Robust stability and Performance

Transcription

1 122 c Perry Y.Li

2 Chapter 5 Robust stability and Performance Topics: ([ author ] is supplementary source) Sensitivities and internal stability (Goodwin ) Modeling Error and Model Uncertainty (Goodwin , Doyle 4.1) Robust stability (Goodwin , Doyle 4.2) Robust performance (Doyle , Goodwin 5.9) Performance limitation (Glad , Doyle , Goodwin [8.6], ) Loop-shaping technique (Doyle 7.1) 5.1 Nominal Sensitivity Functions ( Y (s) T o (s) = D m (s) = Y (s) ) R(s) = Go (s)c(s) 1 + G o (s)c(s) ( ) Y (s) S o (s) = D o (s) = G o (s)c(s) ( ) Y (s) S io (s) = D i (s) = G o (s) 1 + G o (s)c(s) ( ) U(s) S uo (s) = D o (s) = C(s) 1 + G o (s)c(s) (5.1) (5.2) (5.3) (5.4) T o - Complementary sensitivity (goal: small for noise, 1 for command following) S o - Sensitivity (goal: small) S io - Input disturbance sensitivity (goal: small) S uo - Control sensitivity (goal: small) 123

3 124 c Perry Y.Li 5.2 Internal Stability The nominal loop is internally stable if and only if all eight transfer functions below is stable: ( ) Yo (s) = U o (s) ( Go (s)c(s) G o (s) 1 G o (s)c(s) C(s) G o (s)c(s) C(s) C(s) 1 + G o (s)c(s) Let C(s) = P (s)/l(s), and G o (s) = B o (s)/a o (s). ) H(s)R(s) D i (s) D o (s) D m (s) Proposition The system is internally stable if and only if the roots of the nominal closed loop characteristic equation A o (s)l(s) + B o (s)p (s) = 0 all lie in the open left half plane. Example Consider G o (s) = s s + 2, C(s) = (s + 4)( s + 2) s where pole-zero cancellations at s = 2 and s = 0 occur. Complementary sensitivity is stable: T o (s) = Y (s)/d m (s) = G(s)C(s) 1 + G o (s)c(s) = 1 s + 5 Sensitivity is also stable: S(s) = 1 T o (s) = s + 4 s + 5 However, control sensitivity is marginally stable: S uo (s) = U(s) D m (s) = And, input disturbance sensitivity is unstable: S io (s) = Y (s) D i (s) = C(s) ( s + 2)(s + 4) = 1 + G o (s)c(s) (s + 5)s s ( s + 2)(s + 5) The effect of unstable pole at s = 2 shows up in S io as input disturbance will drive the output Y to be unbounded. This effect is not observed by the controller as it is blocked by its zero. The effect of unstable pole at s = 0 shows up in the control sensitivity so that output disturbance or measurement noise will drive the control output to be unbounded. This effect is not observed at the output as it is blocked by the plant s zero. Characteristic equation has unstable and marginally stable poles: ( s + 2)s + (s + 4)( s + 2)s = ( s + 2)(s + 5)s predicting correctly that the system is NOT internally stable.

4 University of Minnesota ME 8281: Advanced Control Systems Design, Modeling Error For linear systems, if G o (s) is the nominal system, and the actual system is G(s), then define: Additive uncertainty: G ɛ (s) = G(s) G o (s) Multiplicative uncertainty: or G o (s)g (s) = G(s) G o (s) G (s) = G(s) G o(s) G o (s) Example: Time delays - τ The transfer function of time delay is e τs. Since it is not rational, one often models it as ( ) τs + 2k k e τs τs + 2k where k = 0, 1, 2,.... The additive modeling error is: The multiplicative modeling error is: [ G (s) = G ɛ (s) = e τs e τs ( ) τs + 2k k τs + 2k ( ) ] τs + 2k k ( ) τs + 2k k /. τs + 2k τs + 2k Multiplicative Modeling error of approximation of various orders for 1s time delay 10 2 MME k=0 k=1 k=2 k=3 k= Other examples: ω rad/s

5 126 c Perry Y.Li Uncertain pole location: 1/(s + A m ) 1/(s + A) where A [A 0, A 1 ] Neglected (possibly structural) dynamics: ω 2 n/(s 2 + 2ζω n s + omega 2 n) 1. Neglected compressibility effect, etc Robust Stability Let G o (s) be the nominal plant. Consider a family of plants characterized by: where P := {G(s) = (1 + W u (s) (s))g o (s)} G (s) = W u (s) (s) is the multiplicative model uncertainty W u (s) is a given stable uncertainty weighting function (s) is the uncertainty itself, which is any stable transfer function with (jω) < 1 for all ω R. The question of robust stability is whether a controller designed for G o (s) also stabilizes the any plant G(s) P. If so, we say that the controller provides robust stability.

6 University of Minnesota ME 8281: Advanced Control Systems Design, Robust Stability Theorem: Assume that the controller C(s) internally stabilizes the nominal plant G o (s). Suppose that G(s)C(s) and G o (s)c(s) have the same number of unstable (open loop) poles on the open right half plane. Then, 1. If G (jω) T o (jω) < 1 for all ω R (in other words, G T o < 1) then the controller C(s) internally stabilizes the perturbed plant G(s). 2. C(s) provides robust stability for the plant set P if and only if where T o (s) is the complementary sensitivity function: where L o (s) = G o (s)c(s). W u T o < 1 (5.5) T o (s) = L o(s) 1 + L o (s) Remark: F := sup ω R F (jω) is the so called infinity norm of the transfer function F (s). Every plant G(s) P corresponds to a G (s) that satisfies G T o < 1. So the extra interesting aspect of robust stability condition is the necessity.

7 128 c Perry Y.Li Before we give a proof of this theorem, recall the Nyquist theorem: Nyquist Theorem: Suppose that L(s) has P unstable poles on the open RHP. Then, the Nyquist plot (i.e. the plot of L(s) on the complex plane, as s traverses the Nyquist contour (i.e. the imaginary axis, indented to the right in case of poles of L(s) on the imaginary axis, and a half circle at infinity encircling the RHP), encircles the ( 1, 0) point clockwise N = Z P times, where Z is the number of unstable poles of the closed loop system: L(s) 1 + L(s) Corollary: The closed loop system is stable if and only if the Nyquist plot of L(s) encircles the ( 1, 0) point P times in the counter-clockwise direction. This is obtained by setting Z = 0. Note: P is the number of open-loop unstable poles. We now return to the robust stability theorem. Notice that the perturbed plant lies in a family of disks of size L o (jω)w u (jω) centered at L o (jω). The perturbed plant is at a distance G (jω)l o (jω) from L o (jω). Geometric interpretation: Since T o W u s=jω = L o (jω)w u (jω) / 1 + L o (jω) and 1 + L o (jω) is the distance between ( 1, 0) and L o (jω), T o W u s=jω < 1 if and only L o (jω)w u (jω) < 1 + L o (jω) Thus, the robust stability theorem states that the family of disks of size L o (jω)w u (jω) centered at L o (jω) should not contain the ( 1, 0) point. Proof of Robust Stability: Item 1 and sufficiency in item 2: Show that the number of encirclement of (-1, 0) does not change The distance from L o (jω) to ( 1, 0) is: 1 + L o (jω) However, since This implies that T o (jω)g (jω) < 1 L o (jω) 1 + L o (jω) G (jω) < 1 G (s)l o (jω) < 1 + L o (jω) = distance to (-1, 0) Hence, the perturbed plant cannot reach (and hence change the encirclement of) the ( 1, 0) point. This is the case because if the perturbed plant does change the encirclement of ( 1, 0) there would be a β < 1 such that the perturbed plant (1 + βg (s))l o (s) touches ( 1, 0) at some s = jω 1. Necessity in item 2:

8 University of Minnesota ME 8281: Advanced Control Systems Design, Suppose that W u T o 1 (i.e. the robust stability condition not satisfied) hence, at some ω 1. W u (jω 1 ) T o (jω 1 ) = γ 1 1 We need to construct a (s) with < 1 such that, with G (s) = W u (s) (s), (1 + G (jω))l o (s) touches the ( 1, 0) point. To do this consider (s) of the form: (s) = ±1 (s β) γ (s + β) Note that (jω) = 1/ γ for all ω (i.e. all pass filter) so that: (jω) = 1 γ ejφ(β,ω) where φ(β, ω) = π 2tan 1 (ω/β). Thus, for any ω, by choice of β, it is possible to achieve any φ(β, ω) (0, π]. Thus, we can choose γ = γ 1 and β such that 1 + L o (jω 1 ) = ±1 γ 1 W u (jω 1 )e jφ(β,ω 1) = L o (jω 1 )G (jω 1 ) This makes the perturbed Nyquist plot touch the (-1, 0) point. QED. Remarks: The perturbed closed loop system can be formulated into a closed loop system between G 1 (s) = G (s) = W u (s) (s) and G 2 (s) = T o (s). Sufficiency part of robust stability is a special case of the Small Gain Theorem (SGT). Small Gain Theorem (SGT): Let G 1 and G 2 be (possibly nonlinear) stable systems with finite input-output gains. Let G 1 and G 2 denote their respective gains, i.e. their induced norms. If G 1 G 2 < 1, then the closed loop system consisting of G 1 and G 2 will also be stable.

9 130 c Perry Y.Li Remarks: If we consider G i : u( ) y( ) then, the induced norm (gain) of G i is defined to be: G i i := sup u( ) y( ) u( ) where u( ) and y( ) are the respective signal norms of the input and output. By using different signal norms, different induced norms of the system can be obtained. For linear systems, it turns out that G is the induced 2-norm. i.e. the input and output are measured using the 2-norm: ( u( ) 2 = u(t) 2 dt Therefore, the sufficiency of (5.5) is exactly what is provided by the Small Gain Theorem. What is interesting for LTI systems is that (5.5) is also necessary condition. 5.5 Robust Performance ) 1 2. We assume that the performance is specified by the smallness of the achieved sensitivity S(s). Nominal performance: S(s) = G(s)C(s) = L(s) W p S o < 1; S o (s) = L o (s) where L o (s) = G o (s)c(s), W p (s) is the sensitivity performance weight. 5.6 How to specify sensitivity weighting W p? For example, output disturbance is where DC disturbance: k [0, 10], AC disturbance: ω [2, 4]rads 1, a A. d o = k + asin(ωt + φ) If we would like the effect of disturbance to be smaller than 1, then, we would choose S(j0) < 1/10 to satisfy DC disturbance; and S(jω) < 1/A for ω [2, 4]rads 1 to satisfy AC disturbance requirements. Similar methodology can be used to specify requirements for S uo, S io etc. Consider a family of plants characterized by: P := {G(s) = (1 + W u (s) (s))g o (s)} G (s) = W u (s) (s) is the multiplicative model error W u (s) is a given stable uncertainty weighting function

10 University of Minnesota ME 8281: Advanced Control Systems Design, (s) is the uncertainty itself, which is any stable transfer function with (jω) < 1 for all ω R. Perturbed sensitivity: Lo(s) 1+L o(s). S(s) = (1 + G )L o (s) = S o 1 + G (s)t o (s) where T o (s) = We are interested in the controller C(s) stabilizing all plants in set P and also satisfying the performance specification W p S < 1. Robust performance means: < 1, W u T o < 1 and W p S o 1 + W u T o < 1 (5.6) Theorem: A necessary and sufficient condition for robust performance is: W p S o + W u T o < 1 (5.7) Sketch of Proof: Sufficiency Suppose that (5.7) is satisfied. Clearly (5.6) implies that W u T o < 1 We need only show that WpSo 1+ W ut o < 1 for any allowable (s). Since W p S o + W u T o < 1, we have W p S o 1 W u T o < 1 It is easy to see that for any (s) < 1, W p S o 1 W u T o W p S o 1 + W u T o (worst case with < 1 is for RHS to equal LHS) thus, we have W p S o 1 + W u T o < 1 Necessity: Assume that (5.6) is true. Construct a such that W p S o 1 W u T o = W ps o 1 + W u T o at the frequency where the LHS is maximized. Since the RHS < 1 by robust performance, W p S o 1 W u T o < 1 and hence (5.7) is true.

11 132 c Perry Y.Li Geometric interpretation: The robust performance theorem states that at all ω, disks of size L o (jω)w u (jω) centered at L o (jω) should not intersect disk with radius W p (jω) the ( 1, 0) point. This is because: Thus, W p S o + W u T o < 1 if and only if: W p (jω)s o (jω) = W p (jω) / 1 + L o (jω) W u (jω)t o (jω) = W u (jω)l o (jω) / 1 + L o (jω) W p (jω) + W u (jω)l o (jω) < 1 + L o (jω) the latter is the distance between ( 1, 0) and L o (jω). Wp(jw) ( 1, 0) Wu(jw) Lo(jw) Lo(jw) Robust performance can be achieved by robust stability and nominal performance Proposition: If a controller provides Nominal performance W p S o < 1 for a performance weighting W p (s). Robust stability W u T o < 1 for an uncertainty weighting W u (s).

12 University of Minnesota ME 8281: Advanced Control Systems Design, then the controller provides for robust performance w.r.t 1 2 W ps o W ut o < 1. i.e. when the performance requirement, and uncertainty requirements are halved. Proof: 1 2 W ps o + 1 }{{} 2 W ut o < 1 }{{} Q.E.D. <0.5 <0.5 We can also run the argument backwards. To solve the robust performance criteria for: we can design a controller that satisfies W ps o + W ut o < 1. Nominal performance with double requirement: (2W p)s o < 1 Robust stability with double uncertainty: (2W u )T < 1 Note: Other ratios are also possible. For example, 0 < β < 1. βw p S o + (1 β)w u T o < 1. Choose β to be large or small depending on which of performance or robust stability can be more easily satisfied. 5.7 Other Performance Specification? Robust performance theorem is stated for performance in terms of S. What about other performance specifications? e.g. Complementary sensitivity T (noise to output) Input-disturbance sensitivity S i Control sensitity S u (output disturbance to control)? Perturbed plant: Error sensitivity: G(s) = (1 + G (s))g o (s); ( ) < 1 S (s) = Perturbed sensitivities in terms of nominal: A Conservative Result: S(s) = S o (s)s (s) T o (s)g (s) T (s) = T o (s)(1 + G (s))s (s) S i (s) = S io (s)(1 + G (s))s (s) S u (s) = S uo (s)s (s) Design the nominal performance to be acceptable

13 134 c Perry Y.Li Try to ensure that achieved performances are similar to the nominal performance. This will be the case if the error sensitivity S (s) 1 + j0. This is the case if T o (jω)g (jω) <<< 1. Hence, T o (jω) should roll off before G (jω) becomes significant, or T o (jω)w u (jω) <<< 1 Notice that this condition is similar, but more stringent than just robust stability. This result means that to achieve robust performance when performance is specified using sensitivities that are not S, one can design the nominal system to have good performance, and then make sure that the complementary sensitivity has rolled off before the uncertainty becomes important. 5.8 Servo-hydraulic Example Consider a servo-valve controlled hydraulic actuator with precisely controlled flow rate. Nominal model based on incompressible fluid is just an integrator: X(s) U(s) = G o(s) = 1 s where x(t) represents the displacement, and u(t) is the flow rate (normalized by the piston area) into the actuator. The fluid in the actual system has compressibility which manifests itself as: mẍ + bẋ + k f (x y) = 0; ẏ = u This gives the transfer function from u to x to be: G(s) = Thus, the multiplicative model error (MME) is: G (s) = G(s) G o(s) G o (s) K s(ms 2 + bs + K) = (ms2 + bs) m 2 + bs + K = (s2 + 2ζw n s) s 2 + 2ζw n s + w 2 n w n and ζ depend on fluid compressibility which is highly variable due to aeration, dirt, temperature, additives etc. Uncertainty: w n [200, 500], ζ [0.1, 0.5]. Let us define bounds for uncertainty weights W u (s) so that G (jω) < W u (jω) This should ensure that for any allowable G (s), we can write it as: for some (s) < 1. G (s) = W u (s) (s);

14 University of Minnesota ME 8281: Advanced Control Systems Design, Performance Weight Wp Wp = 100 (s + 20) 5 Wu = 9 s s Wu = 9 s (s+200) 10 8 (s ) Various G (s) and 2 definitions of W u Since the bound of the uncertainty looks like a 20dB/decade rise flattening out at about 1500 Hz, we first define 9s W u = s The value 9 is chosen so that it bounds the uncertainty sufficiently. The bound is not very tight at low and high frequencies. This may cause performance to be conservative at low frequency. Secondly try: W u = 9s(s + 200) (s ) 2 This is motivated by the desire to lower the size in low frequency portion, and to introduce a boost at the resonance frequency. Thus we added a lead-lag in the weighting. This turns out to be adequate for low frequencies (at least for at least up to 1000 rad/s). Next we define the performance criteria. The specifications are: Bandwidth of ω c = 20 rad/s Output disturbance attenuation of 1/100 within bandwidth. Define performance weighting to be: w c = 20rad/s. W p (s) = 100w5 c (s + w c ) 5

15 136 c Perry Y.Li The 100 is used, so that we have the 1/100 attenuation at D.C.. We used a high order filter to make sure that we don t try too hard to achieve any performance above the bandwidth w c. The performance requirement is: for all ω, W p (jω)s(jω) < 1 where S is sensitivity function. Proportional control: U(s) = K (R(s) X(s)) Nominal complementary sensitivity and sensitivity: T o = K s + K, S o = s s + K. Consider first nominal performance design. Criteria is S o (s)w p (s) < 1. After several iterations, this is achieved by K = 600. Nominal Performance WpS Robust stability WuT 10 2 WuT WpS Nominal design: K = 600 Nominal sensitivity and nominal complementary sensitivity for K = 600 Unfortunately, K = 600 does not provide for robust stability since W u T o > 1. Next, we try different K s and apply the robust performance criterion: W u T o + W p S o < 1 K = 160

16 University of Minnesota ME 8281: Advanced Control Systems Design, Sensitivity S Robust Stability WuT Nominal Performance WpS 10 2 Robust Performance WpS + WuT K = 180 Sensitivity S Proportional Control K=160 Robust Stability WuT Nominal Performance WpS 10 2 Robust Performance WpS + WuT Proportional Control K=180

17 138 c Perry Y.Li K = 200 Sensitivity S 10 2 Robust Stability WuT Nominal Performance WpS 10 2 Robust Performance WpS + WuT Proportional Control K=200 K = 220 Sensitivity S 10 2 Robust Stability WuT Nominal Performance WpS 10 2 Robust Performance WpS + WuT Proportional Control K=220

18 University of Minnesota ME 8281: Advanced Control Systems Design, Gain - K W u T o W p S o W u T o + W p S o Conclusion: Proportional control cannot simultaneously provide robust stability and performance. Options: Reduce performance specification - e.g. reduce bandwidth w c Reduce allowable size of uncertainties - e.g. better system identification to nail down fluid / structure natural frequency and damping ratio ω n, ζ, thus reducing W u. Use a different type controller loop shaping. K = 180.

19 140 c Perry Y.Li Robust Stability WuT Nominal Performance WpS K = 180 Sensitivity peaks at 15 rad/s whereas robust stability peaks at 1500 rad/s. Can we design a controller that only modifies the sensitivities around 5 to 50 rad/s, but have little effect at other frequencies? 5.9 Loop Shaping Principles The robust performance control problem is essentially a tradeoff between minimizing W p S o (performance) and W u T o (robustness). As we saw, at any frequency, either W p or W u must be less than 1. Typically, at low frequency, performance requirement is important and system uncertainty is low; and at high frequency, the reverse is true. Thus, At low frequency, W p >> 1 and W u << 1. At high frequency W p << 1 and W u is large. Thus intuitively, we solve W p S o < 1 at low frequency, and W u T o < 1 at high frequency. Let L o (s) = G o (s)c(s). These requirements translate to:

20 University of Minnesota ME 8281: Advanced Control Systems Design, Low frequency (W p dominates): High frequency (W u >> 1 dominates): W p 1 + L o < 1 L o > W p W u L o 1 + L o < 1 L o < 1 W u This provides some guidelines for designing L o (s) to satisfy robust performance. To be more precise, we need to find bounds based on the robust performance criterion itself. We consider the cases of 1) when W p < 1 (high frequency) and 2) when W u < 1 (low frequency) and determine the necessary and sufficient conditions for robust performance. Case 1: W p < 1: (High frequency - uncertainty is important) Sufficient condition: Necessary condition: L o < 1 W p 1 + W u W ps o + W u T o < 1. L o < 1 W p W u 1 W ps o + W u T o < 1. If W u >> 1, both conditions approach, L o < 1 W p W u (5.8) Case 2: W u < 1: (Low frequency - performance is important) Sufficient condition: Necessary condition: L o > W p W u W ps o + W u T o < 1. L o > W p 1 1 W u W ps o + W u T o < 1. If W p >> 1, both conditions approach, L o > W p 1 W u (5.9) These bounds (5.8)-(5.9) determine the design rule for L o (s) = C(s)G o (s).

21 142 c Perry Y.Li 5.10 Loop Shaping Procedure Setup: Open loop plant G(s) is stable, and minimum phase (no RHP zeros) W u (s) and W p (s) are designed such that min{ W u (jω), W p (jω) } < 1, ω Procedure: 1. Plot on log-log sacle, magnitude versus frequency (5.8)-(5.9): At frequencies where W p > 1 > W u (low frequencies): W p 1 W u, At frequencies where W u > 1 > W p (high frequencies): 1 W p W u 2. Construct L o (s) = G o (s)c(s) such that L o (jω) within the required bounds L o > low frequency bound L o < high frequency bound 3. Choose L o (s) such that at L o (jω) passes through L o (jω) = 1 with gentle slope (-20db/decade or -40dB/decade). This determines the phase margin. 4. Roll off L o (s) (at least) as fast as G o (s) so that C(s) is proper. 5. Check robust performance - W p S o + W u T o < Check nominal stability: roots of 1 + L o (s) = 0 should lie on open LHP. 7. Determine C(s) from the L o (s).

22 University of Minnesota ME 8281: Advanced Control Systems Design, One possibility is to construct a nice looking L o (s) first, and then take C(s) = L o (s)/g o (s) Another possibility is to start with L 0 o(s) = kg(s) and then successively modify, L 1 o(s) = kg o (s) L 2 o(s) = kc 1 (s)g o (s) L 3 o(s) = kc 2 (s)c 1 (s)g o (s)... where C i (s) are typically lead-lag controller, C i (s) = β i α i (s + α i ) s + β i. The controller is then C(s) = kc m (s)c m 1 (s)... C 1 (s). Cross-over region can be tricky to ensure robust performance is achieved Loop shaping example: EH actuator Recall that using proportional control, it is not possible to have both adequate performance and robustness at the same time. However, for C(s) = K = 180, Performance curve W p S o peaks at rad/s Robust stability W u T o peaks at 1500 rad/s with adequate margin. Thus, it seems feasible that if we can improve W p S o at around rad/s without disturbing W u T o at high frequencies too much, robust performane can be achieved. We use loop shaping techniques to guide us. Trial 0: (nominal proportional controller) L(s) = C(s)G o (s) = 180 s

23 144 c Perry Y.Li 10 4 EH actuator example Proportional Control (K=180) As expected, this controller fails in the performance bound. Trial 1: Boost gain between 1 to 10 rad/s s + 1 L o (s) = 180 (s/10 + 1) G s + 1 o(s) = 180 s(s/10 + 1)

24 University of Minnesota ME 8281: Advanced Control Systems Design, EH actuator example Proportional + lead Performance bound is satisfied at the expense of violating the robustness bound. This is verified by plotting the robustness stability and performance curves.

25 146 c Perry Y.Li Sensitivity S 10 2 Robust Stability WuT Nominal Performance WpS 10 1 Robust Performance WpS + WuT Robust stability, nominal performance, robust performance curves for C(s) = 180 s + 1 s/ Trial 2: Reduce gain at high frequency to regain robustness L o (s) = 180 s/ (s/10 + 1) s + 1 s/ G o(s) = 180 s/ s(s/10 + 1) 2

26 University of Minnesota ME 8281: Advanced Control Systems Design, Trial Trial 0 Trial Both performance and robustness bounds are satisfied. Need to check robust performance curve. Sensitivity S Robust Stability WuT Nominal Performance WpS Robust Performance WpS + WuT

27 148 c Perry Y.Li Robust stability, nominal performance, robust performance curves for (s + 1)(s/ ) C(s) = 180 (s/10 + 1) 2 Robust performance is satisfied. Make sure to check that the system is nominally stable. The characteristic polynomial is: 180(s + 1)(s/ ) + s(s/10 + 1) 2 Roots are: 99.5 ± 90.5j and Thus the system is stable. Nyquist plot confirms that the nominal loop is stable. Thus, by robust stability, the system is robustly stable as well. Nyquist Diagram 10 5 Imaginary Axis Real Axis 5.12 Constraints on W p and W u W p (s) and W u (s) specify the desired performance and allowable model uncertainties. However, we cannot define W p and W u to be arbitrarily large and expect that that a controller can be found that solves the the robust performance problem. Here, we show that they must respect each other, and respect the limitations of the open-loop system. Knowledge of these limitations help us define meaningful performance specifications (W p ), and our need to do accurate modeling (W u ). First, some preliminary results...

28 University of Minnesota ME 8281: Advanced Control Systems Design, Preliminary Results Maximum Modulus Theorem (MMT) Let P (s) be a stable, rational transfer function. Then, P := sup ω R P (jω) = sup s with Re(s) 0 P (s) It is obvious that LHS is less than or equal to the RHS. The interesting part of this theorem is that they are equal. Consider L(s) = G(s)C(s). Let p be any pole of L(s) and z be any zero of L(s), i.e. 1/L(p) = 0 and L(z) = 0. We have: 1 S(p) = 1 + L(s) = 0; T (p) = 1 S(p) = 1; s=p T (z) = L(s) 1 + L(s) = 0; S(z) = 1 T (z) = 1; s=z Constraint 1: W u & W p cannot simultaneously be large For the robust performance to be solvable, a necessary condition is: for all ω R, min{ W u (jω), W p (jω) } < 1. Proof: Suppose that W u W p (reverse the argument otherwise). At each ω, W u = W u (S + T ) W u S + W u T W u S + W p T Thus, W u S + W p T < 1 (robust performance) implies that W u < 1. Significance: We cannot simultaneously tolerate uncertainty, and expect good performance at any frequencies. One cannot have better than open-loop performance ( S < 1 as guranteed by W p S 1 and W p > 1) when uncertainty is larger than 100% ( W u > 1). Constraint 2: Right Half Plane poles and zeros limit robustness (W u ) and performance (W p ) Suppose that C(s) internally stabilizes the nominal plant G o (s). Let S o and T o be the nominal sensitivities, and p and z are respectively an unstable pole, and a non-minimum phase zero of L o (s) = G o (s)c(s), i.e. 1/L o (p) = 0 with Re(p) 0; and L o (z) = 0 with Re(z) 0. The nominal (and robust) performance problem cannot be solved if W p (z) 1, since W p S o = sup Re(s) 0 W p S o W p (z)s o (z) = W p (z). The robust stability (and robust performance) problem cannot be solved if W u (p) 1, since W u T o = sup Re(s) 0 W u T o W u (p)t o (p) = W u (p).

29 150 c Perry Y.Li The remedy for this is to reduce performance requirement (smaller W p (z) ), and better system identification (lower uncertainty, and smaller W u (p) ). The problem is more acute if there is a pair of RHP pole and zero close to each other. Consider the open loop system: G(s) = s z s p G 1(s) where z and p are RHP zero and pole, G 1 (s) does not have any RHP poles or zeros. Then, it can be shown that: W p S o W p (z) z + p z p W u T o W u (p) z + p z p Thus, as p z, the lower bounds for W p S and W u T is significantly amplified. This is related to the fact that the unstable pole is nearly canceled out by the non-minimum phase zero. Thus, the unstable mode becomes either nearly uncontrollable or unobservable Constraints due to Bode Integral These constraints are sometimes referred to as the principle of conservation of dirts or the area formula. The general meaning is that for systems that satisfy the conditions of the theorems, it is not possible to improve the performance / robustness at all frequencies. Theorem: (Bode Integral Theorem for Sensitivity) Let the open loop system L(s) = G(s)C(s) have the following properties: It has relative degree (i.e. order of denominator minus order of numerator) n r 1. L(s) has M 0 RHP (unstable) poles (counting multiplicity): p 1, p 2,..., p M, Re(p i ) > 0. Let κ := lim s sl(s). [Note: κ = 0 if n r 2] Sensitivity function is given by S(s) = 1 1+L(s). Then, the sensitivity function S(jω) satisfies: Significance: 0 0 ln S(jω) dω = π M Re(p i ). n r > 1 (5.10) i=1 ln S(jω) dω = κ π M 2 + π Re(p i ). n r = 1. (5.11) L(s) has relative degree n r > 1 if both plant G(s) and controller C(s) are strictly proper. Then, (5.10) applies. Decreasing S(jω) at some frequencies ω will increase it at other frequencies. Hence, the dirt is conserved. The total amount of dirt is increased if the open-loop system L(s) is unstable since the RHS of (5.10)-(5.11) is increased. i=1

30 University of Minnesota ME 8281: Advanced Control Systems Design, When n r = 1, the total amount of dirt is decreased by increasing the high frequency gain (e.g. L(s) = K/s). The sensitivity peak S which is inversely related to robustness may be increased.

31 152 c Perry Y.Li Example: Proportional control of 2nd order plant with no zeros. k L(s) = kg(s) = s 2 + s + 1 ; S(s) = s 2 + s + 1 s 2 + s + (1 + k) 10 1 log S(jw) L(s) = k 1 s 2 + s + 1 k=0 k= ω rad/s Note that as k increases, the sensitivity at low frequencies decreases but the peaks increase at other frequencies. Theorem: (Bode Integral Theorem for Complementary Sensitivity) Let the open loop system L(s) = G(s)C(s) have the following properties: L(s) has at least 1 pole at 0 (i.e. L 1 (0) = 0 or T (0) = 1). L(s) has M 0 RHP (non-minimum phase) zeros (counting multiplicity): c 1, c 2,..., c M, Re(c i ) > 0. k v is the velocity constant - i.e. ( ) dt (s) 1 k v = lim s 0 = lim s 0sL(s) ds Complementary sensitity is given by T (s) = Then, the sensitivity function T (jω) satisfies: 0 L(s) 1+L(s). 1 M ω 2 ln T (jω) dω = π i=1 1 c i π 2k v (5.12)

32 University of Minnesota ME 8281: Advanced Control Systems Design, Significance: If L(s) has at least 2 free integrators (poles at 0), then k v =. This ensures that steady error is 0 for ramp input. Decreasing T (jω) at some frequencies ω will increase it at other frequencies. Hence, the similar to the S(s) story, dirt is conserved. The total amount of dirt is increased if the open-loop system L(s) has non-minimum phase zeros, since the RHS of (5.12) is increased. If L(s) has only 1 free integrator, one can decrease the total dirt by tolerating steady state error due to ramp input. If L(s) does not have free integrators (i.e. L 1 (0) 0 or T (0) 1, then a similar relation T (jω) to (5.12) exists, except that in the integral we have ln T (0). This however, does not pose limitation on making T (jω) small in all frequencies Application: Closed loop bandwidth and Open loop unstable pole Constraints on W p, W u imply the following design rule: The closed loop bandwidth should be larger than the magnitude of the unstable open loop pole Usually a factor of 2 is used. Reasoning 1: (From complementary sensitivity function) Suppose that the uncertainty weighting is chosen so that it is important at high frequency, unimportant at low frequencies. This roughly translates to requirement on T o since W u T o < 1. One possibility is W u (s) = s ω o + 1 T. So that W 1 u = T ω o (s T + ω) ; T o(jω o ) W 1 u (jω o ). Hence, the cross over frequency is ω o, and W u (0) = T 1. Here, we can interpret ω o as the bandwidth of the system, since beyond which, we allow T o (jω) to be small. Let p R be a real unstable pole of the open loop plant G o (s). Then, from the constraint that W u (p) < 1, p ω o + 1 T < 1 ω o > p p. 1 1/ T If T is 2 (50% uncertainty at D.C.), then this gives the rule of thumb with the factor of 2. Reasoning 2: (From sensitivity function) Let G o (s) have a real unstable pole at p R, and L o (s) = G o (s)c(s). As a rough approximation, let the open loop gain L o (jω) 0 when ω > ω o where ω o is the bandwidth of the closed loop system. This implies that S o (jω) 1 (ln S o (jω) 0) when ω > ω o. Let M be the max of S o (jω) (sensitivity peak) From Bode integral (5.10), πp = ωo 0 ln S o (jω) dω + ω o ln S o (jω) dω ω o ln(m)

33 154 c Perry Y.Li This shows that the sensitivity peak M e πp/ωo. Since M should be reasonably small (otherwise, the system will behave much worse than open loop), p/ω o should not be large. In particular, if ω o = p, then M = e π 23 which is unacceptable in most cases. When, ω o = 2p, the estimated lower bound for M is Note: M >> 1 is bad also for robustness, since T = 1 S. Thus, large T requires very good modeling effort to maintain stability. Also, recall S 1 o (jω) is distance of L o (jω) to ( 1, 0) Application: Closed loop bandwidth and Open loop nonminimum phase zero The constraints on W u and W p also imply the following design rule: The closed loop bandwidth should be smaller than the magnitude of the non-minimum phase zero Suppose that the performance weighting W p (s) satisfies: W 1 p = S s s + ω S Thus, the sensitivity becomes important for ω < ω o ; high frequency sensitivity requirement is given by S. Let z be a real non-minimum phase (RHP) zero of G o (s). Then from the necessary condition, W p (z) < 1, z + ω o S < 1 ω o < (1 1/ Sz S)z < z 5.17 Limitations - summary W u and W p cannot be large simultaneously. Open loop RHP poles and zeros limit how W u and W p can be defined Conservation of dirt theorems say that improving sensitivity S o (jω) in some frequencies require payment at others. Similar theorem for T o (jω), especially when infinite open loop D.C. gain (open loop integrators). Design implications: Closed loop pole should be faster than open loop unstable pole; Closed loop pole should be slower than open loop non-minimum phase zero. Problem when open loop pole is fast, and non-minimum phase zero is slow consider changing system architecture Example: Inverted pendulum Balancing a beam on a palm. Let u be the force on the palm and M is its mass; m and l are the mass and the length of the beam. Assume that the beam mass is concentrated at the tip. Let y be the tip position and x be the position of the palm.

34 University of Minnesota ME 8281: Advanced Control Systems Design, Transfer function from u to x: This has an unstable pole at + Transfer function from u to y: This does not have any zeros. Conclusions: G ux (s) = (M+m)g Ml G uy (s) = ls 2 g s 2 (Mls 2 (M + m)g), and a non-minimum phase zero at + g/l. g s 2 (Mls 2 (M + m)g) Control is more difficult as the beam becomes shorter, or if the beam is heavier, as the unstable pole gets larger. This requires faster bandwidth on the part of the human controller. Control by looking at the palm (x) is virtually impossible, especially for short beams, because of the unstable pole is faster than the non-minimum phase zero. Thus, there will inevitably be a large sensitivity peak ( S o ) and hence robustness problems. Control by looking at the tip of the beam (y) is easier since there are no zeros. For l = 0.5m, assume that the palm is much heavier than the beam (M >> m). Using a factor of 2 p the required bandwidth is 1.4 Hz. This is quite reasonable for humans.