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ECE 321  Electromechanical Motion Devices  Purdue Study Resources

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

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 Hfield is constant Point2 H dl = H ( Point2 Point1) Point1 Thus the MMF drop is given by T H (

Hw9
School: Purdue
EE321 Spring 2008 / Homework 9 Problem 37 Discrete winding function The number of conductors in each slot of the aphase 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

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

Hw1
School: Purdue
EE321 Spring 10 Homework 1 Problem 1 Review of lineintegral and application to MMF drop. Consider a Cartesian coordinate system (x,y,z). Suppose a uniform Hfield of 1 A/m exists in the direction of the xaxis. Calculate the MMF drop from the point (1,1

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 Hfield is constant Point2 H dl = H ( Point2 Point1) Point1 Thus the MMF drop is given by T H (

HW2 Solution
School: Purdue
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 := 1.3 Point where saturation occurs 7 u 0 := 4 10 Now let's compute some reluctances. For flux densities

Hw 4
School: Purdue
Course: Electromechanical Motion Devices
EE321 Spring 2013 Homework 4 Problem 1 Calculation of Torque The fluxlinkage 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 ?

HW1
School: Purdue
EE321 Spring 2012 Homework 1 Problem 1 Review of lineintegral and application to MMF drop. Consider a Cartesian coordinate system (x,y,z). Suppose a uniform Hfield of 1 A/m exists in the direction of the xaxis. Calculate the MMF drop from the point (1

Hw8
School: Purdue
EE321 Spring 2008 / Homework 8 Problem 33 Buck converter operation Consider the example on page 11 of the lecture notes. Suppose the dc voltage is changed to 150 V and the speed to 450 rad/s. Find the average armature current, the average switch current,

Hw4
School: Purdue
EE321 Spring 2010 Homework 4 Problem 1 Calculation of Torque The fluxlinkage 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 ?

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 = 5106 Nms. The applied voltage is 6 V and Bm = 0. Calcula

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

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

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

Hw 1
School: Purdue
Course: Electromechanical Motion Devices
EE321 Spring 2012 Homework 1 Problem 1 Review of lineintegral and application to MMF drop. Consider a Cartesian coordinate system (x,y,z). Suppose a uniform Hfield of 1 A/m exists in the direction of the xaxis. Calculate the MMF drop from the point (1

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

Hw13
School: Purdue
Course: Electromechanical Motion Devices
EE321 Spring 2013 Homework 13 Problem 1 Rotating MMF Using the configuration we studied in class (Figure 6.21 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 th

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

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

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 25 satisfies objective 4. Getting 70% of problem 3 or 4 satisfies objective 2. Getting 70% o

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

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 13 satisfies ABET objective 2. Getting 75% of problem 5 satisfies ABET objective 3. Getting

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

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

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

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

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

Hw11_solution
School: Purdue
Course: Electromechanical Motion Devices
EE321 Spring 2013 Homework #11 Problem 1 Machine Parameters rs := 3 m := 0.17 3 Lss := 10 10 N := 3 P := 4 Load 2 r P TL( r) := 2 250 2.3 Source vs := 100 v := 0 vq := () 2 vs cos v () vd := 2 vs sin v Ok  lets solve the problem ( ) rs vq r m r Ls

Hw11
School: Purdue
Course: Electromechanical Motion Devices
EE321 Spring 2013 Homework 11 Problem 1 Brushless DC Operation from a Voltage Source A three phase brushless DC machine has the following parameters: rs = 3 , Lss = 10 mH, m = 0.17 Vs, P = 4 . It is operating from an inverter and the control is such that

Hw10
School: Purdue
Course: Electromechanical Motion Devices
EE321 Spring 2013 Homework 10 Problem 1 Phase Inductance Suppose the winding function of aphase of the stator is given by was = 2 P N s cos sm P 2 and the winding function of the bphase of the rotor is given by wbr = 2 P N r sin rm P 2 Express the mut

Hw10 Solution
School: Purdue
Course: Electromechanical Motion Devices
EE321 Spring 2013 HW#10 Problem 1 Lasar = r L 0 2 2 Ns Nr P g 2 P P P cos sm sin sm rm dsm 2 2 2 0 Lasar = r L 0 2 Ns Nr P 2 g 2 1 2 ( 0 Lasar = 4 r L 0 2 g P Lasar = 4 r L () 1 Ns Nr sin r 2 2 0 2 g P () Ns Nr sin r Problem 2 ( ) n as = 200 sin

Hw7
School: Purdue
Course: Electromechanical Motion Devices
EE321 Spring 2013 Homework 7 Problem 1 Buck converter operation Consider the example on page 68 of the lecture notes. Suppose the dc voltage is changed to 150 V and the speed to 400 rad/s. Find the average armature current, the average switch current, the

Hw7 Solution
School: Purdue
Course: Electromechanical Motion Devices
EE321 Spring 2013 HW7 Problem 1 Buck Converter Operation Consider a machine with the following parameters: ra := 0.1 kv := 0.2 The machine is fed using a buck converter with the following parameters vfsw := 2.4 vfd := 2.0 vdc := 150 d := 0.7 The machine i

Hw12 Solution
School: Purdue
Course: Electromechanical Motion Devices
EE321 Spring 2013 Homework #12 Problem 1 500 e Xphasor := 5+ 1 20 ( 100j) + Xphasor = 17.379 ( j 1 300 ( 100j) 2 Xphasor 2 = 24.577 ) arg Xphasor = 1.967 x=17.4*sqrt(2)*cos(100t1.97) If the cosine term were a sine, I simply would have adjusted the phase

Hw12
School: Purdue
Course: Electromechanical Motion Devices
EE321 Spring 2013 Homework 12 Problem 1 Phasor Analysis Consider the differential equation 5x + 1 dx 1 d 2x + = 500 2 cos(100t + 1) 20 dt 300 dt 2 where all arguments of the cosine term are in radians. Use phasor analysis to find the steadystate solution

Hw13 Solution
School: Purdue
Course: Electromechanical Motion Devices
EE321 Spring 2013 Homework #13 Problem 1 We have: i as = 50 sin( 200 t ) i bs = 50 cos( 200 t ) Now, we will also have ( ) ( ) was = W cos 2 sm wbs = W sin 2 sm Thus, the stator MMF is ( ) Fs = 50 W sin 200 t 2 sm Whereupon the stator MMF is moving at 100

Hw6 Solution
School: Purdue
Course: Electromechanical Motion Devices
EE321 Spring 2013, HW #6 Problem 1 ra := 6 kv := 0.01 6 K := 5 10 va := 6 kv ( va kv r) ra = K r Set Te = Tl va r := kv 2 kv + K ra r = 461.538 rad/s Problem 2 ra := 10 Rf := 50 LAF := 0.5 Va := 30 Vf := 30 Part 1  Stall Torque i f := i a := Vf Rf Va ra

Hw 5
School: Purdue
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 cphase is energized and the position is as indicated (point 1). Then the position is

Hw 5 Solutions
School: Purdue
Course: Electromechanical Motion Devices
EE321 HW #5 Problem 1 Problem 2 Problem 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

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 23 (together) can be used to satisfy ABET Objective 2. Problems 46 (together) can be used to satisfy ABET Objective 1. Bid me run, an

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 openloop poles (i.e., poles of G(s)H (s) Rule 2: The root locus (or the branches) starts at the openloop poles (K = 0) and ends at the openl

11. BodeDiagram4pages
School: Purdue
Course: Electromechanical Motion Devices
Frequency Response Method Frequency Response Method of a system is Frequency response Frequency Response Example defined as the steadystate response of Consider a massdashpotspring example with f (t ) as input and x (t ) as output. Frequency response

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

16. Bisector Method For RL Lead Design
School: Purdue
Course: Electromechanical Motion Devices
ECE 382 Lead Compensator Design (RootLocus) 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 +

17. DesignLeadBisector0
School: Purdue
Course: Electromechanical Motion Devices
ECE 382 Lead Compensator Design (Bisector Method) j A s1 n 2 2 0 C B =cos1 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

18. DesignLeadBodeSpring2012
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 openloop g

19. DesignlagBodefall2011
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 openlo

Pastfinal2010
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 1page crib sheets allowed (to be submitted with exam). No calculators, you may express answers in ter

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

Hw2
School: Purdue
EE321 Spring 2010 Homework 2 Problems 1 UI Inductor Analysis Consider the UI core shown in Figure 1.41 (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

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 cphase is energized and the position is as indicated (point 1). Then the position is

Hw6
School: Purdue
EE321 Spring 2010 / Homework 6 Problem 1 Problem 3.103 from Electromechanical Motion Devices Problem 2 Problem 3.106 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

9. RouthRootLocus4pages
School: Purdue
Course: Electromechanical Motion Devices
Stability Stability Depends on ClosedLoop 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

8. Sensitivitypoles2ndordersystemsslides1
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

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 ycomponent couples the loop := Single turn = 0.02513 Problem 2 w := 1 cm d s := 2 cm g := 0.1 mm

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 bphase stator windings of a machine are given by was = 100 cos(4 s ) and wbs = 250 sin(4 s ) . The aphase current of the machine is given by i as = 10 cos( e t

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

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

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

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 = 5106 Nms. The applied voltage is 6 V and Bm = 0. Calcula

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

1. Spring2012courseintro4pages
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) 4941384 Email: csglee@purdue.edu Ofce Hours: MWF: 10:30 11:30 AM (or by appointment) ECE382: Feedback System Analysis and Design C.

2. Differentialfall2011
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

3. Differentialequations4page
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

4. Blockdiagramsignalflowgraph4pages
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

5. Modelling%20DCmotor4pages
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

6. Analogyrev0
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

7. Analogoussytstemsslides4pages
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; Forcevelocity,

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

Hw14 Solution
School: Purdue
Course: Electromechanical Motion Devices
EE321 Spring 2013 Homework #14 Problem 1 Machine Parameters 3 3 rs := 72.5 10 3 Lls := 1.32 10 3 rr := 41.3 10 P := 4 Llr := 1.32 10 N Lrr := Llr + Lms 2 N := 2 Source vs := 460 e := 2 60 3 Mechanical Load 50 746 = 1800 2 := 3 2 1800 60 60 50 746 1

Hw9
School: Purdue
Course: Electromechanical Motion Devices
ECE321/ECE595 Homework 9 Problem 1 Discrete Winding Function The number of conductors in each slot of the aphase of the stator of the machine are as follows: N as = [0 4 4 0 4 4 0 4 4 0 4 4]T Compute and graph the winding function associated with this wi

Hw6_rachel_pereira
School: Purdue

HW7_rachel_pereira
School: Purdue

Hw8_rachel_pereira
School: Purdue

HW9_rachel_pereira
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HW10_rachel_pereira
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HW11_rachel_pereira
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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

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

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 DIRECTCURRENT MACHINES 3.1 INTRODUCTION T he directcurrent (dc) machine

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