《航空器的稳定与控制》（英文版）Matrix Diagonalization

Matrix Diagonalization Suppose a is diagonizable with independent eigenvectors use similarity transformations to diagonalize dynamics matrix Ax→汇a=A 入 1 V-LAV A=4 Corresponds to change of state from a to rd=v-lz System response given by eAt, look at power series expansion At= atv (At)2=(VMV-1) →(At)2=VA"V I+At+=(At)+ vI+A+5A2t2+.V

Matrix Diagonalization • Suppose A is diagonizable with independent eigenvectors V = [v1, . . . , vn] ⎤ – use similarity transformations to diagonalize dynamics matrix x˙ = Ax ⇒ x˙ d = Adxd ⎡ ⎢ λ1 ⎢ ⎢ ⎢ ⎢ ⎣ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ Δ V = Λ = Ad −1 AV = · λn – Corresponds to change of state from x to xd = V −1x • System response given by eAt, look at power series expansion At = V ΛtV −1 2 (At) 2 = (V ΛtV −1 )V ΛtV −1 = V Λ t 2 V −1 ⇒ (At) n = V Λnt nV −1 1 At e = I + At + (At) 2 + . . . 2 ⎧ ⎨ ⎫ 1 ⎬ 2 2 V −1 = V I + Λ + Λ t + . . . ⎩ ⎭ 2 ⎡ ⎤ λ1t ⎢ e ⎢ ⎢ ⎢ ⎢ ⎣ · λnt e ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ V eΛt V −1 V −1 = = V 1

Taking Laplace transform (SI-A 1 S where the residue Ri= viw, and we define 1 Note that the wi are the left eigenvectors of A associated with the right eigenvectors vi 入 AV=V V-IA where a2A=入

� � � • Taking Laplace transform, ⎡ ⎤ 1 s−λ1 ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ (sI − A) −1 = V V −1 · 1 s−λn Ri n i=1 s − λi where the residue Ri = viwi T , and we define = ⎤⎡ T ⎢ w1 ⎢ ⎢ ⎢ ⎢ ⎣ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ , V −1 = . V = v . 1 . . . vn . T wn • Note that the wi are the left eigenvectors of A associated with the right eigenvectors vi ⎡ ⎤ ⎡ ⎤ ⎢ λ1 λ1 ⎢ ⎢ ⎢ ⎢ ⎣ · ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⇒ V −1 A = ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ V −1 AV = V · λn λn ⎤⎡ ⎡ ⎡⎤ ⎤ T T ⎢ w λ1 w1 ⎢ ⎢ ⎢ ⎢ ⎣ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ A = ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ · ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ 1 . . . . . . T T wn λn wn T where wi A = λiwT i 2

So, if i=Ar, the time domain solution is given (t) civil ao) dyad x(t)=∑[v2x(O)e e the part of the solution v;e ait is called a mode of a system solution is a weighted sum of the system modes weights depend on the components of r(0)along w Can now give dynamics interpretation of left and right eigenvectors A;=入 A=入 so if x(O) 2, then (t)=∑(2x(0)et so right eigenvectors are initial conditions that result in relatively simple motions x(t) With no external inputs, if the initial condition only disturbs one mode, then the response consists of only that mode for

� � • So, if x˙ = Ax, the time domain solution is given by n = eλit T x(t) viwi x(0) dyad i=1 n x(t) = [ T i x(0)]eλit w vi i=1 • The part of the solution vieλit is called a mode of a system – solution is a weighted sum of the system modes – weights depend on the components of x(0) along wi • Can now give dynamics interpretation of left and right eigenvectors: Avi = λivi , wiA = λiwi , wi T vj = δij so if x(0) = vi, then � n x(t) = (wT i x(0))eλit vi i=1 λit = e vi ⇒ so right eigenvectors are initial conditions that result in relatively simple motions x(t). With no external inputs, if the initial condition only disturbs one mode, then the response consists of only that mode for all time. 3

If a has complex conjugate eigenvalues, the process is similar but a little more complicated Consider a 2x2 case with A having eigenvalues a+bi and associated eigenvectors e1, e2, with e2=e1. Then a+b1 0 4 +b1 0 e1 el ≡TDT-1 . now use the transformation matrix M=0.5 Then it follows that A= TDT-=TM(M-IDM(M-T- TM(M DMITMT which has the nice structure A=Re(e1) Im(er) a b -b a Re(e1)|Im(e1)I Where all the matrices are real With complex roots, the diagonalization is to a block diagonal form

� � � � � � � � � � � � • If A has complex conjugate eigenvalues, the process is similar but a little more complicated. • Consider a 2x2 case with A having eigenvalues a ± bi and associated eigenvectors e1, e2, with e2 = e¯1. Then ⎡ ⎤ a + bi 0 −1 ⎢ ⎣ ⎥ A = e ⎦ 1 e2 e1 e2 0 a − bi a + bi 0 ⎡ ⎤ −1 ≡ T DT ⎢ −1 ⎣ ⎥ = e ⎦ 1 e¯1 e1 e¯1 0 a − bi • Now use the transformation matrix ⎡ ⎤ ⎡ ⎤ ⎢ ⎣ 1 −i ⎥ ⎦ M−1 = ⎢ ⎣ 1 1 ⎥ M = 0.5 ⎦ 1 i i −i • Then it follows that A = T DT −1 = (TM)(M−1 DM)(M−1 T −1 ) = (TM)(M−1 DM)(TM) −1 which has the nice structure: ⎡ ⎤ a b −1 ⎢ ⎣ ⎥ A = Re(e ⎦ 1) Im(e1) Re(e1) Im(e1) −b a where all the matrices are real. • With complex roots, the diagonalization is to a block diagonal form. 4

For this case we have that c41=[Re(1)m(e)e cos(bt) sin(bt) sin(bt)cos(bt) Re(e1)|Im(er) Note that Re(e1 Im(e1 is the matrix that inverts eleme Re(e1)Im(e1)[Re(e1) Im(e1)I 10 So for an initial condition to excite just this mode, can pick (0)=Re(ell, or x(0)=[Im(el)] or a linear combination · Example a(0)=[Re(e1) eAr(0)=Re(e1)Im(e /eat cos(bt)sin(bt) Re(e1) Im(e1) [Re(e1)l Re(el) Im(neat cos(bt)sin(bt)「1 sin(bt)cos( bt)0 eat re(e1) Im(er ea(Re(en) cos(bt)-Im(e1) sin(bt)) which would ensure that only this mode is excited in the response 5

� � � � � � � � � � � � � � � � � � � • For this case we have that ⎡ ⎤ ⎢ ⎣ cos(bt) sin(bt) −1 At = Re(e1) Im(e1) eat ⎥ e ⎦ Re(e1) Im(e1) − sin(bt) cos(bt) −1 Note that Re(e1) Re(e1) Im(e1) • Im(e1) is the matrix that inverts ⎡ ⎤ Re(e1) Im(e1) −1 � ⎢ 1 0 ⎣ ⎥ Re(e ⎦ 1) Im(e1) = 0 1 • So for an initial condition to excite just this mode, can pick x(0) = [Re(e1)], or x(0) = [Im(e1)] or a linear combination. • Example x(0) = [Re(e1)] ⎡ ⎤ ⎢ ⎣ cos(bt) sin(bt) x(t) = eAtx(0) = Re(e1) Im(e1) eat ⎥ ⎦ · − sin(bt) cos(bt) −1 Re(e1) Im(e1) [Re(e1)] ⎡ ⎡⎤ ⎤ cos(bt) sin(bt) 1 = Re(e1) Im(e1) eat ⎢ ⎣ ⎢ ⎣ ⎥ ⎦ ⎥ ⎦ ⎤ − sin(bt) cos(bt) cos(bt) 0 ⎡ ⎢ ⎣ ⎥ ⎦ at = e Re(e1) Im(e1) − sin(bt) at = e (Re(e1) cos(bt) − Im(e1)sin(bt)) which would ensure that only this mode is excited in the response 5

Example: Spring Mass System Classic example: spring mass system consider simple case first d ki= 1 M-1K0 M= diag(mi) k1+k2+k5-k5 k 2 k5 k3+k4+hs -k3 k k2+k3 Eigenvalues and eigenvectors of the undamped system 入1=±0.77i入2=±185iA3=2.00i U 1.00 1.00 1.00 1.41 1.41 0.00 ±0.77i 士1.85i ±2.00i ±0.77i ±1.85i 千2.00i 士1.08i 干2.61i

� � Example: Spring Mass System • Classic example: spring mass system consider simple case first: mi = 1, and ki = 1 M1 k1 k2 k3 k4 k5 M3 M2 Z1 Z3 Z2 x = z1 z2 z3 z˙1 z˙2 z˙3 ⎡ ⎤ ⎢ 0 I ⎣ ⎥ A = ⎦ M = diag (mi) −M−1K 0 k1 + k2 + k5 ⎡ ⎤ ⎢ −k5 −k2 ⎢ ⎢ ⎢ ⎢ ⎣ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ K = −k5 k3 + k4 + k5 −k3 −k2 −k3 k2 + k3 • Eigenvalues and eigenvectors of the undamped system λ1 = ±0.77i λ2 = ±1.85i λ3 = ±2.00i v1 v2 v3 1.00 1.00 1.00 1.00 1.00 −1.00 1.41 −1.41 0.00 ±0.77i ±1.85i ±2.00i ±0.77i ±1.85i �2.00i ±1.08i �2.61i 0.00 1

Initial conditions to excite just the three modes Ci0)=a1Relvi)+a2Im(u1) VaER Simulation using a1=1, 02=0 Visualization important for correct physical interpretation Mode1A1=±0.77i M Lowest frequency mode all masses move in same direction Middle mass has higher amplitude motions 23, motions all in phase

- - - • Initial conditions to excite just the three modes: xi(0) = α1Re(vi) + α2Im(v1) ∀αj ∈ R – Simulation using α1 = 1, α2 = 0 • Visualization important for correct physical interpretation • Mode 1 λ1 = ±0.77i M1 M3 M2 – Lowest frequency mode, all masses move in same direction – Middle mass has higher amplitude motions z3, motions all in phase 0 1 2 3 4 5 6 7 8 9 10 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 time displacement V1 z1 z2 z3 2

Mode2入2=±1.85i M M M2 Middle frequency mode has middle mass moving in opposition to two end masses Again middle mass has higher amplitude motions 23 V3

• Mode 2 λ2 = ±1.85i M1 M3 M2 -  - – Middle frequency mode has middle mass moving in opposition to two end masses – Again middle mass has higher amplitude motions z3 0 1 2 3 4 5 6 7 8 9 10 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4 time displacement V3 z1 z2 z3 3

Mode3入3=±2.0i M M3 M2 Highest frequency mode, has middle mass stationary, and other two masses in opposition V2 03 Eigenvectors with that correspond with more constrained motion of the system are associated with higher frequency eigenvalues

• Mode 3 λ3 = ±2.00i M1 M3 M2 -  0 – Highest frequency mode, has middle mass stationary, and other two masses in opposition 0 1 2 3 4 5 6 7 8 9 10 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4 time displacement V2 z1 z2 z3 • Eigenvectors with that correspond with more constrained motion of the system are associated with higher frequency eigenvalues 4