Region in which λ(a) is included. Which region D brings good response?

Méret: px

Mutatás kezdődik a ... oldaltól:

Download "Region in which λ(a) is included. Which region D brings good response?"

Győző Bogdán
4 évvel ezelőtt
Látták:

1 Region in which λ(a) is included Which region D brings good response? λ(a) D LHP 1 / 25

2 Positive-definite matrix symmetry matrix Q = Q T q1 q ex. Q = 3 q 3 q 2 eigenvalues of a symmetry matrix are real. λ(q) = {λ 1, λ 2 } λ 1 = 1 2 (q 1 + q 2 + D), λ 2 = 1 2 (q 1 + q 2 D) D := (q 1 q 2 ) 2 + 4q3 2 > 0 the eigenvectors are orthogonal. Qv i = λ i v i (i = 1, 2) q1 q v 1 = 2 + D q1 q, v 2 = 2 D 2q 3 2q 3 positive-definite matrix Q > 0 x 0 : x T Qx > 0 λ 1, λ 2 > 0 q 1 > 0, q 1 q 2 q3 2 > 0 semipositive-definite matrix Q 0 x 0 : x T Qx 0 λ 1 > 0, λ 2 = 0 q 1 > 0, q 1 q 2 q3 2 = 0 2 / 25

3 Shure complement P M M T < 0 Q P MQ 1 M T < 0, Q < 0 P < 0, Q M T P 1 M < 0 Proof P M } M T Q {{ } R I MQ 1 P MQ = M T 0 I 0 0 I 0 Q Q 1 M T I I 0 P 0 I P = M M T P 1 I 0 Q M T P 1 M 0 I }{{}}{{}}{{} T T S T x T Rx = x} T {{ T T } S }{{} Tx = y T Sy y T y x 0 : x T Rx < 0 y = Tx 0 : y T Sy < 0 3 / 25

4 Examples q1 q 3 q 3 q 2 q 1 q 4 q 6 q 4 q 2 q 5 q 6 q 5 q 3 q1 q 4 q 4 q 2 < 0 q 1 q 1 2 q2 3 < 0, q 2 < 0 < 0 q6 q 5 q 1 q q 4 q 2 q 5 6 q 5 q 3 q2 q q 1 < 0, 5 q 5 q 3 q1 q 4 q 4 q 2 q 1 < 0, q 2 q 1 1 q2 3 < 0 q3 1 q6 q 5 < 0, q3 < 0 1 q4 q2 q 5 q4 q 6 q 1 1 q 6 < 0, q 5 q 3 q4 q 6 < 0 < 0, q 3 q 6 q 5 q 1 q 4 q 4 q 2 1 q6 q 5 < 0 < 0 4 / 25

5 lmisolver in SCILAB problem formulation Minimize f {(X 1,, X n ) Gi (X subject to 1,, X n ) = 0 (i = 1,, p) H j (X 1,, X n ) > 0 (i = 1,, q) decision variables: X 1,, X n or X (:) objective function (linear function): OBJ ex. f (X 1,, X n ) = c 1 X c n X n constraint by linear matrix equality: LME ex. G(X 1,, X n ) = X X T constraint { by linear matrix inequality: LMI H1 (X ex. 1,, X n ) = X H 2 (X 1,, X n ) = XA A T X 5 / 25

6 LME cond. for λ(a) in LHP ẋ = Ax is asymptotically stable: (0) x(t) = exp(at)x(0) 0 (t ) equivalent condtion #1: (1) λ i λ(a) : Re(λ i ) < 0 equivalent condtion #2: (2) X > 0 : XA + A T X = P < 0 where X = 0 exp(a T t)p exp(at) dt > 0 equivalent condtion #3: (3) Y > 0 : AY + YA T = Q < 0 where Y = 0 exp(at)q exp(a T t) dt > 0 6 / 25

7 Proof (2) (1) Let Av = λv. v T (XA + A T X )v = v T X (λv) + (λ v T )Xv = (λ + λ )v T Xv = 2 Re(λ) v }{{}} T {{ Xv} < 0 <0 >0 (0) (2) XA + A T X = 0 exp(a T t)p exp(at) dt A + A T 0 exp(a T t)p exp(at) dt = { 0 exp(a T t)p d dt exp(at) + d dt exp(at t) P exp(at) } dt = d 0 dt exp(at t)p exp(at) dt = exp(a T t)p exp(at) 0 = exp(a T ) P exp(a ) exp(a T 0) P exp(a0) = P }{{}}{{}}{{}}{{} 0 T 0 I n I n 7 / 25

8 Proof (3) (1) Let A T w = λw. w T (AY + YA T )w = (λ w T )Yw + Y (λw) = (λ + λ )w T Yw = 2 Re(λ) w }{{}} T {{ Yw} < 0 <0 >0 (0) (3) AY + YA T = A 0 exp(at)q exp(a T t) dt + 0 exp(at)q exp(a T t) dt A T = { d 0 dt exp(at) Q exp(at t) + exp(at)q d dt exp(at t) } dt = d 0 dt exp(at)q exp(at t) dt = exp(at)q exp(a T t) 0 = exp(a ) Q exp(a ) exp(a0) P exp(a 0) = Q }{{}}{{}}{{}}{{} 0 T 0 I n I n 8 / 25

9 LMI cond. for λ(a) in D 1 λ(a) D 1 = {s = x + jy C : 2α + s + s < 0} X > 0 : 2αX + XA + A T X < 0 Y > 0 : 2αY + AY + YA T < 0 D 1 2α + s + s < 0 2α + (x + jy) + (x jy) < 0 x < α 9 / 25

10 Proof Let Av = λv. v T (2αX + XA + A T X )v = 2αv T Xv + v T X (λv) + (λ v T )Xv = (2α + λ + λ )v T Xv = 2 (α + Re(λ)) v }{{}} T {{ Xv} < 0 <0 >0 Let y = Xx. x T (2αX + XA + A T X )x < 0 ( x 0) x}{{} T X (2α X}{{} 1 +A X}{{} 1 + X}{{} 1 y T Y Y Y A T ) Xx }{{} y < 0 ( y 0) 10 / 25

11 Using SCILAB (analmi1.sce) function LME,LMI,OBJ=analmi1(XLIST) X=XLIST(:); LME=X-X ; LMI0=X; LMI1=-2*alpha*X-X*A-A *X; LMI=list(LMI0,LMI1); OBJ=; endfunction A=0 1;-1-2; alpha=0.5; //A=0 1;-1-2; alpha=1.5; //A=0 1;-1-1; alpha=0.3; Xinit=eye(2,2); XLIST0=list(Xinit); XLIST=lmisolver(XLIST0,analmi1); X=XLIST(:) 11 / 25

12 LMI cond. for λ(a) in D 2 r s λ(a) D 2 = {s = x + jy C : s r rx XA X > 0 : A T < 0 X rx ry AY Y > 0 : YA T < 0 ry < 0} D 2 r s s r < 0 ( r) s 1 ( r) s < 0, r < 0 r 2 (x + jy)(x jy) > 0 x 2 + y 2 < r 2 12 / 25

13 Proof Let Av = λv. v T 0 T rx XA 0 T v T A T X rx rv = T Xv v T X (λv) (λ v T )Xv rv T Xv r λ = λ v r } T {{ Xv} < 0 }{{} >0 <0 Let y 1 = Xx 1, y 2 = Xx 2. x T 1 x2 T rx XA A T X rx v 0 0 v x1 < 0 ( x x 1, x 2 0) 2 x1 T X x 2 T X rx 1 AX 1 Xx1 X 1 A T rx 1 < 0 Xx 2 y1 T y2 T ry AY y1 YA T < 0 ( y ry 1, y 2 0) y 2 13 / 25

14 Using SCILAB (analmi2.sce) function LME,LMI,OBJ=analmi2(XLIST) X=XLIST(:); LME=X-X ; LMI0=X; LMI2=--r*X X*A;A *X -r*x; LMI=list(LMI0,LMI2); OBJ=; endfunction A=0 1;-1-2; r=2; //A=0 1;-1-2; r=0.5; //A=0 1;-1-1; r=1; Xinit=eye(2,2); XLIST0=list(Xinit); XLIST=lmisolver(XLIST0,analmi2); 14 / 25

15 LMI cond. for λ(a) in D 3 sin θ cos θ sin θ λ(a) D 3 = {s = x + jy C : s + cos θ sin θ cos θ sin θ(xa + A T X ) cos θ(xa A T X ) X > 0 : cos θ(xa A T X ) sin θ(xa + A T < 0 X ) sin θ(ay + YA T ) cos θ(ay YA T ) Y > 0 : cos θ(ay YA T ) sin θ(ay + YA T < 0 ) D 3 T sin θ cos θ sin θ cos θ s + s < 0 cos θ sin θ cos θ sin θ (s + s ) sin θ (s s ) cos θ (s s ) cos θ (s + s < 0 ) sin θ 2x sin θ 2jy cos θ < 0 2jy cos θ 2x sin θ x sin θ jy cos θ 1 ( j)y cos θ < 0, x sin θ < 0 x sin θ x 2 sin 2 θ y 2 cos 2 θ > 0, x < 0 tan θ > y, x < 0 x cos θ sin θ T s < 0} 15 / 25

16 Proof = = Let Av = λv. v T 0 T sin θ(xa + A T X ) cos θ(xa A T X ) v 0 0 T v T cos θ(xa A T X ) sin θ(xa + A T X ) 0 v sin θ(v T X (λv) + (λ v T )Xv) cos θ(v T X (λv) (λ v T )Xv) cos θ(v T X (λv) (λ v T )Xv) sin θ(v T X (λv) + (λ v T )Xv) (λ + λ ) sin θ (λ λ ) cos θ (λ λ ) cos θ (λ + λ ) sin θ } v T {{ Xv } < 0 }{{} >0 <0 Let y 1 = Xx 1, y 2 = Xx 2. x T 1 x2 T sin θ(xa + A T X ) cos θ(xa A T X ) x1 cos θ(xa A T X ) sin θ(xa + A T < 0 ( x X ) x 1, x 2 0) 2 x1 T X x2 T X sin θ(ax 1 + X 1 A T ) cos θ(ax 1 X 1 A T ) Xx1 cos θ(ax 1 X 1 A T ) sin θ(ax 1 + X 1 A T < 0 ) Xx 2 y1 T y2 T sin θ(ay + YA T ) cos θ(ay YA T ) y1 cos θ(ay YA T ) sin θ(ay + YA T < 0 ( y 1, y 2 0) ) y 2 16 / 25

17 Using SCILAB (analmi3.sce) function LME,LMI,OBJ=analmi3(XLIST) X=XLIST(:); LME=X-X ; LMI0=X; LMI3=-sth*(X*A+A *X) cth*(x*a-a *X); -cth*(x*a-a *X) sth*(x*a+a *X); LMI=list(LMI0,LMI3) OBJ=; endfunction A=0 1;-1-2; th=pi/4; sth=sin(th); cth=cos(th); //A=0 1;-1-1; th=pi/4; sth=sin(th); cth=cos(th); //A=0 1;-1-1; th=pi/2; sth=sin(th); cth=cos(th); Xinit=eye(2,2); XLIST0=list(Xinit); XLIST=lmisolver(XLIST0,analmi3); X=XLIST(:) 17 / 25

18 LMI cond. for λ(a) in D = D 1 D 2 D 3 λ(a) D = D 1 D 2 D 3 X > 0 : 2αX + XA + A T X < 0 rx XA A T < 0 X rx sin θ(xa + A T X ) cos θ(xa A T X ) cos θ(xa A T X ) sin θ(xa + A T < 0 X ) Y > 0 : 2αY + AY + YA T < 0 ry AY YA T < 0 ry sin θ(ay + YA T ) cos θ(ay YA T ) cos θ(ay YA T ) sin θ(ay + YA T < 0 ) 18 / 25

19 Using SCILAB (analmi4.sce) function LME,LMI,OBJ=analmi123(XLIST) X=XLIST(:); LME=X-X ; LMI0=X; LMI1=-2*alpha*X-X*A-A *X; LMI2=--r*X X*A;A *X -r*x; LMI3=-sth*(X*A+A *X) cth*(x*a-a *X); -cth*(x*a-a *X) sth*(x*a+a *X); LMI=list(LMI0,LMI1,LMI2,LMI3) OBJ=; endfunction A=0 1;-1-2; alpha=0.5; r=2; th=//a=0 1;-1-1; alpha=0.25; r=2; th=//a=0 1;-1-1; alpha=0.25; r=2; th=xlist0=list(xinit); XLIST=lmisolver(XLIST0,analmi123); X=XLIST(:) 19 / 25

20 Kronecker Product Kronecker product: A B a11 a ex. A = 12 a 21 a 22 a11 B a A B = 12 B a 21 B a 22 B mixed-product property (A C)(B D) = (AB) (CD) 20 / 25

21 LMI cond. for D λ(a) D = {s C : Φ + Θs + Θ T s < 0} X > 0 : Φ X + Θ (XA) + Θ T (A T X ) < 0 Y > 0 : Φ Y + Θ (AY ) + Θ T (YA T ) < 0 Let Av = λv. (I n v T ){Φ X + Θ (XA) + Θ T (A T X )}(I n v) = Φ v T Xv + Θ v T XAv + Θ T v T A T Xv = Φ v T Xv + Θ v T X (λv) + Θ T (λ v T )Xv = (Φ + λθ + λ Θ T ) (v T Xv) < 0 }{{}}{{} <0 >0 21 / 25

22 Problem LEC03 (1) Derive an LMI equvalent to the following. λ(a) D = {s = x + jy C : x 2 a 2 + y 2 b 2 < 1} Develope an program to check the above condition for given a matirx A with constants a, b. 0 1 ex. A = with a = 2, b = / 25

23 Problem LEC03 (2) Derive an LMI equvalent to the following. λ(a) D = {s = x + jy C : x 2 a 2 y 2 > 1, x < a} b2 Develope an program to check the above condition for given a matirx A with constants a, b. 0 1 ex. A = with a = 0.5, b = / 25

24 Hint (1) x2 + y 2 a 2 b 2 < 1 ( s+s ) 2 2 a 2 + ( s s ) 2j 2 < 1 b 2 b 2 (s + s ) 2 a 2 (s s ) 2 < 4a 2 b 2 (b(s + s ) + a(s s ))(b(s + s ) a(s s )) < 4a 2 b 2 4a 2 b 2 + ((b + a)s + (b a)s ))((b a)s + (b + a)s )) < 0 4a 2 b 2 (b + a)s + (b a)s (b a)s + (b + a)s < / 25

25 Hint (2) x 2 a y 2 > 1, x < a ( s+s ) 2 2 ( s s ) 2 2j s + s > 1, < a 2 b2 a 2 b 2 2 b 2 (s + s ) 2 + a 2 (s s ) 2 > 4a 2 b 2, s + s + 2a < 0 b 2 (s + s + 2a)(s + s 2a) + a 2 (s s ) 2 > 0, s + s + 2a < 0 b(s + s 2a) a2 (s s ) 2 b(s + s + 2a) < 0, b(s + s + 2a) < 0 b(s + s + 2a) a(s s ) a(s s ) b(s + s < 0 2a) 25 / 25