FREQUENCY-RESPONSE DESIGN

Size: px
Start display at page:

Download "FREQUENCY-RESPONSE DESIGN"

Transcription

1 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 of a closed-loop system based on open-loop response. Now, we look at frequency-response based design methods which primarily aim at improving stability margins. Also, given the relationship between PM and performance, we have some idea of transient response as well. Start thinking of Bode-magnitude and Bode-phase plots as LEGO to make the frequency response we want. PD compensation Compensator D(s) = K ( + TD s). We have seen from root-locus that this has a stabilizing effect. Magnitude and phase effect: K 9 K. T D T D. T D T D Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli T D T D

2 ECE45/ECE55, FREQUENCY-RESPONSE DESIGN 9 2 PD controller increases phase for frequencies over ω./td.so, locate /T D < crossover so that phase margin at crossover is better. Problem: Magnitude response continues to increase as frequency increases. This amplifies high-frequency sensor noise. Lead compensation [ ] Ts + Compensator D(s) = K αts +, α <. Approximate PD control for frequencies up to ω = αt. Magnitude and phase effect: K α φ max K T ω max αt. T The phase contributed at frequency ω is T ω max αt αt φ = tan (T ω) tan (αt ω). To find where the maximum phase contribution is, we take the derivative of φ with respect to ω and set it to zero dφ dω = dtan (T ω) dω = T T 2 ω 2 + dtan (αt ω) dω αt α 2 T 2 ω 2 + = Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli

3 ECE45/ECE55, FREQUENCY-RESPONSE DESIGN 9 3 α 2 T 3 ω 2 + T = αt 3 ω 2 + αt T 3 ω 2 α(α ) = T (α ) ω max = T α. For this frequency, the amount of phase added is φ max = tan (T ω max ) tan (αt ω max ) ( ) ( ) α = tan α tan α }{{}}{{} θ θ 2 To get a simpler expression, consider the triangles to the right. + α φ max θ θ 2 We can solve for φmax using the law of cosines. This gives: α α + α 2 α α ( α) 2 = ( + α) + (α + α 2 ) 2 + α α + α 2 cos (φ max ) 2α + α 2 = + 2α + α α α + α 2 cos (φ max ) 4α cos (φ max ) = 2 + α α + α = 2 α 2 ( + α). Now, we remember that cos 2 (θ) + sin 2 (θ) =, so sin 2 (φ max ) = 4α ( + α) 2 = ( + α)2 4α ( + α) 2 So, we conclude that = ( α)2 ( + α) 2. Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli

4 ECE45/ECE55, FREQUENCY-RESPONSE DESIGN 9 4 ( ) α φ max = sin + α We can invert this to find α to add a certain phase φmax α = sin(φ max) + sin(φ max ). We can show that ωmax occurs mid-way between T and on a log αt scale ( ) ( ) ( ) log(ω max ) = log T = log + log αt T αt. = 2 [ ( ) ( )] log + log. T αt 9 8 How much can we improve phase with a lead network? φmax If we need more phase improvement, we can use a double-lead compensator: D(s) = K [ ] Ts + 2. αts + Need to compromise between good phase margin and good sensor noise rejection at high frequency. /α Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli

5 ECE45/ECE55, FREQUENCY-RESPONSE DESIGN : Design method # for lead controllers Design specification include: Desired steady-state error. Desired phase margin. Compute steady-state error of compensated system. Set equal to desired steady-state error by computing K. Evaluate the PM of the uncompensated system using the value of K from above. ω max initially set to crossover frequency. Add a small amount of PM (5 to 2 )totheneededphaselead. (Lead network moves crossover point, so need to design conservatively): Gives φ max. Determine α = sin(φ max ) + sin(φ max ). Determine T = ω max α. Iterate if necessary (choose a slightly different ωmax or φ max ). EXAMPLE: Plant G(s) = s(s + ) Specification : Want steady-state error for ramp input <.. Specification 2: Want overshoot Mp < 25%. Start by computing steady-state error [ ] e ss = lim s R(s) s + D(s)G(s) [ ] = lim = s s + D(s) D(). (s+) Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli

6 ECE45/ECE55, FREQUENCY-RESPONSE DESIGN 9 6 [ ] Ts + D(s) = K αts + So, D() = K so D() = K. =.... K =. In Bode plot below, red line is original plant. Green line is plant with added gain of K =, butnoleadcompensator. We see that the new crossover is at 3 rad s,soω max = 3rads. Overshoot specification Mp < 25% gives PM > 45. Evaluating (red line) Bode diagram of KG(s) at crossover, we see we have a phase margin of 8.Needabout27 more to meet spec. Note, addition of pole and zero from lead network will shift crossover frequency somewhat. To be safe, design for added phase of 37 instead of 27. α = sin(37 ) + sin(37 ) =.25. T =.25(3) =.667. So, D(s) = [ 2s/3 + 2s/2 + ]. In plots below, the blue line is the Bode plot of the compensated system KD(s)G(s). Also, uncompensated root locus is plotted top-right; compensated root locus is plotted bottom-right. We find that we just missed specifications: PM = 44.Iterateif desired to meet specs. Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli

7 ECE45/ECE55, FREQUENCY-RESPONSE DESIGN 9 7 Magnitude Phase ω (rad/sec) ω (rad/sec) Imag Axis Imag Axis Real Axis Real Axis Amplitude Step Response Amplitude Ramp Response Time (sec.) Summary of design example Time (sec.). Determine dc-gain so that steady-state errors meet spec. 2. Select α and T to achieve an acceptable PM at crossover. Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli

8 ECE45/ECE55, FREQUENCY-RESPONSE DESIGN : Design method #2 for lead controllers Design specifications Desired closed-loop bandwidth requirements (rather than e ss ). Desired PM requirements. Choose open-loop crossover frequency ωc to be half the desired closed-loop bandwidth. Evaluate KG = G( jω c ) and φ G = G( jω c ). Compute phase lead required φmax = PM φ G 8. Compute α = sin(φ max ) + sin(φ max ). Compute T = αωc. Compensation is [ ] Ts + D(s) = K αts + α D( jω c ) =K + α α + = K α + α α + = K. α Open-loop gain at ωc is designed to be : D( jω c ) G( jω c ) = K α G( jω c ) = Our design is now complete. K = α K G. Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli

9 ECE45/ECE55, FREQUENCY-RESPONSE DESIGN 9 9 EXAMPLE: Desired gain crossover frequency of radians per second. Desired PM of 6. G(s) = s(s + ). KG = j( j + ) =., φ G = φmax = = sin(φ max ) α = + sin(φ max ) =.37. T = α =.35. K D = α K G = Matlab code to automate this procedure: % [K,T,alpha]=bod_lead(np,dp,wc,PM) % Computes the lead compensation of the plant np/dp to have % phase margin PM (in degrees) at crossover frequency wc (in radians/sec). % e.g., [K,T,alpha]=bod_lead([],[ ],5,6); function [K,T,alpha]=bod_lead(np,dp,wc,PM) % first compute the plant response at wc. [magc,phc]=bode(np,dp,wc); % now, compute the needed phase lead at wc and convert to radians % for use with "sine" phir=(-8+pm-phc)*pi/8; if abs(phir)pi/2, fprintf('a simple phase lead/lag cannot change the phase by more\n'); fprintf('than +/- 9 degrees.\n'); error('aborting.'); end; % Compute alpha, T, and K for compensator alpha=(-sin(phir))/(+sin(phir)); T=/(wc*sqrt(alpha)); K=sqrt(alpha)/magc; Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli

10 ECE45/ECE55, FREQUENCY-RESPONSE DESIGN 9 % compute the new open-loop system by convolving the plant polynomials % with the compensator polynomials. nol=conv(np,k*[t ]); dol=conv(dp,[alpha*t ]); % check the solution by plotting the Bode plot for the new open-loop % polynomials. Include the frequency w= % to get the full resonance response to show the gain margin. Also, plot % the uncompensated Bode response. w=logspace(-2,)*wc; w(34)=wc; clf; [mag,ph]=bode(np,dp,w); [mag2,ph2]=bode(nol,dol,w); subplot(2); semilogx(w/wc,2*log(mag),'--'); hold on; semilogx(w/wc,2*log(mag2)); grid; plot(w/wc,*w+,'g-'); ylabel('magnitude (db)'); title('phase-lead design (uncompensated=dashed; compensated=solid)'); subplot(22); semilogx(w/wc,ph,'--'); hold on; semilogx(w/wc,ph2); grid; plot(w/wc,*w-8+pm,'g-'); ylabel('phase'); xlabel('frequency/wc (i.e., ""=wc)'); Matlab results: Magnitude (db) Phase Phase lead design (uncompensated=dashed; compensated=solid) Frequency/wc (i.e., ""=wc) Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli

11 ECE45/ECE55, FREQUENCY-RESPONSE DESIGN 9 9.4: PI and lag compensation PI compensation In many problems it is important to keep bandwidth low, and also reduce steady-state error. PI compensation used here. K D(s) = K [ + T I s ]..ωt I ωt I ωt I K 9 ωt I ωt I Infinite gain at zero frequency Reduces steady-state error to step, ramp, etc. But also has integrator anti-windup problems. Adds phase below breakpoint. We want to keep breakpoint frequency very low to keep from destabilizing system. Lag compensation T I ω c. Approximates PI, but without integrator overflow. [ ] Ts + D(s) = α α>. αts + Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli

12 ECE45/ECE55, FREQUENCY-RESPONSE DESIGN 9 2 Primary objective of lag is to add 2 log α db gain to low frequencies without changing PM. α αt T αt T 9 Steady-state response improves with little effect on transient response. Typical process Assumption is that we need to modify (increase) the dc-gain of the loop transfer function. If we apply only a gain, then ω c typically increases and the phase margin decreases. NOT GOOD. Instead, use lag compensation to lower high-frequency gain. Assume plant has gain K (adjustable), or that we insert a gain K into the system. Adjust the open-loop gain K to meet phase margin requirements (plus about 5 slop factor) at crossover without additional compensation. Draw Bode diagram of system using the gain K from above. Evaluate low-frequency gain. Determine α to meet low-frequency gain requirements. Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli

13 ECE45/ECE55, FREQUENCY-RESPONSE DESIGN 9 3 Choose one corner frequency ω = (the zero) to be about one T decade below crossover frequency. (This way, the phase added by the lag compensator will minimally affect PM. Thephaseaddedat crossover will be about 5,henceourpreviousslopfactor). The other corner frequency is at ω = αt. Iterate design to meet spec. EXAMPLE: K G(s) = (.5 s + ) (s + ) ( 2 s + ). Bode plot for plant is red line, below. Uncompensated root locus is top-right plot. We want to design compensator for PM 25, K p = 9.. Set K = 4.5 for PM = 3.Thisgivescrossoveratω c.2 rads/sec. (Green line on Bode plot.) 2. Low-frequency gain now about 3 db or (3/2) = Should be raised by a factor of 2 to get K p = 9. So,α = Choose corner frequency at ω =.2 rads/sec. =.2, ort = 5. T 5. We then know other corner frequency ω = αt =..4 Step Response From: U() Altogether, we now have the compensator [ ] 5s + D(s) = 2. s + Amplitude To: Y() Time (sec.) Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli

14 ECE45/ECE55, FREQUENCY-RESPONSE DESIGN Magnitude Phase ω (rad/sec) ω (rad/sec) Imag Axis Imag Axis Real Axis Real Axis Please understand that the recipes in these notes are not meant to be exhaustive! And, we often require both lead and lag to meet specs. Many other approaches to control design using frequency-response methods exist. In fact, once you are comfortable with Bode plots, you can add poles and zeros like LEGOs to build whatever frequency response you might like (within the limitations of Bode s gain-phase theorem). Our next section gives some guidance on this. Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli

15 ECE45/ECE55, FREQUENCY-RESPONSE DESIGN : Design based on sensitivity Goal: Develop conditions on Bode plot of loop transfer function D(s)G(s) that will ensure good performance with respect to sensitivity, steady-state errors and sensor noise. Steady-state performance = lower bound on system low-freq. gain. Sensor noise = upper bound on high-frequency gain. Consider also the risk of an unmodeled system resonance: Magnitude may go over. Can cause instability. Must ensure that high-frequency gain is low so magnitude does not go over. Magnitude Frequency Together, these help us design a desired loop frequency response: Magnitude must be high at low frequencies, and Low at high frequencies. Magnitude of D(s)G(s) Steady-state Error Boundary Sensor noise and plant sensitivity boundary Sensitivity functions The presence of noises also enters our design considerations. w(t) r(t) D(s) G(s) y(t) v(t) Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli

16 ECE45/ECE55, FREQUENCY-RESPONSE DESIGN 9 6 Y (s) = W (s) + G(s)D(s)[R(s) V (s) Y (s)] [ + G(s)D(s)] Y (s) = W (s) + G(s)D(s)[R(s) V (s)] or, Y (s) = G(s)D(s) W (s) + [R(s) V (s)]. + G(s)D(s) + G(s)D(s) Tracking error = R(s) Y (s) G(s)D(s) E(s) = R(s) W (s) [R(s) V (s)] + G(s)D(s) + G(s)D(s) = + G(s)D(s) [R(s) W (s)] + G(s)D(s)V (s) + G(s)D(s) Define the sensitivity function S(s) to be S(s) = + G(s)D(s) which is the transfer function from r(t) to e(t) and from w(t) to e(t). The complementary sensitivity function T (s) = S(s) S(s) = G(s)D(s) + G(s)D(s) = T (s) which is the transfer function from r(t) to y(t). If V =, then and Y (s) = S(s)W (s) + T (s)r(s) E(s) = S(s)[R(s) W (s)]. The sensitivity function here is related to the one we saw several weeks ago SG T = T G G T = = + D(s)G(s) D(s)G(s) [ + D(s)G(s)] 2 + D(s)G(s) = S(s). G(s)[ + D(s)G(s)] G(s) Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli

17 ECE45/ECE55, FREQUENCY-RESPONSE DESIGN 9 7 So S(s) is the sensitivity of the transfer function to plant perturbations. Recall that T (s) + S(s) =, regardlessofd(s) and G(s). We would like T (s) =. ThenS(s) = ; Disturbanceiscancelled, design is insensitive to plant perturbation, steady-state error. BUT, forphysicalplants,g(s) for high frequencies (which forces S(s) ). Furthermore, the transfer function between V (s) and E(s) is T (s). To reduce high frequency noise effects, T (s) as frequency increases, and S(s). Typical sensitivity and complementary sensitivity (closed-loop transfer) functions are: S( jω) Performance Spec. T ( jω) Stability Spec. Another view of sensitivity: So, + D( jω)g( jω) is the distance between the Nyquist + D( jω)g( jω) curve to the point, and S( jω) =. D( jω)g( jω) + D( jω)g( jω) I(s) R(s) Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli

18 ECE45/ECE55, FREQUENCY-RESPONSE DESIGN 9 8 Alargevalueof S( jω) indicates a nearly unstable Nyquist plot. The maximum value of S is a more accurate measure of stability than PM or GM. So,wewantmax S( jω) small. How small? Note: E( jω) = S( jω)r( jω) E( jω) = S( jω)r( jω) S( jω) R( jω) put a frequency-based error bound Let W (ω) = R( jω)/e b.then, E( jω) S( jω) R( jω) e b S( jω) W (ω). EXAMPLE: Aunity-feedbacksystemistohaveanerrorlessthan.5 for all unity-amplitude sinusoids below 5 rads/sec. Draw W ( jω) for this design. Spectrum of R( jω) is unity for ω 5. Since eb =.5, W (ω) =.5 = 2 5 for this range Magnitude 2 5 Frequency (rads/sec.) We can translate this requirement into a loop-gain requirement. When errors are small, loop gain is high, so S( jω) D( jω)g( jω) and D( jω)g( jω) W (ω) or, D( jω)g( jω) W (ω) Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli

19 ECE45/ECE55, FREQUENCY-RESPONSE DESIGN : Robustness Typically there is some uncertainty in the plant transfer function. We want our design to be robustly stable, and to robustly give good performance (often called H design). Uncertainty often expressed as multiplicative G( jω) = G n ( jω)[ + W 2 ( jω) ( jω)] W2 (ω) is a function of frequency expressing uncertainty, or size of possible error in transfer function as a function of frequency. W2 (ω) is almost always small at low frequencies. W2 (ω) increases at high frequencies as unmodeled structural flexibility is common. Typical W2 Magnitude Frequency (rads/sec.) ( jω) expresses uncertainty in phase. The only restriction is Design ( jω). Assume design for nominal plant Gn (s) is stable. Thus, + D( jω)g n ( jω) = ω. Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli

20 ECE45/ECE55, FREQUENCY-RESPONSE DESIGN 9 2 For robust stability, + D( jω)g n ( jω) + D( jω)g n ( jω) }{{} = by assumption + D( jω)g( jω) = ω + D( jω)g n ( jω)[ + W 2 (ω) ( jω)] = + D( jω)g n( jω) + D( jω)g n ( jω) W 2(ω) ( jω) = recall, T ( jω) = D( jω)g n( jω) + D( jω)g n ( jω), [ + T ( jω)w 2 (ω) ( jω)] = so, T ( jω)w 2 (ω) ( jω) < or, T ( jω) W 2 (ω) <. For high freqencies D( jω)gn ( jω) is typically small, so T ( jω) D( jω)g n ( jω). Thus D( jω)g n ( jω) W 2 (ω) < D( jω)g n ( jω) < W 2 (ω). EXAMPLE: The uncertainty in a plant model is described by a function W 2 (ω) which is zero until ω = 3 rads/sec, and increases linearly from there to a value of at ω =, rads/sec. It remains constant at for higher frequencies. Plot constraint on D( jω)g n ( jω). Where W2 (ω) =, thereisnoconstraintonthemagnitudeoftheloop gain. Above ω = 3, /W 2 (ω) is a hyperbola from to. at,. Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli

21 ECE45/ECE55, FREQUENCY-RESPONSE DESIGN 9 2 Magnitude Frequency (rads/sec.) Combining W (ω) and W 2 (ω) requirements, 8 Log Magnitude Frequency (rads/sec.) Limitation: Crossover needs to be with slope. So,cannotmake constraints too strict, or design will be unstable. Lecture notes prepared by and copyright , Gregory L. Plett and M. Scott Trimboli

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

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

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

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

DESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD

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)

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

Automatic Control 2. Loop shaping. Prof. Alberto Bemporad. University of Trento. Academic year

Automatic Control 2. Loop shaping. Prof. Alberto Bemporad. University of Trento. Academic year Automatic Control 2 Loop shaping Prof. Alberto Bemporad University of Trento Academic year 21-211 Prof. Alberto Bemporad (University of Trento) Automatic Control 2 Academic year 21-211 1 / 39 Feedback

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

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

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

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

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

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

CDS 101/110a: Lecture 8-1 Frequency Domain Design

CDS 101/110a: Lecture 8-1 Frequency Domain Design CDS 11/11a: Lecture 8-1 Frequency Domain Design Richard M. Murray 17 November 28 Goals: Describe canonical control design problem and standard performance measures Show how to use loop shaping to achieve

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

Dr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Review

Dr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Review Week Date Content Notes 1 6 Mar Introduction 2 13 Mar Frequency Domain Modelling 3 20 Mar Transient Performance and the s-plane 4 27 Mar Block Diagrams Assign 1 Due 5 3 Apr Feedback System Characteristics

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

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

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

DIGITAL CONTROLLER DESIGN

DIGITAL CONTROLLER DESIGN ECE4540/5540: Digital Control Systems 5 DIGITAL CONTROLLER DESIGN 5.: Direct digital design: Steady-state accuracy We have spent quite a bit of time discussing digital hybrid system analysis, and some

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

Chapter 2. Classical Control System Design. Dutch Institute of Systems and Control

Chapter 2. Classical Control System Design. Dutch Institute of Systems and Control Chapter 2 Classical Control System Design Overview Ch. 2. 2. Classical control system design Introduction Introduction Steady-state Steady-state errors errors Type Type k k systems systems Integral Integral

More information

Lecture 6 Classical Control Overview IV. Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science - Bangalore

Lecture 6 Classical Control Overview IV. Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science - Bangalore Lecture 6 Classical Control Overview IV Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science - Bangalore Lead Lag Compensator Design Dr. Radhakant Padhi Asst.

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

Exercises for lectures 13 Design using frequency methods

Exercises for lectures 13 Design using frequency methods Exercises for lectures 13 Design using frequency methods Michael Šebek Automatic control 2016 31-3-17 Setting of the closed loop bandwidth At the transition frequency in the open loop is (from definition)

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

ECE317 : Feedback and Control

ECE317 : Feedback and Control ECE317 : Feedback and Control Lecture : Steady-state error Dr. Richard Tymerski Dept. of Electrical and Computer Engineering Portland State University 1 Course roadmap Modeling Analysis Design Laplace

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

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

Plan of the Lecture. Goal: wrap up lead and lag control; start looking at frequency response as an alternative methodology for control systems design.

Plan of the Lecture. Goal: wrap up lead and lag control; start looking at frequency response as an alternative methodology for control systems design. Plan of the Lecture Review: design using Root Locus; dynamic compensation; PD and lead control Today s topic: PI and lag control; introduction to frequency-response design method Goal: wrap up lead and

More information

Raktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Frequency Response-Design Method

Raktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Frequency Response-Design Method .. AERO 422: Active Controls for Aerospace Vehicles Frequency Response- Method Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. ... Response to

More information

Chapter 9: Controller design

Chapter 9: Controller design Chapter 9. Controller Design 9.1. Introduction 9.2. Effect of negative feedback on the network transfer functions 9.2.1. Feedback reduces the transfer function from disturbances to the output 9.2.2. Feedback

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

EECS C128/ ME C134 Final Wed. Dec. 15, am. Closed book. Two pages of formula sheets. No calculators.

EECS C128/ ME C134 Final Wed. Dec. 15, am. Closed book. Two pages of formula sheets. No calculators. Name: SID: EECS C28/ ME C34 Final Wed. Dec. 5, 2 8- am Closed book. Two pages of formula sheets. No calculators. There are 8 problems worth points total. Problem Points Score 2 2 6 3 4 4 5 6 6 7 8 2 Total

More information

FEEDBACK CONTROL SYSTEMS

FEEDBACK CONTROL SYSTEMS FEEDBAC CONTROL SYSTEMS. Control System Design. Open and Closed-Loop Control Systems 3. Why Closed-Loop Control? 4. Case Study --- Speed Control of a DC Motor 5. Steady-State Errors in Unity Feedback Control

More information

= rad/sec. We can find the last parameter, T, from ωcg new

= rad/sec. We can find the last parameter, T, from ωcg new EE572 Solution to HW#22. Keep working on your project!! 1. Consider the following system: W(s) + T s =1 msec G lead (z) G zoh (z) 8 ( s+ 4) - a) Design a lead compensator, G lead (z), which meets the following

More information

FREQUENCY-RESPONSE ANALYSIS

FREQUENCY-RESPONSE ANALYSIS ECE450/550: Feedback Control Systems. 8 FREQUENCY-RESPONSE ANALYSIS 8.: Motivation to study frequency-response methods Advantages and disadvantages to root-locus design approach: ADVANTAGES: Good indicator

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

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

MEM 355 Performance Enhancement of Dynamical Systems

MEM 355 Performance Enhancement of Dynamical Systems MEM 355 Performance Enhancement of Dynamical Systems Frequency Domain Design Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University 5/8/25 Outline Closed Loop Transfer Functions

More information

The loop shaping paradigm. Lecture 7. Loop analysis of feedback systems (2) Essential specifications (2)

The loop shaping paradigm. Lecture 7. Loop analysis of feedback systems (2) Essential specifications (2) Lecture 7. Loop analysis of feedback systems (2). Loop shaping 2. Performance limitations The loop shaping paradigm. Estimate performance and robustness of the feedback system from the loop transfer L(jω)

More information

The requirements of a plant may be expressed in terms of (a) settling time (b) damping ratio (c) peak overshoot --- in time domain

The requirements of a plant may be expressed in terms of (a) settling time (b) damping ratio (c) peak overshoot --- in time domain Compensators To improve the performance of a given plant or system G f(s) it may be necessary to use a compensator or controller G c(s). Compensator Plant G c (s) G f (s) The requirements of a plant may

More information

MAE143a: Signals & Systems (& Control) Final Exam (2011) solutions

MAE143a: Signals & Systems (& Control) Final Exam (2011) solutions MAE143a: Signals & Systems (& Control) Final Exam (2011) solutions Question 1. SIGNALS: Design of a noise-cancelling headphone system. 1a. Based on the low-pass filter given, design a high-pass filter,

More information

Control System Design

Control System Design ELEC ENG 4CL4: Control System Design Notes for Lecture #11 Wednesday, January 28, 2004 Dr. Ian C. Bruce Room: CRL-229 Phone ext.: 26984 Email: ibruce@mail.ece.mcmaster.ca Relative Stability: Stability

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

STABILITY ANALYSIS TECHNIQUES

STABILITY ANALYSIS TECHNIQUES ECE4540/5540: Digital Control Systems 4 1 STABILITY ANALYSIS TECHNIQUES 41: Bilinear transformation Three main aspects to control-system design: 1 Stability, 2 Steady-state response, 3 Transient response

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

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

Introduction. Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31

Introduction. Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31 Introduction Classical Control Robust Control u(t) y(t) G u(t) G + y(t) G : nominal model G = G + : plant uncertainty Uncertainty sources : Structured : parametric uncertainty, multimodel uncertainty Unstructured

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

BASIC PROPERTIES OF FEEDBACK

BASIC PROPERTIES OF FEEDBACK ECE450/550: Feedback Control Systems. 4 BASIC PROPERTIES OF FEEDBACK 4.: Setting up an example to benchmark controllers There are two basic types/categories of control systems: OPEN LOOP: Disturbance r(t)

More information

IC6501 CONTROL SYSTEMS

IC6501 CONTROL SYSTEMS DHANALAKSHMI COLLEGE OF ENGINEERING CHENNAI DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING YEAR/SEMESTER: II/IV IC6501 CONTROL SYSTEMS UNIT I SYSTEMS AND THEIR REPRESENTATION 1. What is the mathematical

More information

Robust Performance Example #1

Robust Performance Example #1 Robust Performance Example # The transfer function for a nominal system (plant) is given, along with the transfer function for one extreme system. These two transfer functions define a family of plants

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

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

EECS C128/ ME C134 Final Thu. May 14, pm. Closed book. One page, 2 sides of formula sheets. No calculators.

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

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

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

Loop shaping exercise

Loop shaping exercise Loop shaping exercise Excerpt 1 from Controlli Automatici - Esercizi di Sintesi, L. Lanari, G. Oriolo, EUROMA - La Goliardica, 1997. It s a generic book with some typical problems in control, not a collection

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

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

Robust Control 3 The Closed Loop

Robust Control 3 The Closed Loop Robust Control 3 The Closed Loop Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University /2/2002 Outline Closed Loop Transfer Functions Traditional Performance Measures Time

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

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

Compensator Design to Improve Transient Performance Using Root Locus

Compensator Design to Improve Transient Performance Using Root Locus 1 Compensator Design to Improve Transient Performance Using Root Locus Prof. Guy Beale Electrical and Computer Engineering Department George Mason University Fairfax, Virginia Correspondence concerning

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

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

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

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

D(s) G(s) A control system design definition

D(s) G(s) A control system design definition R E Compensation D(s) U Plant G(s) Y Figure 7. A control system design definition x x x 2 x 2 U 2 s s 7 2 Y Figure 7.2 A block diagram representing Eq. (7.) in control form z U 2 s z Y 4 z 2 s z 2 3 Figure

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

STABILITY ANALYSIS. Asystemmaybe stable, neutrallyormarginallystable, or unstable. This can be illustrated using cones: Stable Neutral Unstable

STABILITY ANALYSIS. Asystemmaybe stable, neutrallyormarginallystable, or unstable. This can be illustrated using cones: Stable Neutral Unstable ECE4510/5510: Feedback Control Systems. 5 1 STABILITY ANALYSIS 5.1: Bounded-input bounded-output (BIBO) stability Asystemmaybe stable, neutrallyormarginallystable, or unstable. This can be illustrated

More information

Topic # Feedback Control Systems

Topic # Feedback Control Systems Topic #19 16.31 Feedback Control Systems Stengel Chapter 6 Question: how well do the large gain and phase margins discussed for LQR map over to DOFB using LQR and LQE (called LQG)? Fall 2010 16.30/31 19

More information

6.302 Feedback Systems Recitation 16: Compensation Prof. Joel L. Dawson

6.302 Feedback Systems Recitation 16: Compensation Prof. Joel L. Dawson Bode Obstacle Course is one technique for doing compensation, or designing a feedback system to make the closed-loop behavior what we want it to be. To review: - G c (s) G(s) H(s) you are here! plant For

More information

Topic # Feedback Control. State-Space Systems Closed-loop control using estimators and regulators. Dynamics output feedback

Topic # Feedback Control. State-Space Systems Closed-loop control using estimators and regulators. Dynamics output feedback Topic #17 16.31 Feedback Control State-Space Systems Closed-loop control using estimators and regulators. Dynamics output feedback Back to reality Copyright 21 by Jonathan How. All Rights reserved 1 Fall

More information

Performance of Feedback Control Systems

Performance of Feedback Control Systems Performance of Feedback Control Systems Design of a PID Controller Transient Response of a Closed Loop System Damping Coefficient, Natural frequency, Settling time and Steady-state Error and Type 0, Type

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

Intro to Frequency Domain Design

Intro to Frequency Domain Design Intro to Frequency Domain Design MEM 355 Performance Enhancement of Dynamical Systems Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University Outline Closed Loop Transfer Functions

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

Systems Analysis and Control

Systems Analysis and Control Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture : Different Types of Control Overview In this Lecture, you will learn: Limits of Proportional Feedback Performance

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

PID controllers. Laith Batarseh. PID controllers

PID controllers. Laith Batarseh. PID controllers Next Previous 24-Jan-15 Chapter six Laith Batarseh Home End The controller choice is an important step in the control process because this element is responsible of reducing the error (e ss ), rise time

More information

An Internal Stability Example

An Internal Stability Example An Internal Stability Example Roy Smith 26 April 2015 To illustrate the concept of internal stability we will look at an example where there are several pole-zero cancellations between the controller and

More information

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

Course Summary. The course cannot be summarized in one lecture.

Course Summary. The course cannot be summarized in one lecture. Course Summary Unit 1: Introduction Unit 2: Modeling in the Frequency Domain Unit 3: Time Response Unit 4: Block Diagram Reduction Unit 5: Stability Unit 6: Steady-State Error Unit 7: Root Locus Techniques

More information

Last week: analysis of pinion-rack w velocity feedback

Last week: analysis of pinion-rack w velocity feedback Last week: analysis of pinion-rack w velocity feedback Calculation of the steady state error Transfer function: V (s) V ref (s) = 0.362K s +2+0.362K Step input: V ref (s) = s Output: V (s) = s 0.362K s

More information

Lecture 11. Frequency Response in Discrete Time Control Systems

Lecture 11. Frequency Response in Discrete Time Control Systems EE42 - Discrete Time Systems Spring 28 Lecturer: Asst. Prof. M. Mert Ankarali Lecture.. Frequency Response in Discrete Time Control Systems Let s assume u[k], y[k], and G(z) represents the input, output,

More information

Dynamic Compensation using root locus method

Dynamic Compensation using root locus method CAIRO UNIVERSITY FACULTY OF ENGINEERING ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 00/0 CONTROL ENGINEERING SHEET 9 Dynamic Compensation using root locus method [] (Final00)For the system shown in the

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. Control Systems Design Lead-Lag Compensator.

Control Systems. Control Systems Design Lead-Lag Compensator. Design Lead-Lag Compensator hibum@seoulteh.a.kr Outline Lead ompensator design in frequeny domain Lead ompensator design steps. Example on lead ompensator design. Frequeny Domain Design Frequeny response

More information

Positioning Servo Design Example

Positioning Servo Design Example Positioning Servo Design Example 1 Goal. The goal in this design example is to design a control system that will be used in a pick-and-place robot to move the link of a robot between two positions. Usually

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

Chapter 6 - Solved Problems

Chapter 6 - Solved Problems Chapter 6 - Solved Problems Solved Problem 6.. Contributed by - James Welsh, University of Newcastle, Australia. Find suitable values for the PID parameters using the Z-N tuning strategy for the nominal

More information

Analysis of SISO Control Loops

Analysis of SISO Control Loops Chapter 5 Analysis of SISO Control Loops Topics to be covered For a given controller and plant connected in feedback we ask and answer the following questions: Is the loop stable? What are the sensitivities

More information

Introduction to Feedback Control

Introduction to Feedback Control 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

More information

Dr Ian R. Manchester

Dr Ian R. Manchester 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

Systems Analysis and Control

Systems Analysis and Control Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 22: The Nyquist Criterion Overview In this Lecture, you will learn: Complex Analysis The Argument Principle The Contour

More information

Lecture 1: Feedback Control Loop

Lecture 1: Feedback Control Loop Lecture : Feedback Control Loop Loop Transfer function The standard feedback control system structure is depicted in Figure. This represend(t) n(t) r(t) e(t) u(t) v(t) η(t) y(t) F (s) C(s) P (s) Figure

More information

Table of Laplacetransform

Table of Laplacetransform Appendix Table of Laplacetransform pairs 1(t) f(s) oct), unit impulse at t = 0 a, a constant or step of magnitude a at t = 0 a s t, a ramp function e- at, an exponential function s + a sin wt, a sine fun

More information

Automatic Control (TSRT15): Lecture 7

Automatic Control (TSRT15): Lecture 7 Automatic Control (TSRT15): Lecture 7 Tianshi Chen Division of Automatic Control Dept. of Electrical Engineering Email: tschen@isy.liu.se Phone: 13-282226 Office: B-house extrance 25-27 Outline 2 Feedforward

More information