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


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1 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 : frequencydomain uncertainty, unmodeled dynamics, nonlinearity Robust Control Objective : Design a controller satisfying stability and performance for a set of models Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
2 Model Uncertainty and Feedback The aim of feedback is to overcome the model uncertainty r(t) e(t) u(t) K G +  v(t) y(t) Whatever the plant model is, large GK leads to T = GK 1 (good tracking) 1+GK 1 S = 0 (good disturbance rejection) 1+GK For an openloop stable system : K = 0 (robust stability) K (good performance) Loopshaping : G(jω)K(jω) should be large in the frequencies where good performances are desired and small where the stability is critical. Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
3 Basic Definitions Stability : A transfer function G (s) is stable if it is analytic in the closed Right Half Plane RHP (Re s 0). Properness : G(s) isproper if G(j ) is finite (deg den deg num) G(s) isstrictly proper if G (j ) =0(degden> deg num) G(s) isbiproper if (deg den = deg num) Internal Stability : A closedloop system is internally stable if the transfer functions from all external inputs to all internal signals are stable. For a unity feedback system the following four transfer functions should be stable. 1 1+GK G 1+GK K 1+GK GK 1+GK Wellposedness : A closedloop system with unity feedback is wellposed iff 1 + GK 0. This condition is met if GK( ) 1 orifgk is strictly proper (a feedback system with G =1, K = 1 is not wellposed). Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
4 Internal Stability Theorem A unity feedback system is internally stable if and only if there are no zeros in Re s 0 in the characteristic polynomial where N G N K + M G M K =0 G = N G M G, K = N K M K or the following two conditions hold : (a) The transfer function 1+GK has no zeros in Re s 0. (b) There is no polezero cancellation in Re s 0 when the product GK is formed. or the Nyquist plot of GK does not pass through the point 1 and encircles it n times counterclockwise, where n denotes the number of unstable poles of G and K. Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
5 Norms for Signals Consider piecewise continuous signals mapping (, + ) tor. Anorm must have the following four properties : 1 u 0 (positivity) 2 au = a u, a R (homogenity) 3 u =0 u(t) =0 t (positive definiteness) 4 u + v u + v (triangle inequality) 1Norm : u 1 = u(t) dt ( ) 1/2 2Norm : u 2 = u 2 (t)dt ( u 2 2 is the total signal energy) Norm : u =sup u(t) t ( 1/p pnorm : u p = u(t) dt) p 1 p Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
6 Norms for Systems (SISO) Consider linear, timeinvariant, causal and finitedimensional systems. y(t) =g(t) u(t), y(t) = g(t τ)u(τ)dτ, G (s) =L[g(t)] 2Norm : This norm is bounded if G (s) is strictly proper and has no pole on the imaginary axis. ( 1 ) 1/2 G 2 = G (jω) 2 dω 2π Norm : is bounded if G (s) has no pole on the imaginary axis. G =sup G (jω) ω Parseval s theorem : (for stable systems) ( 1 1/2 ( G 2 = G (jω) dω) 2 = g(t) 2 dt 2π ) 1/2 Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
7 Norms for Systems (MIMO) Given G(s) a multiinput multioutput system 2Norm : This norm is defined as ( 1 G 2 = trace [G (jω)g (jω)] dω 2π Norm : The H norm is defined as G =sup ω G (jω) =sup σ[g (jω)] ω ) 1/2 Remark : The infinity norm has an important property (submultiplicative) GH G H Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
8 Computing the Norms How to compute the 2norm : Suppose that G has bounded twonorm, we have : G 2 2 = 1 2π = 1 2πj G (jω) 2 dω = 1 j G( s)g (s)ds 2πj j G ( s)g (s)ds Then by the residue theorem, G 2 2 equals the sum of the residues of G( s)g(s) at its poles in the left halfplane (LHP). Example Compute the two norm of the following transfer function : G (s) = 1 τs +1 Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
9 Computing the Norms How to compute the norm : Choose a fine grid of frequency points {ω 1,...,ω N },then SISO : G max 1 k N G (jω k) MIMO : G max 1 k N σ[g (jω k)] or alternatively, solve d G(jω) 2 dω =0 Example Compute the infinity norm of G(s) = as +1 bs +1 a, b > 0 If a b If a < b Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
10 Inputoutput relationships If we know how big the input is, how big is the output going to be? Proofs : If u(t) =δ(t) theny(t) = Output Norms for Two Inputs u(t) δ(t) sin(ωt) y 2 G 2 y g G(jω) g(t τ)δ(τ)dτ = g(t), so y 2 = g 2 = G 2 If u(t) =δ(t) theny(t) =g(t), so y = g If u(t) =sin(ωt) theny(t) = G(jω) sin(ωt + φ), so y 2 = and y = G(jω) Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
11 Inputoutput relationships Norms for Signals and Systems System Gains : u 2 =1 u =1 y 2 G y G 2 g 1 Entry (1,1) : We have y 2 2 = 1 G(jω) 2 U(jω) 2 dω G 2 1 U(jω) 2 dω 2π 2π = G 2 U 2 2 = G 2 u 2 2 Entry(2,1) : According to the CauchySchwartz inequality ( ) 1/2 ( y(t) = g(t τ)u(τ)dτ g 2 (t τ)dτ u 2 (τ)dτ = g 2 u 2 = G 2 u 2 y G 2 u 2 ) 1/2 Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
12 Asymptotic Tracking Internal Model Principle : For perfect asymptotic tracking of r(t), the loop transfer function L = GK must contain the unstable poles of r(s). Theorem Assume that the feedback system is internally stable and n=d=0. (a) If r(t) is a step, then lim e(t) =r(t) y(t) =0iff t S =(1+L) 1 has at least one zero at the origin. (b) If r(t) is a ramp, then lim e(t) =0iffS has at least two t zerosattheorigin. (c) If r(t) =sin(ωt), then lim e(t) =0iffS has at least one t zero at s = jω. FinalValue Theorem : If y(s) has no poles in Re s 0 except possibly one pole at s =0then: lim y(t) = lim sy(s) t s 0 Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
13 Nominal Performance Tracking performance can be quantified in terms of a weighted norm of the sensitivity function 1 Sensitivity Function : TF from r to tracking error e : S = 1+GK Complementary Sensitivity Function : TF from r to y : T = GK 1+GK S is the relative sensitivity of T with respect to relative perturbations in G : T /T S = lim G 0 G/G = dt G dg T Performance Specification : 2 K (1 + GK ) GK G(1 + GK ) = (1 + GK ) 2 = GK 1 1+GK 1 r(t) is any sinusoid of amplitude 1 filtered by W 1, then the max. amp. of e is W 1 S. 2 In some applications good performance is achieved if S(jω) < W 1 (jω) 1, ω or W 1 S < 1 W 1 (jω) < 1+L(jω), ω Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
14 Nominal Performance In many control applications, the nominal performance can be defined as W 1 S < 1 where W 1 (s) is typically a lowpass filter It guarantees small S(jω) at low frequencies. Graphical interpretation : The nominal performance in the frequencydomain is given by : Im W 1 (jω) W 1 (jω)s(jω) < 1 ω W 1 (jω) 1 1+L(jω) < 1 ω, Re W 1 (jω) < 1+L(jω), ω L(jω) Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
15 Model Uncertainty We cannot exactly model the physical systems so there is always the modeling errors. The best technique is to define a model set which can be structured or unstructured. Structured model set such as parametric uncertainty 1 G = { s 2 + as +1 : a min a a max } or multimodel uncertainty G = {G 0, G 1, G 2, G 3 } Unstructured model set such as unmodeled dynamics or disk uncertainty G = {G 0 + : γ} or frequencydomain uncertainty G = {G (jω) S 1 (jω) < G (jω) < S 2 (jω) } Conservatism : Controller design for a model set greater than the real model set leads to a conservative design. Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
16 Model Uncertainty Unstructured uncertainty : Additive uncertainty G = G + W 2 1 Multiplicative uncertainty Feedback uncertainty G = G = G (1 + W 2 ) 1 G 1+ W 2 G or G = G 1+ W 2 1 G :truemodel : normbounded uncertainty G : nominal model W 2 : weighting filter Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
17 Model Uncertainty Example (Multimodel to multiplicative uncertainty) m frequencyresponse models are identified. Find the multiplicative uncertainty model and the weighting filter. if G = G (1 + W 2 ) G G 1= W 2 G (jω) 1 G (jω) 1 W 2(jω) Let G k (jω) be the frequency response of the model at the kth experiment and G(jω) that of the nominal model (e.g. the mean value). max G k (jω i ) k G (jω i ) 1 W 2(jω i ) i Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
18 Model Uncertainty Example (Parametric to multiplicative uncertainty) { } k Suppose that G(s) = :0.1 k 10. s 2 G(s) = k 0 s 2 G(jω) G(jω) 1 W 2(jω) max k 0.1 k 10 1 k 0 W 2(jω) Thebestvaluefork 0 is 5.05 which gives W 2 (s) =4.95/5.05 Example (Parametric to feedback uncertainty) { } 1 s 2 :0.4 a 0.8 take a = , as +1 So G(s) = where G(s) = 1 s s +0.2 s +1 = 1 s s +1, G (s) 1+ W 2 (s)g (s) W 2(s) =0.2s Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
19 Model Uncertainty Example (Timedelay to multiplicative uncertainty) Assume that G(s) = 1 and G(s) =e τs 1 where 0 τ 0.1. s 2 s 2 G(jω) G(jω) 1 W 2(jω) e τjω 1 W 2 (jω) ω,τ 10 Bode Diagram Using the Bode diagram we can find W 2 (s) = 0.21s 0.1s +1 Magnitude (db) Frequency (rad/s) Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
20 Robust Stability Robustness : A controller is robust with respect to a closedloop characteristic, if this characteristic holds for every plant in G Robust Stability : A controller is robust in stability if it provides internal stability for every plant in G. Stability margin : For a given model set with an associate size, it can be defined as the largest model set stabilized by a controller. Stability margin for an uncertainty model : Given G = G (1 + W 2 ) with β, the stability margin for a controller C is the least upper bound of β. Im L Modulus margin : The distance from 1 to the openloop Nyquist curve. M m = inf 1 L(jω) =inf 1+L(jω) ω ω [ ] 1 1 = sup = S 1 ω 1+L(jω) 1 M m L(jω) 1 Re L Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
21 Robust Stability Theorem (Small Gain) Suppose H is stable and has bounded infinity norm and let γ>0. The following feedback loop is internally stable for all stable (s) with (s) H(s) 1/γ if and only if H <γ Robust stability condition for plants with additive uncertainty : K G = G+ W 2 H = W 2 1+GK Closedloop system is internally stable for all 1 iff W 2 KS < 1 W 2 r(t) + K G y(t)  Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
22 Robust Stability Robust stability condition for plants with multiplicative uncertainty : GK G = G(1+ W 2 ) H = W 2 W 2 1+GK r(t) + K G y(t) Closedloop system is internally  stable for all 1 iff W 2 T < 1. Proof : Assume that W 2 T < 1. We show that the winding number of 1+GK around zero is equal to that of 1 + GK. 1+ GK =1+GK(1+ W 2 )=1+GK+GK W 2 =1+GK+(1+GK)T W 2 1+ GK =(1+GK)(1 + W 2 T ) so wno { (1 + GK)} =wno{(1 + GK)} +wno{(1 + W 2 T )}. But wno {(1 + W 2 T )} = 0 because W 2 T < 1. Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
23 Robust Stability Robust stability condition for plants with feedback uncertainty (1) : G 1 G = H = W 2 1+ W 2 1+GK Closedloop system is internally stable for all 1 iff W 2 S < 1. W 2 r(t) K G  y(t)  Robust stability condition for plants with feedback uncertainty (2) : G G = 1+ W 2 G H = W G 2 1+GK r(t) Closedloop system is internally  stable for all 1 iff W 2 GS < 1. K W 2  G y(t) Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
24 Robust Stability The robust stability condition for systems with multiplicative uncertainty is defined as W 2 T < 1 where W 2 (s) is typically a highpass filter. It guarantees small T (jω) at high frequencies, where unmodelled dynamics are large. Graphical interpretation : The robust stability condition in the frequencydomain is given by : Im W 2 (jω)t (jω) < 1 ω W 2 (jω)l(jω) 1+L(jω) < 1 ω, 1 Re W 2 (jω)l(jω) < 1+L(jω), ω W 2 L L Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
25 Robust Performance Nominal performance condition : W 1 S < 1 Robust stability condition for multiplicative uncertainty : W 2 T < 1 Robust performance for multiplicative uncertainty : W 2 T < 1and W 1 S < 1 where : 1 S = 1+ GK = 1 1+GK (1+ W 2 ) = 1 (1 + GK )(1+ W 2 T ) = S 1+ W 2 T or W 2 T < 1and W 1 S 1+ W 2 T < 1 Theorem A necessary and sufficient condition for robust performance of a plant model with multiplicative uncertainty is W 1 S + W 2 T < 1 Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
26 Robust Performance Graphical interpretation : The robust performance condition for systems with multiplicative uncertainty is given by : W 1 S + W 2 T < 1 Im W 1 (jω) 1+L(jω) + W 2 (jω)l(jω) 1+L(jω) < 1 ω W 1 (jω) + W 2 (jω)l(jω) < 1+L(jω), ω W 11 W 2 L L Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
27 Limit of Performance Algebraic Constraints : S + T =1so S(jω) and T (jω) cannot both be less than 1/2 at the same frequency. A necessary condition for robust performance is that : min{ W 1 (jω), W 2 (jω) } < 1, ω So at every frequency either W 1 or W 2 must be less than 1. Typically W 1 is monotonically decreasing and W 2 is monotonically increasing. If p is a pole and z azeroofl both in Re s 0then: S(p) =0 S(z) =1 T (p) =1 T (z) =0 Analytic Constraints : Bounds on the weights W 1 and W 2 : W 1 S W 1 (z) W 2 T W 2 (p) Proof from the Maximum Modulus Theorem : F = sup F (s) Re s>0 Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
28 Analytic Constraints Definition (AllPass and MinimumPhase Transfer Functions) Defin M as the set of stable transfer functions with bounded infinity norm. F (s) Mis allpass if F (jω) =1 ω G(s) Mis minimumphase if it has no zeros in Re s > 0. Every function G Mcan be presented as G = G ap G mp Suppose that z and p are the only zero and pole of G in the closed RHP and K has neither poles nor zeros there. Then : S ap (s) = s p s + p S(z) =1 S mp (z) =Sap 1 + p (z) =z z p W 1(z) z + p Then : W 1 S = W 1 S mp W 1 (z)s mp (z) = Similarly : T ap (s) = s z s + z, T (p) =1 W 2T z p W 2(p) p + z p z Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
29 Analytic Constraints Example Consider the inverse pendulum problem. y m (M + m)ẍ + ml( θ cos θ θ 2 sin θ) = u m(ẍ cos θ + l θ g sin θ) = d x u l Linearized model : ( ) x = θ 1 s 2 [Mls 2 (M + m)g] ls 2 g ls 2 s 2 M+m m s2 M ( ) u d Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
30 Analytic Constraints Example Measuring x : T ux = ls 2 g s 2 [Mls 2 (M + m)g] RHP poles and zeros : z = (M + m)g g/l p =0, 0, Ml If m M W 2 T 1( W 2 (p) is an increasing function) the system is difficult to control. The best case is m/m and l large. Measuring y or θ : T uθ = 1 Mls 2 (M + m)g T uy = g s 2 [Mls 2 (M + m)g] Since there is no RHP zero a larger l gives a smaller p so the system is easier to stabilize. Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
31 Analytic Constraints Theorem (The Waterbed Effect) Suppose that G has a zero at z with Re z > 0 and : M 1 := max S(jω) ω 1 ω ω 2 M 2 := S Then there exist positive constants c 1 and c 2, depending only on ω 1,ω 2 and z, such that : Theorem (The Area Formula) c 1 log M 1 + c 2 log M 2 log S 1 ap (z) 0 Assume that the relative degree of L is at least 2. Then 0 log S(jω) dω = π(log e) i Re p i where {p i } denotes the set of poles of L in Re s > 0. Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
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