Topic #19 16.31 Feedback Control Closed-loop system analysis . Bounded gain theorem Copyright 2001 by Jonathan How
Topic #19 16.31 Feedback Control Closed-loop system analysis • Bounded Gain Theorem Copyright 2001 by Jonathan How. 1
Fall 2001 16.3119-1 Bounded gain There exist very easy ways of testing(analytically) whether S Gu)< SISO Bounded Gain Theorem: Gain of generic stable system A r+ Bu C r+ Du is bounded in the sense that Gmax=sup GG)=sup C(jwI-A)B+D <y if and only if (iff 2. The hamiltonian matrix A+b(I-D'D)-D C B(- DID)-IBT C(+D(I-DD-DCT-A-CDOI-D'D-B has no eigenvalues on the imaginary axis Note that with D=0. the Hamiltonian matrix is 4 BB Eigenvalues of this matrix are symmetric about the real and imaginary axis (related to the SRL
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 Gmax = sup ω |G(jω)| = sup ω |C(jωI − A) −1 B + D| < γ if and only if (iff) 1. |D| < γ 2. The Hamiltonian matrix H = A + B(γ2I − DTD)−1DTC B(γ2I − DT D)−1BT −CT (I + D(γ2I − DTD)−1DT )C −AT − CTD(γ2I − DTD)−1BT has no eigenvalues on the imaginary axis. • Note that with D = 0, the Hamiltonian matrix is H = A 1 γ2BBT −CTC −AT – Eigenvalues of this matrix are symmetric about the real and imaginary axis (related to the SRL)
Fall 2001 163119-2 So supu G lw)0 such that AX+XA+CC+-XBBX=0 and A+BBX 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≈0 sInce e=-Sr+ Sufficient to discuss the magnitude of s because the only require- ment is that it be small Direct approach is to develop an upperbound for S and then test if S is below this bound Sgo or equivalently, whether Ws(ja)S Gu)< 1, Ww Note: The state-space tests can also be used for MIMo systems but in that case, we need diffferent Frequency Domain tests
Fall 2001 16.31 19–2 • So supω |G(jω)| < γ iff H has no eigenvalues on the jω-axis. • An equivalent test is if there exists a X ≥ 0 such that ATX + XA + CTC + 1 γ2XBBTX = 0 and A + 1 γ2BBTX 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 ≈ 0 since e = −Sr + ... – Sufficient to discuss the magnitude of S because the only requirement 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 |Ws(jω)| ∀ω? or equivalently, whether |Ws(jω)S(jω)| < 1, ∀ω • Note: The state-space tests can also be used for MIMO systems – but in that case, we need diffferent Frequency Domain tests
Fall 2001 6.3119-3 Typically pick simple forms for weighting functions(first or second order), and then cascade them as necessary. Basic one W(Ss)=s/+uB +Wba M A Freq(rad/sec) Figure 1: Example of a standard performance weighting filter. Typically have A≤1,M>1,and|1/W≈1atuB Thus we can test whether Ws(jw)(jw)<1, Vw by Forming a state space model of the combined system Ws(s)S(s) Use the bounded gain theorem with y=1 Typically use a bisection section of y to find Ws(ja)S(ja)lmax
Fall 2001 16.31 19–3 • Typically pick simple forms for weighting functions (first or second order), and then cascade them as necessary. Basic one: Ws(s) = s/M + ωB s + ωBA 10−2 10−1 100 101 102 10−2 10−1 100 101 Freq (rad/sec) |1/W s | A ωb M Figure 1: Example of a standard performance weighting filter. Typically have A 1, M > 1 , and |1/Ws| ≈ 1at ωB • Thus we can test whether |Ws(jω)S(jω)| < 1, ∀ω by: – Forming a state space model of the combined system Ws(s)S(s) – Use the bounded gain theorem with γ = 1 – Typically use a bisection section of γ to find |Ws(jω)S(jω)|max
Fall 2001 16.3119-4 Example: Simple system 150 (10s+1)(0.05s+1) ith G= 1 Require wB 5, a slope of 1, low frequency value less than A=0 and a high frequency peak less than M=5 s/M+wB s+WBA Sensitivity and Inverse of Performance Weight SW 1 Y1.0219 Frequency Figure 2: Want Wss< 1, so we just fail the test
Fall 2001 16.31 19–4 • Example: Simple system G(s) = 150 (10s + 1)(0.05s +1)2 with Gc = 1 • Require ωB ≈ 5, a slope of 1, low frequency value less than A = 0.01 and a high frequency peak less than M = 5. Ws = s/M + ωB s + ωBA 10−2 10−1 100 101 102 10−3 10−2 10−1 100 101 Frequency Magnitude Sensitivity and Inverse of Performance Weight S W s 1/W s S γ= 1.0219 Figure 2: Want |WsS| < 1, so we just fail the test
Fall 2001 163119-5 Sketch of Proof Sufficiency: consider y= 1, and assume D=0 for simplicity G(s) AB AT-CT G(s) and G(s)=G(S) B Note that /r=S(s)=[1 Now find the state space representation of S (s) C1= A 1+b(r+92)=A C1+BB 2+ Br 2 AT2-Cu y=-A2-CTCE u=r+Bx ABB11「x B CTC -A 0 B 2 poles of S(s) are contained in the eigenvalues of the matrix H
Fall 2001 16.31 19–5 Sketch of Proof • Sufficiency: consider γ = 1, and assume D = 0 for simplicity G(s) GT (−s) ✲ ✻ ru y y2 + G(s) := A B C 0 and G˜(s) = GT (−s) := −AT −CT BT 0 • Note that u/r = S(s) = [1 − G˜G] −1 • Now find the state space representation of S(s) x˙ 1 = Ax1 + B(r + y2) = Ax1 + BBT x2 + Br x˙ 2 = −AT x2 − CT y = −AT x2 − CTCx u = r + BT x2 ⇒ x˙ 1 x˙ 2 = A BBT −CTC −AT x1 x2 + B 0 r u = 0 BT x1 x2 + r ⇒ poles of S(s) are contained in the eigenvalues of the matrix H.
Fall 2001 16.31196 Now assume that H has no eigenvalues on the jw-axis has no poles there I-g g has no zeros there So 1-G g has no zeros on the jw-axis, and we also know that I-G*G→I>0asu→( since d=0 Together, these imply that I-G(juG(jw)=I-GG>0 V For a SISo system, this condition(I-G*G>0 is equivalent to G o<1 which is true iff Gmax= max G(j)<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
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 − GG → I > 0 as ω → ∞ (since D = 0). – Together, these imply that I − GT (−jω)G(jω) = I − G G > 0 ∀ ω • For a SISO system, this condition (I − GG > 0) is equivalent to |G(jω)| < 1 ∀ ω which is true iff Gmax = 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 163119-7 ssues Note that it is actually not easy to find gmax directly using the state space techniques It is easy to check if Gmax < r So we just keep changing y to find the smallest value for which we can show that Gmax y(called ?min Bisection search algorithm Bisection search algorithm(see web 1. Select y, O so that≤Gmax≤mu 2. Test (u -nu/y< TOL Yes→Stop(Gmnk≈(7n+) No→g o to step 3 3. With y=5(n +Yu), test if Gmax y using Ai(H) 4. If Ai H)E jR, then set y(test value too low), otherwise set fu=y and go to step 2 This is the basis of Hoo control theory
Fall 2001 16.31 19–7 Issues • Note that it is actually not easy to find Gmax directly using the state space techniques – It is easy to check if Gmax < γ – So we just keep changing γ to find the smallest value for which we can show that Gmax < γ (called γmin) ⇒ Bisection search algorithm. • Bisection search algorithm (see web) 1. Select γu, γl so that γl ≤ Gmax ≤ γu 2. Test (γu − γl)/γl < TOL. Yes ⇒ Stop (Gmax ≈ 1 2(γu + γl)) No ⇒ go to step 3. 3. With γ = 1 2(γl + γu), test if Gmax < γ using λi(H) 4. If λi(H) ∈ jR, then set γl = γ (test value too low), otherwise set γu = γ and go to step 2. • This is the basis of H∞ control theory
