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2 EE 343 Spring 206 Exam II Page of. (20 total points) For the system shown in the above figure, assume that G(s) = (a) (2 points) What is the closed-loop transfer function (CLTF)? (s + 2) 2. G CLTF (s) = G(s) + G(s) = (s + 2) 2 + = s 2 + 4s + 5. (b) (2 points) What are the poles of the CLTF? Poles are: p,2 = 2 ± j. (c) (3 points) Find the settling time via the 5% criterion (t s = 3 ζω n ). ω n = 5,ζ = 2/ 5. Settling time is t s = 3 ζω n = 3 2 =.5sec (d) (3 points) Compute the steady-state error corresponding to a unit step input you can (should) use the provided SSE table. System is Type 0, then the SSE is e ss = + G(0) = 0.8. (e) (0 points) We want to design the system such that t s = sec and damping ration ζ = 3 2. Find the corresponding desired CLTF poles given these design objectives. Given that the settling time is second, we can find ζ d ω nd since t s = 3. ζω n Hence, ζ d ω dn = 3 = σ d. Also, we re given that the damping ratio is ζ d = 3 2 ω dn = 6/ 3. Therefore, the desired closed loop poles are generated from the desired CLTF denominator (s 2 + 6s + 2), and they are: p,2 d = 3 ± 3j.

3 EE 343 Spring 206 Exam II Page 2 of 2. (20 total points) For the system shown in the above figure, assume that: G(s) = (a) (5 points) Find the CLTF. s 3 + s 2 + 2s 0.5, C(s) = + K s, K 0. G CLTF (s) = C(s)G(s) + C(s)G(s) = s + K s 4 + s 3 + 2s s + K (b) (5 points) Using the Routh array criterion, find the range of K such that the system is stable. s 4 2 K s s 2.5 K s s K.5 Hence, the range of K s.t. the system is stable is 0 < K < K (c) (5 points) In terms of K, obtain the SSE of the above system given that the input is a unit ramp function. You ll have to figure out the System Type and use the SSE table (see last page of your exam booklet). The system is of Type, hence the steady state error for this system given a ramp input is e ss = lim s 0 sc(s)g(s) = 2K. Note that this expression only holds if 0 < K < (d) (5 points) By varying the value of K, what is the smallest absolute value of such tracking error one can achieve (this part is related to (c) above)?

4 EE 343 Spring 206 Exam II Page 3 of We want to be as small as possible. Hence, we want K to be as close to K as possible, but not 0.75 (since K = 0.75 would make the system marginally stable). Therefore, the smallest value would be e min ss = /(.5) = 0.66.

5 EE 343 Spring 206 Exam II Page 4 of 3. (25 total points) The characteristic polynomial (CP) of a system is given as follows: + KH(s)G(s) = + K (s 2 + 2s + 2)(s ). We want to sketch the root locus in this problem of the given CP. (a) (5 points) Find the part of the real axis on the root locus. How many branches does the root locus have? The poles are p,2,3 = ± j,+. The points on the real axis to the left of p 3 = belong to the root locus. (b) (5 points) Find the asymptotes of the root locus, including their angles (φ q ) and point of intersection (σ a ). n p n z = 3, hence we have 3 branches. The angles of asymptotes are: φ q = 60,80, 60deg and they all intersect at σ a = (c) (5 points) Find the break-in/breakaway points of the root locus. Setting the CP to zero, we obtain: K(s) = (s 3 + s 2 2) dk/ds = (3s 2 + 2s) = 0 s,2 = 2/3,0. Both of these points belong to the root locus: s is a break-in point and s 2 is a breakaway point. (d) (5 points) Find the angle of departure from the complex poles (you can only find one of these angles, since the RL is symmetric). You are also given that tan (/2) = 26.56deg. Angle of departure from p : φ p = ( ) = 63.4deg. Similarly, φ p2 = +63.4deg. (e) (5 points) Find the crossings with the jω axis, and sketch the root locus. Crossings occur at ω = 0 and K = 2. Root locus plot:

6 EE 343 Spring 206 Exam II Page 5 of 3 Root Locus 2 Imaginary Axis (seconds - ) Real Axis (seconds - )

7 EE 343 Spring 206 Exam II Page 6 of 4. (5 total points) For the above system, match the four given step-response plots given in the below figure with the four transfer functions given in the below table. Justify your answer in clear sentences. Lazy justifications get lazy credits. Transfer Function G(s) 0 (s + )(s + 2)(s + 3) 0 (s 2 + 2s + 2)(s + 3) (s )(s + ) 0 s 2 + 2s + 2 Step Response Plot #????????

8 EE 343 Spring 206 Exam II Page 7 of Your solution here: 0 : This transfer function corresponds to time response plot (s + )(s + 2)(s + 3) 3, as the poles are real and in the LHP, which signifies no overshoot and the step response approaches the final value. 0 (s 2 : This transfer function corresponds to time response plot + 2s + 2)(s + 3) 2, as the poles are complex and in the LHP, which signifies an overshoot and the step response approaches the final value. To determine the final value, we can use the final value theorem which gives y( ) = lim s 0 su(s)g(s) = lim s 0 G(s) = 0/6 =.6 which is shown in time response 2. (s )(s + ) : This system is unstable with a pole p = in the RHP. Hence the step-response will be unstable, which is only shown in time response plot. 0 s 2 : Similar to the solution of time-response 2, the solution for this + 2s + 2 transfer function corresponds to time response plot 4: y( ) = lim s 0 su(s)g(s) = lim s 0 G(s) = 0/2 = 5. Transfer Function G(s) 0 (s + )(s + 2)(s + 3) 0 (s 2 + 2s + 2)(s + 3) (s )(s + ) 0 s 2 + 2s + 2 Step Response Plot # 3 2 4

9 EE 343 Spring 206 Exam II Page 8 of 5. (20 total points) For the above system, consider that G(s) = K (s + 3)(s + ), K 0. (a) (5 points) Find the closed loop transfer function. G CLTF (s) = K s 2 + 4s + K + 3. (b) (5 points) What are ω n K? and ζ for this standard second order system in terms of ω n = K + 3,ζ = 2 K + 3. (c) (0 points) Find the value of K is the maximum overshoot is equal to 0.. The maximum overshoot is given by the following formula: M p = e ζ ζ 2 π, and ln(0.) π Your solution (K) can be approximate, but you have to clearly show all the steps involved. M p = e ζ ζ 2 π = 0. ζ ζ 0.6 = 0.6 K 8.5. ζ 2 K + 3

10 EE 343 Spring 206 Exam II Page 9 of 6. (20 total points) The characteristic polynomial of a closed-loop system is given by: ( + K)s 2 + (2 2K)s + 2K = 0. (a) (20 points) Plot the root locus. You should follow all the steps we discussed in class. A simple sketch will not be graded. You are also given that tan () = 45deg and tan (/3) = 8.43deg (0) Using the hint, we can write the given characteristic polynomial as a standard + KG(s) = 0: ( + K)s 2 + (2 2K)s + 2K = 0 s 2 + 2s + K(s 2 2s + 2) = 0 + K s2 2s + 2 s 2 + 2s = 0. Hence, the new G(s), is equal to G(s) = s2 2s + 2 s s. Poles: p,2 = 0, 2. Zeros: z,2 = ± j. n p = 2,n z = 2 2. Asymptotes: None, since n p n z 0 3. Breakaway points: dk ds = 0 (2s + 2)(s2 2s + 2) = 0 s,2,3 =, 0.62,.62 s 2 = 0.62 is the breakaway point 4. Angles of arrival at the complex zeros: φ z = arctan(/3) 90 = 53.43deg φ z2 = 53.43deg 5. jω axis crossing: K 0.95,ω ± 6. Plot:

11 EE 343 Spring 206 Exam II Page 0 of.5 Root Locus Imaginary Axis (seconds - ) Real Axis (seconds - )

12 EE 343 Spring 206 Exam II Page of Rule 6 Asymptotes angles: RL branches ending at OL zeros at approach the asymptotic lines with angles: ( + 2q)80 φ q = deg, q = 0,,2,...,n p n z n p n z Rule 7 Real-axis intercept of asymptotes: σ A = n p i= Re(p i) n z j= Re(z j) n p n z Rule 9 Angle of Departure (AoD): defined as the angle from a complex pole or Angle of Arrival (AoA) at a complex zero: AoD from a complex pole : φ p = 80 i AoA at a complex zero : φ z = 80 + i p i + z j, j p i z j j i p i is the sum of all angles of vectors to a complex pole in question from other poles, j z j is the sum of all angles of vectors to a complex pole in question from other zeros denotes the angle of a complex number Rule 0 Determine whether the RL crosses the imaginary y-axis by setting: + KG(s = jω)h(s = jω) = 0 + 0i and finding the ω and K that solves the above equation. The value of ω you get is the frequency at which the RL crosses the imaginary y-axis and the K you get is the associated gain for the controller. You should obtain two equations (real = 0 and imaginary = 0) with two unknowns (K,ω). From there, you solve for K,ω pairs

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MAE143 B - Linear Control - Spring 2018 Midterm, May 3rd MAE143 B - Linear Control - Spring 2018 Midterm, May 3rd Instructions: 1. This exam is open book. You can consult any printed or written material of your liking. 2. You have 70 minutes. 3. Most questions

EECS C128/ ME C134 Final Thu. May 14, pm. Closed book. One page, 2 sides of formula sheets. No calculators. Name: SID: EECS C28/ ME C34 Final Thu. May 4, 25 5-8 pm Closed book. One page, 2 sides of formula sheets. No calculators. There are 8 problems worth points total. Problem Points Score 4 2 4 3 6 4 8 5 3

GEORGIA INSTITUTE OF TECHNOLOGY SCHOOL of ELECTRICAL & COMPUTER ENGINEERING FINAL EXAM. COURSE: ECE 3084A (Prof. Michaels) GEORGIA INSTITUTE OF TECHNOLOGY SCHOOL of ELECTRICAL & COMPUTER ENGINEERING FINAL EXAM DATE: 09-Dec-13 COURSE: ECE 3084A (Prof. Michaels) NAME: STUDENT #: LAST, FIRST Write your name on the front page

DESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD 206 Spring Semester ELEC733 Digital Control System LECTURE 7: DESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD For a unit ramp input Tz Ez ( ) 2 ( z ) D( z) G( z) Tz e( ) lim( z) z 2 ( z ) D( z)

Course roadmap. Step response for 2nd-order system. Step response for 2nd-order system ME45: Control Systems Lecture Time response of nd-order systems Prof. Clar Radcliffe and Prof. Jongeun Choi Department of Mechanical Engineering Michigan State University Modeling Laplace transform Transfer

6.1 Sketch the z-domain root locus and find the critical gain for the following systems K., the closed-loop characteristic equation is K + z 0. 6. Sketch the z-domain root locus and find the critical gain for the following systems K (i) Gz () z 4. (ii) Gz K () ( z+ 9. )( z 9. ) (iii) Gz () Kz ( z. )( z ) (iv) Gz () Kz ( + 9. ) ( z. )( z 8. ) (i)

Dr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Root Locus Week Content Notes 1 Introduction 2 Frequency Domain Modelling 3 Transient Performance and the s-plane 4 Block Diagrams 5 Feedback System Characteristics Assign 1 Due 6 Root Locus 7 Root Locus 2 Assign

AN INTRODUCTION TO THE CONTROL THEORY Open-Loop controller An Open-Loop (OL) controller is characterized by no direct connection between the output of the system and its input; therefore external disturbance, non-linear dynamics and parameter

EE 3CL4: Introduction to Control Systems Lab 4: Lead Compensation EE 3CL4: Introduction to Control Systems Lab 4: Lead Compensation Tim Davidson Ext. 27352 davidson@mcmaster.ca Objective To use the root locus technique to design a lead compensator for a marginally-stable

Exercise 1 (A Non-minimum Phase System) Prof. Dr. E. Frazzoli 5-59- Control Systems I (Autumn 27) Solution Exercise Set 2 Loop Shaping clruch@ethz.ch, 8th December 27 Exercise (A Non-minimum Phase System) To decrease the rise time of the system,

Transient response via gain adjustment. Consider a unity feedback system, where G(s) = 2. The closed loop transfer function is. s 2 + 2ζωs + ω 2 n Design via frequency response Transient response via gain adjustment Consider a unity feedback system, where G(s) = ωn 2. The closed loop transfer function is s(s+2ζω n ) T(s) = ω 2 n s 2 + 2ζωs + ω 2

EE302 - Feedback Systems Spring Lecture KG(s)H(s) = KG(s) EE3 - Feedback Systems Spring 19 Lecturer: Asst. Prof. M. Mert Ankarali Lecture 1.. 1.1 Root Locus In control theory, root locus analysis is a graphical analysis method for investigating the change of

Control Systems I Lecture 10: System Specifications Control Systems I Lecture 10: System Specifications Readings: Guzzella, Chapter 10 Emilio Frazzoli Institute for Dynamic Systems and Control D-MAVT ETH Zürich November 24, 2017 E. Frazzoli (ETH) Lecture - - - - - - - - - - - - - - - - - - DISCLAIMER - - - - - - - - - - - - - - - - - - General Information: This midterm is a sample midterm. This means: The sample midterm contains problems that are of similar,

Final exam (practice) UCLA: Math 31B, Spring 2017 Instructor: Noah White Date: Final exam (practice) UCLA: Math 31B, Spring 2017 This exam has 8 questions, for a total of 80 points. Please print your working and answers neatly. Write your solutions in

Root Locus Design Example #3 Root Locus Design Example #3 A. Introduction The system represents a linear model for vertical motion of an underwater vehicle at zero forward speed. The vehicle is assumed to have zero pitch and roll

Outline. Classical Control. Lecture 5 Outline Outline Outline 1 What is 2 Outline What is Why use? Sketching a 1 What is Why use? Sketching a 2 Gain Controller Lead Compensation Lag Compensation What is Properties of a General System Why use?

EXAMPLE PROBLEMS AND SOLUTIONS Similarly, the program for the fourth-order transfer function approximation with T = 0.1 sec is [num,denl = pade(0.1, 4); printsys(num, den, 'st) numlden = sa4-2o0sa3 + 1 80O0sA2-840000~ + 16800000 sa4

Engraving Machine Example Engraving Machine Example MCE44 - Fall 8 Dr. Richter November 24, 28 Basic Design The X-axis of the engraving machine has the transfer function G(s) = s(s + )(s + 2) In this basic example, we use a proportional

Exercise 1 (A Non-minimum Phase System) Prof. Dr. E. Frazzoli 5-59- Control Systems I (HS 25) Solution Exercise Set Loop Shaping Noele Norris, 9th December 26 Exercise (A Non-minimum Phase System) To increase the rise time of the system, we Introduction to Feedback Control Control System Design Why Control? Open-Loop vs Closed-Loop (Feedback) Why Use Feedback Control? Closed-Loop Control System Structure Elements of a Feedback Control System Overview of Bode Plots Transfer function review Piece-wise linear approximations First-order terms Second-order terms (complex poles & zeros) J. McNames Portland State University ECE 222 Bode Plots Ver.