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8–1ECE4510/5510: Feedback Control Systems.FREQUENCY-RESPONSE ANALYSIS8.1: Motivation to study frequency-response methods Advantages and disadvantages to root-locus design approach:ADVANTAGES : Good indicator of transient response. Explicitly shows location of closed-loop poles. Tradeoffs areclear.DISADVANTAGES : Requires transfer function of plant be known. Difficult to infer all performance values. Hard to extract steady-state response (sinusoidal inputs). Frequency-response methods can be used to supplement root locus: Can infer performance and stability from same plot. Can use measured data when no model is available. Design process is independent of system order (# poles). Time delays handled correctly (e sτ). Graphical techniques (analysis/synthesis) are “quite simple.”What is a frequency response? We want to know how a linear system responds to sinusoidal input, insteady state.c 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝
8–2ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSIS Consider system Y (s) G(s)U (s) with input u(t) u 0 cos(ωt), sos.U (s) u 0 2s ω2With zero initial conditions,Y (s) u 0 G(s) s.s 2 ω2Do a partial-fraction expansion (assume distinct roots)α2αnα0α1α0 ··· Y (s) s a1 s a2s an s jω s jωty(t) !α1ea1t α2ea2"# · · · αn ean t α0e j ωt α0 e j ωt .If stable, these decay to zero.yss (t) α0e j ωt α0 e j ωt . Let α0 Ae j φ . Then,yss Ae j φ e j ωt Ae j φ e j ωt%& A e j (ωt φ) e j (ωt φ) 2A cos (ωt φ) .We find α0 via standard partial-fraction-expansion means:'α0 (s jω)Y (s) s j ω)(u 0 sG(s) )) (s jω) )s j ω u 0( jω)G( jω) u 0 G( jω) .(2 jω)2Substituting into our prior resultyss u 0 G( jω) cos (ωt ̸ G( jω)) .c 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝
8–3ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSIS Important LTI-system fact: If the input to an LTI system is a sinusoid,the “steady-state” output is a sinusoid of the same frequency butdifferent amplitude and phase.Transfer function at s jω tells us response to asinusoid.but also about stability as jω-axis is stability boundary!FORESHADOWING :EXAMPLE :Suppose that we have a system with transfer function2G(s) .3 s Then, the system’s frequency response is)2 ))2G( jω) . )3 s s j ω 3 jωThe magnitude response is))) 2 ) 2 22) .A( jω) )) 23 jω ) 3 jω (3 jω)(3 jω)9 ω The phase response isω *2φ( jω) ̸3 jωθ ̸ (2) ̸ (3 jω)123 0 tan 1 (ω/3) . Now that we know the amplitude and phase response, we can findthe amplitude gain and phase change caused by the system for anyspecific frequency. For example, if ω 3 rad s 1,A( j3) 29 9 23φ( j3) tan 1(3/3) π/4.c 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝
8–4ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSIS8.2: Plotting a frequency response There are two common ways to plot a frequency response themagnitude and phase for all frequencies.EXAMPLE :RU (s) C11 RCsG(s) Y (s)Frequency response1(let RC 1)1 jω RC1 1 jω1̸ tan 1 (ω). 1 ω2G( jω) We will need to separate magnitude and phase information fromrational polynomials in jω. Magnitude magnitude of numerator / magnitude of denominator,R(num)2 I(num)2,R(den)2 I(den)2. Phase phase of numerator phase of denominatortan 1*I(num)R(num) tan 1*I(den)R(den) .Plot method #1: Polar plot in complex plane Evaluate G( jω) at each frequency for 0 ω .Result will be a complex number at each frequency: a jb or Ae j φ .c 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝
ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSIS Plot each point on the complex plane at (a jb) or Ae j φ for eachfrequency-response value. Result polar plot. We will later call this a “Nyquist plot”.ω00.51.01.52.03.05.010.0 G( jω)1.000̸ 0.0 0.894̸ 26.6 0.707̸ 45.0 0.555̸ 56.3 0.447̸ 63.4 0.316̸ 71.6 0.196̸ 78.7 0.100̸ 84.3 0.000̸ 90.0 The polar plot is parametric in ω, so it is hard to read thefrequency-response for a specific frequency from the plot. We will see later that the polar plot will help us determine stabilityproperties of the plant and closed-loop system.Plot method #2: Magnitude and phase plots We can replot the data by separating the plots for magnitude andphase making two plots versus frequency.c 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝8–5
8–6ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSIS10 30G( jω) 20 G( jω) 100.80.60.4 40 50̸ 60 700.2 8000 12345Frequency, (rads/sec.)6 90012345Frequency, (rads/sec.)6The above plots are in a natural scale, but usually a log-log plot ismade This is called a “Bode plot” or “Bode diagram.”Reason for using a logarithmic scale Simplest way to display the frequency response of arational-polynomial transfer function is to use a Bode Plot.Logarithmic G( jω) versus logarithmic ω, and logarithmic ̸ G( jω)versus ω.REASON :* ablog10 log10 a log10 b log10 c log10 d .cd The polynomial factors that contribute to the transfer function canbe split up and evaluated separately.(s 1)G(s) (s/10 1)( jω 1)G( jω) ( jω/10 1) jω 1 G( jω) jω/10 1 . ω /2,log10 G( jω) log10 1 ω2 log10 1 .10c 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝
8–7ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSIS Consider:log10 For ω ωn ,log100For ω ωn ,KEY POINT: log1000*ω1 ωn* 2* 2ω1 ωnω1 ωn 2. log10(1) 0. log10* ω.ωnTwo straight lines on a log-log plot; intersect at ω ωn .Typically plot 20 log10 G( jω) ; that is, in dB.20dBApproximationExact0.1ωn ωn10ωnA transfer function is made up of first-order zeros and poles, complexzeros and poles, constant gains and delays. We will see how to makestraight-line (magnitude- and phase-plot) approximations for all these,and combine them to form the appropriate Bode diagram.c 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝
8–8ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSIS8.3: Bode magnitude diagrams (a) The log10(·) operator lets us break a transfer function up into pieces. If we know how to plot the Bode plot of each piece, then we simplyadd all the pieces together when we’re done.Bode magnitude: Constant gain dBdB 20 log10 K . K 1Not a function of frequency.Horizontal straight line. If K 1, then negative, elsepositive.0.1101 K 1Bode magnitude: Zero or pole at origin For a zero at the origin,20 dB20 dB perdecadeG(s) s0.1dB 20 log10 G( jω) 20 log10 jω dB. For a pole at the origin,1G(s) sdB 20 log10 G( jω) 20 log10 jω dB.110 20 dB20 dB 20 dB perdecade0.11 20 dBc 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝10
8–9ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSIS Both are straight lines, slope 20 dB per decade of frequency. Line intersects ω-axis at ω 1. For an nth-order pole or zero at the origin,dB 20 log10 ( jω)n 20 log10 ωn 20n log10 ω. Still straight lines. Still intersect ω-axis at ω 1. But, slope 20n dB per decade.Bode magnitude: Zero or pole on real axis, but not at origin For a zero on the real axis, (LHP or RHP), the standard Bode form is* sG(s) 1 ,ωnwhich ensures unity dc-gain. If you start out with something likeG(s) (s ωn ),then factor as s 1 .G(s) ωnωnDraw the gain term (ωn ) separately from the zero term (s/ωn 1). *In general, a LHP or RHP zero has standard Bode form *s 1G(s) ωnc 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝
8–10ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSISG( jω) 1 j20 log10 G( jω) 20 log10 0*ωωn0 1 *ωωn 2* ω 220 log10 1 20 log10 1 0.ωn0* * 2ωω. 20 log1020 log10 1 ωnωnFor ω ωn ,For ω ωn ,Two straight lines on a log scale which intersect at ω ωn .For a pole on the real axis, (LHP or RHP) standard Bode form is 1*s 1G(s) ωn0* 2ω20 log10 G( jω) 20 log10 1 .ωnThis is the same except for a minus sign.0.1ωn1ωn 20 dB perdecade20 dB 20 dB20 dB perdecade0.1ωn1ωn10ωnc 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝10ωn
8–11ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSIS8.4: Bode magnitude diagrams (b)Bode magnitude: Complex zero pair or complex pole pair For a complex-zero pair (LHP or RHP) standard Bode form is* * 2ss 1, 2ζωnωnwhich has unity dc-gain. If you start out with something likes 2 2ζ ωn s ωn2 ,which we have seen before as a “standard form,” the dc-gain is ωn2 . Convert forms by factoring out ωn221* * 2ss 1 . 2ζs 2 2ζ ωn s ωn2 ωn2ωnωnComplex zeros do not lend themselves very well to straight-lineapproximation. 2*s If ζ 1, then this is 1 .ωn Double real zero at ω slope of 40 dB/decade.n For ζ ̸ 1, there will be overshoot or undershoot at ω ωn .For other values of ζ : Dip frequency: ωd ωn Value of H ( jωd ) is:,20 log10(2ζ (1 ζ 2)).,1 2ζ 2 Note: There is no dip unless 0 ζ 1/ 2 0.707 .Dip amount for 0 ζ 0.7070-10Dip (dB) -20-30-40-50-6000.10.20.3c 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝0.4ζ0.50.60.70.8
8–12ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSIS We write complex poles (LHP or RHP) as2 11* * 2ss 2ζ 1.G(s) ωnωn, The resonant peak frequency is ωr ωn 1 2ζ 2, Value of H ( jωr ) is 20 log10 (2ζ (1 ζ 2)).Same graph as for “dip” for complex-conjugate zeros. Note that there is no peak unless 0 ζ 1/ 2 0.707 . For ω ωn , magnitude 0 dB. For ω ωn , magnitude slope 40 dB/decade.Bode Mag: Complex zeros40 dB20 dB20 dBζ 0.90.70.50 dB0.30.20.10.050 dB 20 dBωn0.1ωn 20 dB10ωnBode Mag: Complex polesζ 0.050.10.20.30.50.7ζ 0.9 40 dB0.1ωnωnBode magnitude: Time delay G(s) e sτ. G( jω) 1.20 log10 1 0 dB.Does not change magnitude response.EXAMPLE :Sketch the Bode magnitude plot for2000(s 0.5).G(s) s(s 10)(s 50)c 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝10ωn
8–13ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSIS The first step is to convert the terms of the transfer function into“Bode standard form”.%&2000·0.5 s 12000(s 0.5)0.5&%& % 10·50G(s) sss(s 10)(s 50) s 10 1 50 1/.jω2 0.5 1././.G( jω) jωjωjω 10 1 50 1We can see that the components of the transfer function are: DC gain of 20 log10 2 6 dB; Pole at origin; One real zero not at origin, and Two real poles not at origin.806040200 20 40 60 210 110010110210c 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝310
8–14ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSIS8.5: Bode phase diagrams (a) Bode diagrams consist of the magnitude plots we have seen so far, BUT, also phase plots. These are just as easy to draw. BUT, they differ depending on whether the dynamics are RHP or LHP.Finding the phase of a complex number Plot the location of the number as a vector in the complex plane. Use trigonometry to find the phase. For numbers with positive real part,* I(#)̸ (#) tan 1.R(#) Ip2̸p1( p2 )For numbers with negative real part, *I(#)̸ (#) 180 tan 1. R(#) ̸( p1 )RIf you are lucky enough to have the “atan2(y, x)” function, then̸(#) atan2(I(#), R(#))for any complex number. Also note,̸*abcd ̸ (a) ̸ (b) ̸ (c) ̸ (d).Finding the phase of a complex function of ω This is the same as finding the phase of a complex number, if specificvalues of ω are substituted into the function.c 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝
8–15ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSISBode phase: Constant gain G(s) K .3(K ) 0 ,K 0;. 180 , K 0. ̸ Constant phase of 0 or 180 .Bode phase: Zero or pole at originZero: G(s) s, . . . G( jω) jω ω̸ 90 .11 j1 Pole: G(s) , . . . G( jω) ̸ 90 .sjωωω Constant phase of 90 . Bode phase: Real LHP zero or pole *s Zero: G(s) 1 .ωn *ω̸ G( jω) ̸ 1jωn* ω tan 1.ωnPole: G(s) . ̸1sωn 1*/,ω 1ωn* ω. tan 1ωnG( jω) ̸ (1) ̸j90 45 perdecade0.1ωn0.1ωnωnωn10ωn10ωn 45 perdecade 90 c 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝
8–16ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSISBode phase: Real RHP zero or pole* s180 Zero: G(s) 1 .ωn* ω̸ G( jω) ̸j 190 ωn* ω. 180 tan 1ωn0.1ωn1 Pole: G(s) ./,0.1ωns 1ωn* ω̸ G( jω) ̸ (1) ̸j 1ωn 90 ** ω 180 tan 1ωn* 180 ω 180 tan 1.ωnc 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝ 45 perdecadeωn10ωnωn10ωn45 perdecade
8–17ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSIS8.6: Bode phase diagrams (b)Bode phase: Complex LHP zero pair or pole pair Complex LHP zeros cause phase to go from 0 to 180 . Complex LHP poles cause phase to go from 180 to 0 . Transition happens in about ζ decades, centered at ωn .Bode Phase: Complex LHP zerosζ 0.051600.11400.21200.31000.5800.70.960180Bode Phase: Complex LHP poles0 20 40 60 80 100 12040 14020 16000.1ωn 180ωn10ωnζ 0.90.70.50.30.20.10.050.1ωnωn10ωnBode phase: Complex RHP zero pair or pole pair Complex RHP zeros cause phase to go from 360 to 180 . Complex RHP poles cause phase to go from 360 to 180 .Bode Phase: Complex RHP zerosBode Phase: Complex RHP poles3500300 50250 100200 150150 200100 25050 3000 3500.1ωnωn10ωn0.1ωnωnc 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝10ωn
8–18ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSISBode phase: Time delay G(s) e sτ ,0.1ωiωi10ωiG( jω) e j ωτ 1̸ ωτ̸G( jω) ωτ in radians. 56.3ωτ in degrees.Note: Line curve in log scale.EXAMPLE : Sketch the Bode phase plot for2000(s 0.5)2 ( jω/0.5 1)G(s) ,or G( jω) s(s 10)(s 50)jω ( jω/10 1) ( jω/50 1)where we converted to “Bode standard form” in a prior example. Constant: K 2. Zero phase contribution.Pole at origin: Phase contribution of 90 .Two real LHP poles: Phase from 0 to 90 , each.One real LHP zero: Phase from 0 to 90 .900 90 180 210 110010110210c 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝310
8–19ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSISEXAMPLE : Sketch the Bode magnitude and phase plots for1200 (s 3).G(s) s (s 12) (s 50)First, we convert to Bode standard form, which gives&%1200 (3) 1 3s%&%&G(s) sss (12) (50) 1 121 50./jω6 1 3/./.G( jω) jωjωjω 1 12 1 50Positive gain, one real LHP zero, one pole at origin, two real LHPpoles.3020100 10 20 11001011021031090450 45 90 135 180 110010110210c 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝310
8–20ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSIS8.7: Some observations based on Bode plotsNonminimum-phase systems A system is called a nonminimum-phase if it has pole(s) or zero(s) inthe RHP. ConsiderG 1(s) 10G 2(s) 10s 1s 104s 1s 104zero at 1pole at 10zero at 1pole at 105minimumphase5nonminimumphase The magnitude responses of these two systems are: ω2 1 jω 1 G 1( jω) 10 10 jω 10 ω2 100 ω2 1 jω 1 10 G 2( jω) 10 jω 10 ω2 100which are the same! The phase responses are very different:(G 1 ( jω)), ̸ (G 2 ( jω))Bode-Magnitude Plot G 1 ( jω) , G 2 ( jω) 15015Non-minimumPhase, G 250 210 110010110210Frequency, (rads/sec.)310̸12010 Bode-Phase Plot180209060MinimumPhase, G 1300 210 110010110210Frequency, (rads/sec.)310Note that the change in phase of G 1 is much smaller than change ofphase in G 2. Hence G 1 is “minimum phase” and G 2 is“nonminimum-phase”c 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝
8–21ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSIS Non-minimum phase usually associated with delay.s 1G 2(s) G 1(s) 1 !s "#Delay s 1is very similar to a first-order Padé approximation to as 1delay. It is the same when evaluated at s jω.Note: Consider using feedback to control a nonminimum-phase system.What do the root-locus plotting techniques tell us? Consequently, nonminimum-phase systems are harder to designcontrollers for; step response often tends to “go the wrong way,” atleast initially.Steady-state errors from Bode magnitude plot Recall our discussion of steady-state errors to step/ramp/parabolicinputs versus “system type” (summarized on pg. 4–24) Consider a unity-feedback system. If the open-loop plant transfer function has N poles at s 0 then thesystem is “type N ” K p is error constant for type 0. K v is error constant for type 1. K a is error constant for type 2. For a unity-feedback system, K p lim G(s).s 0 At low frequency, a type 0 system will have G(s) K p . We can read this off the Bode-magnitude plot directly!c 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝
8–22ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSIS Horizontal y-intercept at low frequency K p . ess 11 Kpfor step input.K v lim sG(s), and is nonzero for a type 1 system.s 0 At low frequency, a type 1 system will have G(s) Kv.sKv. Slope of 20 dB/decade.ω Use the above approximation to extend the low-frequencyasymptote to ω 1. The asymptote (NOT THE ORIGINAL G( jω) ) evaluated at ω 1 is K v .1for ramp input. ess Kv At low frequency, G( jω) K a lim s 2 G(s), and is nonzero for a type 2 system.s 0 At low frequency, a type 2 system will have G(s) Ka.s2Ka. Slope of 40 dB/decade.ω2 Again, use approximation to extend low-frequency asymptote toω 1. The asymptote evaluated at ω 1 is K a .1for parabolic input. ess Ka At low frequency, G( jω) Similar for higher-order systems.c 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝
8–23ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSISExample 12050MagnitudeMagnitude100 10 20 3040302010 40 210 110010Frequency, (rads/sec.)EXAMPLEExample 2601100 210 110010110Frequency, (rads/sec.)1: Horizontal as ω 0, so we know this is type 0. Intercept 6 dB. . .K p 6 dB 2 [linear units].EXAMPLE2: Slope 20 dB/decade as ω 0, so we know this is type 1. Extend slope at low frequency to ω 1. Intercept 20 dB. . . K v 20 dB 10 [linear units].c 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝
8–24ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSIS8.8: Stability revisited If we know the closed-loop transfer function of a system in rationalpolynomial form, we can use Routh to find stable ranges for K . Motivation: What if we only have open-loop frequency response?A simple example Consider, for now, that we know the transfer-function of the system,and can plot the root-locus.EXAMPLE :I(s)r (t) K1s(s 1)2y(t)K 2R(s)We see neutral stability at K 2. The system is stable for K 2 andunstable for K 2.Recall that a point is on the root locus if K G(s) 1 and̸ G(s) 180 .If system is neutrally stable, jω-axis will have a point (points) where K G( jω) 1 and ̸ G( jω) 180 .c 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝
8–25ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSIS Consider the Bode plot ofK G(s). . .40 A neutral-stability condition fromBode plot is: K G( jωo) 1 AND̸ K G( jωo ) 180 at the samefrequency ωo .In this case, increasingK instability K G( jω) 1at ̸ K G( jω) 180 stability.200 20 40 60 80 210K 0.1K 2K 10 110010110 90 120 150 180 210In some cases, decreasing 240K instability K G( jω) 1 27010101010 ̸at K G( jω) 180 stability.KEY POINT: We can find neutral stability point on Bode plot, but don’t(yet) have a way of determining if the system is stable or not. Nyquistfound a frequency-domain method to do so. 2 101Nyquist stability Poles of closed-loop transfer function in RHP—the system is unstable. Nyquist found way to count closed-loop poles in RHP. If count is greater than zero, system is unstable. Idea: First, find a way to count closed-loop poles inside a contour. Second, make the contour equal to the RHP. Counting is related to complex functional mapping.c 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝
8–26ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSIS8.9: Interlude: Complex functional mapping Nyquist technique is a graphical method to determine system stability,regions of stability and MARGINS of stability. Involves graphing complex functions of s as a polar plot.EXAMPLE :Plotting f (x), a real function of a real variable x.f (x)x This can be done.EXAMPLE :Plotting F(s), a complex function of a complex variable s.F(s)?NO! This is wrong!s Must draw mapping of points or lines from s-plane to F(s)-plane.jωF(s)F(s0)mappingσs0EXAMPLE :I(F)sR(F)F(s) 2s 1 . . . “map the four points: A, B, C, D”D 1CAjI(s)R(s)1 jBc 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝
8–27ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSISEXAMPLE :Map a square contour (closed path) by F(s) jD(s)j IF(D)A 1 jCF( A)F(B)1 R(s) 11s.s 2 j F(C)BFORESHADOWING :By drawing maps of a specific contour, using amapping function related to the plant open-loop frequency-response,we will be able to determine closed-loop stability of systems.Mapping function: Poles of the function When we map a contour containing (encircling) poles and zeros ofthe mapping function, this map will give us information about howmany poles and zeros are encircled by the contour. Practice drawing maps when we know poles and zeros. EvaluateEXAMPLE :G(s) s so G(so) ⃗v e j α66̸ (poles).̸ (zeros) α F(c1)I(s)c1αR(s)c 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝
8–28ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSIS In this example, there are no zeros or poles inside the contour. Thephase α increases and decreases, but never undergoes a net changeof 360 (does not encircle the origin).EXAMPLE :I(s)F(c2)c2αR(s) One pole inside contour. Resulting map undergoes 360 net phasechange. (Encircles the origin).EXAMPLE :I(s)F(c3)c3R(s) In this example, there are two poles inside the contour, and the mapencircles the origin twice.c 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝
8–29ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSIS8.10: Cauchy’s theorem and Nyquist’s rule These examples give heuristic evidence of the general rule: Cauchy’stheorem“Let F(s) be the ratio of two polynomials in s. Let the closedcurve C in the s-plane be mapped into the complex planethrough the mapping F(s). If the curve C does not pass throughany zeros or poles of F(s) as it is traversed in the CW direction,the corresponding map in the F(s)-plane encircles the originN Z P times in the CW direction,” whereZ # of zeros of F(s) in C,P # of poles of F(s) in C. Consider the following feedback system:r (t)D(s)G(s)y(t)H (s) T (s) D(s)G(s).1 D(s)G(s)H (s)For closed-loop stability, no poles of T (s)in RHP.I(s)R No zeros of 1 D(s)G(s)H (s) in RHP. Let F(s) 1 D(s)G(s)H (s). Count zeros in RHP using Cauchytheorem! (Contour entire RHP). The Nyquist criterion simplifies Cauchy’s criterion for feedbacksystems of the above form.c 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝R(s)
8–30ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSIS Cauchy: F(s) 1 D(s)G(s)H (s).Nyquist: F(s) D(s)G(s)H (s).I(s)N # of encirclements of origin.N # of encirclements of 1.CauchyNyquistI(s)R(s) Simple? YES!!! Think of Nyquist path as four parts:R(s)I(s)I. Origin. Sometimes a special case (laterexamples).II. jω-axis. FREQUENCY-response ofO.L. system! Just plot it as a polar plot.IIIIIR(s)IIVIII. For physical systems 0.IV. Complex conjugate of II. So, for most physical systems, the Nyquist plot, used to determineCLOSED-LOOP stability, is merely a polar plot of LOOP frequencyresponse D( jω)G( jω)H ( jω). We don’t even need a mathematical model of the system. Measureddata of G( jω) combined with our known D( jω) and H ( jω) areenough to determine closed-loop stability.c 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝
ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSISTHE TEST: N # encirclements of 1 point when F(s) D(s)G(s)H (s). P # poles of 1 F(s) in RHP # of open-loop unstable poles. (assuming that H (s) is stable—reasonable). Z # of zeros of 1 F(s) in RHP # of closed-loop unstable poles.Z N PThe system is stable iff Z 0. Be careful counting encirclements!Draw line from 1 in any direction.Count # crossings of line and diagram.N #CW crossings #CCW crossings.Changing the gain K of F(s) MAGNIFIES the entire plot.ENHANCED TEST:Loop transfer function is K D(s)G(s)H (s). N # encirclements of 1/K point when F(s) D(s)G(s)H (s). Rest of test is the same. Gives ranges of K for stability.c 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝8–31
8–32ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSIS8.11: Nyquist test exampleD(s) H (s) 1.5G(s) (s 1)25or, G( jω) ( jω 1)2EXAMPLE :ωR(G( jω))I(G( 460-0.0005-0.0000500.0000-0.0000-0.0000At s 0, G(s) 5.I:5.(1 jω)2III : At s , G(s) 0.II :At s jω, G( jω) IV :At s 5.(1 jω)2 jω, G(s) I(s)R(s) No encirclements of 1, N 0.No open-loop unstable poles P 0.Z N P 0. Closed-loop system is stable.No encirclements of 1/K for any K 0. So, system is stable for any K 0.c 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝
8–33ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSIS Confirm by checking Routh array. Routh array: a(s) 1 K G(s) s 2 2s 1 5K .11 5Ks2s12s 0 1 5K Stable for any K 0.50.(s 1)2(s 10)I: G(0) 50/10 5.50II: G( jω) .2( jω 1) ( jω 10)III: G( ) 0.EXAMPLE :G(s) IV: G( jω) G( jω) . Note loop to left of origin. Systemis NOT stable for all K 0.I(s)ωR(G( jω))I(G( -0.00000.0000ZoomR(s)c 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝
8–34ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSIS8.12: Nyquist test example with pole on jω-axis1.s(τ s 1) WARNING! We cannot blindly follow procedure!EXAMPLE :Pole(s) at origin. G(s) Nyquist path goes through pole at zero! (Remember from Cauchy’stheorem that the path cannot pass directly through a pole or zero.) Remember: We want to count closed-loop poles inside a “box” thatencompasses the RHP. So, we use a slightly-modified Nyquist path.I(s)I(s)IIIρ 0IIθR(s)IZoomR(s)IV The bump at the origin makes a detour around the offending pole. Bump defined by curve: s lim ρe j θ ,ρ 0 From above,G(s) s ρe j θ 0 θ 90 .1,ρe j θ (τρe j θ 1)Consider magnitude as ρ 0lim G(s) s ρe j θ ρ 00 θ 90 11 .ρ τρe j θ 1 ρc 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝
8–35ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSIS Consider phase as ρ 0lim ̸ G(s) s ρe j θ θ ̸ (τρe j θ 1).ρ 0 So,1̸ θ ρ 0 ρlim G(ρe j θ ) limρ 0 This is an arc of infinite radius, sweeping from 0 to 90 (a littlemore than 90 because of contribution from WE CANNOT DRAW THIS TO SCALE! 1term).(τ s 1)Z N P.N # encirclements of 1. N 0.P # Loop transfer function polesinside MODIFIED contour. P 0.Z 0. Closed-loop system is stable.EXAMPLE :G(s) Use modified Nyquist path again1s 2(s 1)I: Near origin1.ρ 2e j 2θ (1 ρe j θ )11jθ Magnitude: lim G(ρe .) 2ρ 0ρ 1 ρe j θ ρ 2jθ Phase: lim ̸ G(ρe) 0 [2θ ̸ (1 ρe j θ )] 2θ . So,G(s) s ρe j θ ρ 01̸ 2θ 2ρ 0 ρlim G(ρe j θ ) limρ 00 θ 90 .c 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝
ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSIS8–36Infinite arc from 0 to 180 (a little1term.)more than 180 because of1 s Z N P 2 0 2. Unstable forK 1. In fact, unstable for any K 0! Matlab for aboveG(s) num [0 0 0 1];1s 3 s 2 0s 0den [1 1 0 0];nyquist1(num,den);axis([xmin xmax ymin ymax]); “nyquist1.m” is available on course web site. It repairs the standard Matlab “nyquist.m” program, which doesn’twork when poles are on imaginary axis. “nyquist2.m” is also available. It draws contours around poles on theimaginary axis in the opposite way to “nyquist1.m”. Counting isdifferent.c 1998–2017, Gregory L. Plett and M. Scott TrimboliLecture notes prepared by and copyright ⃝
8–37ECE4510/ECE5510, FREQUENCY-RESPONSE ANALYSIS8.13: Stability (gain and phase) margins A large fraction of systems to be controlled are stable for small gainbut become unstable if gain is increased beyond a certain point. The distance between the current (stable) system and an unstablesystem is called a “stability margin.” Can have a gain margin and a phase margin.GAIN MARGIN :Factor by which the gain is less than the neutral stabilityvalue. Gain margin measures “How much can we increase the gain of theloop transfer function L(s) D(s)G(s)H (s) and still have a stablesystem?” Many Nyquist plots are like this one.Increasing loop gain magnifies theplot. GM 1/(distance between origin andplace where Nyquist map crossesreal axis).1GMPM If we increase gain, Nyquist map“stretches” and we may encircle 1. For a stable system, GM 1 (linear units) or GM 0 dB.
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