Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: Ph:

Size: px
Start display at page:

Download "Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: Ph:"

Transcription

1 Serial :. PT_EE_A+C_Control Sytem_798 Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubanewar olkata Patna Web: Ph: CLASS TEST 8-9 ELECTRICAL ENGINEERING Subject : Control Sytem Date of tet : 7/9/8 Anwer ey. (d) 7. (b) 3. (a) 9. (d) 5. (c). (c) 8. (b) 4. (b). (c) 6. (a) 3. (d) 9. (b) 5. (c). (a) 7. (d) 4. (c). (b) 6. (b). (c) 8. (d) 5. (b). (a) 7. (b) 3. (b) 9. (c) 6. (b). (b) 8. (b) 4. (c) 3. (b)

2 8 Electrical Engineering Detailed Explanation. (d) From log -magnitude plot, corner frequency at log ω ; or ω.; Hence pole at ω., and gain; log G, G, Hence tranfer fuction TF ( + /.). (c) +. T D the delay time, i not a deirable feature. It i the time taken by the ytem before tarting to repond. Hence. T S /T D cannot be le than a the ytem i untable in thi cae. So the ratio T S /T D hould be made a large a poible to make the ytem controllable.therefore, the anwer at (c) (greater than ) i appropriate. 3. (d) ( j4) ( + 3 j4) ( + 3) (j4) ; ; ω n 5 ; ω n 5 rad/ec; ζω n 6 ζ (c) y(t) AM in (t + φ) Where, A, & M jω + 5. (b) At ω, M and φ ω tan tan (/ ) 45 π/4 y(t) in( t π/4) Here, A Now, [I A] Now, characteritic equation i I A +

3 CT8 EE Control Sytem 9 or, ( + ) + or, + + Now, on comparing with tandard form we get, ω n (natural frequency) rad/ec and ξ ω n ξ ω ξ.5 Here ξ and ω n are repectively.5 and rad/ec. n 6. (b) From Bode plot of the given ytem, we can ee that magnitude plot croe db axi before the phae plot croe the 8 axi. Hence, gain cro over frequency i le than phae cro over frequency, i.e. ω gc < ω pc For ω gc < ω pc, obtained gain margin i poitive and phae margin i alo poitive. Therefore, the ytem i table. 7. (b) C V i () R R V () Thi i the circuit of phae lead compenator and it tranfer function i given a: V () V () i α ( + T) ( +αt)...(i) Where α R R + R, (α < ) R Here, α R + R + 3 and T time contant R C For given tranfer function maximum phae i given by φ m φ m in in α +α 3 in in (/) φ m 3

4 Electrical Engineering 8. (b) The open loop tranfer function i G()H() ( + 6) ( + )( + 4) So, we have the open loop pole and zero a Pole: and 4 (P ) Zero: 6 (Z ) Therefore, the number of aymptote i P Z So, the characteritic () i correct. Now, we have the characteritic equation for the ytem given by + G() H() or, ( + )( + 4)+ ( + 6) or, + (6 + ) For the characteritic equation, we form the Routh array a Root locu i plotted for to. i.e. >. Here, for > root locu doe not interect jω axi becaue row will not be zero. Thu, characteritic () i incorrect. For the given ytem, we have two pole and one zero. So, one imaginary zero lie at infinite. Therefore, the characteritic (4) i incorrect. Hence, (b) mut be a correct option. But, we check further for characteritic (3) a follow. We ketch the root locu for the given ytem a hown below. Img 6 4 Re It ha two real axi interection. So, characteritic (3) i correct. 9. (b) The overall tranfer function i, C() R () T e + + e Now, e T T [we can approximate] Hence, the characteritic equation i, + + ( T) or, + ( T) + The Routh array i a hown below. T

5 CT8 EE Control Sytem T For tability, T > Hence, T < and T (time) hould be greater than. Therefore, < T <.. (b) For phae lead compenator, zero i nearer to origin than pole. Hence tranfer function G c () α ( + T) ( + αt) αt α < T. (a) G() T ( + T ) TF T + T ( + T ) ( + T ) T + + ( T ) T TT + + TT + + T TT ω n TT ; ξ T T for ξ <<, T << T. (b) From the equation, we have [A] [C] [ ] Check for Controllability: [AB] [B AB] ; [B] whoe determinant i zero. Hence the ytem i not controllable.

6 Electrical Engineering Tet for Obervability: [C] [ ] C T A T C T 4 Obervability O whoe determinant, Hence, obervable. T T T C A C 4 3. (a) Zero Pole: + j, j The tranfer function i G() ( + ) ( + j)( + + j) where multiplication of vector length drawn from all pole multiplication of vector length drawn from all zero G( j) Hence choice (a) i correct. 5 ( j+ ) ( + j) ( j+ j)( j+ + j) ( j)(+ j3) (b) Before the witch i cloed the TF, C () R () ( + a) ( + a) a + ( + a) ω n, ζω n a, Steady tate error i ζ a e R lim. ( ) + GH ( ) ( ) lim. + ( + a)

7 CT8 EE Control Sytem 3 After the witch i cloed, ( + a) lim. (+ a) + + a a lim + a + TF C () ( + a) + R () ( + a) + +. T T ( + a) + + ( a + ) + ω n, ζω n a + T, T ζ a + t a T + Damping factor i increaed by Steady tate error e T R lim. ( ) + GH ( ) ( ) lim + [( + a) + ] T lim + [( + a) + T] a + T a + Steady tate error will increae by T. Hence, both ζ and e are increaed. T lim + + ( + a) + a + T T 5. (c) A we know, The characteritic equation of the ytem become. I F (+ 3) Here, ω n and ξω n 3

8 4 Electrical Engineering or ξ.5.6 (over damped) 6. (b) Equivalent impedance circuit diagram i hown in figure. Z v t i () Z v t () Here, Z () R + + C 6 and Z () 5 R + + C + + R Given op-amp circuit i an inverting amplifier, therefore tranfer function i V () V () i 5 + Z() + Z() 5 + ( + ) ( + ) 7. (b) The characteritic equation i 3 + b ( + ) or 3 + b + ( + 3) + ( + ) The Routh array i b ( + 3) + ( + ) b + The ytem will have utained ocillation if row i zero i.e. b ( + 3) ( + ) b

9 CT8 EE Control Sytem 5 i.e. b The auxiliary equation of row i b + ( + ) 4 or (jω) (j) (becaue ω rad/ec)...(i) b Putting b + 4 +, in equation (i), we get or or Alo, b (b) Corner frequencie are, ω.5; ω ; ω 3 5 Change in magnitude in db Slope (Number of decade between two frequencie) (log.5 log ) log db Magnitude (in db) db 4.5 db (Magnitude at ω rad/ec) From initial line equation, log 4.5 or.63 Since initial line ha a lope of db/ec and tart from a point 4.5 db at ω rad/ec, the factor contributing i.63 Plot between ω.5 and ω i having a lope of db/dec. At ω.5 the lope ha changed from db/dec to db/dec and thi can only happen due to a factor in the numerator and i + (+.4 ).5 At ω, the lope ha changed from db/dec to + db/dec and i due to a factor in the numerator and i + (+. ) At ω 5, the lope ha changed from + db/dec to db/dec and i due to a factor in the denominator and i + (+.4 ) 5

10 6 Electrical Engineering The open-loop tranfer function i thu given by G().63( +.4 )( +. ) ( +.4 ).63( +.5)( + ) ( + 5) 9. (d) ( + T ) OLTF i given by G() d ( + T ) putting jω ( + jωt ) G(jω) d ω ( + jωt ) Magnitude, Gjω ( ) ( ) ( T ) + ωt d ω + ω Phae angle G(jω) 8 + tan (ωt d ) tan (ωt ). Gj ( ω ) 8...(a) ω Gj ( ω ) 8...(b) ω Since at ω, G(jω) 9, which i poible only if tan (ωt d ) tan (ωt ) > or, tan (ωt d ) > tan (ωt ) T d > T. (c) Forward path tranfer function of the given ytem i: G() ( +α )( + ) ( ) 3 ; and α > So, the characteritic equation of the ytem i obtained a + G() H() or, ( +α )( + ) 3 + ( ) or, 3 + [ + (α + 3) + 3α]...() or, ( + + 9) + α ( + 3) α ( + 3) or, + ( + + 9) So, we have the open loop gain a α i varied given by G()H()...() ( ) α + 3 ( + + 9) Therefore, the number of aymptote are P Z 3...(3)

11 CT8 EE Control Sytem 7 So, two root branche will go to infinite along aymptote a α. Now, from equation (), we have: (9 + α) + 3α So, we form the Routh array a 3 9+ α 3α 9 + 7α 3α For α >, row can not be zero. Hence, root locu doe not interect jω axi for α >. Only option (c) atifie thee condition.. (a) The open loop tranfer function i G() ( + )( + ) (a) Finite pole are at,,, (P 3) Finite zero are nil (z ) (b) Number of aymptote P Z 3 The centroid σ (or the meeting point of the aympote) i at σ 3 The angle of the aymptote are given by θ ( + ) π,,..., ( P Z ) P Z ( + ) π,,, 6, 8, 3 ( 6 ) 3 (c) The breakaway point are given by d d d or [ ( )( )] d d d + + or ( ) or 6± ,.577 (invalid) Thu,.4 i only valid break away point (d) The number of branche of the root loci i the greater of P and Z, viz, 3. Uing all the above information, we plot the root loci, jω -plane 6.4 σ Hence choice (a) i correct.

12 8 Electrical Engineering. (c) Approximated root locu for the given tranfer function Im x 4 x Re Interection with imaginary axi: characteritic equation or 48 Auxiliary equation 6 + or ω 8j ±.8j 3. (b) G() 4 ( + 8) e due to parabolic input A/ a where a 4 lim G () H () lim ( + 8) 4 8 e 3 4 / (c) Calculating ω pc 8 9 tan (. ω ) pc tan (. ω ) pc 9 or, (. ω pc ) tan ω pc 3 or, ω pc.36 rad/ec.ω pc +.ω (..) ω pc pc GM log.36 ( + (. ω pc ) )( + (. ωpc ) )

13 CT8 EE Control Sytem 9.4 db. 5. (c) G(jω) 4 jω + + jω jω+ + ( jω) 4 jω ω jω 4 + jω jω Separating real and imaginary part, we get G(jω) ( ω ) 4 j ω...(i) Gjω ( ) ( ω ) 4+ 4ω 4 4 ω ω ω 4 ω 4 ω a, ω and from equation (i) a, ω Gjω ( ) G(jω) tan ω ω G(jω) 8 tan (a) All finite pole ha poitive real part mean G() H() ( a)( b)( c) at ω ω (d) ẋ x + u ẋ x + u x + x u x + u + x y [ ] x x Obervability T T T Q o C : A C Q o Unobervable Alo I A ( ) ( )

14 Electrical Engineering Pole lie on RHS of -plane hence untable. 8. (d) Characteritic matrix I + G()H() ( + 3) ( + ) T() G () + G ( ) H ( ) G() [ + G()H()] Here, T() ( + ) ( + ) ( + ) + 5+ ( + ) (c) Uing Routh table for openloop ytem The number of ign change in the firt column i. One pole located at RHS, hence untable. Now, for cloed loop ytem + G()H() + ( + )

15 CT8 EE Control Sytem Uing Routh table The number of ign change in the firt column of Routh table i zero Hence, the cloed loop ytem i table. 3. (b) d Impule repone [tep repone] dt Now, a we know at t ; c(t) Here, we can write c(t) ( e 3t + 3e t ) u(t) Now, t ; c(t) d Hence, h(t) [ ct ( )] dt h(t) δ(t) e 3t δ(t) + 3e 3t u(t) + 3e t δ(t) 3e t. u(t) δ(t) δ(t) + 3δ(t) + (3e 3t 3e t )u(t) h(t) 3δ(t) + (3e 3t 3e t ) u(t) H() Thu, Gain 3 3 3( ) + 3( + ) 3( + 3) ( + ) ( + 3)( + ) 3( ) 6 3( ) ( + 3)( + ) 3 + ( + ) 3

The state variable description of an LTI system is given by 3 1O. Statement for Linked Answer Questions 3 and 4 :

The state variable description of an LTI system is given by 3 1O. Statement for Linked Answer Questions 3 and 4 : CHAPTER 6 CONTROL SYSTEMS YEAR TO MARKS MCQ 6. The tate variable decription of an LTI ytem i given by Jxo N J a NJx N JN K O K OK O K O xo a x + u Kxo O K 3 a3 OKx O K 3 O L P L J PL P L P x N K O y _

More information

CONTROL SYSTEMS. Chapter 5 : Root Locus Diagram. GATE Objective & Numerical Type Solutions. The transfer function of a closed loop system is

CONTROL SYSTEMS. Chapter 5 : Root Locus Diagram. GATE Objective & Numerical Type Solutions. The transfer function of a closed loop system is CONTROL SYSTEMS Chapter 5 : Root Locu Diagram GATE Objective & Numerical Type Solution Quetion 1 [Work Book] [GATE EC 199 IISc-Bangalore : Mark] The tranfer function of a cloed loop ytem i T () where i

More information

Root Locus Diagram. Root loci: The portion of root locus when k assume positive values: that is 0

Root Locus Diagram. Root loci: The portion of root locus when k assume positive values: that is 0 Objective Root Locu Diagram Upon completion of thi chapter you will be able to: Plot the Root Locu for a given Tranfer Function by varying gain of the ytem, Analye the tability of the ytem from the root

More information

MODERN CONTROL SYSTEMS

MODERN CONTROL SYSTEMS MODERN CONTROL SYSTEMS Lecture 1 Root Locu Emam Fathy Department of Electrical and Control Engineering email: emfmz@aat.edu http://www.aat.edu/cv.php?dip_unit=346&er=68525 1 Introduction What i root locu?

More information

ME2142/ME2142E Feedback Control Systems

ME2142/ME2142E Feedback Control Systems Root Locu Analyi Root Locu Analyi Conider the cloed-loop ytem R + E - G C B H The tranient repone, and tability, of the cloed-loop ytem i determined by the value of the root of the characteritic equation

More information

Chapter 7. Root Locus Analysis

Chapter 7. Root Locus Analysis Chapter 7 Root Locu Analyi jw + KGH ( ) GH ( ) - K 0 z O 4 p 2 p 3 p Root Locu Analyi The root of the cloed-loop characteritic equation define the ytem characteritic repone. Their location in the complex

More information

Lecture 12: Examples of Root Locus Plots. Dr. Kalyana Veluvolu. Lecture 12: Examples of Root Locus Plots Dr. Kalyana Veluvolu

Lecture 12: Examples of Root Locus Plots. Dr. Kalyana Veluvolu. Lecture 12: Examples of Root Locus Plots Dr. Kalyana Veluvolu ROOT-LOCUS ANALYSIS Example: Given that G( ) ( + )( + ) Dr. alyana Veluvolu Sketch the root locu of 1 + G() and compute the value of that will yield a dominant econd order behavior with a damping ratio,

More information

Function and Impulse Response

Function and Impulse Response Tranfer Function and Impule Repone Solution of Selected Unolved Example. Tranfer Function Q.8 Solution : The -domain network i hown in the Fig... Applying VL to the two loop, R R R I () I () L I () L V()

More information

Lecture 4. Chapter 11 Nise. Controller Design via Frequency Response. G. Hovland 2004

Lecture 4. Chapter 11 Nise. Controller Design via Frequency Response. G. Hovland 2004 METR4200 Advanced Control Lecture 4 Chapter Nie Controller Deign via Frequency Repone G. Hovland 2004 Deign Goal Tranient repone via imple gain adjutment Cacade compenator to improve teady-tate error Cacade

More information

Analysis of Stability &

Analysis of Stability & INC 34 Feedback Control Sytem Analyi of Stability & Steady-State Error S Wonga arawan.won@kmutt.ac.th Summary from previou cla Firt-order & econd order ytem repone τ ωn ζω ω n n.8.6.4. ζ ζ. ζ.5 ζ ζ.5 ct.8.6.4...4.6.8..4.6.8

More information

SKEE 3143 CONTROL SYSTEM DESIGN. CHAPTER 3 Compensator Design Using the Bode Plot

SKEE 3143 CONTROL SYSTEM DESIGN. CHAPTER 3 Compensator Design Using the Bode Plot SKEE 3143 CONTROL SYSTEM DESIGN CHAPTER 3 Compenator Deign Uing the Bode Plot 1 Chapter Outline 3.1 Introduc4on Re- viit to Frequency Repone, ploang frequency repone, bode plot tability analyi. 3.2 Gain

More information

Root Locus Contents. Root locus, sketching algorithm. Root locus, examples. Root locus, proofs. Root locus, control examples

Root Locus Contents. Root locus, sketching algorithm. Root locus, examples. Root locus, proofs. Root locus, control examples Root Locu Content Root locu, ketching algorithm Root locu, example Root locu, proof Root locu, control example Root locu, influence of zero and pole Root locu, lead lag controller deign 9 Spring ME45 -

More information

EE Control Systems LECTURE 14

EE Control Systems LECTURE 14 Updated: Tueday, March 3, 999 EE 434 - Control Sytem LECTURE 4 Copyright FL Lewi 999 All right reerved ROOT LOCUS DESIGN TECHNIQUE Suppoe the cloed-loop tranfer function depend on a deign parameter k We

More information

The Root Locus Method

The Root Locus Method The Root Locu Method MEM 355 Performance Enhancement of Dynamical Sytem Harry G. Kwatny Department of Mechanical Engineering & Mechanic Drexel Univerity Outline The root locu method wa introduced by Evan

More information

Feedback Control Systems (FCS)

Feedback Control Systems (FCS) Feedback Control Sytem (FCS) Lecture19-20 Routh-Herwitz Stability Criterion Dr. Imtiaz Huain email: imtiaz.huain@faculty.muet.edu.pk URL :http://imtiazhuainkalwar.weebly.com/ Stability of Higher Order

More information

Control Systems Analysis and Design by the Root-Locus Method

Control Systems Analysis and Design by the Root-Locus Method 6 Control Sytem Analyi and Deign by the Root-Locu Method 6 1 INTRODUCTION The baic characteritic of the tranient repone of a cloed-loop ytem i cloely related to the location of the cloed-loop pole. If

More information

MEM 355 Performance Enhancement of Dynamical Systems Root Locus Analysis

MEM 355 Performance Enhancement of Dynamical Systems Root Locus Analysis MEM 355 Performance Enhancement of Dynamical Sytem Root Locu Analyi Harry G. Kwatny Department of Mechanical Engineering & Mechanic Drexel Univerity Outline The root locu method wa introduced by Evan in

More information

ME 375 FINAL EXAM Wednesday, May 6, 2009

ME 375 FINAL EXAM Wednesday, May 6, 2009 ME 375 FINAL EXAM Wedneday, May 6, 9 Diviion Meckl :3 / Adam :3 (circle one) Name_ Intruction () Thi i a cloed book examination, but you are allowed three ingle-ided 8.5 crib heet. A calculator i NOT allowed.

More information

Figure 1: Unity Feedback System

Figure 1: Unity Feedback System MEM 355 Sample Midterm Problem Stability 1 a) I the following ytem table? Solution: G() = Pole: -1, -2, -2, -1.5000 + 1.3229i, -1.5000-1.3229i 1 ( + 1)( 2 + 3 + 4)( + 2) 2 A you can ee, all pole are on

More information

Automatic Control Systems. Part III: Root Locus Technique

Automatic Control Systems. Part III: Root Locus Technique www.pdhcenter.com PDH Coure E40 www.pdhonline.org Automatic Control Sytem Part III: Root Locu Technique By Shih-Min Hu, Ph.D., P.E. Page of 30 www.pdhcenter.com PDH Coure E40 www.pdhonline.org VI. Root

More information

1 Routh Array: 15 points

1 Routh Array: 15 points EE C28 / ME34 Problem Set 3 Solution Fall 2 Routh Array: 5 point Conider the ytem below, with D() k(+), w(t), G() +2, and H y() 2 ++2 2(+). Find the cloed loop tranfer function Y () R(), and range of k

More information

EE/ME/AE324: Dynamical Systems. Chapter 8: Transfer Function Analysis

EE/ME/AE324: Dynamical Systems. Chapter 8: Transfer Function Analysis EE/ME/AE34: Dynamical Sytem Chapter 8: Tranfer Function Analyi The Sytem Tranfer Function Conider the ytem decribed by the nth-order I/O eqn.: ( n) ( n 1) ( m) y + a y + + a y = b u + + bu n 1 0 m 0 Taking

More information

Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: Ph:

Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web:     Ph: Serial : 0. LS_D_ECIN_Control Systems_30078 Delhi Noida Bhopal Hyderabad Jaipur Lucnow Indore Pune Bhubaneswar Kolata Patna Web: E-mail: info@madeeasy.in Ph: 0-4546 CLASS TEST 08-9 ELECTRONICS ENGINEERING

More information

Given the following circuit with unknown initial capacitor voltage v(0): X(s) Immediately, we know that the transfer function H(s) is

Given the following circuit with unknown initial capacitor voltage v(0): X(s) Immediately, we know that the transfer function H(s) is EE 4G Note: Chapter 6 Intructor: Cheung More about ZSR and ZIR. Finding unknown initial condition: Given the following circuit with unknown initial capacitor voltage v0: F v0/ / Input xt 0Ω Output yt -

More information

Department of Mechanical Engineering Massachusetts Institute of Technology Modeling, Dynamics and Control III Spring 2002

Department of Mechanical Engineering Massachusetts Institute of Technology Modeling, Dynamics and Control III Spring 2002 Department of Mechanical Engineering Maachuett Intitute of Technology 2.010 Modeling, Dynamic and Control III Spring 2002 SOLUTIONS: Problem Set # 10 Problem 1 Etimating tranfer function from Bode Plot.

More information

NAME (pinyin/italian)... MATRICULATION NUMBER... SIGNATURE

NAME (pinyin/italian)... MATRICULATION NUMBER... SIGNATURE POLITONG SHANGHAI BASIC AUTOMATIC CONTROL June Academic Year / Exam grade NAME (pinyin/italian)... MATRICULATION NUMBER... SIGNATURE Ue only thee page (including the bac) for anwer. Do not ue additional

More information

376 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD. D(s) = we get the compensated system with :

376 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD. D(s) = we get the compensated system with : 376 CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Therefore by applying the lead compenator with ome gain adjutment : D() =.12 4.5 +1 9 +1 we get the compenated ytem with : PM =65, ω c = 22 rad/ec, o

More information

Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: Ph:

Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web:     Ph: Serial : LS_B_EC_Network Theory_0098 CLASS TEST (GATE) Delhi Noida Bhopal Hyderabad Jaipur Lucknow ndore Pune Bhubanewar Kolkata Patna Web: E-mail: info@madeeay.in Ph: 0-4546 CLASS TEST 08-9 ELECTRONCS

More information

ME 375 FINAL EXAM SOLUTIONS Friday December 17, 2004

ME 375 FINAL EXAM SOLUTIONS Friday December 17, 2004 ME 375 FINAL EXAM SOLUTIONS Friday December 7, 004 Diviion Adam 0:30 / Yao :30 (circle one) Name Intruction () Thi i a cloed book eamination, but you are allowed three 8.5 crib heet. () You have two hour

More information

Lecture 5 Introduction to control

Lecture 5 Introduction to control Lecture 5 Introduction to control Tranfer function reviited (Laplace tranform notation: ~jω) () i the Laplace tranform of v(t). Some rule: ) Proportionality: ()/ in () 0log log() v (t) *v in (t) () * in

More information

Stability Criterion Routh Hurwitz

Stability Criterion Routh Hurwitz EES404 Fundamental of Control Sytem Stability Criterion Routh Hurwitz DR. Ir. Wahidin Wahab M.Sc. Ir. Arie Subiantoro M.Sc. Stability A ytem i table if for a finite input the output i imilarly finite A

More information

Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: Ph:

Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web:     Ph: Serial : LS_N_A_Network Theory_098 Delhi Noida Bhopal Hyderabad Jaipur Lucknow ndore Pune Bhubanewar Kolkata Patna Web: E-mail: info@madeeay.in Ph: 0-4546 CLASS TEST 08-9 NSTRUMENTATON ENGNEERNG Subject

More information

Homework 12 Solution - AME30315, Spring 2013

Homework 12 Solution - AME30315, Spring 2013 Homework 2 Solution - AME335, Spring 23 Problem :[2 pt] The Aerotech AGS 5 i a linear motor driven XY poitioning ytem (ee attached product heet). A friend of mine, through careful experimentation, identified

More information

ROOT LOCUS. Poles and Zeros

ROOT LOCUS. Poles and Zeros Automatic Control Sytem, 343 Deartment of Mechatronic Engineering, German Jordanian Univerity ROOT LOCUS The Root Locu i the ath of the root of the characteritic equation traced out in the - lane a a ytem

More information

Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: Ph:

Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web:     Ph: Serial : Ch1_EE_C_Power Electronic_6818 Delhi Noida Bhopal Hyderabad Jaipur ucknow Indore Pune Bhubanewar Kolkata Patna Web: E-mail: info@madeeay.in Ph: 11-451461 CASS ES 18-19 EECRICA ENGINEERING Subject

More information

March 18, 2014 Academic Year 2013/14

March 18, 2014 Academic Year 2013/14 POLITONG - SHANGHAI BASIC AUTOMATIC CONTROL Exam grade March 8, 4 Academic Year 3/4 NAME (Pinyin/Italian)... STUDENT ID Ue only thee page (including the back) for anwer. Do not ue additional heet. Ue of

More information

G(s) = 1 s by hand for! = 1, 2, 5, 10, 20, 50, and 100 rad/sec.

G(s) = 1 s by hand for! = 1, 2, 5, 10, 20, 50, and 100 rad/sec. 6003 where A = jg(j!)j ; = tan Im [G(j!)] Re [G(j!)] = \G(j!) 2. (a) Calculate the magnitude and phae of G() = + 0 by hand for! =, 2, 5, 0, 20, 50, and 00 rad/ec. (b) ketch the aymptote for G() according

More information

CISE302: Linear Control Systems

CISE302: Linear Control Systems Term 8 CISE: Linear Control Sytem Dr. Samir Al-Amer Chapter 7: Root locu CISE_ch 7 Al-Amer8 ١ Learning Objective Undertand the concept of root locu and it role in control ytem deign Be able to ketch root

More information

Control Systems. Root locus.

Control Systems. Root locus. Control Sytem Root locu chibum@eoultech.ac.kr Outline Concet of Root Locu Contructing root locu Control Sytem Root Locu Stability and tranient reone i cloely related with the location of ole in -lane How

More information

Lecture 6: Resonance II. Announcements

Lecture 6: Resonance II. Announcements EES 5 Spring 4, Lecture 6 Lecture 6: Reonance II EES 5 Spring 4, Lecture 6 Announcement The lab tart thi week You mut how up for lab to tay enrolled in the coure. The firt lab i available on the web ite,

More information

Digital Control System

Digital Control System Digital Control Sytem - A D D A Micro ADC DAC Proceor Correction Element Proce Clock Meaurement A: Analog D: Digital Continuou Controller and Digital Control Rt - c Plant yt Continuou Controller Digital

More information

Control Systems Engineering ( Chapter 7. Steady-State Errors ) Prof. Kwang-Chun Ho Tel: Fax:

Control Systems Engineering ( Chapter 7. Steady-State Errors ) Prof. Kwang-Chun Ho Tel: Fax: Control Sytem Engineering ( Chapter 7. Steady-State Error Prof. Kwang-Chun Ho kwangho@hanung.ac.kr Tel: 0-760-453 Fax:0-760-4435 Introduction In thi leon, you will learn the following : How to find the

More information

Control Systems. Root locus.

Control Systems. Root locus. Control Sytem Root locu chibum@eoultech.ac.kr Outline Concet of Root Locu Contructing root locu Control Sytem Root Locu Stability and tranient reone i cloely related with the location of ole in -lane How

More information

Lecture 10 Filtering: Applied Concepts

Lecture 10 Filtering: Applied Concepts Lecture Filtering: Applied Concept In the previou two lecture, you have learned about finite-impule-repone (FIR) and infinite-impule-repone (IIR) filter. In thee lecture, we introduced the concept of filtering

More information

ECE-320 Linear Control Systems. Spring 2014, Exam 1. No calculators or computers allowed, you may leave your answers as fractions.

ECE-320 Linear Control Systems. Spring 2014, Exam 1. No calculators or computers allowed, you may leave your answers as fractions. ECE-0 Linear Control Sytem Spring 04, Exam No calculator or computer allowed, you may leave your anwer a fraction. All problem are worth point unle noted otherwie. Total /00 Problem - refer to the unit

More information

into a discrete time function. Recall that the table of Laplace/z-transforms is constructed by (i) selecting to get

into a discrete time function. Recall that the table of Laplace/z-transforms is constructed by (i) selecting to get Lecture 25 Introduction to Some Matlab c2d Code in Relation to Sampled Sytem here are many way to convert a continuou time function, { h( t) ; t [0, )} into a dicrete time function { h ( k) ; k {0,,, }}

More information

EE105 - Fall 2005 Microelectronic Devices and Circuits

EE105 - Fall 2005 Microelectronic Devices and Circuits EE5 - Fall 5 Microelectronic Device and ircuit Lecture 9 Second-Order ircuit Amplifier Frequency Repone Announcement Homework 8 due tomorrow noon Lab 7 next week Reading: hapter.,.3. Lecture Material Lat

More information

Chapter 9: Controller design. Controller design. Controller design

Chapter 9: Controller design. Controller design. Controller design Chapter 9. Controller Deign 9.. Introduction 9.2. Eect o negative eedback on the network traner unction 9.2.. Feedback reduce the traner unction rom diturbance to the output 9.2.2. Feedback caue the traner

More information

MM1: Basic Concept (I): System and its Variables

MM1: Basic Concept (I): System and its Variables MM1: Baic Concept (I): Sytem and it Variable A ytem i a collection of component which are coordinated together to perform a function Sytem interact with their environment. The interaction i defined in

More information

Stability. ME 344/144L Prof. R.G. Longoria Dynamic Systems and Controls/Lab. Department of Mechanical Engineering The University of Texas at Austin

Stability. ME 344/144L Prof. R.G. Longoria Dynamic Systems and Controls/Lab. Department of Mechanical Engineering The University of Texas at Austin Stability The tability of a ytem refer to it ability or tendency to eek a condition of tatic equilibrium after it ha been diturbed. If given a mall perturbation from the equilibrium, it i table if it return.

More information

Question 1 Equivalent Circuits

Question 1 Equivalent Circuits MAE 40 inear ircuit Fall 2007 Final Intruction ) Thi exam i open book You may ue whatever written material you chooe, including your cla note and textbook You may ue a hand calculator with no communication

More information

Chapter #4 EEE Automatic Control

Chapter #4 EEE Automatic Control Spring 008 EEE 00 Chapter #4 EEE 00 Automatic Control Root Locu Chapter 4 /4 Spring 008 EEE 00 Introduction Repone depend on ytem and controller parameter > Cloed loop pole location depend on ytem and

More information

ECE 3510 Root Locus Design Examples. PI To eliminate steady-state error (for constant inputs) & perfect rejection of constant disturbances

ECE 3510 Root Locus Design Examples. PI To eliminate steady-state error (for constant inputs) & perfect rejection of constant disturbances ECE 350 Root Locu Deign Example Recall the imple crude ervo from lab G( ) 0 6.64 53.78 σ = = 3 23.473 PI To eliminate teady-tate error (for contant input) & perfect reection of contant diturbance Note:

More information

Design of Digital Filters

Design of Digital Filters Deign of Digital Filter Paley-Wiener Theorem [ ] ( ) If h n i a caual energy ignal, then ln H e dω< B where B i a finite upper bound. One implication of the Paley-Wiener theorem i that a tranfer function

More information

Lecture 8. PID control. Industrial process control ( today) PID control. Insights about PID actions

Lecture 8. PID control. Industrial process control ( today) PID control. Insights about PID actions Lecture 8. PID control. The role of P, I, and D action 2. PID tuning Indutrial proce control (92... today) Feedback control i ued to improve the proce performance: tatic performance: for contant reference,

More information

Lecture #9 Continuous time filter

Lecture #9 Continuous time filter Lecture #9 Continuou time filter Oliver Faut December 5, 2006 Content Review. Motivation......................................... 2 2 Filter pecification 2 2. Low pa..........................................

More information

CONTROL SYSTEMS. Chapter 2 : Block Diagram & Signal Flow Graphs GATE Objective & Numerical Type Questions

CONTROL SYSTEMS. Chapter 2 : Block Diagram & Signal Flow Graphs GATE Objective & Numerical Type Questions ONTOL SYSTEMS hapter : Bloc Diagram & Signal Flow Graph GATE Objective & Numerical Type Quetion Quetion 6 [Practice Boo] [GATE E 994 IIT-Kharagpur : 5 Mar] educe the ignal flow graph hown in figure below,

More information

ECE Linear Circuit Analysis II

ECE Linear Circuit Analysis II ECE 202 - Linear Circuit Analyi II Final Exam Solution December 9, 2008 Solution Breaking F into partial fraction, F 2 9 9 + + 35 9 ft δt + [ + 35e 9t ]ut A 9 Hence 3 i the correct anwer. Solution 2 ft

More information

EE C128 / ME C134 Problem Set 1 Solution (Fall 2010) Wenjie Chen and Jansen Sheng, UC Berkeley

EE C128 / ME C134 Problem Set 1 Solution (Fall 2010) Wenjie Chen and Jansen Sheng, UC Berkeley EE C28 / ME C34 Problem Set Solution (Fall 200) Wenjie Chen and Janen Sheng, UC Berkeley. (0 pt) BIBO tability The ytem h(t) = co(t)u(t) i not BIBO table. What i the region of convergence for H()? A bounded

More information

Wolfgang Hofle. CERN CAS Darmstadt, October W. Hofle feedback systems

Wolfgang Hofle. CERN CAS Darmstadt, October W. Hofle feedback systems Wolfgang Hofle Wolfgang.Hofle@cern.ch CERN CAS Darmtadt, October 9 Feedback i a mechanim that influence a ytem by looping back an output to the input a concept which i found in abundance in nature and

More information

Chapter 13. Root Locus Introduction

Chapter 13. Root Locus Introduction Chapter 13 Root Locu 13.1 Introduction In the previou chapter we had a glimpe of controller deign iue through ome imple example. Obviouly when we have higher order ytem, uch imple deign technique will

More information

( 1) EE 313 Linear Signals & Systems (Fall 2018) Solution Set for Homework #10 on Laplace Transforms

( 1) EE 313 Linear Signals & Systems (Fall 2018) Solution Set for Homework #10 on Laplace Transforms EE 33 Linear Signal & Sytem (Fall 08) Solution Set for Homework #0 on Laplace Tranform By: Mr. Houhang Salimian & Prof. Brian L. Evan Problem. a) xt () = ut () ut ( ) From lecture Lut { ()} = and { } t

More information

EE Control Systems LECTURE 6

EE Control Systems LECTURE 6 Copyright FL Lewi 999 All right reerved EE - Control Sytem LECTURE 6 Updated: Sunday, February, 999 BLOCK DIAGRAM AND MASON'S FORMULA A linear time-invariant (LTI) ytem can be repreented in many way, including:

More information

CHE302 LECTURE V LAPLACE TRANSFORM AND TRANSFER FUNCTION. Professor Dae Ryook Yang

CHE302 LECTURE V LAPLACE TRANSFORM AND TRANSFER FUNCTION. Professor Dae Ryook Yang CHE3 ECTURE V APACE TRANSFORM AND TRANSFER FUNCTION Profeor Dae Ryook Yang Fall Dept. of Chemical and Biological Engineering Korea Univerity CHE3 Proce Dynamic and Control Korea Univerity 5- SOUTION OF

More information

ECE-202 FINAL December 13, 2016 CIRCLE YOUR DIVISION

ECE-202 FINAL December 13, 2016 CIRCLE YOUR DIVISION ECE-202 Final, Fall 16 1 ECE-202 FINAL December 13, 2016 Name: (Pleae print clearly.) Student Email: CIRCLE YOUR DIVISION DeCarlo- 8:30-9:30 Talavage-9:30-10:30 2021 2022 INSTRUCTIONS There are 35 multiple

More information

CONTROL SYSTEMS. Chapter 7 : Bode Plot. 40dB/dec 1.0. db/dec so resultant slope will be 20 db/dec and this is due to the factor s

CONTROL SYSTEMS. Chapter 7 : Bode Plot. 40dB/dec 1.0. db/dec so resultant slope will be 20 db/dec and this is due to the factor s CONTROL SYSTEMS Chapter 7 : Bode Plot GATE Objective & Numerical Type Solutio Quetio 6 [Practice Book] [GATE EE 999 IIT-Bombay : 5 Mark] The aymptotic Bode plot of the miimum phae ope-loop trafer fuctio

More information

Figure 1 Siemens PSSE Web Site

Figure 1 Siemens PSSE Web Site Stability Analyi of Dynamic Sytem. In the lat few lecture we have een how mall ignal Lalace domain model may be contructed of the dynamic erformance of ower ytem. The tability of uch ytem i a matter of

More information

Digital Control System

Digital Control System Digital Control Sytem Summary # he -tranform play an important role in digital control and dicrete ignal proceing. he -tranform i defined a F () f(k) k () A. Example Conider the following equence: f(k)

More information

Chapter 8. Root Locus Techniques

Chapter 8. Root Locus Techniques Chapter 8 Rt Lcu Technique Intrductin Sytem perfrmance and tability dt determined dby cled-lp l ple Typical cled-lp feedback cntrl ytem G Open-lp TF KG H Zer -, - Ple 0, -, -4 K 4 Lcatin f ple eaily fund

More information

Gain and Phase Margins Based Delay Dependent Stability Analysis of Two- Area LFC System with Communication Delays

Gain and Phase Margins Based Delay Dependent Stability Analysis of Two- Area LFC System with Communication Delays Gain and Phae Margin Baed Delay Dependent Stability Analyi of Two- Area LFC Sytem with Communication Delay Şahin Sönmez and Saffet Ayaun Department of Electrical Engineering, Niğde Ömer Halidemir Univerity,

More information

R. W. Erickson. Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder

R. W. Erickson. Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder R. W. Erickon Department of Electrical, Computer, and Energy Engineering Univerity of Colorado, Boulder ZOH: Sampled Data Sytem Example v T Sampler v* H Zero-order hold H v o e = 1 T 1 v *( ) = v( jkω

More information

MAE140 Linear Circuits Fall 2012 Final, December 13th

MAE140 Linear Circuits Fall 2012 Final, December 13th MAE40 Linear Circuit Fall 202 Final, December 3th Intruction. Thi exam i open book. You may ue whatever written material you chooe, including your cla note and textbook. You may ue a hand calculator with

More information

SIMON FRASER UNIVERSITY School of Engineering Science ENSC 320 Electric Circuits II. Solutions to Assignment 3 February 2005.

SIMON FRASER UNIVERSITY School of Engineering Science ENSC 320 Electric Circuits II. Solutions to Assignment 3 February 2005. SIMON FRASER UNIVERSITY School of Engineering Science ENSC 320 Electric Circuit II Solution to Aignment 3 February 2005. Initial Condition Source 0 V battery witch flip at t 0 find i 3 (t) Component value:

More information

Solutions. Digital Control Systems ( ) 120 minutes examination time + 15 minutes reading time at the beginning of the exam

Solutions. Digital Control Systems ( ) 120 minutes examination time + 15 minutes reading time at the beginning of the exam BSc - Sample Examination Digital Control Sytem (5-588-) Prof. L. Guzzella Solution Exam Duration: Number of Quetion: Rating: Permitted aid: minute examination time + 5 minute reading time at the beginning

More information

Chapter #4 EEE8013. Linear Controller Design and State Space Analysis. Design of control system in state space using Matlab

Chapter #4 EEE8013. Linear Controller Design and State Space Analysis. Design of control system in state space using Matlab EEE83 hapter #4 EEE83 Linear ontroller Deign and State Space nalyi Deign of control ytem in tate pace uing Matlab. ontrollabilty and Obervability.... State Feedback ontrol... 5 3. Linear Quadratic Regulator

More information

Linear System Fundamentals

Linear System Fundamentals Linear Sytem Fundamental MEM 355 Performance Enhancement of Dynamical Sytem Harry G. Kwatny Department of Mechanical Engineering & Mechanic Drexel Univerity Content Sytem Repreentation Stability Concept

More information

Homework 7 - Solutions

Homework 7 - Solutions Homework 7 - Solutions Note: This homework is worth a total of 48 points. 1. Compensators (9 points) For a unity feedback system given below, with G(s) = K s(s + 5)(s + 11) do the following: (c) Find the

More information

R. W. Erickson. Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder

R. W. Erickson. Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder R. W. Erickon Department of Electrical, Computer, and Energy Engineering Univerity of Colorado, Boulder Cloed-loop buck converter example: Section 9.5.4 In ECEN 5797, we ued the CCM mall ignal model to

More information

8.1.6 Quadratic pole response: resonance

8.1.6 Quadratic pole response: resonance 8.1.6 Quadratic pole response: resonance Example G(s)= v (s) v 1 (s) = 1 1+s L R + s LC L + Second-order denominator, of the form 1+a 1 s + a s v 1 (s) + C R Two-pole low-pass filter example v (s) with

More information

Module 4: Time Response of discrete time systems Lecture Note 1

Module 4: Time Response of discrete time systems Lecture Note 1 Digital Control Module 4 Lecture Module 4: ime Repone of dicrete time ytem Lecture Note ime Repone of dicrete time ytem Abolute tability i a baic requirement of all control ytem. Apart from that, good

More information

Lecture 8 - SISO Loop Design

Lecture 8 - SISO Loop Design Lecture 8 - SISO Loop Deign Deign approache, given pec Loophaping: in-band and out-of-band pec Fundamental deign limitation for the loop Gorinevky Control Engineering 8-1 Modern Control Theory Appy reult

More information

ROOT LOCUS. Consider the system. Root locus presents the poles of the closed-loop system when the gain K changes from 0 to. H(s) H ( s) = ( s)

ROOT LOCUS. Consider the system. Root locus presents the poles of the closed-loop system when the gain K changes from 0 to. H(s) H ( s) = ( s) C1 ROOT LOCUS Consider the system R(s) E(s) C(s) + K G(s) - H(s) C(s) R(s) = K G(s) 1 + K G(s) H(s) Root locus presents the poles of the closed-loop system when the gain K changes from 0 to 1+ K G ( s)

More information

6.302 Feedback Systems Recitation 6: Steady-State Errors Prof. Joel L. Dawson S -

6.302 Feedback Systems Recitation 6: Steady-State Errors Prof. Joel L. Dawson S - 6302 Feedback ytem Recitation 6: teadytate Error Prof Joel L Dawon A valid performance metric for any control ytem center around the final error when the ytem reache teadytate That i, after all initial

More information

Introduction to Laplace Transform Techniques in Circuit Analysis

Introduction to Laplace Transform Techniques in Circuit Analysis Unit 6 Introduction to Laplace Tranform Technique in Circuit Analyi In thi unit we conider the application of Laplace Tranform to circuit analyi. A relevant dicuion of the one-ided Laplace tranform i found

More information

5.5 Application of Frequency Response: Signal Filters

5.5 Application of Frequency Response: Signal Filters 44 Dynamic Sytem Second order lowpa filter having tranfer function H()=H ()H () u H () H () y Firt order lowpa filter Figure 5.5: Contruction of a econd order low-pa filter by combining two firt order

More information

ME 375 EXAM #1 Tuesday February 21, 2006

ME 375 EXAM #1 Tuesday February 21, 2006 ME 375 EXAM #1 Tueday February 1, 006 Diviion Adam 11:30 / Savran :30 (circle one) Name Intruction (1) Thi i a cloed book examination, but you are allowed one 8.5x11 crib heet. () You have one hour to

More information

Main Topics: The Past, H(s): Poles, zeros, s-plane, and stability; Decomposition of the complete response.

Main Topics: The Past, H(s): Poles, zeros, s-plane, and stability; Decomposition of the complete response. EE202 HOMEWORK PROBLEMS SPRING 18 TO THE STUDENT: ALWAYS CHECK THE ERRATA on the web. Quote for your Parent' Partie: 1. Only with nodal analyi i the ret of the emeter a poibility. Ray DeCarlo 2. (The need

More information

What lies between Δx E, which represents the steam valve, and ΔP M, which is the mechanical power into the synchronous machine?

What lies between Δx E, which represents the steam valve, and ΔP M, which is the mechanical power into the synchronous machine? A 2.0 Introduction In the lat et of note, we developed a model of the peed governing mechanim, which i given below: xˆ K ( Pˆ ˆ) E () In thee note, we want to extend thi model o that it relate the actual

More information

EE 4443/5329. LAB 3: Control of Industrial Systems. Simulation and Hardware Control (PID Design) The Inverted Pendulum. (ECP Systems-Model: 505)

EE 4443/5329. LAB 3: Control of Industrial Systems. Simulation and Hardware Control (PID Design) The Inverted Pendulum. (ECP Systems-Model: 505) EE 4443/5329 LAB 3: Control of Indutrial Sytem Simulation and Hardware Control (PID Deign) The Inverted Pendulum (ECP Sytem-Model: 505) Compiled by: Nitin Swamy Email: nwamy@lakehore.uta.edu Email: okuljaca@lakehore.uta.edu

More information

6.447 rad/sec and ln (% OS /100) tan Thus pc. the testing point is s 3.33 j5.519

6.447 rad/sec and ln (% OS /100) tan Thus pc. the testing point is s 3.33 j5.519 9. a. 3.33, n T ln(% OS /100) 2 2 ln (% OS /100) 0.517. Thu n 6.7 rad/ec and the teting point i 3.33 j5.519. b. Summation of angle including the compenating zero i -106.691, The compenator pole mut contribute

More information

EECS2200 Electric Circuits. RLC Circuit Natural and Step Responses

EECS2200 Electric Circuits. RLC Circuit Natural and Step Responses 5--4 EECS Electric Circuit Chapter 6 R Circuit Natural and Step Repone Objective Determine the repone form of the circuit Natural repone parallel R circuit Natural repone erie R circuit Step repone of

More information

( ) 2. 1) Bode plots/transfer functions. a. Draw magnitude and phase bode plots for the transfer function

( ) 2. 1) Bode plots/transfer functions. a. Draw magnitude and phase bode plots for the transfer function ECSE CP7 olution Spring 5 ) Bode plot/tranfer function a. Draw magnitude and phae bode plot for the tranfer function H( ). ( ) ( E4) In your magnitude plot, indicate correction at the pole and zero. Step

More information

Massachusetts Institute of Technology Dynamics and Control II

Massachusetts Institute of Technology Dynamics and Control II I E Maachuett Intitute of Technology Department of Mechanical Engineering 2.004 Dynamic and Control II Laboratory Seion 5: Elimination of Steady-State Error Uing Integral Control Action 1 Laboratory Objective:

More information

S_LOOP: SINGLE-LOOP FEEDBACK CONTROL SYSTEM ANALYSIS

S_LOOP: SINGLE-LOOP FEEDBACK CONTROL SYSTEM ANALYSIS S_LOOP: SINGLE-LOOP FEEDBACK CONTROL SYSTEM ANALYSIS by Michelle Gretzinger, Daniel Zyngier and Thoma Marlin INTRODUCTION One of the challenge to the engineer learning proce control i relating theoretical

More information

1 (s + 3)(s + 2)(s + a) G(s) = C(s) = K P + K I

1 (s + 3)(s + 2)(s + a) G(s) = C(s) = K P + K I MAE 43B Linear Control Prof. M. Krstic FINAL June 9, Problem. ( points) Consider a plant in feedback with the PI controller G(s) = (s + 3)(s + )(s + a) C(s) = K P + K I s. (a) (4 points) For a given constant

More information

Homework Assignment No. 3 - Solutions

Homework Assignment No. 3 - Solutions ECE 6440 Summer 2003 Page 1 Homework Aignment o. 3 Problem 1 (10 point) Aume an LPLL ha F() 1 and the PLL parameter are 0.8V/radian, K o 100 MHz/V, and the ocillation frequency, f oc 500MHz. Sketch the

More information

ECE382/ME482 Spring 2004 Homework 4 Solution November 14,

ECE382/ME482 Spring 2004 Homework 4 Solution November 14, ECE382/ME482 Spring 2004 Homework 4 Solution November 14, 2005 1 Solution to HW4 AP4.3 Intead of a contant or tep reference input, we are given, in thi problem, a more complicated reference path, r(t)

More information

CHAPTER # 9 ROOT LOCUS ANALYSES

CHAPTER # 9 ROOT LOCUS ANALYSES F K א CHAPTER # 9 ROOT LOCUS ANALYSES 1. Introduction The basic characteristic of the transient response of a closed-loop system is closely related to the location of the closed-loop poles. If the system

More information

Bogoliubov Transformation in Classical Mechanics

Bogoliubov Transformation in Classical Mechanics Bogoliubov Tranformation in Claical Mechanic Canonical Tranformation Suppoe we have a et of complex canonical variable, {a j }, and would like to conider another et of variable, {b }, b b ({a j }). How

More information

SIMON FRASER UNIVERSITY School of Engineering Science ENSC 320 Electric Circuits II. R 4 := 100 kohm

SIMON FRASER UNIVERSITY School of Engineering Science ENSC 320 Electric Circuits II. R 4 := 100 kohm SIMON FRASER UNIVERSITY School of Engineering Science ENSC 320 Electric Circuit II Solution to Aignment 3 February 2003. Cacaded Op Amp [DC&L, problem 4.29] An ideal op amp ha an output impedance of zero,

More information