Chapter 8 Objectives
|
|
- Lorin Johnston
- 5 years ago
- Views:
Transcription
1 haper 8 Engr8 ircui Analyi Dr uri Nelon haper 8 Objecive Be able o eermine he naural an he ep repone of parallel circui; Be able o eermine he naural an he ep repone of erie circui. Engr8 haper 8, Nilon e
2 an ircui evie Tranien, naural, or homogeneou repone: Fae over ime; ei change. Force, eayae, paricular repone: Follo he inpu; Inepenen of ime pae. The oal repone ill be of he form: k k e The ircui circui conain boh an inucor an a capacior; Thee circui have a ie range of applicaion incluing ocillaor, frequency filer, fligh imulaion, moeling auomobile upenion, an more; The repone of circui ih D ource an iche ill coni of a naural repone an a force repone: v v f v n The complee repone mu aify boh he iniial coniion an he final coniion of he force repone. Engr8 haper 8, Nilon e
3 SourceFree Parallel ircui We ill fir uy he naural repone of econorer circui by looking a a ourcefree parallel circui: i i v i i i i v v ò v v v v Seconorer Differenial equaion Thi econorer ifferenial equaion can be olve by auming he form of a oluion: hich mean SourceFree Parallel ircui v Ae v v v A e Ae Ae Ae Thi i knon a he characeriic equaion. Engr8 haper 8, Nilon e 3
4 SourceFree Parallel ircui Uing he quaraic formula, e ge Define reonan frequency: Define amping facor: æ ç è æ ç è ö ø ö ø a Then:, a ± a SeconOrer Differenial Equaion Soluion We ill no ivie he circui repone ino hree cae accoring o he ign of he erm uner he raical., ± æ ö ç è ø, a ± a α > ω overampe:!# & ' * &. α ω criically ampe: α < ω unerampe: v Ae a v e B co B A e in Engr8 haper 8, Nilon e 4
5 Type of ircui epone Overampe Example Fin v in he circui a he righ. Ignore he curren arro. Given iniial coniion: v c, i A a α > ω herefore hi i an overampe cae, 6, a ± a Engr8 haper 8, Nilon e 5
6 Overampe ae coninue The oluion i of he form: v Ae Ue iniial coniion o fin A an A From v c a : v Ae Ae A A From K aken a : A e 6 i i v v 4 A 6A 4 i 6 A e 6A e Overampe ae coninue Solving he o equaion e ge A 84 an A 84 The oluion i: v 84e 84e 84 e e 6 6 V v v 84 e e 6 Engr8 haper 8, Nilon e 6
7 Example: Overampe ircui Fin v for >. v 8e 5, e, V for > riically Dampe ae α ω Fin v in he circui a he righ. Given iniial coniion: v c, i A /4f i v i 8.573Ω 7H i a.45 riically ampe hen α ω.45 The complee oluion i of he form: v Ae A e Engr8 haper 8, Nilon e 7
8 riically Dampe ae coninue Ue iniial coniion o fin A an A From v c a : v A e Ae A Therefore A an he oluion i reuce o v Ae Fin A from K a : i i v v i 4 4 A A.45 e A e.45 riically Dampe ae coninue Solving he equaion: A 4 The oluion i: v 4e.45 V v v 4e.45 Engr8 haper 8, Nilon e 8
9 riically Dampe Example Fin uch ha he circui i criically ampe for > an uch ha vv. Aner: 3.63 kω,.4ω Unerampe ae α < ω, a ± a For he unerampe cae, he erm inie he bracke ill be negaive an ill be a complex number. Define a Then, a ± j v A e v e a j a A e j A e a j A e j Engr8 haper 8, Nilon e 9
10 Engr8 haper 8, Nilon e j j e A A e e v a Uing Euler Ieniy q q q co j in e j in co in co in co in co B B e v A A j A A e v ja A ja A e v a a a in co B B e v a Unerampe ae coninue ooking a he magniue: A penulum i an example of an unerampe econorer mechanical yem. iplacemen Mechanical Analogue
11 Unerampe ae Example Fin v in he circui a he righ. Given iniial coniion: v c, i A /4f i v i.5ω 7H i a 6 α < ω herefore, hi i an unerampe cae a v i of he form: v e B co B in Unerampe ae coninue Ue iniial coniion o fin B an B From v c a : v e B co B in B Therefore B an he oluion i reuce o v e B in Fin B from K a : i i v v i 4 4 B B e co B e in Engr8 haper 8, Nilon e
12 Unerampe ae coninue Solving: B 97 The oluion i: v 97e in V v v 97e in V Unerampe Example Fin i for >. i e..7 co in 4.75 A Engr8 haper 8, Nilon e
13 Engr8 haper 8, Nilon e 3 Summary of Tranien epone Source Free Serie ircui ò i i i i i i v v v v v i v
14 omparing Serie an Parallel ircui Parallel v v v v A e Ae, ± æ ö ç è ø Serie i i i i A e Ae, ± æ ö ç è ø, a ± a, a ± a a a Serie ircui Soluion a a a a a If: α > ω overampe: α ω criically ampe: α < ω unerampe: i A e Ae A i e a A B co B in a i e Engr8 haper 8, Nilon e 4
15 Texbook Problem 8.49 Nilon h The circui conain no iniial energy. Fin v o for. v o 6 6e 4 co3.33e 4 in3 V Summary: Solving ircui. Ienify he erie or parallel circui;. Fin α an ω ; 3. Deermine heher he circui i overampe, criically ampe, or unerampe; 4. Aume a oluion naural repone force repone: A e Ae A e A e e V a B co B in V V f f f Overampe riically ampe Unerampe 5. Fin A, B, an V f uing iniial an final coniion. Engr8 haper 8, Nilon e 5
16 haper 8 Summary Shoe ho o eermine he naural an he ep repone of parallel circui; Shoe ho o eermine he naural an he ep repone of erie circui. Engr8 haper 8, Nilon e 6
CHAPTER 7: SECOND-ORDER CIRCUITS
EEE5: CI RCUI T THEORY CHAPTER 7: SECOND-ORDER CIRCUITS 7. Inroducion Thi chaper conider circui wih wo orage elemen. Known a econd-order circui becaue heir repone are decribed by differenial equaion ha
More informationEE202 Circuit Theory II
EE202 Circui Theory II 2017-2018, Spring Dr. Yılmaz KALKAN I. Inroducion & eview of Fir Order Circui (Chaper 7 of Nilon - 3 Hr. Inroducion, C and L Circui, Naural and Sep epone of Serie and Parallel L/C
More informationPackage Inductance and RLC Circuit Analysis
Package Inucance an ircui Analyi FATO Jawel, ik a beken al fenomeen Tenmine oner mahemaici En in wikunig aangelege kringen Mijn naam, al u he ween wil i i Ik zi in allerlei berekeningen En aan he eine
More informations-domain Circuit Analysis
Domain ircui Analyi Operae direcly in he domain wih capacior, inducor and reior Key feaure lineariy i preerved c decribed by ODE and heir I Order equal number of plu number of Elemenbyelemen and ource
More information8. Basic RL and RC Circuits
8. Basic L and C Circuis This chaper deals wih he soluions of he responses of L and C circuis The analysis of C and L circuis leads o a linear differenial equaion This chaper covers he following opics
More informationPackage Inductance and RLC Circuit Analysis. New Today. Bondwires. Package Parasitics. Model for Invertor with Inductance. Inductance (Inductantie)
Package Inucance an ircui Analyi New Toay FATO Jawel, ik a beken al fenomeen Tenmine oner mahemaici En in wikung aangelege kringen Mijn naam, al u he ween wil i i Ik zi in allerlei berekeningen En aan
More information6.8 Laplace Transform: General Formulas
48 HAP. 6 Laplace Tranform 6.8 Laplace Tranform: General Formula Formula Name, ommen Sec. F() l{ f ()} e f () d f () l {F()} Definiion of Tranform Invere Tranform 6. l{af () bg()} al{f ()} bl{g()} Lineariy
More informationUT Austin, ECE Department VLSI Design 5. CMOS Gate Characteristics
La moule: CMOS Tranior heory Thi moule: DC epone Logic Level an Noie Margin Tranien epone Delay Eimaion Tranior ehavior 1) If he wih of a ranior increae, he curren will ) If he lengh of a ranior increae,
More informationSample Final Exam (finals03) Covering Chapters 1-9 of Fundamentals of Signals & Systems
Sample Final Exam Covering Chaper 9 (final04) Sample Final Exam (final03) Covering Chaper 9 of Fundamenal of Signal & Syem Problem (0 mar) Conider he caual opamp circui iniially a re depiced below. I LI
More informationLecture 13 RC/RL Circuits, Time Dependent Op Amp Circuits
Lecure 13 RC/RL Circuis, Time Dependen Op Amp Circuis RL Circuis The seps involved in solving simple circuis conaining dc sources, resisances, and one energy-sorage elemen (inducance or capaciance) are:
More informationSingle Phase Line Frequency Uncontrolled Rectifiers
Single Phae Line Frequency Unconrolle Recifier Kevin Gaughan 24-Nov-03 Single Phae Unconrolle Recifier 1 Topic Baic operaion an Waveform (nucive Loa) Power Facor Calculaion Supply curren Harmonic an Th
More informationTo become more mathematically correct, Circuit equations are Algebraic Differential equations. from KVL, KCL from the constitutive relationship
Laplace Tranform (Lin & DeCarlo: Ch 3) ENSC30 Elecric Circui II The Laplace ranform i an inegral ranformaion. I ranform: f ( ) F( ) ime variable complex variable From Euler > Lagrange > Laplace. Hence,
More informationChapter 9 - The Laplace Transform
Chaper 9 - The Laplace Tranform Selece Soluion. Skech he pole-zero plo an region of convergence (if i exi) for hee ignal. ω [] () 8 (a) x e u = 8 ROC σ ( ) 3 (b) x e co π u ω [] ( ) () (c) x e u e u ROC
More informationVoltage/current relationship Stored Energy. RL / RC circuits Steady State / Transient response Natural / Step response
Review Capaciors/Inducors Volage/curren relaionship Sored Energy s Order Circuis RL / RC circuis Seady Sae / Transien response Naural / Sep response EE4 Summer 5: Lecure 5 Insrucor: Ocavian Florescu Lecure
More informationR.#W.#Erickson# Department#of#Electrical,#Computer,#and#Energy#Engineering# University#of#Colorado,#Boulder#
.#W.#Erickson# Deparmen#of#Elecrical,#Compuer,#and#Energy#Engineering# Universiy#of#Colorado,#Boulder# Chaper 2 Principles of Seady-Sae Converer Analysis 2.1. Inroducion 2.2. Inducor vol-second balance,
More informationEECE 301 Signals & Systems Prof. Mark Fowler
EECE 31 Signal & Syem Prof. Mark Fowler Noe Se #27 C-T Syem: Laplace Tranform Power Tool for yem analyi Reading Aignmen: Secion 6.1 6.3 of Kamen and Heck 1/18 Coure Flow Diagram The arrow here how concepual
More informationMath 2214 Solution Test 1 B Spring 2016
Mah 14 Soluion Te 1 B Spring 016 Problem 1: Ue eparaion of ariable o ole he Iniial alue DE Soluion (14p) e =, (0) = 0 d = e e d e d = o = ln e d uing u-du b leing u = e 1 e = + where C = for he iniial
More informationLet. x y. denote a bivariate time series with zero mean.
Linear Filer Le x y : T denoe a bivariae ime erie wih zero mean. Suppoe ha he ime erie {y : T} i conruced a follow: y a x The ime erie {y : T} i aid o be conruced from {x : T} by mean of a Linear Filer.
More informationMath 2214 Solution Test 1A Spring 2016
Mah 14 Soluion Tes 1A Spring 016 sec Problem 1: Wha is he larges -inerval for which ( 4) = has a guaraneed + unique soluion for iniial value (-1) = 3 according o he Exisence Uniqueness Theorem? Soluion
More information13.1 Circuit Elements in the s Domain Circuit Analysis in the s Domain The Transfer Function and Natural Response 13.
Chaper 3 The Laplace Tranform in Circui Analyi 3. Circui Elemen in he Domain 3.-3 Circui Analyi in he Domain 3.4-5 The Tranfer Funcion and Naural Repone 3.6 The Tranfer Funcion and he Convoluion Inegral
More informationCONTROL SYSTEMS. Chapter 10 : State Space Response
CONTROL SYSTEMS Chaper : Sae Space Repone GATE Objecive & Numerical Type Soluion Queion 5 [GATE EE 99 IIT-Bombay : Mark] Conider a econd order yem whoe ae pace repreenaion i of he form A Bu. If () (),
More informationECE 2100 Circuit Analysis
ECE 1 Circui Analysis Lesson 35 Chaper 8: Second Order Circuis Daniel M. Liynski, Ph.D. ECE 1 Circui Analysis Lesson 3-34 Chaper 7: Firs Order Circuis (Naural response RC & RL circuis, Singulariy funcions,
More informationLAB 5: Computer Simulation of RLC Circuit Response using PSpice
--3LabManualLab5.doc LAB 5: ompuer imulaion of RL ircui Response using Ppice PURPOE To use a compuer simulaion program (Ppice) o invesigae he response of an RL series circui o: (a) a sinusoidal exciaion.
More informationChapter 6. Laplace Transforms
Chaper 6. Laplace Tranform Kreyzig by YHLee;45; 6- An ODE i reduced o an algebraic problem by operaional calculu. The equaion i olved by algebraic manipulaion. The reul i ranformed back for he oluion of
More informationChapter 7 Response of First-order RL and RC Circuits
Chaper 7 Response of Firs-order RL and RC Circuis 7.- The Naural Response of RL and RC Circuis 7.3 The Sep Response of RL and RC Circuis 7.4 A General Soluion for Sep and Naural Responses 7.5 Sequenial
More informationEEEB113 CIRCUIT ANALYSIS I
9/14/29 1 EEEB113 CICUIT ANALYSIS I Chaper 7 Firs-Order Circuis Maerials from Fundamenals of Elecric Circuis 4e, Alexander Sadiku, McGraw-Hill Companies, Inc. 2 Firs-Order Circuis -Chaper 7 7.2 The Source-Free
More information, the. L and the L. x x. max. i n. It is easy to show that these two norms satisfy the following relation: x x n x = (17.3) max
ecure 8 7. Sabiliy Analyi For an n dimenional vecor R n, he and he vecor norm are defined a: = T = i n i (7.) I i eay o how ha hee wo norm aify he following relaion: n (7.) If a vecor i ime-dependen, hen
More informationChapter 6. Laplace Transforms
6- Chaper 6. Laplace Tranform 6.4 Shor Impule. Dirac Dela Funcion. Parial Fracion 6.5 Convoluion. Inegral Equaion 6.6 Differeniaion and Inegraion of Tranform 6.7 Syem of ODE 6.4 Shor Impule. Dirac Dela
More informationDesign of Controller for Robot Position Control
eign of Conroller for Robo oiion Conrol Two imporan goal of conrol: 1. Reference inpu racking: The oupu mu follow he reference inpu rajecory a quickly a poible. Se-poin racking: Tracking when he reference
More informationMath 2214 Solution Test 1B Fall 2017
Mah 14 Soluion Tes 1B Fall 017 Problem 1: A ank has a capaci for 500 gallons and conains 0 gallons of waer wih lbs of sal iniiall. A soluion conaining of 8 lbsgal of sal is pumped ino he ank a 10 galsmin.
More informationChapter Three Systems of Linear Differential Equations
Chaper Three Sysems of Linear Differenial Equaions In his chaper we are going o consier sysems of firs orer orinary ifferenial equaions. These are sysems of he form x a x a x a n x n x a x a x a n x n
More information(b) (a) (d) (c) (e) Figure 10-N1. (f) Solution:
Example: The inpu o each of he circuis shown in Figure 10-N1 is he volage source volage. The oupu of each circui is he curren i( ). Deermine he oupu of each of he circuis. (a) (b) (c) (d) (e) Figure 10-N1
More informationChapter 7: Inverse-Response Systems
Chaper 7: Invere-Repone Syem Normal Syem Invere-Repone Syem Baic Sar ou in he wrong direcion End up in he original eady-ae gain value Two or more yem wih differen magniude and cale in parallel Main yem
More informationSome Basic Information about M-S-D Systems
Some Basic Informaion abou M-S-D Sysems 1 Inroducion We wan o give some summary of he facs concerning unforced (homogeneous) and forced (non-homogeneous) models for linear oscillaors governed by second-order,
More informationChapter 10. Optimization: More than One Choice Variable
Chaper Opimiaion: More han One Choice Variable William Sanle Jevons 85-88 Carl Menger 8 9. Opimiaion Problems Chaper 9: ma u one choice variable: consumpion Chaper : ma u o choice variables: leisure Chaper
More information8.022 (E&M) Lecture 16
8. (E&M) ecure 16 Topics: Inducors in circuis circuis circuis circuis as ime Our second lecure on elecromagneic inducance 3 ways of creaing emf using Faraday s law: hange area of circui S() hange angle
More informationu(t) Figure 1. Open loop control system
Open loop conrol v cloed loop feedbac conrol The nex wo figure preen he rucure of open loop and feedbac conrol yem Figure how an open loop conrol yem whoe funcion i o caue he oupu y o follow he reference
More information18.03SC Unit 3 Practice Exam and Solutions
Sudy Guide on Sep, Dela, Convoluion, Laplace You can hink of he ep funcion u() a any nice mooh funcion which i for < a and for > a, where a i a poiive number which i much maller han any ime cale we care
More informationDirect Current Circuits. February 19, 2014 Physics for Scientists & Engineers 2, Chapter 26 1
Direc Curren Circuis February 19, 2014 Physics for Scieniss & Engineers 2, Chaper 26 1 Ammeers and Volmeers! A device used o measure curren is called an ammeer! A device used o measure poenial difference
More informationElectrical Circuits. 1. Circuit Laws. Tools Used in Lab 13 Series Circuits Damped Vibrations: Energy Van der Pol Circuit
V() R L C 513 Elecrical Circuis Tools Used in Lab 13 Series Circuis Damped Vibraions: Energy Van der Pol Circui A series circui wih an inducor, resisor, and capacior can be represened by Lq + Rq + 1, a
More informationChapter 8 The Complete Response of RL and RC Circuits
Chaper 8 The Complee Response of RL and RC Circuis Seoul Naional Universiy Deparmen of Elecrical and Compuer Engineering Wha is Firs Order Circuis? Circuis ha conain only one inducor or only one capacior
More informationINDEX. Transient analysis 1 Initial Conditions 1
INDEX Secion Page Transien analysis 1 Iniial Condiions 1 Please inform me of your opinion of he relaive emphasis of he review maerial by simply making commens on his page and sending i o me a: Frank Mera
More informationEECE251. Circuit Analysis I. Set 4: Capacitors, Inductors, and First-Order Linear Circuits
EEE25 ircui Analysis I Se 4: apaciors, Inducors, and Firs-Order inear ircuis Shahriar Mirabbasi Deparmen of Elecrical and ompuer Engineering Universiy of Briish olumbia shahriar@ece.ubc.ca Overview Passive
More informationWEEK-3 Recitation PHYS 131. of the projectile s velocity remains constant throughout the motion, since the acceleration a x
WEEK-3 Reciaion PHYS 131 Ch. 3: FOC 1, 3, 4, 6, 14. Problems 9, 37, 41 & 71 and Ch. 4: FOC 1, 3, 5, 8. Problems 3, 5 & 16. Feb 8, 018 Ch. 3: FOC 1, 3, 4, 6, 14. 1. (a) The horizonal componen of he projecile
More informationChapter 2: Principles of steady-state converter analysis
Chaper 2 Principles of Seady-Sae Converer Analysis 2.1. Inroducion 2.2. Inducor vol-second balance, capacior charge balance, and he small ripple approximaion 2.3. Boos converer example 2.4. Cuk converer
More informationCHAPTER 12 DIRECT CURRENT CIRCUITS
CHAPTER 12 DIRECT CURRENT CIUITS DIRECT CURRENT CIUITS 257 12.1 RESISTORS IN SERIES AND IN PARALLEL When wo resisors are conneced ogeher as shown in Figure 12.1 we said ha hey are conneced in series. As
More informationAlgorithms and Data Structures 2011/12 Week 9 Solutions (Tues 15th - Fri 18th Nov)
Algorihm and Daa Srucure 2011/ Week Soluion (Tue 15h - Fri 18h No) 1. Queion: e are gien 11/16 / 15/20 8/13 0/ 1/ / 11/1 / / To queion: (a) Find a pair of ube X, Y V uch ha f(x, Y) = f(v X, Y). (b) Find
More informationAlgorithmic Discrete Mathematics 6. Exercise Sheet
Algorihmic Dicree Mahemaic. Exercie Shee Deparmen of Mahemaic SS 0 PD Dr. Ulf Lorenz 7. and 8. Juni 0 Dipl.-Mah. David Meffer Verion of June, 0 Groupwork Exercie G (Heap-Sor) Ue Heap-Sor wih a min-heap
More informationSection 3.5 Nonhomogeneous Equations; Method of Undetermined Coefficients
Secion 3.5 Nonhomogeneous Equaions; Mehod of Undeermined Coefficiens Key Terms/Ideas: Linear Differenial operaor Nonlinear operaor Second order homogeneous DE Second order nonhomogeneous DE Soluion o homogeneous
More informationdy dx = xey (a) y(0) = 2 (b) y(1) = 2.5 SOLUTION: See next page
Assignmen 1 MATH 2270 SOLUTION Please wrie ou complee soluions for each of he following 6 problems (one more will sill be added). You may, of course, consul wih your classmaes, he exbook or oher resources,
More informationECE 2100 Circuit Analysis
ECE 1 Circui Analysis Lesson 37 Chaper 8: Second Order Circuis Discuss Exam Daniel M. Liynski, Ph.D. Exam CH 1-4: On Exam 1; Basis for work CH 5: Operaional Amplifiers CH 6: Capaciors and Inducor CH 7-8:
More informationReading from Young & Freedman: For this topic, read sections 25.4 & 25.5, the introduction to chapter 26 and sections 26.1 to 26.2 & 26.4.
PHY1 Elecriciy Topic 7 (Lecures 1 & 11) Elecric Circuis n his opic, we will cover: 1) Elecromoive Force (EMF) ) Series and parallel resisor combinaions 3) Kirchhoff s rules for circuis 4) Time dependence
More informationFirst Order RC and RL Transient Circuits
Firs Order R and RL Transien ircuis Objecives To inroduce he ransiens phenomena. To analyze sep and naural responses of firs order R circuis. To analyze sep and naural responses of firs order RL circuis.
More informationAC Circuits AC Circuit with only R AC circuit with only L AC circuit with only C AC circuit with LRC phasors Resonance Transformers
A ircuis A ircui wih only A circui wih only A circui wih only A circui wih phasors esonance Transformers Phys 435: hap 31, Pg 1 A ircuis New Topic Phys : hap. 6, Pg Physics Moivaion as ime we discovered
More informationV L. DT s D T s t. Figure 1: Buck-boost converter: inductor current i(t) in the continuous conduction mode.
ECE 445 Analysis and Design of Power Elecronic Circuis Problem Se 7 Soluions Problem PS7.1 Erickson, Problem 5.1 Soluion (a) Firs, recall he operaion of he buck-boos converer in he coninuous conducion
More informationES 250 Practice Final Exam
ES 50 Pracice Final Exam. Given ha v 8 V, a Deermine he values of v o : 0 Ω, v o. V 0 Firs, v o 8. V 0 + 0 Nex, 8 40 40 0 40 0 400 400 ib i 0 40 + 40 + 40 40 40 + + ( ) 480 + 5 + 40 + 8 400 400( 0) 000
More informationStudy of simple inductive-capacitive series circuits using MATLAB software package
ecen Advance in ircui, Syem and Auomaic onrol Sudy of imple inducive-capaciive erie circui uing MAAB ofware package NIUESU IU, PĂSUESU DAGOŞ Faculy of Mechanical and Elecrical Engineering Univeriy of Peroani
More informationMath 2142 Exam 1 Review Problems. x 2 + f (0) 3! for the 3rd Taylor polynomial at x = 0. To calculate the various quantities:
Mah 4 Eam Review Problems Problem. Calculae he 3rd Taylor polynomial for arcsin a =. Soluion. Le f() = arcsin. For his problem, we use he formula f() + f () + f ()! + f () 3! for he 3rd Taylor polynomial
More informationME 391 Mechanical Engineering Analysis
Fall 04 ME 39 Mechanical Engineering Analsis Eam # Soluions Direcions: Open noes (including course web posings). No books, compuers, or phones. An calculaor is fair game. Problem Deermine he posiion of
More information( ) = Q 0. ( ) R = R dq. ( t) = I t
ircuis onceps The addiion of a simple capacior o a circui of resisors allows wo relaed phenomena o occur The observaion ha he ime-dependence of a complex waveform is alered by he circui is referred o as
More information2.4 Cuk converter example
2.4 Cuk converer example C 1 Cuk converer, wih ideal swich i 1 i v 1 2 1 2 C 2 v 2 Cuk converer: pracical realizaion using MOSFET and diode C 1 i 1 i v 1 2 Q 1 D 1 C 2 v 2 28 Analysis sraegy This converer
More informationMore on ODEs by Laplace Transforms October 30, 2017
More on OE b Laplace Tranfor Ocober, 7 More on Ordinar ifferenial Equaion wih Laplace Tranfor Larr areo Mechanical Engineering 5 Seinar in Engineering nali Ocober, 7 Ouline Review la cla efiniion of Laplace
More informationTopics in Combinatorial Optimization May 11, Lecture 22
8.997 Topics in Combinaorial Opimizaion May, 004 Lecure Lecurer: Michel X. Goemans Scribe: Alanha Newman Muliflows an Disjoin Pahs Le G = (V,E) be a graph an le s,,s,,...s, V be erminals. Our goal is o
More informationLaplace Transform. Inverse Laplace Transform. e st f(t)dt. (2)
Laplace Tranform Maoud Malek The Laplace ranform i an inegral ranform named in honor of mahemaician and aronomer Pierre-Simon Laplace, who ued he ranform in hi work on probabiliy heory. I i a powerful
More informationMath 333 Problem Set #2 Solution 14 February 2003
Mah 333 Problem Se #2 Soluion 14 February 2003 A1. Solve he iniial value problem dy dx = x2 + e 3x ; 2y 4 y(0) = 1. Soluion: This is separable; we wrie 2y 4 dy = x 2 + e x dx and inegrae o ge The iniial
More informationUniversity of Cyprus Biomedical Imaging and Applied Optics. Appendix. DC Circuits Capacitors and Inductors AC Circuits Operational Amplifiers
Universiy of Cyprus Biomedical Imaging and Applied Opics Appendix DC Circuis Capaciors and Inducors AC Circuis Operaional Amplifiers Circui Elemens An elecrical circui consiss of circui elemens such as
More informationHomework 2 Solutions
Mah 308 Differenial Equaions Fall 2002 & 2. See he las page. Hoework 2 Soluions 3a). Newon s secon law of oion says ha a = F, an we know a =, so we have = F. One par of he force is graviy, g. However,
More information6.01: Introduction to EECS I Lecture 8 March 29, 2011
6.01: Inroducion o EES I Lecure 8 March 29, 2011 6.01: Inroducion o EES I Op-Amps Las Time: The ircui Absracion ircuis represen sysems as connecions of elemens hrough which currens (hrough variables) flow
More informationMA 214 Calculus IV (Spring 2016) Section 2. Homework Assignment 1 Solutions
MA 14 Calculus IV (Spring 016) Secion Homework Assignmen 1 Soluions 1 Boyce and DiPrima, p 40, Problem 10 (c) Soluion: In sandard form he given firs-order linear ODE is: An inegraing facor is given by
More informationPhys1112: DC and RC circuits
Name: Group Members: Dae: TA s Name: Phys1112: DC and RC circuis Objecives: 1. To undersand curren and volage characerisics of a DC RC discharging circui. 2. To undersand he effec of he RC ime consan.
More information6.302 Feedback Systems Recitation : Phase-locked Loops Prof. Joel L. Dawson
6.32 Feedback Syem Phae-locked loop are a foundaional building block for analog circui deign, paricularly for communicaion circui. They provide a good example yem for hi cla becaue hey are an excellen
More informationEE202 Circuit Theory II , Spring. Dr. Yılmaz KALKAN & Dr. Atilla DÖNÜK
EE202 Circui Theory II 2018 2019, Spring Dr. Yılmaz KALKAN & Dr. Ailla DÖNÜK 1. Basic Conceps (Chaper 1 of Nilsson - 3 Hrs.) Inroducion, Curren and Volage, Power and Energy 2. Basic Laws (Chaper 2&3 of
More informationPhysics 1402: Lecture 22 Today s Agenda
Physics 142: ecure 22 Today s Agenda Announcemens: R - RV - R circuis Homework 6: due nex Wednesday Inducion / A curren Inducion Self-Inducance, R ircuis X X X X X X X X X long solenoid Energy and energy
More informationMath 2214 Sol Test 2B Spring 2015
Mah 14 Sol Tes B Sring 015 roblem 1: An objec weighing ounds sreches a verical sring 8 fee beond i naural lengh before coming o res a equilibrium The objec is ushed u 6 fee from i s equilibrium osiion
More informationWall. x(t) f(t) x(t = 0) = x 0, t=0. which describes the motion of the mass in absence of any external forcing.
MECHANICS APPLICATIONS OF SECOND-ORDER ODES 7 Mechanics applicaions of second-order ODEs Second-order linear ODEs wih consan coefficiens arise in many physical applicaions. One physical sysems whose behaviour
More informationDifferential Equations
Mah 21 (Fall 29) Differenial Equaions Soluion #3 1. Find he paricular soluion of he following differenial equaion by variaion of parameer (a) y + y = csc (b) 2 y + y y = ln, > Soluion: (a) The corresponding
More information= + t ] can be used to calculate the angular
WEEK-7 Reciaion PHYS 3 Mar 8, 8 Ch. 8 FOC:, 4, 6,, 3 &5. () Using Equaion 8. (θ = Arc lengh / Raius) o calculae he angle (in raians) ha each objec subens a your eye shows ha θ Moon = 9. 3 ra, θ Pea = 7.
More informationDiscussion Session 2 Constant Acceleration/Relative Motion Week 03
PHYS 100 Dicuion Seion Conan Acceleraion/Relaive Moion Week 03 The Plan Today you will work wih your group explore he idea of reference frame (i.e. relaive moion) and moion wih conan acceleraion. You ll
More informationChapter 3 Kinematics in Two Dimensions
Chaper 3 KINEMATICS IN TWO DIMENSIONS PREVIEW Two-dimensional moion includes objecs which are moing in wo direcions a he same ime, such as a projecile, which has boh horizonal and erical moion. These wo
More informationA dynamic AS-AD Model
A ynamic AS-AD Moel (Lecure Noes, Thomas Seger, Universiy of Leipzig, winer erm 10/11) This file escribes a ynamic AS-AD moel. The moel can be employe o assess he ynamic consequences of macroeconomic shocks
More informationWhen analyzing an object s motion there are two factors to consider when attempting to bring it to rest. 1. The object s mass 2. The object s velocity
SPH4U Momenum LoRuo Momenum i an exenion of Newon nd law. When analyzing an ojec moion here are wo facor o conider when aeming o ring i o re.. The ojec ma. The ojec velociy The greaer an ojec ma, he more
More informationENGI 9420 Engineering Analysis Assignment 2 Solutions
ENGI 940 Engineering Analysis Assignmen Soluions 0 Fall [Second order ODEs, Laplace ransforms; Secions.0-.09]. Use Laplace ransforms o solve he iniial value problem [0] dy y, y( 0) 4 d + [This was Quesion
More informationSolutions to Assignment 1
MA 2326 Differenial Equaions Insrucor: Peronela Radu Friday, February 8, 203 Soluions o Assignmen. Find he general soluions of he following ODEs: (a) 2 x = an x Soluion: I is a separable equaion as we
More information5.2 GRAPHICAL VELOCITY ANALYSIS Polygon Method
ME 352 GRHICL VELCITY NLYSIS 52 GRHICL VELCITY NLYSIS olygon Mehod Velociy analyi form he hear of kinemaic and dynamic of mechanical yem Velociy analyi i uually performed following a poiion analyi; ie,
More informationExponential Sawtooth
ECPE 36 HOMEWORK 3: PROPERTIES OF THE FOURIER TRANSFORM SOLUTION. Exponenial Sawooh: The eaie way o do hi problem i o look a he Fourier ranform of a ingle exponenial funcion, () = exp( )u(). From he able
More informationHomework-8(1) P8.3-1, 3, 8, 10, 17, 21, 24, 28,29 P8.4-1, 2, 5
Homework-8() P8.3-, 3, 8, 0, 7, 2, 24, 28,29 P8.4-, 2, 5 Secion 8.3: The Response of a Firs Order Circui o a Consan Inpu P 8.3- The circui shown in Figure P 8.3- is a seady sae before he swich closes a
More informationCONTROL SYSTEMS. Chapter 3 Mathematical Modelling of Physical Systems-Laplace Transforms. Prof.Dr. Fatih Mehmet Botsalı
CONTROL SYSTEMS Chaper Mahemaical Modelling of Phyical Syem-Laplace Tranform Prof.Dr. Faih Mehme Boalı Definiion Tranform -- a mahemaical converion from one way of hinking o anoher o make a problem eaier
More informationCHAPTER. Forced Equations and Systems { } ( ) ( ) 8.1 The Laplace Transform and Its Inverse. Transforms from the Definition.
CHAPTER 8 Forced Equaion and Syem 8 The aplace Tranform and I Invere Tranform from he Definiion 5 5 = b b {} 5 = 5e d = lim5 e = ( ) b {} = e d = lim e + e d b = (inegraion by par) = = = = b b ( ) ( )
More informationA Theoretical Model of a Voltage Controlled Oscillator
A Theoreical Model of a Volage Conrolled Ocillaor Yenming Chen Advior: Dr. Rober Scholz Communicaion Science Iniue Univeriy of Souhern California UWB Workhop, April 11-1, 6 Inroducion Moivaion The volage
More informationSignal and System (Chapter 3. Continuous-Time Systems)
Signal and Sysem (Chaper 3. Coninuous-Time Sysems) Prof. Kwang-Chun Ho kwangho@hansung.ac.kr Tel: 0-760-453 Fax:0-760-4435 1 Dep. Elecronics and Informaion Eng. 1 Nodes, Branches, Loops A nework wih b
More informationMon Apr 2: Laplace transform and initial value problems like we studied in Chapter 5
Mah 225-4 Week 2 April 2-6 coninue.-.3; alo cover par of.4-.5, EP 7.6 Mon Apr 2:.-.3 Laplace ranform and iniial value problem like we udied in Chaper 5 Announcemen: Warm-up Exercie: Recall, The Laplace
More informationGraphs III - Network Flow
Graph III - Nework Flow Flow nework eup graph G=(V,E) edge capaciy w(u,v) 0 - if edge doe no exi, hen w(u,v)=0 pecial verice: ource verex ; ink verex - no edge ino and no edge ou of Aume every verex v
More informationln 2 1 ln y x c y C x
Lecure 14 Appendi B: Some sample problems from Boas Here are some soluions o he sample problems assigned for Chaper 8 8: 6 Soluion: We wan o find he soluion o he following firs order equaion using separaion
More informationMTH Feburary 2012 Final term PAPER SOLVED TODAY s Paper
MTH401 7 Feburary 01 Final erm PAPER SOLVED TODAY s Paper Toal Quesion: 5 Mcqz: 40 Subjecive quesion: 1 4 q of 5 marks 4 q of 3 marks 4 q of marks Guidelines: Prepare his file as I included all pas papers
More informationChallenge Problems. DIS 203 and 210. March 6, (e 2) k. k(k + 2). k=1. f(x) = k(k + 2) = 1 x k
Challenge Problems DIS 03 and 0 March 6, 05 Choose one of he following problems, and work on i in your group. Your goal is o convince me ha your answer is correc. Even if your answer isn compleely correc,
More informationMA Study Guide #1
MA 66 Su Guide #1 (1) Special Tpes of Firs Order Equaions I. Firs Order Linear Equaion (FOL): + p() = g() Soluion : = 1 µ() [ ] µ()g() + C, where µ() = e p() II. Separable Equaion (SEP): dx = h(x) g()
More information1 1 + x 2 dx. tan 1 (2) = ] ] x 3. Solution: Recall that the given integral is improper because. x 3. 1 x 3. dx = lim dx.
. Use Simpson s rule wih n 4 o esimae an () +. Soluion: Since we are using 4 seps, 4 Thus we have [ ( ) f() + 4f + f() + 4f 3 [ + 4 4 6 5 + + 4 4 3 + ] 5 [ + 6 6 5 + + 6 3 + ]. 5. Our funcion is f() +.
More information1 Motivation and Basic Definitions
CSCE : Deign and Analyi of Algorihm Noe on Max Flow Fall 20 (Baed on he preenaion in Chaper 26 of Inroducion o Algorihm, 3rd Ed. by Cormen, Leieron, Rive and Sein.) Moivaion and Baic Definiion Conider
More informationProblem Set #1. i z. the complex propagation constant. For the characteristic impedance:
Problem Se # Problem : a) Using phasor noaion, calculae he volage and curren waves on a ransmission line by solving he wave equaion Assume ha R, L,, G are all non-zero and independen of frequency From
More informationChapter 8 The Complete Response of RL and RC Circuits
Chaper 8 he Complee Response of R and RC Ciruis Exerises Ex 8.3-1 Before he swih loses: Afer he swih loses: 2 = = 8 Ω so = 8 0.05 = 0.4 s. 0.25 herefore R ( ) Finally, 2.5 ( ) = o + ( (0) o ) = 2 + V for
More information