State Space Design. MEM 355 Performance Enhancement of Dynamical Systems

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1 State Space Design MEM 355 Performance Enhancement of Dynamical Systems Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University

2 Outline State space techniques emerged around They are direct and exploit the efficient computations of linear algebra. State space models, modes & similarity transformations Controllability & Observability Special forms of state equations State feedback & pole placement Observers Design via separation principle Design examples

3 Similarity Transformations xx = AAAA + BBBB xx RR yy = CCCC + DDDD nn, uu RR mm, yy RR pp xx = AAAA + bbbb yy = cccc + dddd Now consider the transformation to new states zz, defined by xx = TTTT zz = TT 1 xx TTzz = AAAAAA + BBBB yy = CCCCCC + DDDD zz = TT 1 AAAAAA + TT 1 BBBB yy = CCCCCC + DDDD so that, zz = AA zz + BB uu yy = CC zz + DD uu, AA = TT 1 AAAA, BB = TT 1 BB, CC = CCCC, DD = DD

4 Diagonal Form eigen system of AA: λλ 1 λλ 2 λλ nn eigenvalues h 1 h 2 h nn independent eigenvectors TT h 1 h 2 h nn AA = h 1 h 2 h 1 AA h 1 nn h 2 h nn λλ = 0 λλ λλ nn zz ii = λλ ii zz ii + bb ii uu, ii = 1,, nn A decoupled system of n 1 st order ode s

5 Example >> A=[3 2 1;4 5 6;1 2 3]; >> [V,D]=eig(A) V = Define A D = Compute eigensystem Check similarity trans Use linear solve rather than inv >> inv(v)*a*v ans = >> V\A*V ans =

6 3 Mass (1) friction, c m k m k m 1 x m 0 0 x1 x m x 2 mx 1 = k ( x2 x1) cx 1 1 d x mx3 mx 2 = k( x2 x1) + k( x3 x2) cx 2 = dt p1 k k 0 c m 0 0 p1 mx 3 = k ( x3 x2) cx 3 p 2 k 2k k 0 c m 0p 2 p 0 k k 0 0 p c 3 m 3

7 3 Mass (2) EDU» A=[ ; ; ; ; ; ]; EDU» [v,e]=eig(a) v = Columns 1 through i i i i i i i i i i i i i i i i i i i i i i i i Columns 5 through

8 3 Mass (3) e = Columns 1 through i i i i Columns 5 through

9 3 Mass (4) Translation Slow exponential decay Slow damped oscillation Fast damped oscillation m k Low frequency m k m m k m k m high frequency

10 Controllability & Observability xx = AAAA + BBBB yy = CCCC + DDDD Controllability: The system is (completely) controllable if there exists a control input uu tt defined on a finite time interval 0, TT that steers the system from any initial state xx 0 to any final state xx 1. Observability: The system is (completely) observable if the intial state xx 0 can be determined from knowledge of the input uu tt and the measurement of the output yy tt over a finite time interval 0, TT.

11 Controllability / Observability Tests Controllability Matrix: C = BB AAAA CC AA nn 1 BB Observability Matrix: O = CCCC CCAA nn 1 Controllable rank C = nn (for SI det C 0) Observable rank O = nn (for SO det O 0)

12 Combustion Example air fuel microphone air speaker controller ( ) s+ z ( 2 2 s 2ρω ) z zs ωz ( 2 2)( 2 2 2ρω ) 1 1 ω1 2ρω 2 2 ω2 + + f G s = K s p f s s s s flame acoustics

13 Combustion Example >> s=tf('s'); zf = 1.500; pf = 1.000; rhoz = ; omegaz = 4.500; rho1 = 0.5; omega1 = 1.0; rho2 = 0.3; omega2 = 3.500; Gp=((s + zf)*(s^2 + 2*rhoz*omegaz*s + omegaz^2))/((s + pf)*(s^2-2*rho1*omega1*s + omega1^2)*(s^2 + 2*rho2*omega2*s + omega2^2)); Gpss=ss(Gp,'min'); [A,B,C,D]=ssdata(Gpss) A = B = C = D = 0

14 Combustion Example Cont d >> Co=ctrb(A,B) Co = >> rank(co) ans = 5 >> Ob=obsv(A,C) Ob = >> rank(ob) ans = 5

15 Special Forms Consider a SISO controllable & observable system c n 1 ca C = b Ab A b, det 0 C O =, deto 0 n 1 ca q1 1 1 C O [ p1 pn ] q n We will consider four state transformations defined by T C, T O, T q n qa n n 1, T4 pn Apn A p n qa n n 1

16 Controllability Form for SISO Systems x = Ax + bu, y = cx n 1 C = b Ab A b, detc 0 = = n 1 n 1 T b Ab A b T A b Ab b 0 0 a0 1 an a 0 0 = + = + 0 a an 1 0 a y = ctz 1 z z u z z u Note special structure of A, b

17 Controllability Form the transformation zz = TT 1 AAAA zz + TT 1 bb uu TT 1 TT = II TT 1 bb TT 1 AAAA TT 1 AA nn 1 bb = II TT 1 AAAA = TT 1 AAAA TT 1 AA 2 bb TT 1 AA nn bb 1 TT YY bb = 1 1 YY = 2, YY = 0 0 TT 1AAnn bb 0 1 YY nn suppose det λλλλ AA = λλ nn + aa nn 1 λλ nn aa 0, C H Thm AA nn + aa nn 1 AA nn aa 0 II = 0 YY = TT 1 AA nn bb = aa nn 1 TT 1 AA nn 1 bb aa 0 TT 1 bb = aa 0 aa 1 aa nn 1

18 Observability Form x = Ax + bu, y = cx c ca O =, deto 0 n 1 ca T = O ( 1 z = z+ T buy ), = [ 1 0 0] z 0 1 an 1 a1 a0 Note special structure of A, c

19 Observability Form the transformation yy = cccc yy = cccccc, cccc = 0 yy = ccaa 2 xx, cccccc = 0 yy nn 1 = ccaa nn 1 xx, ccaa nn 2 bb = 0 yy nn = ccaa nn xx + uu, ccaa nn 1 bb = 1 zz zz 1 = cccc 1 = zz 2 zz zz 2 = cccccc 2 = zz 3 zz nn = ccaa nn 1 zz xx nn 1 = zz nn zz nn = ccaa nn SS 1 zz + uu zz = SSSS

20 Observability Form the transformation, cont d 1 c cs ca cas = = = n 1 n 1 1 ca ca S 1 1 S, SS I I n n 1 A an 1A aa 1 ai = n 1 n ca S = an 1cA S a1cas a0cs [ a a a ] = n

21 Summary of Forms Observability T 0 0 a a 1 0 z z u z = + = z + bu an 1 0 an 1 a1 a 0 y = cz y = z Controllability [ 1 0 0] an z z u z = = + z + bu a1 1 a0 a1 a 1 1 a0 0 0 n y = cz y = z T 4 [ ] T T 3 Controller/phase variable Observer

22 Transfer Functions from State Space x = Ax + Bu y = Cx + Du take Laplace transform: ( ) = ( ) + ( ) [ ] ( ) = ( ) ( ) = ( ) + ( ) sx s AX s BU s si A X s BU s Y s CX s DU s { 1 } ( ) ( ) [ ] Y s = C si A B + D U s ( ) [ ] 1 G s = C si A B + D

23 Example x =, [ 1 0] 0 4 x+ u y = x C = [ B AB] = not controllable 1 4 s s G s = C si A B = ( ) [ 1 ] [ 1 0] ( s 2 )( s 4 ) ( s + 2) 1 = = ( s+ 2)( s+ 4) s+ 4

24 Characteristic Polynomial of Companion Form det λ + an λ 1 0 λ λ ( λ an 1 ) det = + + a a 0 λ 0 0 λ ( λ a ) ( n ) ( n ) λ n 1 = + n 1 + n n 1 λ an 1λ a1λ a0 = Expand along 1 st column Expand along 1 st column

25 Pole Placement Problem Given a linear system: x = Ax + Bu find a state feedback control: u = Kx such that the closed loop system: ( ) x = Ax + BKx = A + BK x { p p p } has a specified (self-conjugate) set of poles,,,. 1 2 n

26 Pole Placement Sol n: SISO Case Convert x = Ax + bu to controller form (phase variable form) using x = Tz: z z = + u a0 a1 an 1 1 [ ] Set u k k k z and obtain closed loop: z = 1 2 n = k a k a k a Expand desired closed loop characteristic polynomial and compare coefficients, and solve for k,, k : ( ) ( )( ) ( ) n n 1 φ λ = λ p λ p λ p = λ + α λ + + α α = a k, α = a k,, α = a k n n 1 cl 1 2 n n n 1 n 1 n 1 1 Convert back to x-coordinates: Kz KT = x u = KT x n ( ) controller form z

27 Pole Place Design: The Easy Way PLACE Pole placement technique K = PLACE(A,B,P) computes a state-feedback matrix K such that the eigenvalues of A-B*K are those specified in vector P. No eigenvalue should have a multiplicity greater than the number of inputs. Warning!! Notice the sign difference.

28 Ackermann s Formula [ 0 0 1] C φ ( ) φ ( ) K = A L= A 1 1 cl cl ACKER Pole placement gain selection using Ackermann's formula. K = ACKER(A,B,P) calculates the feedback gain matrix K such that the single input system. x = Ax + Bu with a feedback law of u = -Kx has closed loop poles at the values specified in vector P, i.e., P = eig(a-b*k). O

29 Full-State Estimator Plant u B x C y A B ˆx C ŷ - A L Estimator

30 Estimator Error Dynamics x = Ax + Bu, y = Cx x ˆ = Axˆ+ Bu + L yˆ y, yˆ = Cxˆ ( A + LC) ( ) ( A + BK ),( A, B) ( + ) e : = x xˆ e = Ae + LCe e = A LC e One approach is to select of problems are equivalent: L so as to place the poles. Notice that the following two pole placement ( T T T A + C L ) ( AC) controllable,, observable

31 Closed Loop Dynamics x = Ax + Bu x ˆ = Axˆ+ Bu + L Cxˆ Cx u = Kxˆ ( ) x = Ax + BKxˆ = A + BK x BKe ( ) e = Ae + LCe = A + LC ( ) x A + BK BK x e = 0 A + LC e λ e closed loop poles A+ BK + λ A+ LC ( ) ( )

32 Compensator xˆ = Axˆ+ Bu + L yˆ y A + BK + LC xˆ Ly ( ) ( ) u = Kxˆ ( ) = ( + + ) compensator 1 Gc s K si A BK LC L Equivalent compensator y - Estimator K x = Ax + Bu y = Cx + Du y

33 Implementation compensator manual system y - Estimator K x = Ax + Bu y = Cx + Du y design a state feedback controller u = Kx design a state estimator/observer to produce xˆ implement the control Kxˆ

34 Examples In the following examples we will Examine open loop modes Design controller using pole placement Compute equivalent compensator Perform root locus analysis of feedback loop

35 Example: F-16 landing approach longitudinal dynamics u u 0 α α = + δ E q q θ θ 0 u α y = [ ] q θ phugoid: λ = ± j h= ± j short period: λ = , h = , Open loop modes

36 ) F-16: PI Control G G p c ( s) ( s) = s + 5 = s ( )( s ) s s 2 ( s )( s )( s s ) Root Locus Design via root locus Imaginary Axis (seconds Real Axis (seconds -1 )

37 Example: F-16 state feedback Desired poles - short period: λ = 1.25 ± j ω = 2.5, ρ = 0.5 3,4 1,2 ( ) phugoid: λ = 0.01± j ω = 0.1, ρ = 0.1 ( ) K = [ ] Design via pole placement requires an observer

38 Example: F-16 Rynaski robust observer "place observer poles at LHP plant zeros, remainder are placed arbitrarily" L T G G λ = 0, , , 1 p c [ ] = ( s) ( s) = ( )( s ) s s ( s )( s )( s s ) 2 ( s )( s s ) = s s s s ( )( )( ) Equivalent compensator

39 Root Locus G ( s) G p c ( s) = ( )( s ) s s 2 ( s )( s )( s s ) 2 ( s )( s s ) = s s s s ( )( )( )

40 Root Locus 2

41 Margins

42 Boeing altitude hold controller Compensator h - Gc ( s) δ e Aircraft h K q q Inner loop K θ θ

43 Boeing 747 Dynamics (cruise) u u 0 w w 32.7 q = q+ 2.08δ e θ θ 0 h h 0 h = [ ] u w q θ h

44 Boeing 747 Open Loop Longitudinal Modes-1 >> A=[ ; ; ; ; ]; >> eig(a) ans = i i Short period i i Phugoid i Vertical translation

45 Boeing 747 Open Loop Longitudinal Modes-2 >> [V,D]=eig(A) Short period V = i i i i i i i i i i i i i i i i i i i i Vertical translation Phugoid

46 Boeing 747 Inner Loop Design A=[ ; ; ; ; ]; B=[0;-32.7;-2.08;0;0]; C=[ ]; poles=[0, i, i, , ]; Kinner=place(A,B,poles) Kinner = eig(a-b*kinner) ans = i i Small contribution, so we ll drop these two terms. Thus, the implementation does not need an observer. [0, i, i, i, i] K q K θ original poles

47 Boeing 747 cont d RHP zero e ( ) δ h: G s = 32.7( s )( s )( s 5.61) ( ± )( ± ) phugoid Choose: K = , K = p q s s j s j [ ] A A = A+ b θ short-period = ( s )( s )( s 5.61) G Gp ( s) = s s ± j2.99 s s ( )( )( ) Inner loop improves stability New A matrix Note zeros are unchanged

48 Outer Loop Design Computations-1 >> A=[ ; ; ; ; ] A = >> b=[0;-32.7;-2.08;0;0] b = >> p=[-.0045;-.145;-.513;-2.25+i*2.98;-2.25-i*2.98];

49 Computations 2 >> K=place(A,b,p) K = >> eig(a-b*k) ans = i i >> c=[0,0,0,0,1]; >> poles=[ ,-5.645,-9,-10,-11]; >> L=place(A',c',poles)' L =

50 Computations 3 >> eig(a-l*c) ans = >> Ac=A-b*K-L*c; >> Bc=L; >> Cc=K; h - Altitude Hold Compensator Stabilized Airframe >> Gcss=ss(Ac,Bc,Cc,0); >> Gc=tf(Gcss); >> zpk(gc) Zero/pole/gain: (s ) (s ) (s^ s ) (s+13.22) (s+5.644) (s ) (s^ s ) h

51 Computations Summary

52 Root Locus

53 Margins

54 Step response

55 Sensitivity Function

56 Pole Locations: Degree of Stability & Damping Ratio Im α degree of stability, decay rate 1/α ideal region for closed loop poles θ damping ratio θ = sin 1 ρ Re If specifications are given in terms of degree of stability and damping ratio, then simply choose poles in the shaded region.

57 Pole Locations: Settling Time and Overshoot We can translate settling time and overshoot specifications into degree of stability and damping ratio specifications. * 1 c1 c2 c2 c E( s) = S( s) = s s s s ( ) ( ) ( * + λ ) 1 + λ2 s + λ2 λ1 φ2 σ2 ω2 φ2 σ2 ω2 ( ) ρ ρ e t c = ce + e e e + e e e + t j t j t j t j t λ1t σ2t ce e cos t ( ) = + ρ ω + φ ( ) αts If λ, σ < α then e < ε α > ln ε / T i i s s 0

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