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)< y iff H has no eigenvalues on the ju-axis An equivalent test is if there exists a X >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