ECE382/ME482 Spring 2005 Homework 6 Solution April 17, (s/2 + 1) s(2s + 1)[(s/8) 2 + (s/20) + 1]

Size: px
Start display at page:

Download "ECE382/ME482 Spring 2005 Homework 6 Solution April 17, (s/2 + 1) s(2s + 1)[(s/8) 2 + (s/20) + 1]"

Transcription

1 ECE382/ME482 Spring 25 Homework 6 Solution April 17, 25 1 Solution to HW6 P8.17 We are given a system with open loop transfer function G(s) = 4(s/2 + 1) s(2s + 1)[(s/8) 2 + (s/2) + 1] (1) and unity negative feedback. We are asked to determine (a) sketch the Bode diagram and (b) find the frequency ω 1 at which the magnitude is db and the frequency ω 2 where the phase is 18 degrees. Solution: (a) In order to sketch the Bode diagram we first rewrite the open loop transfer function as G(s) = = 4(1/2)(s + 2) s(2)(s + 1/2)(1/64)[s s/5 + 64] 64(s + 2) s(s + 1/2)[s s/5 + 64] and note that for the quadratic factor in the denominator ω n = 8 rad/s and ζ = (1/2)(16/5)(1/8) =.2. For the magnitude plot we thus have corner frequencies.5, 2, and 8. We start with magnitude 2log 1 4 (we must use the gain from (1) not (2) see the form of equation (8.26) in section 8.2 of Dorf and Bishop). The slope will be 2 db/decade for frequencies less than.5. 4 db/decade for.5 < ω < 2, 2 db/decade for 2 < ω < 8 and 4 db/ decade for frequencies above 8 rad/s. From Figure 8.1 of the text, we see that we should have a peak of about 8 db corresponding to the value ζ =.2. For the phase plot, we see that the phase angle for low frequencies will be 9 degrees because of the pole at the origin, while for high frequencies the phase angle will ten to 27 degrees because the transfer function has three more poles than zeros. Because the corner frequencies are so close together it would be extremely difficult to accurately draw the phase angle by hand. The Bode diagram generated by Matlab is shown in Figure 1. (b) From the Bode diagram generated by Matlab we determine that ω rad/s and ω 2 = The margin command >> margin(tf(64*[1 2],conv([1 1/2 ],[1 16/5 64]))) gives us the values exactly as shown in Figure 2. AP8.1 We are given a mechanical system in which a mass is suspended from a fixed surface by a spring and supported by a damper. We are also given a Bode diagram. We are asked to determine the spring, damper, and mass parameters k, b, and m, respectively. Solution: (2) (3) (4)

2 ECE382/ME482 Spring 25 Homework 6 Solution April 17, 25 2 Figure 1: Bode Diagram for P8.17a 1 Bode Diagram for P A force balance on the mass yields so the transfer function of the system is ky + bẏ + mÿ = r (5) G(s) = 1 ms 2 + bs + k. (6) From the diagram we obtain the following estimates: low frequency gain 26 db, corner frequency ω n = 3 rad/s, and difference between low frequency gain and peak gain 4 db. From Figure 8.1a of the text we estimate the value of ζ to be.3. Then G() = 1/k so 2log 1 k 26 db, yielding k 2 N/m. The natural frequency ω 2 n = k/m then yields m 2/9 2.2 kg. Finally, b = 2ω n ζ 1.8 Ns/m. DP8.1 We are given a system with unity negative feedback and forward path transfer function We are asked to G c (s)g(s) = K(s + 2) s 2 (s + 12). (7)

3 ECE382/ME482 Spring 25 Homework 6 Solution April 17, Figure 2: Bode Diagram with gain and phase margins for P8.17b Bode Diagram Gm = 9.33 db (at 7.7 rad/sec), Pm = 51.2 deg (at 1.59 rad/sec) (a,b) For K = 1, sketch the Bode diagrams for the open and closed loop systems, respectively. (c) Repeat (a,b) for K = 5. (d) Choose K so that M pω 2 and the bandwidth is maximized. (e) Find the steady state error for a ramp input. Solution: The closed loop transfer function is T(s) = K(s + 2) s s 2 + K(s + 2). (8) When K = 1, the poles obtained using the roots function of Matlab are , and.349±.48j. When K = 5, the poles are and ±2.822j. To find the corner frequencies we would factor s 2 + 2ζω n s + ωn 2 to find that the real part of a complex root is ζω n and the imaginary part is ±ω n 1 ζ 2, then solve for ω n and ζ.

4 ECE382/ME482 Spring 25 Homework 6 Solution April 17, 25 4 (a) For the open loop transfer function we have corner frequencies 2 and 12 rad/s. The phase will start at 18 degrees due to the double pole at the origin. There will be an increase of 9 degrees due to the zero at 2 rad/s over the range 2/1 to 1(2) rad/s and a decrease of 9 degrees due to the pole at 12 over the range 12/1 to 1(12) rad/s. It is clear that these ranges overlap so we expect the maximum phase to be less than 9 degrees. The magnitude will decrease by 4 db/decade for frequencies less than 2 rad/s, decrease by only 2 db/dec for frequencies between the corner frequencies, and decrease by 4 db/dec for frequencies above 12 rad/s. The Bode plot is shown in Figure 3. 5 Figure 3: Bode Diagram for DP8.1a Bode plot for DP8.1 Open Loop System with K= (b) The Bode plot for the closed loop system is shown in Figure 4. (c) The Bode plot for the open loop system with K = 5 is shown in Figure 5. The Bode plot for the closed loop system with K = 5 is shown in Figure 6. (d) Since M pω is given by M pω = 1 2ζ ζ <.77, (9) 1 ζ2,

5 ECE382/ME482 Spring 25 Homework 6 Solution April 17, Figure 4: Bode Diagram for DP8.1b Bode plot for DP8.1 Closed Loop System with K= constraining its value constrains the value of ζ. Solving (9) for ζ in terms of M pω yields ζ 4 ζ M 2 p ω = (1) or 1 ± 1 1 ζ = 2 where we will choose the appropriate square roots to make the damping coefficient real and positive. For M pω = 2 we obtain ζ =.9659 or Since, in fact, we are trying to obtain bounds on ζ, we test the value of M pω resulting for ζ = 1/2 which is between these and find that the resulting M pω = so we have determined that we may safely choose any value for the damping coefficient between.26 and.96. Now by doing long division of s 3 +12s 2 +Ks+2K by s 2 +2ζω n s+ω 2 n and constraining the remainder to be zero, I was able to obtain an expression for ω n in terms of ζ and K. However, it was not a very tractable expression so instead we will do a root locus plot to find a reasonable value of K and see whether that helps. M 2 pω (11)

6 ECE382/ME482 Spring 25 Homework 6 Solution April 17, Figure 5: Open Loop Bode Diagram for DP8.1c Bode plot for DP8.1 Open Loop System with K= Checking the values of the damping coefficient as we travel along the loci corresponding to the complex conjugate pair of roots we find that K = 9 corresponds roughly to ζ =.26 but that for the default root locus plot we never obtain ζ =.96. We enlarge the region shown by specifying a set of gains in the second argument to the rlocus command, rlocus(oltf,[.1:.1:1]) and then check further along the loci. We discover that the value of ζ decreases as we increase the gain further and that around K = 38 we again have ζ =.26. Since bandwidth should increase with K, we test values of K near the large end of the range to find the largest K that meets the constraint M pω < 2. I used the following command for K = [25:1:4]; [mag,phase,w]=bode(feedback(tf(k*[1 2],[1 12 ]),1)); if max(mag)> 2 disp([ K =,num2str(k-1)]) stop

7 ECE382/ME482 Spring 25 Homework 6 Solution April 17, Figure 6: Closed Loop Bode Diagram for DP8.1c Bode plot for DP8.1 Closed Loop System with K= end; end; to determine that K = 337 is the maximum allowable gain. (This is sloppy code. There is no Matlab command stop so an error is generated when the command is encountered. This stops the loop since we have the answer we need.) Finally, we determine the resulting system bandwidth from the Bode plot shown in Figure 7 to be 26.7 rad/s. (e) Applying the Final Value Theorem, we obtain ( ) 1 e ss = lim s (1 T(s)) (12) s 1 = lim s s s 2 ( s s 2 ) + K(s + 2) K(s + 2) s s 2 + K(s + 2) (13) s 2 (s + 12) = lim s s(s s 2 =. (14) + Ks + 2K) DP8.3 We are given a system having transfer function G(s) in the forward path and transfer

8 ECE382/ME482 Spring 25 Homework 6 Solution April 17, Figure 7: Closed Loop Bode Diagram for DP8.1d Bode Diagram for DP8.1 Closed Loop System with K= System: untitled1 : 26.7 : function H(s) in the (negative) feedback path where and G(s) = K s 2 + 2s + 2 (15) H(s) = 5 s + 5. (16) We are asked to (a) choose a value for K for which the system is stable and plot the Bode diagram, and (b) adjust the gain K to obtain 2log 1 M pω = 3 db, then (c) with this gain, determine the bandwidth ω B and the steady state error e ss for a step input. Solution: The closed loop transfer function is T(s) = K(s + 5) s 3 + 7s s + (1 + 5K) (17)

9 ECE382/ME482 Spring 25 Homework 6 Solution April 17, 25 9 (a) The corresponding Routh array is given in Table 1. From the fourth row we determine that we need K > 2 for stability. The third row simplifies to the constraint K < or K < 74/5 so the range of values of K for which the system is stable is 2 < K < (18) Table 1: Routh array for DP8.3a s 3 : 1 12 s 2 : K s 1 1 : 7 (1 + 5K 84) s : 1 + 5K We choose K = 1 and obtain the Bode plot shown in Figure 8. Figure 8: Closed Loop Bode Diagram for DP8.3a Bode plot for DP8.3 Closed Loop System with K= (b) To determine the value of K within the stable range that yields M pω = 3 db I used the following Matlab commands which identified the appropriate gain to be K = 4.13.

10 ECE382/ME482 Spring 25 Homework 6 Solution April 17, 25 1 for K = [1:.1:14]; [mag,phase,w]=bode(tf(k*[1 5],[ (1+5*K)])); if 2*log1(max(mag))> 3 disp([ K =,num2str(k-.1)]) stop end; end; (c) We obtain the Bode plot shown in Figure 9 with bandwidth 3 rad/s (obtained by clicking on magnitude curve and finding the frequency at which the magnitude is 3 db). 2 Figure 9: Closed Loop Bode Diagram for DP8.3b with K=4.13 Bode plot for DP8.3 Closed Loop System with K= Applying the final value theorem we obtain ( 1 e ss = lim s s s ) (1 T(s)) = 1 T() = K (19) so for K = 4.13, e ss =.3263.

11 ECE382/ME482 Spring 25 Homework 6 Solution April 17, MP8.5 We are given a closed loop transfer function T(s) = 1 s 2 + 6s + 1 (2) and asked to (a) determine ω r, M pω, and ω B from the Bode plot, (b) estimate ζ and ω n from the Bode plot, (c) find ω n and ζ analytically, then predict ω r and M pω from them and compare the results. In part (a), we are asked to use the logspace function to generate a frequency vector with log spacing and frequency range [.1,1] rad/s. Solution: Part of the purpose of this exercise is to familiarize the reader with the units used in the bode and logspace commands. logspace takes two arguments: the exponents of the power of 1 corresponding to the minimum and maximum frequencies for the vector to be generated. When used without return variables, bodegenerates plots of magnitude in db and phase in deg. However if return variables are used, it generates a vector of magnitudes, which are not in db, and a vector of phases in deg. (a) I used the Matlab script shown below to find the requested values M pω = 4.8 db, ω r = 8.7 rad/s, and bandwidth 15.2 rad/s. % % MP8_5.m solves part of problem MP8_5 from Dorf and Bishop 1th ed. % % 17 April 5 --sk % cltf = tf([1],[1 6 1]) % (a) closed loop Bode plot figure(5) w = logspace(-1,3,1); bode(cltf,w) grid title( Bode plot for MP8.5 Closed Loop System ) print -deps mp8_5a [mag,phase,w]=bode(cltf,w); smag = squeeze(mag); % for some reason mag is 1 x 1 x 1. make smag 1 x 1. [Mpw,loc] = max(smag); MpwdB = 2*log1(Mpw) wr = w(loc) [foo,loc]=min(abs(2*log1(smag)-(-3*ones(size(smag))))) wb = w(loc) The Bode plot is shown in Figure 1. (b) From (11) above (in DP8.1), using the value of M pω (not in db) obtained above we find that ζ =.3. We reject the option ζ =.95 resulting using the positive square root because the formula (8.37) is only valid for ζ <.77. Then inserting this value

12 ECE382/ME482 Spring 25 Homework 6 Solution April 17, Figure 1: Closed Loop Bode Diagram for MP8.5a Bode plot for MP8.5 Closed Loop System for ζ and the value determined earlier for for ω r into (8.36) and solving for ω n yields ω n = Here are the Matlab commands I used. % (b) invert (8.36) and (8.37) zetap = sqrt(1/2+1/2*(sqrt(1-1/mpw^2))) zetam = sqrt(1/2-1/2*(sqrt(1-1/mpw^2))) wn = wr/sqrt(1-2*zetam^2) %%% zetap >.77 so eqn used isn t valid (c) From the denominator of T(s) we see that ω n = 1 = 1 and ζ = 6/(2ω n ) =.3. This is quite good agreement. It s possible we would have gotten even better agreement if we had use more points in our log-spaced frequency vector. (The default is 5 points. I used 1.) MP8.9 We are asked to design a filter G(s) such that the magnitude of the frequency response satisfies the design specifications given in Table 2. We are also asked to maximize the peak magnitude close to 4 rad/s. Solution: This is rather an open-ended question. We are not told how complicated we

13 ECE382/ME482 Spring 25 Homework 6 Solution April 17, Table 2: Design Specs for magnitude of frequency response of filter of MP8.9 Frequency Range Magnitude Requirement ω < 1 G(jω) < db 1 < ω < 1 G(jω) db ω > 1 G(jω) < db are allowed to make the filter. We also are not told what additional criteria we should use to decide whether a design is good or not. Obviously, the more poles and zeros we use, the sharper the cutoffs we can obtain (at the cost of some ripple in the middle of the range). Keeping in mind that simplicity is desirable, we note that we will need at least one zero to increase the magnitude so that it goes through db at 1 rad/s and at least two poles to decrease it so that it goes back through db at 1 rad/s. I played around with the gain and zero and pole locations to get the filter whose Bode plot is shown in Figure 11. The transfer function of my filter is G(s) = 1(s +.9) s s (21)

14 ECE382/ME482 Spring 25 Homework 6 Solution April 17, Figure 11: Closed Loop Bode Diagram for MP8.9 Bode Diagram

ECE382/ME482 Spring 2005 Homework 7 Solution April 17, K(s + 0.2) s 2 (s + 2)(s + 5) G(s) =

ECE382/ME482 Spring 2005 Homework 7 Solution April 17, K(s + 0.2) s 2 (s + 2)(s + 5) G(s) = ECE382/ME482 Spring 25 Homework 7 Solution April 17, 25 1 Solution to HW7 AP9.5 We are given a system with open loop transfer function G(s) = K(s +.2) s 2 (s + 2)(s + 5) (1) and unity negative feedback.

More information

ECE382/ME482 Spring 2005 Homework 8 Solution December 11,

ECE382/ME482 Spring 2005 Homework 8 Solution December 11, ECE382/ME482 Spring 25 Homework 8 Solution December 11, 27 1 Solution to HW8 P1.21 We are given a system with open loop transfer function G(s) = K s(s/2 + 1)(s/6 + 1) and unity negative feedback. We are

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

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

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

More information

Homework Assignment 3

Homework Assignment 3 ECE382/ME482 Fall 2008 Homework 3 Solution October 20, 2008 1 Homework Assignment 3 Assigned September 30, 2008. Due in lecture October 7, 2008. Note that you must include all of your work to obtain full

More information

ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 2010/2011 CONTROL ENGINEERING SHEET 5 Lead-Lag Compensation Techniques

ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 2010/2011 CONTROL ENGINEERING SHEET 5 Lead-Lag Compensation Techniques CAIRO UNIVERSITY FACULTY OF ENGINEERING ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 00/0 CONTROL ENGINEERING SHEET 5 Lead-Lag Compensation Techniques [] For the following system, Design a compensator such

More information

MAE 143B - Homework 9

MAE 143B - Homework 9 MAE 143B - Homework 9 7.1 a) We have stable first-order poles at p 1 = 1 and p 2 = 1. For small values of ω, we recover the DC gain K = lim ω G(jω) = 1 1 = 2dB. Having this finite limit, our straight-line

More information

EE C128 / ME C134 Fall 2014 HW 8 - Solutions. HW 8 - Solutions

EE C128 / ME C134 Fall 2014 HW 8 - Solutions. HW 8 - Solutions EE C28 / ME C34 Fall 24 HW 8 - Solutions HW 8 - Solutions. Transient Response Design via Gain Adjustment For a transfer function G(s) = in negative feedback, find the gain to yield a 5% s(s+2)(s+85) overshoot

More information

Due Wednesday, February 6th EE/MFS 599 HW #5

Due Wednesday, February 6th EE/MFS 599 HW #5 Due Wednesday, February 6th EE/MFS 599 HW #5 You may use Matlab/Simulink wherever applicable. Consider the standard, unity-feedback closed loop control system shown below where G(s) = /[s q (s+)(s+9)]

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

Systems Analysis and Control

Systems Analysis and Control Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 21: Stability Margins and Closing the Loop Overview In this Lecture, you will learn: Closing the Loop Effect on Bode Plot Effect

More information

MAS107 Control Theory Exam Solutions 2008

MAS107 Control Theory Exam Solutions 2008 MAS07 CONTROL THEORY. HOVLAND: EXAM SOLUTION 2008 MAS07 Control Theory Exam Solutions 2008 Geir Hovland, Mechatronics Group, Grimstad, Norway June 30, 2008 C. Repeat question B, but plot the phase curve

More information

a. Closed-loop system; b. equivalent transfer function Then the CLTF () T is s the poles of () T are s from a contribution of a

a. Closed-loop system; b. equivalent transfer function Then the CLTF () T is s the poles of () T are s from a contribution of a Root Locus Simple definition Locus of points on the s- plane that represents the poles of a system as one or more parameter vary. RL and its relation to poles of a closed loop system RL and its relation

More information

1 (20 pts) Nyquist Exercise

1 (20 pts) Nyquist Exercise EE C128 / ME134 Problem Set 6 Solution Fall 2011 1 (20 pts) Nyquist Exercise Consider a close loop system with unity feedback. For each G(s), hand sketch the Nyquist diagram, determine Z = P N, algebraically

More information

INTRODUCTION TO DIGITAL CONTROL

INTRODUCTION TO DIGITAL CONTROL ECE4540/5540: Digital Control Systems INTRODUCTION TO DIGITAL CONTROL.: Introduction In ECE450/ECE550 Feedback Control Systems, welearnedhow to make an analog controller D(s) to control a linear-time-invariant

More information

Frequency Response Techniques

Frequency Response Techniques 4th Edition T E N Frequency Response Techniques SOLUTION TO CASE STUDY CHALLENGE Antenna Control: Stability Design and Transient Performance First find the forward transfer function, G(s). Pot: K 1 = 10

More information

Radar Dish. Armature controlled dc motor. Inside. θ r input. Outside. θ D output. θ m. Gearbox. Control Transmitter. Control. θ D.

Radar Dish. Armature controlled dc motor. Inside. θ r input. Outside. θ D output. θ m. Gearbox. Control Transmitter. Control. θ D. Radar Dish ME 304 CONTROL SYSTEMS Mechanical Engineering Department, Middle East Technical University Armature controlled dc motor Outside θ D output Inside θ r input r θ m Gearbox Control Transmitter

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 3F2: Systems and Control EXAMPLES PAPER 2 ROOT-LOCUS. Solutions

Module 3F2: Systems and Control EXAMPLES PAPER 2 ROOT-LOCUS. Solutions Cambridge University Engineering Dept. Third Year Module 3F: Systems and Control EXAMPLES PAPER ROOT-LOCUS Solutions. (a) For the system L(s) = (s + a)(s + b) (a, b both real) show that the root-locus

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

MAE 143B - Homework 7

MAE 143B - Homework 7 MAE 143B - Homework 7 6.7 Multiplying the first ODE by m u and subtracting the product of the second ODE with m s, we get m s m u (ẍ s ẍ i ) + m u b s (ẋ s ẋ u ) + m u k s (x s x u ) + m s b s (ẋ s ẋ u

More information

Active Control? Contact : Website : Teaching

Active Control? Contact : Website :   Teaching Active Control? Contact : bmokrani@ulb.ac.be Website : http://scmero.ulb.ac.be Teaching Active Control? Disturbances System Measurement Control Controler. Regulator.,,, Aims of an Active Control Disturbances

More information

ECE382/ME482 Spring 2005 Homework 1 Solution February 10,

ECE382/ME482 Spring 2005 Homework 1 Solution February 10, ECE382/ME482 Spring 25 Homework 1 Solution February 1, 25 1 Solution to HW1 P2.33 For the system shown in Figure P2.33 on p. 119 of the text, find T(s) = Y 2 (s)/r 1 (s). Determine a relationship that

More information

CHAPTER 7 : BODE PLOTS AND GAIN ADJUSTMENTS COMPENSATION

CHAPTER 7 : BODE PLOTS AND GAIN ADJUSTMENTS COMPENSATION CHAPTER 7 : BODE PLOTS AND GAIN ADJUSTMENTS COMPENSATION Objectives Students should be able to: Draw the bode plots for first order and second order system. Determine the stability through the bode plots.

More information

Boise State University Department of Electrical Engineering ECE461 Control Systems. Control System Design in the Frequency Domain

Boise State University Department of Electrical Engineering ECE461 Control Systems. Control System Design in the Frequency Domain Boise State University Department of Electrical Engineering ECE6 Control Systems Control System Design in the Frequency Domain Situation: Consider the following block diagram of a type- servomechanism:

More information

Outline. Classical Control. Lecture 5

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?

More information

Frequency (rad/s)

Frequency (rad/s) . The frequency response of the plant in a unity feedback control systems is shown in Figure. a) What is the static velocity error coefficient K v for the system? b) A lead compensator with a transfer

More information

Overview of Bode Plots Transfer function review Piece-wise linear approximations First-order terms Second-order terms (complex poles & zeros)

Overview of Bode Plots Transfer function review Piece-wise linear approximations First-order terms Second-order terms (complex poles & zeros) 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.

More information

16.30/31, Fall 2010 Recitation # 2

16.30/31, Fall 2010 Recitation # 2 16.30/31, Fall 2010 Recitation # 2 September 22, 2010 In this recitation, we will consider two problems from Chapter 8 of the Van de Vegte book. R + - E G c (s) G(s) C Figure 1: The standard block diagram

More information

Exercise 1 (A Non-minimum Phase System)

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

More information

ECE 486 Control Systems

ECE 486 Control Systems ECE 486 Control Systems Spring 208 Midterm #2 Information Issued: April 5, 208 Updated: April 8, 208 ˆ This document is an info sheet about the second exam of ECE 486, Spring 208. ˆ Please read the following

More information

PM diagram of the Transfer Function and its use in the Design of Controllers

PM diagram of the Transfer Function and its use in the Design of Controllers PM diagram of the Transfer Function and its use in the Design of Controllers Santiago Garrido, Luis Moreno Abstract This paper presents the graphical chromatic representation of the phase and the magnitude

More information

Exercise 1 (A Non-minimum Phase System)

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,

More information

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall 2007

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall 2007 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering.4 Dynamics and Control II Fall 7 Problem Set #9 Solution Posted: Sunday, Dec., 7. The.4 Tower system. The system parameters are

More information

Digital Control Systems

Digital Control Systems Digital Control Systems Lecture Summary #4 This summary discussed some graphical methods their use to determine the stability the stability margins of closed loop systems. A. Nyquist criterion Nyquist

More information

VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur

VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur 603 203. DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING SUBJECT QUESTION BANK : EC6405 CONTROL SYSTEM ENGINEERING SEM / YEAR: IV / II year

More information

Systems Analysis and Control

Systems Analysis and Control Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 23: Drawing The Nyquist Plot Overview In this Lecture, you will learn: Review of Nyquist Drawing the Nyquist Plot Using the

More information

KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING

KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING QUESTION BANK SUB.NAME : CONTROL SYSTEMS BRANCH : ECE YEAR : II SEMESTER: IV 1. What is control system? 2. Define open

More information

Engraving Machine Example

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

More information

Systems Analysis and Control

Systems Analysis and Control Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 23: Drawing The Nyquist Plot Overview In this Lecture, you will learn: Review of Nyquist Drawing the Nyquist Plot Using

More information

100 (s + 10) (s + 100) e 0.5s. s 100 (s + 10) (s + 100). G(s) =

100 (s + 10) (s + 100) e 0.5s. s 100 (s + 10) (s + 100). G(s) = 1 AME 3315; Spring 215; Midterm 2 Review (not graded) Problems: 9.3 9.8 9.9 9.12 except parts 5 and 6. 9.13 except parts 4 and 5 9.28 9.34 You are given the transfer function: G(s) = 1) Plot the bode plot

More information

K(s +2) s +20 K (s + 10)(s +1) 2. (c) KG(s) = K(s + 10)(s +1) (s + 100)(s +5) 3. Solution : (a) KG(s) = s +20 = K s s

K(s +2) s +20 K (s + 10)(s +1) 2. (c) KG(s) = K(s + 10)(s +1) (s + 100)(s +5) 3. Solution : (a) KG(s) = s +20 = K s s 321 16. Determine the range of K for which each of the following systems is stable by making a Bode plot for K = 1 and imagining the magnitude plot sliding up or down until instability results. Verify

More information

EECE 301 Signals & Systems Prof. Mark Fowler

EECE 301 Signals & Systems Prof. Mark Fowler EECE 301 Signals & Systems Prof. Mark Fowler C-T Systems: Bode Plots Note Set #36 1/14 What are Bode Plots? Bode Plot = Freq. Resp. plotted with H() in db on a log frequency axis. Its easy to use computers

More information

MAE 143B - Homework 9

MAE 143B - Homework 9 MAE 43B - Homework 9 7.2 2 2 3.8.6.4.2.2 9 8 2 2 3 a) G(s) = (s+)(s+).4.6.8.2.2.4.6.8. Polar plot; red for negative ; no encirclements of, a.s. under unit feedback... 2 2 3. 4 9 2 2 3 h) G(s) = s+ s(s+)..2.4.6.8.2.4

More information

Poles and Zeros and Transfer Functions

Poles and Zeros and Transfer Functions Poles and Zeros and Transfer Functions Transfer Function: Considerations: Factorization: A transfer function is defined as the ratio of the Laplace transform of the output to the input with all initial

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. Erickson Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder 8.1. Review of Bode plots Decibels Table 8.1. Expressing magnitudes in decibels G db = 0 log 10

More information

Problem Weight Score Total 100

Problem Weight Score Total 100 EE 350 EXAM IV 15 December 2010 Last Name (Print): First Name (Print): ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO Problem Weight Score 1 25 2 25 3 25 4 25 Total

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

EE3CL4: Introduction to Linear Control Systems

EE3CL4: Introduction to Linear Control Systems 1 / 30 EE3CL4: Introduction to Linear Control Systems Section 9: of and using Techniques McMaster University Winter 2017 2 / 30 Outline 1 2 3 4 / 30 domain analysis Analyze closed loop using open loop

More information

Software Engineering 3DX3. Slides 8: Root Locus Techniques

Software Engineering 3DX3. Slides 8: Root Locus Techniques Software Engineering 3DX3 Slides 8: Root Locus Techniques Dr. Ryan Leduc Department of Computing and Software McMaster University Material based on Control Systems Engineering by N. Nise. c 2006, 2007

More information

Professor Fearing EE C128 / ME C134 Problem Set 7 Solution Fall 2010 Jansen Sheng and Wenjie Chen, UC Berkeley

Professor Fearing EE C128 / ME C134 Problem Set 7 Solution Fall 2010 Jansen Sheng and Wenjie Chen, UC Berkeley Professor Fearing EE C8 / ME C34 Problem Set 7 Solution Fall Jansen Sheng and Wenjie Chen, UC Berkeley. 35 pts Lag compensation. For open loop plant Gs ss+5s+8 a Find compensator gain Ds k such that the

More information

APPLICATIONS FOR ROBOTICS

APPLICATIONS FOR ROBOTICS Version: 1 CONTROL APPLICATIONS FOR ROBOTICS TEX d: Feb. 17, 214 PREVIEW We show that the transfer function and conditions of stability for linear systems can be studied using Laplace transforms. Table

More information

Root Locus Methods. The root locus procedure

Root Locus Methods. The root locus procedure Root Locus Methods Design of a position control system using the root locus method Design of a phase lag compensator using the root locus method The root locus procedure To determine the value of the gain

More information

AMME3500: System Dynamics & Control

AMME3500: System Dynamics & Control Stefan B. Williams May, 211 AMME35: System Dynamics & Control Assignment 4 Note: This assignment contributes 15% towards your final mark. This assignment is due at 4pm on Monday, May 3 th during Week 13

More information

(b) A unity feedback system is characterized by the transfer function. Design a suitable compensator to meet the following specifications:

(b) A unity feedback system is characterized by the transfer function. Design a suitable compensator to meet the following specifications: 1. (a) The open loop transfer function of a unity feedback control system is given by G(S) = K/S(1+0.1S)(1+S) (i) Determine the value of K so that the resonance peak M r of the system is equal to 1.4.

More information

Stability of CL System

Stability of CL System Stability of CL System Consider an open loop stable system that becomes unstable with large gain: At the point of instability, K( j) G( j) = 1 0dB K( j) G( j) K( j) G( j) K( j) G( j) =± 180 o 180 o Closed

More information

Course roadmap. Step response for 2nd-order system. Step response for 2nd-order system

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

More information

Classify a transfer function to see which order or ramp it can follow and with which expected error.

Classify a transfer function to see which order or ramp it can follow and with which expected error. Dr. J. Tani, Prof. Dr. E. Frazzoli 5-059-00 Control Systems I (Autumn 208) Exercise Set 0 Topic: Specifications for Feedback Systems Discussion: 30.. 208 Learning objectives: The student can grizzi@ethz.ch,

More information

If you need more room, use the backs of the pages and indicate that you have done so.

If you need more room, use the backs of the pages and indicate that you have done so. EE 343 Exam II Ahmad F. Taha Spring 206 Your Name: Your Signature: Exam duration: hour and 30 minutes. This exam is closed book, closed notes, closed laptops, closed phones, closed tablets, closed pretty

More information

Frequency Response Analysis

Frequency Response Analysis Frequency Response Analysis Consider let the input be in the form Assume that the system is stable and the steady state response of the system to a sinusoidal inputdoes not depend on the initial conditions

More information

Control of Electromechanical Systems

Control of Electromechanical Systems Control of Electromechanical Systems November 3, 27 Exercise Consider the feedback control scheme of the motor speed ω in Fig., where the torque actuation includes a time constant τ A =. s and a disturbance

More information

Lecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types

Lecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types Lecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types Venkata Sonti Department of Mechanical Engineering Indian Institute of Science Bangalore, India, 562 This

More information

LINEAR CONTROL SYSTEMS. Ali Karimpour Associate Professor Ferdowsi University of Mashhad

LINEAR CONTROL SYSTEMS. Ali Karimpour Associate Professor Ferdowsi University of Mashhad LINEAR CONTROL SYSTEMS Ali Karimpour Associate Professor Ferdowsi University of Mashhad Controller design in the frequency domain Topics to be covered include: Lag controller design 2 Dr. Ali Karimpour

More information

Control Systems I. Lecture 9: The Nyquist condition

Control Systems I. Lecture 9: The Nyquist condition Control Systems I Lecture 9: The Nyquist condition adings: Guzzella, Chapter 9.4 6 Åstrom and Murray, Chapter 9.1 4 www.cds.caltech.edu/~murray/amwiki/index.php/first_edition Emilio Frazzoli Institute

More information

2.010 Fall 2000 Solution of Homework Assignment 8

2.010 Fall 2000 Solution of Homework Assignment 8 2.1 Fall 2 Solution of Homework Assignment 8 1. Root Locus Analysis of Hydraulic Servomechanism. The block diagram of the controlled hydraulic servomechanism is shown in Fig. 1 e r e error + i Σ C(s) P(s)

More information

EE C128 / ME C134 Fall 2014 HW 6.2 Solutions. HW 6.2 Solutions

EE C128 / ME C134 Fall 2014 HW 6.2 Solutions. HW 6.2 Solutions EE C28 / ME C34 Fall 24 HW 6.2 Solutions. PI Controller For the system G = K (s+)(s+3)(s+8) HW 6.2 Solutions in negative feedback operating at a damping ratio of., we are going to design a PI controller

More information

10ES-43 CONTROL SYSTEMS ( ECE A B&C Section) % of Portions covered Reference Cumulative Chapter. Topic to be covered. Part A

10ES-43 CONTROL SYSTEMS ( ECE A B&C Section) % of Portions covered Reference Cumulative Chapter. Topic to be covered. Part A 10ES-43 CONTROL SYSTEMS ( ECE A B&C Section) Faculty : Shreyus G & Prashanth V Chapter Title/ Class # Reference Literature Topic to be covered Part A No of Hours:52 % of Portions covered Reference Cumulative

More information

Controls Problems for Qualifying Exam - Spring 2014

Controls Problems for Qualifying Exam - Spring 2014 Controls Problems for Qualifying Exam - Spring 2014 Problem 1 Consider the system block diagram given in Figure 1. Find the overall transfer function T(s) = C(s)/R(s). Note that this transfer function

More information

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall K(s +1)(s +2) G(s) =.

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall K(s +1)(s +2) G(s) =. MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering. Dynamics and Control II Fall 7 Problem Set #7 Solution Posted: Friday, Nov., 7. Nise problem 5 from chapter 8, page 76. Answer:

More information

The Frequency-response Design Method

The Frequency-response Design Method Chapter 6 The Frequency-response Design Method Problems and Solutions for Section 6.. (a) Show that α 0 in Eq. (6.2) is given by α 0 = G(s) U 0ω = U 0 G( jω) s jω s= jω 2j and α 0 = G(s) U 0ω = U 0 G(jω)

More information

Today (10/23/01) Today. Reading Assignment: 6.3. Gain/phase margin lead/lag compensator Ref. 6.4, 6.7, 6.10

Today (10/23/01) Today. Reading Assignment: 6.3. Gain/phase margin lead/lag compensator Ref. 6.4, 6.7, 6.10 Today Today (10/23/01) Gain/phase margin lead/lag compensator Ref. 6.4, 6.7, 6.10 Reading Assignment: 6.3 Last Time In the last lecture, we discussed control design through shaping of the loop gain GK:

More information

INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad

INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad - 500 043 Electrical and Electronics Engineering TUTORIAL QUESTION BAN Course Name : CONTROL SYSTEMS Course Code : A502 Class : III

More information

Systems Analysis and Control

Systems Analysis and Control Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 24: Compensation in the Frequency Domain Overview In this Lecture, you will learn: Lead Compensators Performance Specs Altering

More information

EE 4343/ Control System Design Project LECTURE 10

EE 4343/ Control System Design Project LECTURE 10 Copyright S. Ikenaga 998 All rights reserved EE 4343/5329 - Control System Design Project LECTURE EE 4343/5329 Homepage EE 4343/5329 Course Outline Design of Phase-lead and Phase-lag compensators using

More information

16.31 Homework 2 Solution

16.31 Homework 2 Solution 16.31 Homework Solution Prof. S. R. Hall Issued: September, 6 Due: September 9, 6 Problem 1. (Dominant Pole Locations) [FPE 3.36 (a),(c),(d), page 161]. Consider the second order system ωn H(s) = (s/p

More information

] [ 200. ] 3 [ 10 4 s. [ ] s + 10 [ P = s [ 10 8 ] 3. s s (s 1)(s 2) series compensator ] 2. s command pre-filter [ 0.

] [ 200. ] 3 [ 10 4 s. [ ] s + 10 [ P = s [ 10 8 ] 3. s s (s 1)(s 2) series compensator ] 2. s command pre-filter [ 0. EEE480 Exam 2, Spring 204 A.A. Rodriguez Rules: Calculators permitted, One 8.5 sheet, closed notes/books, open minds GWC 352, 965-372 Problem (Analysis of a Feedback System) Consider the feedback system

More information

Response to a pure sinusoid

Response to a pure sinusoid Harvard University Division of Engineering and Applied Sciences ES 145/215 - INTRODUCTION TO SYSTEMS ANALYSIS WITH PHYSIOLOGICAL APPLICATIONS Fall Lecture 14: The Bode Plot Response to a pure sinusoid

More information

Control Systems I Lecture 10: System Specifications

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

More information

SECTION 5: ROOT LOCUS ANALYSIS

SECTION 5: ROOT LOCUS ANALYSIS SECTION 5: ROOT LOCUS ANALYSIS MAE 4421 Control of Aerospace & Mechanical Systems 2 Introduction Introduction 3 Consider a general feedback system: Closed loop transfer function is 1 is the forward path

More information

Dynamic circuits: Frequency domain analysis

Dynamic circuits: Frequency domain analysis Electronic Circuits 1 Dynamic circuits: Contents Free oscillation and natural frequency Transfer functions Frequency response Bode plots 1 System behaviour: overview 2 System behaviour : review solution

More information

MAE 143B - Homework 8 Solutions

MAE 143B - Homework 8 Solutions MAE 43B - Homework 8 Solutions P6.4 b) With this system, the root locus simply starts at the pole and ends at the zero. Sketches by hand and matlab are in Figure. In matlab, use zpk to build the system

More information

FREQUENCY-RESPONSE DESIGN

FREQUENCY-RESPONSE DESIGN ECE45/55: Feedback Control Systems. 9 FREQUENCY-RESPONSE DESIGN 9.: PD and lead compensation networks The frequency-response methods we have seen so far largely tell us about stability and stability margins

More information

Notes for ECE-320. Winter by R. Throne

Notes for ECE-320. Winter by R. Throne Notes for ECE-3 Winter 4-5 by R. Throne Contents Table of Laplace Transforms 5 Laplace Transform Review 6. Poles and Zeros.................................... 6. Proper and Strictly Proper Transfer Functions...................

More information

Bangladesh University of Engineering and Technology. EEE 402: Control System I Laboratory

Bangladesh University of Engineering and Technology. EEE 402: Control System I Laboratory Bangladesh University of Engineering and Technology Electrical and Electronic Engineering Department EEE 402: Control System I Laboratory Experiment No. 4 a) Effect of input waveform, loop gain, and system

More information

ECSE 4962 Control Systems Design. A Brief Tutorial on Control Design

ECSE 4962 Control Systems Design. A Brief Tutorial on Control Design ECSE 4962 Control Systems Design A Brief Tutorial on Control Design Instructor: Professor John T. Wen TA: Ben Potsaid http://www.cat.rpi.edu/~wen/ecse4962s04/ Don t Wait Until The Last Minute! You got

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 :. PT_EE_A+C_Control Sytem_798 Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubanewar olkata Patna Web: E-mail: info@madeeay.in Ph: -4546 CLASS TEST 8-9 ELECTRICAL ENGINEERING Subject

More information

Frequency methods for the analysis of feedback systems. Lecture 6. Loop analysis of feedback systems. Nyquist approach to study stability

Frequency methods for the analysis of feedback systems. Lecture 6. Loop analysis of feedback systems. Nyquist approach to study stability Lecture 6. Loop analysis of feedback systems 1. Motivation 2. Graphical representation of frequency response: Bode and Nyquist curves 3. Nyquist stability theorem 4. Stability margins Frequency methods

More information

PD, PI, PID Compensation. M. Sami Fadali Professor of Electrical Engineering University of Nevada

PD, PI, PID Compensation. M. Sami Fadali Professor of Electrical Engineering University of Nevada PD, PI, PID Compensation M. Sami Fadali Professor of Electrical Engineering University of Nevada 1 Outline PD compensation. PI compensation. PID compensation. 2 PD Control L= loop gain s cl = desired closed-loop

More information

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.

6.1 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)

More information

Second-order filters. EE 230 second-order filters 1

Second-order filters. EE 230 second-order filters 1 Second-order filters Second order filters: Have second order polynomials in the denominator of the transfer function, and can have zeroth-, first-, or second-order polynomials in the numerator. Use two

More information

CHAPTER 7 STEADY-STATE RESPONSE ANALYSES

CHAPTER 7 STEADY-STATE RESPONSE ANALYSES CHAPTER 7 STEADY-STATE RESPONSE ANALYSES 1. Introduction The steady state error is a measure of system accuracy. These errors arise from the nature of the inputs, system type and from nonlinearities of

More information

Some special cases

Some special cases Lecture Notes on Control Systems/D. Ghose/2012 87 11.3.1 Some special cases Routh table is easy to form in most cases, but there could be some cases when we need to do some extra work. Case 1: The first

More information

NADAR SARASWATHI COLLEGE OF ENGINEERING AND TECHNOLOGY Vadapudupatti, Theni

NADAR SARASWATHI COLLEGE OF ENGINEERING AND TECHNOLOGY Vadapudupatti, Theni NADAR SARASWATHI COLLEGE OF ENGINEERING AND TECHNOLOGY Vadapudupatti, Theni-625531 Question Bank for the Units I to V SE05 BR05 SU02 5 th Semester B.E. / B.Tech. Electrical & Electronics engineering IC6501

More information

Problem Set #7 Solutions Due: Friday June 1st, 2018 at 5 PM.

Problem Set #7 Solutions Due: Friday June 1st, 2018 at 5 PM. EE102B Spring 2018 Signal Processing and Linear Systems II Goldsmith Problem Set #7 Solutions Due: Friday June 1st, 2018 at 5 PM. 1. Laplace Transform Convergence (10 pts) Determine whether each of the

More information

SAMPLE SOLUTION TO EXAM in MAS501 Control Systems 2 Autumn 2015

SAMPLE SOLUTION TO EXAM in MAS501 Control Systems 2 Autumn 2015 FACULTY OF ENGINEERING AND SCIENCE SAMPLE SOLUTION TO EXAM in MAS501 Control Systems 2 Autumn 2015 Lecturer: Michael Ruderman Problem 1: Frequency-domain analysis and control design (15 pt) Given is a

More information

Controller Design using Root Locus

Controller Design using Root Locus Chapter 4 Controller Design using Root Locus 4. PD Control Root locus is a useful tool to design different types of controllers. Below, we will illustrate the design of proportional derivative controllers

More information

Radar Dish. Armature controlled dc motor. Inside. θ r input. Outside. θ D output. θ m. Gearbox. Control Transmitter. Control. θ D.

Radar Dish. Armature controlled dc motor. Inside. θ r input. Outside. θ D output. θ m. Gearbox. Control Transmitter. Control. θ D. Radar Dish ME 304 CONTROL SYSTEMS Mechanical Engineering Department, Middle East Technical University Armature controlled dc motor Outside θ D output Inside θ r input r θ m Gearbox Control Transmitter

More information

Frequency Response part 2 (I&N Chap 12)

Frequency Response part 2 (I&N Chap 12) Frequency Response part 2 (I&N Chap 12) Introduction & TFs Decibel Scale & Bode Plots Resonance Scaling Filter Networks Applications/Design Frequency response; based on slides by J. Yan Slide 3.1 Example

More information

Module 5: Design of Sampled Data Control Systems Lecture Note 8

Module 5: Design of Sampled Data Control Systems Lecture Note 8 Module 5: Design of Sampled Data Control Systems Lecture Note 8 Lag-lead Compensator When a single lead or lag compensator cannot guarantee the specified design criteria, a laglead compensator is used.

More information

Dr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Root Locus

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

More information