MEM 355 Performance Enhancement of Dynamical Systems

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1 MEM 355 Performance Enhancement of Dynamical Systems State Space Design Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University 11/8/2016

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 x = Ax + Bu n m p x = Ax + bu x R, u R, y R y = Cx + Du y = cx + du Now consider the transformation to new states z, defined by 1 x Tz z T x = + = + y = CTz + Du y = CTz + Du 1 1 Tz ATz Bu z T ATz T Bu so that, = = * * z Az Bu = + * * y Cz Du = +, =, = * 1 * 1 A T AT B T B,, * * C = CT D = D

4 Diagonal Form eigen-system of : 1 2 n A h 1 h 2 h n [ 1 2 n ] * 1 A [ h h h ] A[ h h h ] T h h h = 1 2 n = λ λ λ eigenvalues λ λ λn independent eigenvectors n * i i i i A decoupled system of n 1 st order ode s z = λ z + bu, i= 1,, n

5 Example >> A=[3 2 1;4 5 6;1 2 3]; >> [V,D]=eig(A) V = D = Define A 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 p3 0 k k 0 0 c m p3

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 x = Ax + Bu y = Cx + Du Controllability: The system is (completely) controllable if there ( ) [ T] exists a control input u t defined on a finite time interval 0, that steers the system from any initial state x Observability: 0 0 to any final state x. The system is (completely) observable if the intial u( t) y( t ) over a finite time i [ T ] state x can be determined from knowledge of the input and the measurement of the output 1 nterval 0,.

11 Controllability / Observability Tests Controllability Matrix: Observability Matrix: C B AB A n 1 = C CA O = n 1 CA Controllable rank C = n (for SI detc 0) Observable rank O = n (for SO deto 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 controllabile & 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 q n A 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 1 1 ( ) ( ) z = T AT z + T b u n 1 T T I T b T Ab T A b I 1 T b = = n T AT = 1 T Ab T A b T A b 0 0 Y1 0 = 1 Y 2 1 n =, Y= T Ab Yn 1 n Y T Ab ( ) suppose det λi A = λ + a λ + + a, C-H Thm A + a A + + ai= 0 = n n 1 n n 1 n 1 0 n 1 0 a a a 1 n = an 1T A b at 0 b= 0 n 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 y = cx y = cax, cb = 0 2 y = ca x, cab = 0 ( ) n 1 n 1 n 2 y ca x ca b 1 ( ) 2 =, = 0 y = ca x + u ca b = z z n n n 1, 1 = cx = cax = n 1 zn ca x z = Sx z z z = 1 2 = 2 3 n 1 z z = z n = + n 1 zn ca S z u

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 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 [ ] Observability 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 + = + a + + a + a n n 1 λ n 1λ 1λ 0 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 xˆ = 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 ( ) = ( + + ) 1 Gc s K si A BK LC L compensator Equivalent compensator y - Estimator K x = Ax + Bu y = Cx + Du y

33 Implementation compensator manual y - Estimator K x = Ax + Bu y = Cx + Du y

34 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

35 F-16: PI Control G G p c ( s) ( s) = s + 5 = s ( )( s ) s s 2 ( s )( s )( s s ) Design via root locus

36 Example: F-16 state feedback Desired poles - short period: λ = 1.25 ± j ,4 1,2 phugoid: λ = 0.01± j K = [ ] Design via pole placement requires an observer

37 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

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

39 Root Locus 2

40 Margins

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

42 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

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

44 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

45 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] original poles

46 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 Note zeros are unchanged

47 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];

48 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 =

49 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

50 Computations Summary

51 Root Locus

52 Margins

53 Step response

54 Sensitivity Function

55 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.

56 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 * ( + λ1) ( + 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 + ρ e cos ω t+ φ ( ) αts If λ, σ < α then e < ε α > ln ε / T i i s s 0

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