Topic #3 16.31 Feedback Control Frequency response methods Analysis e Synthes ● Performance Stability Copyright 2001 by Jonathan How
Topic #3 16.31 Feedback Control Frequency response methods • Analysis • Synthesis • Performance • Stability Copyright 2001 by Jonathan How. 1
Fall 2001 16.313-1 Introduction . Root locus methods have Advantages k Good indicator if transient response k Explicity shows location of all closed-loop poles Trade-offs in the design are fairly clear Disadvantages k Requires a transfer function model(poles and zeros) k Difficult to infer all performance metrics k Hard to determine response to steady-state(sinusoids Frequency response methods are a good complement to the root locus techniques Can infer performance and stability from the same plot Can use measured data rather than a transfer function model The design process can be independent of the system order Time delays are handled correctly Graphical techniques(analysis and synthesis) are quite simple
Fall 2001 16.31 3–1 Introduction • Root locus methods have: – Advantages: ∗ Good indicator if transient response; ∗ Explicity shows location of all closed-loop poles; ∗ Trade-offs in the design are fairly clear. – Disadvantages: ∗ Requires a transfer function model (poles and zeros); ∗ Difficult to infer all performance metrics; ∗ Hard to determine response to steady-state (sinusoids) • Frequency response methods are a good complement to the root locus techniques: – Can infer performance and stability from the same plot – Can use measured data rather than a transfer function model – The design process can be independent of the system order – Time delays are handled correctly – Graphical techniques (analysis and synthesis) are quite simple
Fall 2001 16.313-2 Frequency response Function Given a system with a transfer function G(s), we call the G(jw) ∈0,∞) the frequency response function(FRF) GGjw)=G(jw)l arg G ljw The frf can be used to find the steady-state response of a system to a sinusoidal input. If t)→(G(s)→y(t) and e(t)=sin 2t, G(2j)=0.3, arg G(2j)=800, then the steady-state output is y(t)=0.3sin(2t-80°) The FRF clearly shows the magnitude(and phase) of the response of a system to sinusoidal input e a variety of ways to display this 1. Polar(Nyquist) plot-Re vs. Im of G w) in complex plane Hard to visualize, not useful for synthesis, but gives definitive tests for stability and is the basis of the robustness analysis 2. Nichols Plot-GGjw) vs. arg Gw), which is very handy for systems with lightly damped poles 3. Bode Plot-Log G(jw and arg G(jw) vs Log frequency Simplest tool for visualization and synthesis Typically plot 20log G which is given the symbol dB
Fall 2001 16.31 3–2 Frequency response Function • Given a system with a transfer function G(s), we call the G(jω), ω ∈ [0, ∞) the frequency response function (FRF) G(jω) = |G(jω)| arg G(jω) – The FRF can be used to find the steady-state response of a system to a sinusoidal input. If e(t) → G(s) → y(t) and e(t) = sin 2t, |G(2j)| = 0.3, arg G(2j) = 80◦ , then the steady-state output is y(t)=0.3 sin(2t − 80◦ ) ⇒ The FRF clearly shows the magnitude (and phase) of the response of a system to sinusoidal input • A variety of ways to display this: 1. Polar (Nyquist) plot – Re vs. Im of G(jω) in complex plane. – Hard to visualize, not useful for synthesis, but gives definitive tests for stability and is the basis of the robustness analysis. 2. Nichols Plot – |G(jω)| vs. arg G(jω), which is very handy for systems with lightly damped poles. 3. Bode Plot – Log |G(jω)| and arg G(jω) vs. Log frequency. – Simplest tool for visualization and synthesis – Typically plot 20log |G| which is given the symbol dB
Fall 2001 16.313-3 Use logarithmic since if (s+1)(s+2) pg s+3)(s+4 log s+1|+log s+2-log s+3-log s+4 and each of these factors can be calculated separately and then added to get the total FRF Can also split the phase plot since (s+1)(s+2) arg (s+3)(s+4) arg(s +1)+arg(s+ 2) arg(s +3)-arg(s+4) The keypoint in the sketching of the plots is that good straightline approximations exist and can be used to obtain a good prediction of the system response
Fall 2001 16.31 3–3 • Use logarithmic since if log |G(s)| = (s + 1)(s + 2) (s + 3)(s + 4) = log |s + 1| + log |s + 2| − log |s + 3| − log |s + 4| and each of these factors can be calculated separately and then added to get the total FRF. • Can also split the phase plot since arg (s + 1)(s + 2) (s + 3)(s + 4) = arg(s + 1) + arg(s + 2) − arg(s + 3) − arg(s + 4) • The keypoint in the sketching of the plots is that good straightline approximations exist and can be used to obtain a good prediction of the system response
Fall 2001 16.313-4 Example ● Draw bode for S+ /10+ G Gw) j+1 /10+1 log|G(ju)}=log1+(u/1)2]12-log1+(a/102 ●A pproximation log1 +(w/ 211 logw/wil w>wi Two straightline approximations that intersect at w=w Error at wi obvious, but not huge and the straightline approxima tions are very easy to work with
Fall 2001 16.31 3–4 Example • Draw Bode for G(s) = s + 1 s/10 + 1 |G(jω)| = |jω + 1| |jω/10 + 1| log |G(jω)| = log[1 + (ω/1)2] 1/2 − log[1 + (ω/10)2] 1/2 • Approximation log[1 + (ω/ωi) 2 ] 1/2 ≈ 0 ω ωi log[ω/ωi] ω ωi Two straightline approximations that intersect at ω ≡ ωi • Error at ωi obvious, but not huge and the straightline approximations are very easy to work with. 10−2 10−1 100 101 102 100 101 102 Freq |G|
Fall 2001 16.313-5 To form the composite sketch Arrange representation of transfer function so that dC gain of each element is unity(except for parts that have poles or zeros lt the origin)-absorb the gain into the overall plant gain Draw all component sketches Start at low frequency(DC) with the component that has the lowest frequency pole or zero (i.e. s-=0) Use this component to draw the sketch up to the frequency of the next pole/zero Change the slope of the sketch at this point to account for the new dynamics:-1 for pole, +1 for zero, -2 for double poles Scale by overall dC gain 0° Figure 1: G(s)=10(s+1)/(s+10) which is a"lead
Fall 2001 16.31 3–5 To form the composite sketch, – Arrange representation of transfer function so that DC gain of each element is unity (except for parts that have poles or zeros at the origin) – absorb the gain into the overall plant gain. – Draw all component sketches – Start at low frequency (DC) with the component that has the lowest frequency pole or zero (i.e. s=0) – Use this component to draw the sketch up to the frequency of the next pole/zero. – Change the slope of the sketch at this point to account for the new dynamics: -1 for pole, +1 for zero, -2 for double poles, . . . – Scale by overall DC gain 10−2 10−1 100 101 102 103 10−1 100 101 102 Freq |G| Figure 1: G(s) = 10(s + 1)/(s + 10) which is a “lead
Fall 2001 16.313-6 Since arg Gga) +ju)-arg(1+j/10 we can construct phase plot for complete system in a similar fashion Know that arg(1+jw/wi)=tan(w/wi Can use straightline approximations 0 arg(1+j/u)≈{90/≥10 Draw the components using breakpoints that are at wi /10 and 10wi igure 2: Phase plot for(s+1)
Fall 2001 16.31 3–6 • Since arg G(jω) = arg(1 + jω) − arg(1 + jω/10), we can construct phase plot for complete system in a similar fashion – Know that arg(1 + jω/ωi) = tan−1(ω/ωi) • Can use straightline approximations arg(1 + jω/ωi) ≈ 0 ω/ωi ≤ 0.1 90◦ ω/ωi ≥ 10 45◦ ω/ωi = 1 • Draw the components using breakpoints that are at ωi/10 and 10ωi 10−2 10−1 100 101 102 103 0 10 20 30 40 50 60 70 80 90 100 Freq Arg G Figure 2: Phase plot for (s + 1)
Fall 2001 16.313-7 Then add them up starting from zero frequency and changing the Slope as→ Figure 3: Phase plot G(s)=10(s+1)/(s+ 10)which is a"lead
Fall 2001 16.31 3–7 • Then add them up starting from zero frequency and changing the slope as ω → ∞ 10−2 10−1 100 101 102 103 −80 −60 −40 −20 0 20 40 60 80 Freq Arg G Figure 3: Phase plot G(s) = 10(s + 1)/(s + 10) which is a “lead
Fall 2001 16.313-8 Freq(Hz) ode for G(s)=s+0.18182-31288-454 The poles are at(-0.892,0.886,-0.0227
Fall 2001 16.31 3–8 10−4 10−3 10−2 10−1 100 101 10−3 10−2 10−1 100 Freq (Hz) Magnitude Actual LF MF HF +1 0 −2 −2 +1 −1 10−4 10−3 10−2 10−1 100 101 102 −180 −160 −140 −120 −100 −80 −60 −40 −20 0 20 Freq (Hz) Phase (deg) Actual LF MF HF Bode for G(s) = 4.54s s3 + 0.1818s2 − 31.1818s − 4.4545. The poles are at (-0.892, 0.886, -0.0227)
Fall 2001 Non-minimum Phase Systems Bode plots are particularly complicated when we have non-minimum tems A system that has a pole/zero in the RhP is called non-minimum ase The reason is clearer once you have studied the Bode Gain Phase relationship Key point: We can construct two(and many more) systems that have identical magnitude plots, but very different phase diagrams Consider Gi(s)=s+2 and G2(s)=s s+2 --=-“""--- Freq Figure 4: Magnitude plots are identical, but the phase plots are dramatically different. NMP has a 180 deg phase loss over this frequency range
Fall 2001 Non-minimum Phase Systems • Bode plots are particularly complicated when we have non-minimum phase systems – A system that has a pole/zero in the RHP is called non-minimum phase. – The reason is clearer once you have studied the Bode GainPhase relationship – Key point: We can construct two (and many more) systems that have identical magnitude plots, but very different phase diagrams. • Consider G1(s) = s+1 s+2 and G2(s) = s−1 s+2 10−1 100 101 102 10−1 100 Freq |G| MP NMP 10−1 100 101 102 0 50 100 150 200 Freq Arg G MP NMP Figure 4: Magnitude plots are identical, but the phase plots are dramatically different. NMP has a 180 deg phase loss over this frequency range
