Pro Ed (11) cfw_595P Differences 321

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###### Hw 6

School: Purdue

Course: Electromechanical Motion Devices

ECE321/ECE595 Spring 2013 Homework 6 Problem 1 Permanent Magnet DC Machine A permanent magnet dc machine has ra = 6 and kv = 0.01 Vs/rad. The shaft load torque is approximated as TL = Kr, where K = 510-6 Nms. The applied voltage is 6 V and Bm = 0. Calcula

• 7 Pages
• ###### 1. Spring2012-course-intro-4pages
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###### 1. Spring2012-course-intro-4pages

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Course: Electromechanical Motion Devices

Instructor and TA Prerequisite: ECE 301 or equivalent. Instructor: Professor C. S. George Lee Ofce: MSEE 256 Phone: (765) 494-1384 Email: csglee@purdue.edu Ofce Hours: MWF: 10:30 -11:30 AM (or by appointment) ECE382: Feedback System Analysis and Design C.

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###### Practice Exam3_solution

School: Purdue

Course: Electromechanical Motion Devices

ECE321 Spring 2008 Exam 3 Solution Outline Problem 1 T Nas := ( 3 3 6 3 3 6 3 3 6 3 3 6 ) Nslts := 12 P := 4 Nslts P =3 1 Was := [ 3 + ( 3 ) + ( 6 ) ] 1 2 j := 2 . 12 k := 1 . 11 Was := Was Nas j j 1 j 1 T 1 Was = 2 -3 1 -6 3 -3 Problem 2 () was = 100 co

• 4 Pages
###### Practice Exam2

School: Purdue

Course: Electromechanical Motion Devices

EE321 Exam 2 Spring 2008 Write your name on the bluebook, and on the last sheet of this exam (which has a figure you will need). Turn in the bluebook and the last page of the exam. Assume that they will be separated so put your name on both. Notes: 1.) Yo

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###### Practice Exam2_solution

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Course: Electromechanical Motion Devices

EE321 Exam 2 Problem 1 3 O 4 O2 Problem 2 A. No B. Yes C. Yes D. Yes E. No Problem 3 3 ra := 10 10 Laf := 150 10 3 ifd_mx := 10 ia_mx := 200 v a_mx := 400 Kl := 0.01 Assume we are against the armature voltage and armature current limits: if = v a_mx ra ia

• 2 Pages
###### Practice Exam1

School: Purdue

Course: Electromechanical Motion Devices

EE321 Exam 1 Spring 2008 Write your name and student ID on the bluebook. Notes: You must show work for credit. Problems 2-3 (together) can be used to satisfy ABET Objective 2. Problems 4-6 (together) can be used to satisfy ABET Objective 1. Bid me run, an

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###### Hw6[1]

School: Purdue

Course: Electromechanical Motion Devices

ECE 321 Homework Set 6 Due Monday. Nov. 8 Must be turned in at beginning of class. Staple this page to front of your solutions. Homework will be collected at beginning of class. If not submitted in time, it will not be graded. Name: Student ID: 1. Conside

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###### Practice Exam1_solution

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Course: Electromechanical Motion Devices

ECE321. Fall 2008 Exam 1 Solution Outline Handy Stuff 2 cm := 1 10 mm := 1.0 10 3 7 0 := 4 10 Problem 1 Hy := 20 A := 100 10 By := Hy 0 := By A Only the y-component couples the loop := Single turn = 0.02513 Problem 2 w := 1 cm d s := 2 cm g := 0.1 mm

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###### Hw 9 Solutions

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Course: Electromechanical Motion Devices

ECE321/ECE595 Spring 2013 HW#9 Problem 1 ( ) ( ) was = 100 cos 4 s wbs = 250 sin 4 s ias = 10 cos e t + B = 1.2 cos e t + 8 8 4 s 7 0 := 4 10 Expanding B we have B = 1.2 cos e t + Also B= F= F g 0 g 0 B cos( 4 s) + 1.2 sin e t + 8 sin( 4 s) 8 Finally F

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###### HW6_Solution[1]

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Course: Electromechanical Motion Devices

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###### Hw 8

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Course: Electromechanical Motion Devices

ECE321/ECE595 Homework 8 Problem 1 Hysteresis Current Control Consider a machine with an armature resistance of 1 , a voltage constant of 0.05 Vs, and an armature inductance of 2 mH. Suppose it is fed from a dc source of 20 V, using a chopper circuit with

• 11 Pages
###### Hw 8 Solutions

School: Purdue

Course: Electromechanical Motion Devices

EE321/595 HW#8 Problem 1 ra := 1 kv := 0.05 3 Laa := 2 10 ia := Tedes kv vdc := 20 vfsw := 1 Tedes := 0.1 vfd := 0.8 fmxdes := 30 10 3 ia = 2 Manipulating the expression in the notes we have ( ) fsw r , h := (vfd + ra ia + kv r) (vdc vfsw ra ia kv r) 2 h

• 1 Page
###### Hw 9

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Course: Electromechanical Motion Devices

EE321/ECE595 Spring 2013 Homework 9 Problem 1 Rotating MMF The winding function of the a- and b-phase stator windings of a machine are given by was = 100 cos(4 s ) and wbs = 250 sin(4 s ) . The a-phase current of the machine is given by i as = 10 cos( e t

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###### Hw 5 Solutions

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Course: Electromechanical Motion Devices

EE321 HW #5 Problem 1 Problem 2 Problem 3

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###### Hw 4

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Course: Electromechanical Motion Devices

EE321 Spring 2013 Homework 4 Problem 1 Calculation of Torque The flux-linkage of a certain rotational electromechanical device may be expressed = (5 + 2sin 4 rm )i where rm is the rotor position and i is the current. What is the electromagnetic torque ?

• 2 Pages
###### Hw 3

School: Purdue

Course: Electromechanical Motion Devices

ECE321/ECE595 Spring 2013 Homework 3 Problem 1 UI Inductor Analysis Consider the UI inductor design we did in class. Recall we had N = 260 Turns d = 8.4857 cm g = 13.069 mm w = 1.813 cm aw = 21.5181 mm2 ds = 8.94 cm ws = 8.94 cm In our design, we assumed

• 6 Pages
###### Hw 3 Solutions

School: Purdue

Course: Electromechanical Motion Devices

EE321, Homework 3 Problem 1 From minimum cost solution in class N := 260 2 d := 8.4857 10 2 d s := 8.94 10 ws := 8.94 10 3 i := 40 3 g := 13.069 10 Ldes := 5 10 2 2 w := 1.813 10 7 0 := 4 10 r := 2000 Recomputing the reluctance R := 2 ( ws + 2w) + 2d s 2

• 2 Pages
###### Hw 2

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Course: Electromechanical Motion Devices

EE321 Spring 2013 Homework 2 Problems 1 UI Inductor Analysis Consider the UI core below. Consider the following parameters: w = 1 cm; ws = 5 cm; d s = 2 cm; d = 5 cm; g = 1.5 mm; N = 100 . Suppose the material used is such that for a flux density less tha

• 11 Pages
###### Hw 2 Solutions

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Course: Electromechanical Motion Devices

Problem 1 - Simple UI Core Analyiss Dimensions, etc 2 3 cm := 1 10 mm := 1.0 10 w := 1 cm d s := 2 cm g := 1.5 mm ws := 5 cm d := 5 cm N := 100 B sat := 1.5 Point where saturation occurs 7 u 0 := 4 10 Now let's compute some reluctances. For flux densities

• 1 Page
###### Hw 1

School: Purdue

Course: Electromechanical Motion Devices

EE321 Spring 2012 Homework 1 Problem 1 Review of line-integral and application to MMF drop. Consider a Cartesian co-ordinate system (x,y,z). Suppose a uniform H-field of 1 A/m exists in the direction of the x-axis. Calculate the MMF drop from the point (1

• 6 Pages
###### Hw 1 Solutions

School: Purdue

Course: Electromechanical Motion Devices

Homework 1, Problem 1 Let the elements of the vector correspond to the x, y, and z components 1 H := 0 0 Now 1 Point1 := 1 1 5 Point2 := 2 1 Since the H-field is constant Point2 H dl = H ( Point2 Point1) Point1 Thus the MMF drop is given by T H (

• 13 Pages
###### Exam 3 Solutions

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Course: Electromechanical Motion Devices

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###### Exam 2 Solutions

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Course: Electromechanical Motion Devices

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###### Exam 1 Solutions

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Course: Electromechanical Motion Devices

• 2 Pages
###### Practice Exam5

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Course: Electromechanical Motion Devices

EE321 Exam 5 Spring 2008 Write your name and student ID on the bluebook. Only turn in the bluebook. Notes: You must show work for credit. Getting 70% on problems 1 or 2 satisfies objective 1 and 2. Getting 70% on problems 4 or 5 satisfies objective 4. Get

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###### Practice Exam5_solution

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Course: Electromechanical Motion Devices

EE321 Exam 5 Solution Outline Problem 1 cm := 0.01 mm := 0.001 w := 1 cm d := 5 cm ws := 5 cm d s := 2 cm g := 1 mm N := 100 r := 1000 7 0 := 4 10 Fe := 25 A := w d w ds + 2 R 45 := A 0r R 56 := ws + w 0 r A R 45 = 39.789 10 3 R 56 = 95.493 10 3 w R 81

• 2 Pages
###### Practice Exam4

School: Purdue

Course: Electromechanical Motion Devices

EE321 Exam 4 Spring 2008 Write your name and student ID on the bluebook. Only turn in the bluebook. Notes: You must show work for credit. Getting 70% of problems 2-5 satisfies objective 4. Getting 70% of problem 3 or 4 satisfies objective 2. Getting 70% o

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###### Practice Exam4_solution

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Course: Electromechanical Motion Devices

EE321 Exam 4 Solution Outline Problem 1 By inspection, the secondary to primary turns ratio is 10. Nsp := 10 Lm := 500 Referred to secondary rs := 2 From secondary rp := 1 Nsp 2 0 rp = 100 10 From secondary The leakage inductances are zero. Problem 2 Lls

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###### Practice Exam3

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Course: Electromechanical Motion Devices

EE321 Exam 3 Spring 2008 Write your name and student ID on the bluebook. Only turn in the bluebook. Notes: You must show work for credit. Getting 75% of problems 1-3 satisfies ABET objective 2. Getting 75% of problem 5 satisfies ABET objective 3. Getting

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###### 2. Differential-fall2011

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Course: Electromechanical Motion Devices

ECE 382 Review of Solutions of Linear Ordinary Differential Equations with Constant Coefficients We shall consider an nth -order, linear, ordinary differential equation with constant coefficients, and discuss some physical problems giving rise to such equ

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Course: Electromechanical Motion Devices

Linear Ordinary Differential Equations Linear Ordinary Differential Equations Why study linear ordinary differential equations? Consider an nth -order ordinary differential equation of the form: Use ODE to model or describe the behavior of a physical syst

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###### 4. Block-diagram-signal-flow-graph-4pages

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Course: Electromechanical Motion Devices

Transfer Functions Block Diagrams Transfer function is dened as: L cfw_output variable Transfer function = L cfw_input variable initial conditions are zero For example, nd the transfer function Eo (s) of an RC circuit Ei (s) A block diagram of a system i

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###### Hw4

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EE321 Spring 2010 Homework 4 Problem 1 Calculation of Torque The flux-linkage of a certain rotational electromechanical device may be expressed = (5 + 2 sin 4 rm )i where rm is the rotor position and i is the current. What is the electromagnetic torque ?

• 1 Page
###### Hw5

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EE321 Spring 2010 Homework 5 Problem 1 Torque Versus Position Trajectory Consider the torque versus position characteristics of a VR stepper shown below. Initialize the c-phase is energized and the position is as indicated (point 1). Then the position is

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###### Hw6

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EE321 Spring 2010 / Homework 6 Problem 1 Problem 3.10-3 from Electromechanical Motion Devices Problem 2 Problem 3.10-6 from Electromechanical Motion Devices Problem 3 PM DC Machine Performance A PM DC machine has a back emf constant of 0.1 Vs, and an arma

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###### Hw7

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EE321 Spring 2010 Homework 7 Problem 1 Buck converter operation Consider the example on page 55 of the lecture notes. Suppose the dc voltage is changed to 125 V and the speed to 400 rad/s. Find the average armature current, the average switch current, the

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###### Hw8

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EE321 Spring 2010 Homework 8 Problem 1 Hysteresis Current Control Consider a machine with an armature resistance of 1 , a voltage constant of 0.05 Vs, and an armature inductance of 3 mH. Suppose it is fed from a dc source of 20 V, using a chopper circuit

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###### Hw9

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EE321 Spring 2010 Homework 9 Problem 1 Winding Functions Find the winding function for nas ( sm ) = N s sin( Psm / 2) + N s 3 sin(3Psm / 2) Problem 2 Rotating MMF The winding function of the a- and b-phase stator windings of a machine are given by was = 1

• 2 Pages
###### Hw10

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EE321 Spring 2010 Homework 10 Problem 1 Electrical and mechanical rotor speed The electrical frequency applied to an synchronous machine (an AC machine in which the rotor travels at the same speed of the MMF) is 60 Hz. The mechanical rotor speed is 900 RP

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###### Hw11

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EE321 Spring 2010 Homework 11 Problem 1 Brushless DC Operation from a Voltage Source The flux linkage equations for a certain PMSM may be expressed abcs L ss = 0 0 0 L ss 0 0 cos r cos(3 r ) i cos( 2 / 3) cos(3 ) 0 abcs + m r m3 r cos( r + 2 / 3) cos(

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###### Hw12

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EE321 Spring 2010 Homework 12 For this homework, consider a transformer. The primary side resistance and leakage inductance are 2 and 1 mH, respectively. The magnetizing inductance is 100 mH. The (referred) secondary resistance and leakage inductance are

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###### Hw13

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EE321 Spring 2010 Homework 13 Problem 1 Rotating MMF Using the configuration we studied in class (Figure 5.2-1 in text), the stator currents of a 4 pole machine are given by ias = 50 sin( 200t ) ibs = 50 cos(200t ) The speed of the machine is 500 rpm in t

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###### Hw14

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EE321 Spring 2010 Homework 14 Problem 1 Steady-State Operation Consider a 2-phase machine with the following parameters: rs = 72.5 m , Lls = L'lr = 1.32 mH, Lm = 20.1 mH, rr' = 41.3 m , and P = 4 . The load torque varies with the speed cubed, and is such

• 8 Pages
###### Exam5

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EE321 Exam 5 Spring 2011 Notes: You must show work for credit on Problems 1-3. Good luck! Handy Facts 0 = 4 107 H/m Taken from, Continuous and Discrete Signal and Systems Analysis, 2nd Edition, by McGillem & Cooper, 1984, CBS College Publishing, and one h

• 8 Pages
###### Exam4

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EE321 Exam 4 Spring 2011 Notes: You must show work for credit on Problems 1-3. Good luck! Handy Facts 0 = 4 107 H/m Taken from, Continuous and Discrete Signal and Systems Analysis, 2nd Edition, by McGillem & Cooper, 1984, CBS College Publishing, and one h

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###### Exam1_solution

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###### Hw2

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EE321 Spring 2010 Homework 2 Problems 1 UI Inductor Analysis Consider the UI core shown in Figure 1.4-1 (or Lecture Set 1, slide 34). Consider the following parameters: w = 1 cm; ws = 5 cm; d s = 2 cm; d = 5 cm; g = 1 mm; N = 100 . Suppose the material us

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Course: Electromechanical Motion Devices

Analogous Systems Mechanical Elements - Inertial Element Inertial elements - masses and moments of inertia. The change in Force (Torque) required to make a unit change in acceleration (angular acceleration). Units: N /m/s2 or kg for mass; Force-velocity,

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###### Pastfinal-2010

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Course: Electromechanical Motion Devices

Fall, 2010 Final Exam ECE321 Last Name:_ First Name:_ User ID (login):_ Work problems and provide answers in space provided - do not unstaple pages Four 1-page crib sheets allowed (to be submitted with exam). No calculators, you may express answers in ter

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Course: Electromechanical Motion Devices

Modeling DC Motors Motor Example Puma Robot - Joint one Modeling Write differential equations to describe the dynamic behavior of a physical system. The differential equations are then used to analyze the expected performance of the physical system. Two c

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###### 6. Analogy-rev0

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Course: Electromechanical Motion Devices

ECE 382 NOTES ON ANALOGOUS SYSTEMS Systems that are governed by the same type of differential equations are called analogous systems. If the response of one physical system to a given excitation is found, then the response of all other systems that are de

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###### 8. Sensitivity-poles-2nd-order-systems-slides1

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Course: Electromechanical Motion Devices

Feedback Control System Characteristics Sensitivity Analysis For a system to perform well, it must be less sensitive to parameter variation. We would like to analyze how the variation of a parameter will affect the performance of the overall system. Why s

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###### 9. Routh-Root-Locus-4pages

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Course: Electromechanical Motion Devices

Stability Stability Depends on Closed-Loop Poles A system is stable if a bounded input always produces a bounded output (BIBO). Bounded means bounded in magnitude. The system response to a bounded input will result in either a decreasing, neutral, or incr

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• ###### 10. Rules for Plotting Root Locus and Bode Plot
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###### 10. Rules For Plotting Root Locus And Bode Plot

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Course: Electromechanical Motion Devices

ECE 382 ROOT LOCUS CONSTRUCTION RULES FOR K > 0 Rule 1: The root locus has n branches, where n is the number of open-loop poles (i.e., poles of G(s)H (s) Rule 2: The root locus (or the branches) starts at the open-loop poles (K = 0) and ends at the open-l

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###### 11. Bode-Diagram-4pages

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Course: Electromechanical Motion Devices

Frequency Response Method Frequency Response Method of a system is Frequency response Frequency Response Example defined as the steady-state response of Consider a mass-dashpot-spring example with f (t ) as input and x (t ) as output. Frequency response

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###### 12. Bode-Plot-Examples

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Course: Electromechanical Motion Devices

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• ###### 15. Introduction to Compensator Design
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###### 15. Introduction To Compensator Design

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Course: Electromechanical Motion Devices

School of Electrical and Computer Engineering Introduction to Compensators Process of Compensation of Electrical and Computer Engineering School Compensators R(s) + R(s) + C(s) - C(s) - G(s) T (s) = G(s) G (s) 1 + G (s) Uncompensated system If T(s) does n

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• ###### 16. Bisector Method for RL Lead Design
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###### 16. Bisector Method For RL Lead Design

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Course: Electromechanical Motion Devices

ECE 382 Lead Compensator Design (Root-Locus) G(s) = 4 s(s + 2) Design Objective: = 0.5 and n = 4 rad/sec. (Interpret the design objective in terms of performance specication.) Procedure: 1) General form of a lead compensator Gc (s) = Kc (s + 1 ) , 1 (s +

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Course: Electromechanical Motion Devices

ECE 382 Lead Compensator Design (Bisector Method) j A s1 n 2 2 0 C B =cos-1 To determine the location of B (zero) and C (pole) analytically 180 = = 90 2 2 22 = 180 = 90 + . 22 OA OB Using the sine law: sin = sin (Note that OA n ) (1) (2) n sin n sin(9

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Course: Electromechanical Motion Devices

ECE 382 Lead Compensator Design (Frequency domain) G(s) = K s(s + 2) Design Objective: Kv = 20 sec1 ; d 50 PM ; Gain margin, Gm 10 dB. Procedure: 1) General form of a lead compensator Gc (s) = Kc (s + 1 ) , 1 (s + ) Kc , > 0, 0<1 Determine the open-loop g

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###### 19. Design-lag-Bode-fall2011

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Course: Electromechanical Motion Devices

ECE 382 Lag Compensator Design (Frequency domain) G(s) = K s(s + 1)(s + 2) Design Objective: Kv = 5 sec1 ; Gd 40 PM Gain margin, Gm 10 dB. ; Procedure: 1) General form of a lag compensator Gc (s) = Kc (s + 1 ) 1 (s + ) , Kc , > 0, >1 Determine the open-lo

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• ###### pastfinal-2009 solutions
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###### Pastfinal-2009 Solutions

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Course: Electromechanical Motion Devices

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###### Pastfinal-2009

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Course: Electromechanical Motion Devices

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###### Exam 3 Solution

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###### Hw 5

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Course: Electromechanical Motion Devices

EE321 Spring 2013 Homework 5 Problem 1 Torque Versus Position Trajectory Consider the torque versus position characteristics of a VR stepper shown below. Initialize the c-phase is energized and the position is as indicated (point 1). Then the position is

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###### Exam2

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1. Consider the following variable-reluctance stepper motor. rm rm cs bs as as rm bs cs cs bs as as cs bs Maximum self-inductance = 6 H, minimum = 2 H. (a) SL = o (b) Lbsbs = + cos [ o (rm )] (c) If ibs = 2 A, and ias = ics = 0, Te = sin [ (rm o )] (d)

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###### Ch2

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Electromechanical Motion Devices, Second Edition by Paul Krause, Oleg Wasynczuk and Steven Pekarek Copyright 2012 Institute of Electrical and Electronics Engineers, Inc. C hapter 2 ELECTROMECHANICAL ENERGY CONVERSION 2.1 INTRODUCTION T he theory of electr

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###### Ch3

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Electromechanical Motion Devices, Second Edition by Paul Krause, Oleg Wasynczuk and Steven Pekarek Copyright 2012 Institute of Electrical and Electronics Engineers, Inc. C hapter 3 DIRECT-CURRENT MACHINES 3.1 INTRODUCTION T he direct-current (dc) machine

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###### Ch4

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Electromechanical Motion Devices, Second Edition by Paul Krause, Oleg Wasynczuk and Steven Pekarek Copyright 2012 Institute of Electrical and Electronics Engineers, Inc. C hapter 4 WINDINGS AND ROTATING MAGNETOMOTIVE FORCE 4.1 INTRODUCTION In the previous

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###### Ch5

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Electromechanical Motion Devices, Second Edition by Paul Krause, Oleg Wasynczuk and Steven Pekarek Copyright 2012 Institute of Electrical and Electronics Engineers, Inc. C hapter 5 INTRODUCTION TO REFERENCE-FRAME THEORY 5.1 INTRODUCTION In recent years, t

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###### Ch6

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Electromechanical Motion Devices, Second Edition by Paul Krause, Oleg Wasynczuk and Steven Pekarek Copyright 2012 Institute of Electrical and Electronics Engineers, Inc. C hapter 6 SYMMETRICAL INDUCTION MACHINES 6.1 INTRODUCTION Although the induction mac

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###### Ch7

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Electromechanical Motion Devices, Second Edition by Paul Krause, Oleg Wasynczuk and Steven Pekarek Copyright 2012 Institute of Electrical and Electronics Engineers, Inc. C hapter 7 SYNCHRONOUS MACHINES 7.1 INTRODUCTION Nearly all electric power is generat

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###### Ch8

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Electromechanical Motion Devices, Second Edition by Paul Krause, Oleg Wasynczuk and Steven Pekarek Copyright 2012 Institute of Electrical and Electronics Engineers, Inc. C hapter 8 PERMANENT-MAGNET ac MACHINE 8.1 INTRODUCTION T he permanent-magnet ac mach

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###### App1

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Electromechanical Motion Devices, Second Edition by Paul Krause, Oleg Wasynczuk and Steven Pekarek Copyright 2012 Institute of Electrical and Electronics Engineers, Inc. A ppendix A A BBREVIATIONS, C ONSTANTS, CONVERSIONS, AND I DENTITIES 477 480 APPENDIX

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###### App2

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Electromechanical Motion Devices, Second Edition by Paul Krause, Oleg Wasynczuk and Steven Pekarek Copyright 2012 Institute of Electrical and Electronics Engineers, Inc. A ppendix B MATRIX ALGEBRA B asic Definitions A rectangular array of numbers or funct

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###### App3

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Electromechanical Motion Devices, Second Edition by Paul Krause, Oleg Wasynczuk and Steven Pekarek Copyright 2012 Institute of Electrical and Electronics Engineers, Inc. A ppendix C THREE-PHASE SYSTEMS In a three-phase system, there are two types of conne

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###### ABET Exam

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EE321 ABET Exam Spring 2009 You may establish credit for ABET objectives (by answering the following questions). 1.) Objective 1. Ability to Analyze / Design Electromagnetic Devices. A solenoid is an electromechanical device used for actuation. In this si

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###### Exam1

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EE321 Exam 1 Spring 2009 Notes: Write your name and ID on blue book. This part of exam will be recycled. You must show work for credit, except for problem 6. Achieving a score of 60% or above on this exam satisfies ABET Objective 1 and Objective 2. Good l

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###### Hw10

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EE321 Spring 2008 / Homework 10 Problem 42 Unbalanced MMF The conductor turns density of a two phase machine is given by n as = 100 cos 2 s nbs = 100 sin 2 s The a- and b-phase currents are given by i as = 5 cos(400t ) ibs = 4 sin( 400t ) Express the tota

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###### Hw10_solution

School: Purdue

EE321 Spring 2008 HW#10 Problem 42 The winding functions are () was = 50 sin 2 s () wbs = 50 cos 2 s but now the currents are ias = 5 cos( 400t) ibs = 4 sin( 400t) So the total MMF is given by () () F = 250 cos( 400 t) sin 2 s 200 sin( 400 t) cos 2 s ( )

• 1 Page
###### Hw11

School: Purdue

EE321 Spring 2008 Homework 11 Problem 47 QD Transformation Starting with P = vas ias + vbs ibs + vcs ics = v T i abcs abcs Show that P= ( 3 rr rr vqs iqs + vds ids + 2v0 s i0 s 2 ) Problem 48 Brushless DC Operation from a Voltage Source A three phase brus

• 6 Pages
###### Hw11_solution

School: Purdue

EE321 Spring 2008 Homework #11 Problem 47 T T 1 1 r r r v K r i P = v abcs iabcs = qd0s T Ks s qd0s T 3 r c cn cp c s 1 iqs n n 2 v r v r v n p c s 1 r = v r v r v P = qs ds 0s s s s i qs ds 0s 0 p p ds 1 1 1 c s 1 i0s 0 3 i r 2 qs 3 r r

• 48 Pages
###### Ch1

School: Purdue

Electromechanical Motion Devices, Second Edition by Paul Krause, Oleg Wasynczuk and Steven Pekarek Copyright 2012 Institute of Electrical and Electronics Engineers, Inc. C hapter 1 MAGNETIC AND MAGNETICALLY COUPLED CIRCUITS 1.1 INTRODUCTION Before diving

• 5 Pages
###### ABETexam

School: Purdue

• 5 Pages
###### HW6(1)

School: Purdue

ECE 321 Homework Set 6 Due Wed. Oct. 9 Work each problem on attached sheets. First page blank with only your name and should be stapled. Homework will be collected promptly at 2:30 pm. If not submitted in time, it will not be graded. Not all problems will

• 3 Pages
###### Hw4

School: Purdue

ECE 202: Linear Circuit Analysis II Fall2013 HOMEWORK SET 4: DUE TUESDAY, SEPTEMBER 10, 5 PM IN MSEE 180 ALWAYS CHECK THE ERRATA on the web. Main Topics: Equivalent circuits for L and C with initial conditions; transfer functions; H(s). Suggestion: Do wha

• 2 Pages
###### HW5_rachel_pereira

School: Purdue

• 3 Pages
###### HW6_rachel_pereira

School: Purdue

• 3 Pages
###### Hw6_rachel_pereira

School: Purdue

• 4 Pages
###### HW7_rachel_pereira

School: Purdue

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