• 6 Pages HW1 Solution
    HW1 Solution

    School: Purdue

    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 (

  • 1 Page hw9
    Hw9

    School: Purdue

    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]
    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

  • 1 Page hw1
    Hw1

    School: Purdue

    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

  • 2 Pages hw3
    Hw3

    School: Purdue

    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

  • 6 Pages hw 4 solutions
    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

  • 3 Pages HW5 Solution
    HW5 Solution

    School: Purdue

    EE321 HW #5 Problem 1 Problem 2 Problem 3

  • 2 Pages HW6
    HW6

    School: Purdue

    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
    Abet

    School: Purdue

    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
    Lecture Set 0

    School: Purdue

    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

  • 4 Pages practice exam1_solution
    Practice Exam1_solution

    School: Purdue

    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

  • 2 Pages practice exam1
    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

  • 4 Pages practice exam2
    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

  • 4 Pages practice exam2_solution
    Practice Exam2_solution

    School: Purdue

    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

  • 1 Page hw 9
    Hw 9

    School: Purdue

    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

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

    School: Purdue

    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.

  • 2 Pages hw6[1]
    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

  • 2 Pages HW6_Solution[1]
    HW6_Solution[1]

    School: Purdue

    Course: Electromechanical Motion Devices

  • 4 Pages 18. Design-Lead-Bode-Spring2012
    18. Design-Lead-Bode-Spring2012

    School: Purdue

    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

  • 1 Page hw 6
    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

  • 11 Pages hw 8 solutions
    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 8
    Hw 8

    School: Purdue

    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

  • 8 Pages hw 9 solutions
    Hw 9 Solutions

    School: Purdue

    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

  • 3 Pages 2. differential-fall2011
    2. Differential-fall2011

    School: Purdue

    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

  • 11 Pages hw 2 solutions
    Hw 2 Solutions

    School: Purdue

    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

  • 2 Pages hw 4
    Hw 4

    School: Purdue

    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
    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
    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
    Hw 2

    School: Purdue

    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

  • 1 Page hw 1
    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
    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
    Exam 3 Solutions

    School: Purdue

    Course: Electromechanical Motion Devices

  • 12 Pages exam 2 solutions
    Exam 2 Solutions

    School: Purdue

    Course: Electromechanical Motion Devices

  • 8 Pages exam 1 solutions
    Exam 1 Solutions

    School: Purdue

    Course: Electromechanical Motion Devices

  • 2 Pages practice exam5
    Practice Exam5

    School: Purdue

    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

  • 5 Pages practice exam5_solution
    Practice Exam5_solution

    School: Purdue

    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
    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

  • 4 Pages practice exam4_solution
    Practice Exam4_solution

    School: Purdue

    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

  • 2 Pages practice exam3
    Practice Exam3

    School: Purdue

    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

  • 6 Pages practice exam3_solution
    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

  • 3 Pages hw 5 solutions
    Hw 5 Solutions

    School: Purdue

    Course: Electromechanical Motion Devices

    EE321 HW #5 Problem 1 Problem 2 Problem 3

  • 1 Page 3. differential-equations-4page
    3. Differential-equations-4page

    School: Purdue

    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

  • 8 Pages 4. block-diagram-signal-flow-graph-4pages
    4. Block-diagram-signal-flow-graph-4pages

    School: Purdue

    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

  • 1 Page hw5
    Hw5

    School: Purdue

    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

  • 1 Page hw6
    Hw6

    School: Purdue

    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

  • 1 Page hw7
    Hw7

    School: Purdue

    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

  • 1 Page hw8
    Hw8

    School: Purdue

    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

  • 1 Page hw9
    Hw9

    School: Purdue

    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
    Hw10

    School: Purdue

    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

  • 1 Page hw11
    Hw11

    School: Purdue

    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(

  • 1 Page hw12
    Hw12

    School: Purdue

    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

  • 1 Page hw13
    Hw13

    School: Purdue

    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

  • 1 Page hw14
    Hw14

    School: Purdue

    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
    Exam5

    School: Purdue

    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
    Exam4

    School: Purdue

    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

  • 5 Pages exam1_solution
    Exam1_solution

    School: Purdue

  • 5 Pages exam 3 solution
    Exam 3 Solution

    School: Purdue

  • 2 Pages hw4
    Hw4

    School: Purdue

    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 ?

  • 2 Pages hw2
    Hw2

    School: Purdue

    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

  • 10 Pages 8. sensitivity-poles-2nd-order-systems-slides1
    8. Sensitivity-poles-2nd-order-systems-slides1

    School: Purdue

    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

  • 6 Pages 5. Modelling%20DC-motor-4pages
    5. Modelling%20DC-motor-4pages

    School: Purdue

    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

  • 8 Pages 6. analogy-rev0
    6. Analogy-rev0

    School: Purdue

    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

  • 6 Pages 7. Analogous-sytstems-slides-4pages
    7. Analogous-sytstems-slides-4pages

    School: Purdue

    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,

  • 12 Pages 9. Routh-Root-Locus-4pages
    9. Routh-Root-Locus-4pages

    School: Purdue

    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

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

    School: Purdue

    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

  • 7 Pages 11. Bode-Diagram-4pages
    11. Bode-Diagram-4pages

    School: Purdue

    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

  • 4 Pages 12. Bode-Plot-Examples
    12. Bode-Plot-Examples

    School: Purdue

    Course: Electromechanical Motion Devices

  • 4 Pages 15. Introduction to Compensator Design
    15. Introduction To Compensator Design

    School: Purdue

    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

  • 3 Pages 16. Bisector Method for RL Lead Design
    16. Bisector Method For RL Lead Design

    School: Purdue

    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 +

  • 1 Page 17. Design-Lead-Bisector0
    17. Design-Lead-Bisector0

    School: Purdue

    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

  • 4 Pages 19. design-lag-Bode-fall2011
    19. Design-lag-Bode-fall2011

    School: Purdue

    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

  • 2 Pages pastfinal-2009 solutions
    Pastfinal-2009 Solutions

    School: Purdue

    Course: Electromechanical Motion Devices

  • 3 Pages pastfinal-2009
    Pastfinal-2009

    School: Purdue

    Course: Electromechanical Motion Devices

  • 4 Pages pastfinal-2010
    Pastfinal-2010

    School: Purdue

    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

  • 7 Pages exam 2 solution
    Exam 2 Solution

    School: Purdue

  • 4 Pages exam2
    Exam2

    School: Purdue

    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)

  • 47 Pages ch2
    Ch2

    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 2 ELECTROMECHANICAL ENERGY CONVERSION 2.1 INTRODUCTION T he theory of electr

  • 47 Pages ch3
    Ch3

    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 3 DIRECT-CURRENT MACHINES 3.1 INTRODUCTION T he direct-current (dc) machine

  • 40 Pages ch4
    Ch4

    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 4 WINDINGS AND ROTATING MAGNETOMOTIVE FORCE 4.1 INTRODUCTION In the previous

  • 28 Pages ch5
    Ch5

    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 5 INTRODUCTION TO REFERENCE-FRAME THEORY 5.1 INTRODUCTION In recent years, t

  • 73 Pages ch6
    Ch6

    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 6 SYMMETRICAL INDUCTION MACHINES 6.1 INTRODUCTION Although the induction mac

  • 57 Pages ch7
    Ch7

    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 7 SYNCHRONOUS MACHINES 7.1 INTRODUCTION Nearly all electric power is generat

  • 69 Pages ch8
    Ch8

    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 8 PERMANENT-MAGNET ac MACHINE 8.1 INTRODUCTION T he permanent-magnet ac mach

  • 4 Pages app1
    App1

    School: Purdue

    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

  • 8 Pages app2
    App2

    School: Purdue

    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

  • 4 Pages app3
    App3

    School: Purdue

    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

  • 3 Pages ABET Exam
    ABET Exam

    School: Purdue

    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

    School: Purdue

    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

  • 1 Page hw10
    Hw10

    School: Purdue

    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

  • 4 Pages hw10_solution
    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
    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
    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
    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
    ABETexam

    School: Purdue

  • 5 Pages HW6(1)
    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
    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
    HW5_rachel_pereira

    School: Purdue

  • 3 Pages HW6_rachel_pereira
    HW6_rachel_pereira

    School: Purdue

  • 3 Pages hw6_rachel_pereira
    Hw6_rachel_pereira

    School: Purdue

  • 4 Pages HW7_rachel_pereira
    HW7_rachel_pereira

    School: Purdue

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