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


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1 Plan of the Lecture Review: design using Root Locus; dynamic compensation; PD and lead control Today s topic: PI and lag control; introduction to frequencyresponse design method Goal: wrap up lead and lag control; start looking at frequency response as an alternative methodology for control systems design. Reading: FPE, Sections , 6.
2 Recap: Lead & Lag Compensators Consider a general controller of the form K s + z s + p K, z, p > 0 are design parameters Depending on the relative values of z and p, we call it: a lead compensator when z < p a lag compensator when z > p Why the name lead/lag? think frequency response jω + z = (jω + z) (jω + p) = ψ φ jω + p if z < p, then ψ φ > 0 (phase lead) if z > p, then ψ φ < 0 (phase lag) z p!
3 Approximate PI via Dynamic Compensation PI control achieves the objective of stabilization and perfect steadystate tracking of constant references; however, just as with PD earlier, we want a stable controller. Here s an idea: replace K s + s More generally, if z = K I /K P, then by K s +, where p is small s + p replace K s + z s by K s + z s + p, where p < z This is lag compensation (or lag control)! We use lag controllers as dynamic compensators for approximate PI control.
4 Approximate PI via Lag Compensation G c (s) = K s + z s + p, p < z G p(s) = s How good is this controller? Tracking a constant reference: assuming closedloop stability, the FVT gives e( ) = = + G c (s)g p (s) s=0 + K s+z = s=0 (s+p)(s ) Kz p Check for stability: no RHP poles for + G c (s)g p (s) (s + p)(s ) + K(s + z) = 0 s 2 + (K + p )s + Kz p = 0 Conditions for stability: K > p, Kz > p
5 Approximate PI via Lag Compensation Tracking a constant reference: if the stability conditions K > p, Kz > p are satisfied, then the steadystate error is e( ) = Kz p this will be close to zero (and negative) if Kz p is large. Lag compensation does not give perfect tracking (indeed, it does not change system type), but we can get as good a tracking as we want by playing with K, z, p. On the other hand, unlike PI, lag compensation gives a stable controller.
6 Effect of Lag Compensation on Root Locus L(s) = s + (s + p)(s ) Intuition: By choosing p very close to zero, we can make the root locus arbitrarily close to PI root locus (stable for large enough K). Let s check: Try p = 0. Compare to PI root locus: What do we see? Compared to PD vs. lead, there is no qualitative change in the shape of RL, since we are not changing #(poles) or #(zeros).
7 More Pole Placement As before, we can choose z lag for a fixed p lag (or vice versa) based on desired pole locations. The procedure is exactly the same as the one we used with lead. (In fact, depending on the pole locations, we may end up with either lead or lag.) Main technique: select parameters to satisfy the phase condition (points on RL must be such that L(s) = 80 ). Caveat: may not always be possible!
8 Pole Placement via RL Let G p (s) = s, G c(s) = K s + z s + p Problem: given p = 2, find K and z to place poles at 2 ± 3j. Desired characteristic polynomial: (s + 2) = s 2 + 4s + 3, damping ratio ζ = Im s given 3 Must have o z (to be selected) ' 2 = 90 x 2 0 x ' = 35 Re ψ }{{} angle from i s to zero ϕ }{{} i = 80 angles from s to poles So, we want ψ = 80 + i ϕ i
9 Pole Placement via RL Im o z (to be selected) s given ' 2 = 90 x x ' = 35 Re We have ϕ = 35, ϕ 2 = 90 We want ψ = 80 + i ϕ i Must have ψ = = 405 = 45 mod 360 Thus, we should have z = 5 Im s given 3 = 45 ' 2 = 90 ' = 35 o x z = 5 0 x 2 Re
10 Let G p (s) = s, G c(s) = K s + z s + p Problem: given p = 2, find z to place poles at 2 ± 3j. Solution: we already found that we need z = 5 resulting characteristic polynomial: (s )(s + 2) + K(s + 5) s 2 + (K + )s + 5K 2 compare against desired characteristic polynomial: s 2 + 4s + 3 = K + = 4, 5K 2 = 3 so we need K = 3 compute s.s. tracking error: Kz p = 6.5 5%
11 Story So Far PD control: provides stability, allows to shape transient response specs replace noncausal Dcontroller Ks with a causal, stable lead controller K s + z s + p, where p > z this introduces a zero in LHP (at z), pulls the root locus into LHP shape of RL differs depending on how large p is PI control: provides stability and perfect steadystate tracking of constant references replace unstable Icontroller K/s with a stable lag controller K s + z s + p, where p < z this does not change the shape of RL compared to PI
12 What About PID Control? Obvious solution combine lead and lag compensation. We will develop this further in homework and later in the course using frequencyresponse design methods which are the subject of several lectures, starting with today s.
13 The FrequencyResponse Design Method Recall the frequencyresponse formula: sin(!t) M sin(!t + ) G(s) where M = M(ω) = G(jω) and φ = φ(ω) = G(jω) Derivation:. u(t) = e st y(t) = G(s)e st 2. Euler s formula: sin(ωt) = ejωt e jωt 3. By linearity, 2j sin(ωt) G(jω)ejωt G( jω)e jωt 2j = M(ω)ej(ωt+φ(ω)) M(ω)e j(ωt+φ(ω)) 2j = M(ω) sin(ωt + φ(ω)) G(jω) = M(ω)e jφ(ω)
14 The FrequencyResponse Design Method sin(!t) M sin(!t + ) G(s) where M = M(ω) = G(jω) and φ = φ(ω) = G(jω) Let s apply this formula to our prototype 2ndorder system: ω 2 n G(s) = s 2 + 2ζω n s + ωn 2 M(ω) = G(jω) = ω 2 n ω 2 + 2jζω n ω + ωn 2 = ( ) ω 2 ω n + 2ζ ω ω n j = [ ( ) ] ω 2 2 ( ) ω n + 4ζ 2 ω 2 ω n
15 The FrequencyResponse Design Method For our prototype 2ndorder system: G(s) = M(ω) = ω 2 n s 2 + 2ζω n s + ωn 2 [ ( ) ] ω 2 2 ( = ) ω n + 4ζ 2 ω 2 ω n + (4ζ 2 2) ( ) ω 2 ( ω n + ω ) 4 ω n MHwL z=ê2 z=ê 2 z= w w n
16 Frequency Response Parameters Here is a typical frequency response magnitude plot: MHwL.0 M r 0.8 / p ! r! BW w ω r resonant frequency M r resonant peak ω BW bandwidth
17 Frequency Response Parameters MHwL.0 M r 0.8 / p ! r! BW w We can get the following formulas using calculus: ω r = ω n 2ζ 2 M r = 2ζ ζ (valid for ζ < ; for ζ, ω r = 0) ω BW = ω n ( 2ζ 2 ) + ( 2ζ 2 ) 2 + }{{} = for ζ=/ 2 so, if we know ω r, M r, ω BW, we can determine ω n, ζ and hence the timedomain specs (t r, M p, t s )
18 Frequency Response & TimeDomain Specs All information about time response is also encoded in frequency response!! MHwL.0 M r 0.8 / p ! r! BW w small M r better damping large ω BW large ω n smaller t r
19 FrequencyResponse Design Method: Main Idea R + K G(s) Y Twostep procedure:. Plot the frequency response of the openloop transfer function KG(s) [or, more generally, D(s)G(s)], at s = jω 2. See how to relate this openloop frequency response to closedloop behavior. We will work with two types of plots for KG(jω):. Bode plots: magnitude KG(jω) and phase KG(jω) vs. frequency ω (could have seen it earlier, in ECE 342) 2. Nyquist plots: Im ( KG(jω) ) vs. Re ( K(jω) ) [Cartesian plot in splane] as ω ranges from to +
20 Note on the Scale Horizontal (ω) axis: we will use logarithmic scale (base 0) in order to display a wide range of frequencies. Note: we will still mark the values of ω, not log 0 ω, on the axis, but the scale will be logarithmic: ! Equal intervals on log scale correspond to decades in frequency.
21 Note on the Scale Vertical axis on magnitude plots: Reason: we will also use logarithmic scale, just like the frequency axis. (M e jφ )(M 2 e jφ 2 ) = M M 2 log(m M 2 ) = log M + log M 2 this means that we can simply add the graphs of log M (ω) and log M 2 (ω) to obtain the graph of log ( M (ω)m 2 (ω) ), and graphical addition is easy. Decibel scale: (M) db = 20 log 0 M (one decade = 20 db)
22 Note on the Scale Vertical axis on phase plots: we will plot the phase on the usual (linear) scale. Reason: ( ) (M e jφ )(M 2 e jφ 2 ) = (M ) M 2 e j(φ +φ 2 ) = φ + φ 2 this means that we can simply add the phase plots for two transfer functions to obtain the phase plot for their product.
23 Scale Convention for Bode Plots magnitude phase horizontal scale log log vertical scale log linear Advantage of the scale convention: we will learn to do Bode plots by starting from simple factors and then building up to general transfer functions by considering products of these simple factors.
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