Topic #19 16.31 Feedback Control Closed-loop system analysis ? Bounded Gain Theorem Copy right 2001 by Jon at h an H ow. 1 Fall 2001 16.31 19–1 Bounded Gain ? There exist very easy ways of testing (analytically) whether |S(jω)| <γ, ?ω ? SISO Bounded Gain Theorem: Gain of generic stable system ˙x = Ax+Bu y = Cx+Du is bounded in the sense that G max =sup ω |G(jω)| =sup ω |C(jωI ?A) ?1 B +D| <γ if and only if (i?) 1. |D| <γ 2. The Hamiltonian matrix H = bracketleftbigg A+ B(γ 2 I ?D T D) ?1 D T C B(γ 2 I ?D T D) ?1 B T ?C T (I + D(γ 2 I ?D T D) ?1 D T )C ?A T ?C T D(γ 2 I ?D T D) ?1 B T bracketrightbigg has no eigenvalues on the imaginary axis. ? Note that with D =0,theHamiltonian matrix is H = bracketleftBigg A 1 γ 2 BB T ?C T C ?A T bracketrightBigg – Eigenvalues of this matrix are symmetric about the real and imaginary axis (related to the SRL) Fall 2001 16.31 19–2 ? So sup ω |G(jω)| <γi? H has no eigenvalues on the jω-axis. ? An equivalent test is if there exists a X ≥ 0 such that A T X +XA+C T C + 1 γ 2 XBB T X =0 and A+ 1 γ 2 BB T X is stable. – This is an Algebraic Riccati Equation (ARE) ? Typical appplication: since e = y?r, then for perfect tracking, we want e ≈ 0 ? want S ≈ 0sincee = ?Sr+... – Su?cient todiscussthemagnitudeofS becausetheonlyrequire- ment is that it be small. ? Direct approach is to develop an upperbound for |S| and then test if |S| is below this bound. |S(jω)| < 1 |W s (jω)| ?ω? or equivalently, whether |W s (jω)S(jω)| < 1, ?ω ? Note: The state-space tests can also be used for MIMO systems – but in that case, we need di?ferent Frequency Domain tests. Fall 2001 16.31 19–3 ? Typically pick simple forms for weighting functions (?rst or second order), and then cascade them as necessary. Basic one: W s (s)= s/M +ω B s+ω B A 10 ?2 10 ?1 10 0 10 1 10 2 10 ?2 10 ?1 10 0 10 1 Freq (rad/sec) |1/W s | A ω b M Figure1: Exampleofastandardperformanceweighting?lter. Typically have A lessmuch 1, M>1,and|1/W s |≈1at ω B ? Thus we can test whether |W s (jω)S(jω)| < 1, ?ω by: – Forming a state space model of the combined system W s (s)S(s) – Use the bounded gain theorem with γ =1 – Typically use a bisection section of γ to ?nd |W s (jω)S(jω)| max Fall 2001 16.31 19–4 ? Example: Simple system G(s)= 150 (10s+1)(0.05s+1) 2 with G c =1 ? Requireω B ≈ 5, aslopeof1, lowfrequencyvaluelessthanA =0.01 and a high frequency peak less than M =5. W s = s/M +ω B s+ω B A 10 ?2 10 ?1 10 0 10 1 10 2 10 ?3 10 ?2 10 ?1 10 0 10 1 Frequency Magnitude Sensitivity and Inverse of Performance Weight S W s 1/W s S γ= 1.0219 Figure 2: Want |W s S| < 1, so we just fail the test Fall 2001 16.31 19–5 Sketch of Proof ? Su?ciency: consider γ = 1, and assume D = 0 for simplicity G(s) G T (?s) ? ? ru y y 2 + G(s):= bracketleftbigg A B C 0 bracketrightbigg and G?(s)=G T (?s):= bracketleftbigg ?A T ?C T B T 0 bracketrightbigg ? Note that u/r = S(s)=[1?G?G] ?1 ? Now ?nd the state space representation of S(s) ˙x 1 = Ax 1 +B(r +y 2 )=Ax 1 +BB T x 2 +Br ˙x 2 = ?A T x 2 ?C T y = ?A T x 2 ?C T Cx u = r +B T x 2 ? bracketleftbigg ˙x 1 ˙x 2 bracketrightbigg = bracketleftbigg ABB T ?C T C ?A T bracketrightbiggbracketleftbigg x 1 x 2 bracketrightbigg + bracketleftbigg B 0 bracketrightbigg r u = bracketleftbig 0 B T bracketrightbig bracketleftbigg x 1 x 2 bracketrightbigg +r ? poles of S(s) are contained in the eigenvalues of the matrix H. Fall 2001 16.31 19–6 ? Now assume that H has no eigenvalues on the jω-axis, ?S=[I ?G?G] ?1 has no poles there ? I ?G?G has no zeros there ? So I ?G?G has no zeros on the jω-axis, and we also know that I ?G star G → I>0asω →∞(since D = 0). – Together, these imply that I ?G T (?jω)G(jω)=I ?G star G>0 ? ω ? For a SISO system, this condition (I ?G star G>0) is equivalent to |G(jω)| < 1 ? ω which is true i? G max =max ω |G(jω)| < 1 ? Can use state-space tools to test if a generic system has a gain less that 1, and can easily re-do this analysis to include the bound γ. Fall 2001 16.31 19–7 Issues ? Note that it is actually not easy to ?ndG max directly using the state space techniques – It is easy to check if G max <γ – So we just keep changing γ to ?nd the smallest value for which we can show that G max <γ(called γ min ) ? Bisection search algorithm. ? Bisection search algorithm (see web) 1. Select γ u , γ l so that γ l ≤ G max ≤ γ u 2. Test (γ u ?γ l )/γ l < TOL. Yes ? Stop (G max ≈ 1 2 (γ u +γ l )) No ? go to step 3. 3. With γ = 1 2 (γ l +γ u ), test if G max <γusing λ i (H) 4. If λ i (H) ∈ jR,thensetγ l = γ (test value too low), otherwise set γ u = γ andgotostep2. ? This is the basis of H ∞ control theory.