Chris Beattie and Serkan Gugercin

Transcription

1 Krylov-based model reduction of second-order systems with proportional damping Chris Beattie and Serkan Gugercin Department of Mathematics, Virginia Tech, USA 44th IEEE Conference on Decision and Control and European Control Conference CDC-ECC 2005 Seville, Spain, December

2 Introduction Consider the second-order system of the form Mẍ(t) + Gẋ(t) + Kx(t) = Bu(t), y(t) = Cx(t) H(s) = C(Ms 2 + Gs + K) 1 B M,G,K R n n, and B,C T R n. In many cases, n is too large for efficient simulation and control Generate, for some r n, an r th order reduced system M r ẍ r (t) + G r ẋ r (t) + K r x r (t) = B r u(t), y r (t) = C r x r (t) M r,g r,k r R r r, and B r,c T r R r y r (t) approximates y(t) for a wide range of inputs u(t). 2

3 H(s) := Krylov-based model reduction E q(t) = Aq(t) + Bu(t) H(s) = C(sE A) 1 B y(t) = Cq(t) Find H r (s) = C r (se r A r ) 1 B r so that d j H(s) ds j = dj H r (s) s=σk ds j, s=σk K : Number of interpolation points J : Number of interpolation conditions at each σ k. k = 1,...,K j = 0,...,J 1. E r := Z T EV, A r := Z T AV, B r := Z T B, C r := CV 3

4 Given F C N N and g C N, define the Krylov subspace: K J (F,g) = span{g, Fg, F 2 g,, F J 1 g } Construct V and Z such that Ran(V) = span {K J1 (F 1,g 1 ),, K JK (F K,g K )}, and Ran(Z) = span { K JK+1 (F K+1,g K+1 ),, K J2K (F 2K,g 2K ) }, where F i = (σ i E A) 1 E, g i = (σ i E A) 1 B, for i = 1,...,K F i = (σ i E A) T E T, g i = (σ i E A) T C T, for i = K + 1,...,2K H r (s) satisfies the interpolation conditions 4

5 Krylov reduction of second-order systems Convert Mẍ(t) + Gẋ(t) + Kx(t) = Bu(t), y(t) = Cx(t) into E q = Aq(t) + Bu(t), y(t) = Cq(t) where E = I 0 0 M, A = K 0 I αm βk, B = 0 B, C T = CT 0. Apply first-order Krylov reduction and transfer back. Not always possible and structure is lost = Apply reduction directly in the second-order system framework Find W R n r such that W T W = I r and M r = W T MW, G r = W T GW, K r = W T KW, B r = W T B, and C r = CW 5

6 Second-order Krylov Subspaces (Bai[2003]) Given F 1, F 2 C n n and g C n, second-order Krylov subspace: K (2) J (F 1, F 2,g) = span{r 0, r 1,, r J 1 } where r 0 = g, r 1 = F 1 r 0, r j = F 1 r j 1 + F 2 r j 2, for j 2. For second-order reduction with interpolation point σ: W r = K (2) r ( K G, KM,r 0 ) where K = σ 2 M + σg + K, D = 2σM + G, r0 = K 1 B Reduction directly in the second-order system framework Reduced model matches the first r moments at σ. 6

7 Second-order systems with proportional damping Mẍ(t) + (αm + βk)ẋ(t) + Kx(t) = Bu(t), where α, β > 0, and αβ < 1. y(t) = Cx(t) E = First-order equivalent system: H(s) = C (se A) 1 B with I 0 0 M, A = K 0 I αm βk, B = 0 B, C T = CT 0. To match the first r moments at σ, construct V 2n r = span { g, Fg,, F r 1 g } where F = (σe A) 1 E and g = (σe A) 1 B. 7

8 Theorem: Given the above set-up, V 2n r W n r W n r where W n r = K r (K 1 σ M, K 1 σ B) with K σ = σ 2 M + σ(αm + βk) + K. Model reduction directly in the second-order framework using Ran(W r ) = W with W T r W r = I r, W r R n r H r (s) matches r moments at σ. Second-order reduction using regular first-order Krylov subspace W r If σ = 0, W n r = span{k 1 B,, (K 1 M) r 1 K 1 B} If σ =, W n r = span{m 1 B,, (M 1 K) r 1 M 1 B} 8

9 Proof: g = (σe A) 1 B = K 1 σ B σk 1 σ B and K 1 σ B W. Let v = v 1 and v 1,v 2 K p (K 1 σ M, K 1 σ B). v 2 The next vector in W is v = (σe A) 1 Ev = v 1 v 2 = κ 1v 1 + κ 2 K 1 σ Mv 1 + κ 3 K 1 σ Mv 2 κ 4 v 1 + κ 5 K 1 σ Mv 1 + κ 6 K 1 σ Mv 2 v 1, v 2 K p+1 (K 1 σ M, K 1 σ B) Moment matching: Second-order reduction using Ran(W) = W is equivalent to first-order reduction with V = Z = W 0. 0 W 9

10 Approximation by Interpolation (Beattie[2004]) Best uniform approximation H r (s) that minimizes max ω R H(jω) H r (jω) should make H(jω) H r (jω) constant as r ( near-circularity of best uniform rational approximation error (Trefethen, 1981) related to Chebyshev equioscillation theorem. Best uniform approximations are hard to calculate. Interpolants are easy to calculate. How to choose the shifts? Pick interpolation points carefully to recover a good uniform approximant. log H(z) H r (z) has positive singularities at system eigenvalues. negative singularities at interpolation points. 10

11 Pick interpolation points to balance the contours of log H(z) H r (z) (makes log H(z) H r (z) nearly constant along the imaginary axis) Interpolation at Ritz values mirrored across the imaginary axis is good. Mirror Ritz values: optimal choice for H 2 minimization as well (G./Antoulas/Beattie [2004]) If system spectra is structured (e.g., circular), interpolation at equivalent lumped charge locations will have much the same effect. 11

12 Shift Selection for Proportional Damping Mẍ + (αm + βk)ẋ + Kx = p. Proposition: All damped eigenvalues (the system poles) are on circle with center: 1 β, radius: 1 αβ β and on ray (, 1 β ]. Distribution depends on undamped natural frequencies, but usual elastic vibration models lead to distributions that are close to equilibrium condenser distributions Interpret log H(z) H r (z) as potential function associated with charge distribution (poles have net charge of +1, interpolation points have net charge of 1). 12

13 Only ONE shift is necessary - replace aggregate of interpolation points (negative charge distribution) with single α shift (an equivalent lumped charge) at σ = β. Optimal choice for condenser distribution of system poles; pretty good choice for most K and M). α Regular first-order Krylov subspace using a single shift: β. 13

14 Pick α, β (0, 1) Exact Condenser Distribution K = α β 2 1 αβ 1 αβ αβ αβ αβ 1 αβ M = 2+ 1 αβ 1 αβ αβ αβ αβ 1 αβ. G = αm + β K, α = β = 0.05 B = C T = [ ] T. 14

15 Reduction from n = 2000 to r = 30 using a single shift 20 Pole locations for exact condenser distribution 10 1 H error vs interpolation point Imag 0 5 H error Real σ σ = α β = 1 is the optimal shift. 15

16 A 1-D Beam Model: n = α = 1/10, β = 1/500, B = e 1, C = e T 200. H error vs interpolation point 500 Distributiion of Observed and Condenser Poles Observed Poles Condenser Poles 10 1 Imag G(s) G r (s) r=5 r=10 r= Real σ σ = α β = : Very close to being optimal. 16

17 Another 1-D Beam Model: n = 200 and α = β = 1/300, B = C T = e 1. Compare with balanced truncation and other shift selections Reduction done in the first-order framework to r (1) = 40 (r = 20) Amplitude bode plots for the beam model 10 4 Amplitude bode plots of the error systems for the beam model H H BT 10 4 H σ* 10 5 H( jw ) H BT H(s) H error ( jw ) H H σ* freq (rad/sec) freq (rad/sec) 17

18 Amplitude bode plots for the beam model 3.6 Beam with n=200, α = β = 1/ σ = σ = 5 H( jw ) σ = 1 σ i, i=1:40 log 10 H er Hinf Observed Convergence rate: Expected Convergence rate: σ = freq (rad/sec) r 18

19 Conclusions Considered Mẍ + (αm + βk)ẋ + Kx = Bu(t), y(t) = Cx(t) Second-order Krylov reduction using first-order Krylov subspaces Pole locations for proportional damping Equivalent single shift (lumped charge) for aggregate of interpolation points (negative charge distribution) Optimal single shift σ = for condenser distribution Close to optimal in general Future work: Extensions to other types of damping. β α 19