Loading DocumentA Control System is represented as: x˙=Ax Where A is an n x n matrix
Show that x(t)=eAtx(0)=L−1[SI−A]−1x(0)
let x=a0+a1t+a2t2+a3t3+⋯
x˙=a1+2a2t+3a3t2+⋯=A(a0+a1t+a2t2+a3t3+⋯)
comparing coefficients we get
a1=Aa0a2=2Aa1=2A2a0a3=3Aa2=6A3a0 and so on
x=a0(1+At+2!A2t2+3!A3t3+⋯)=a0×eAt=x(0)eAt
x˙=Ax take LaPlace transform
Sx(S)−x(0)=Ax(S) solve for x
Sx(S)−Ax(S)=x(0)
x(S)[SI−A]=x(0)
x(S)=[SI−A]−1×x(0)take inverse LaPlace transform
x(t)=L−1[SI−A]−1x(0)
Find The Solution of The State Equation: x˙=⎣⎡032−1⎦⎤⎣⎡x1x2⎦⎤
given that x1(0)=3,x2(0)=−2 then sketch x1(t)&x2(t)
x(t)=L−1([SI−A]−1x(0))
[SI−A]=⎣⎡S−3−2S+1⎦⎤
[SI−A]−1=S2+S−61⎣⎡S+132S⎦⎤=(S−2)(S+3)1⎣⎡S+132S⎦⎤
partial fraction
[SI−A]−1=⎣⎡S−23/5+S+32/5S−23/5−S+33/5S−22/5−S+32/5S−22/5+S+33/5⎦⎤
take inverse LaPlace
[SI−A]−1=51⎣⎡3e2t+2e−3t3e2t−3e−3t2e2t−2e−3t2e2t+3e−3t⎦⎤
x1(t)=53(3e2t+2e−3t)−52(2e2t−2e−3t)=e2t+2e−3t
x2(t)=53(3e2t−3e−3t)−52(2e2t+3e−3t)=e2t−3e−3t
Exercise: given a system represented as: x˙=⎣⎡0−31−4⎦⎤⎣⎡x1x2⎦⎤
given that x1(0)=−3,x2(0)=2 find the solution then sketch x1(t)&x2(t)
x˙=Ax
Sx−x(0)=Ax
Sx−Ax=x(0)
x[SI−A]=x(0)
x(S)=[SI−A]−1x(0)
[SI−A]=⎣⎡S3−1S+4⎦⎤
[SI−A]−1=S2+4S+31⎣⎡S+4−31S⎦⎤=(S+1)(S+3)1⎣⎡S+4−31S⎦⎤
[SI−A]−1=21⎣⎡S+13−S+31S+1−3+S+33S+11−S+31S+1−1+S+33⎦⎤
[SI−A]−1=21⎣⎡3e−t−e−3t−3e−t+3e−3te−t−e−3t−e−t+3e−3t⎦⎤
x1(t)=2−3(3e−t−e−3t)+22(e−t−e−3t)=−3.5e−t+0.5e−3t
x2(t)=2−3(−3e−t+3e−3t)+22(−e−t+3e−3t)=3.5e−t−1.5e−3t
Prove that the solution of the state equation x˙=Ax+Buis given by
x(t)=eAtx(0)+∫0teA(t−τ)Bu(τ)dτ
x˙=Ax+Bu multiply both sides by e−At
e−Atx˙=e−AtAx+e−AtBu rearrange
e−At−e−AtAx=e−AtBubut e−At−e−AtAx=dtde−Atx
dtde−Atx=e−AtBu integrate
e−Atx=x(0)+∫0te−AτBu(τ)dτ multiply both sides by eAt
x=eAtx(0)+∫0teA(t−τ)Bu(τ)dτ but eAt=L−1[SI−A]−1=ϕ(t)
x(t)=eAtx(0)+∫0teA(t−τ)Bu(t−τ)dτ=ϕ(t)x(0)+∫0tϕ(t−τ)Bu(τ)dτ
Exercise: given a system represented as: x˙=⎣⎡0−31−4⎦⎤⎣⎡x1x2⎦⎤+⎣⎡04⎦⎤u
given that x1(0)=−3,x2(0)=2 find the solution given that u is a unit step input then sketch x1(t)&x2(t)
x˙=Ax+Bu
Sx−x(0)=Ax+Bu
Sx−Ax=x(0)+Bu
x[SI−A]=x(0)+Bu
x(S)=[SI−A]−1[x(0)+Bu]
[SI−A]=⎣⎡S3−1S+4⎦⎤
[SI−A]−1=S2+4S+31⎣⎡S+4−31S⎦⎤=(S+1)(S+3)1⎣⎡S+4−31S⎦⎤
x0=⎣⎡−32⎦⎤,Bu=⎣⎡04/S⎦⎤
[SI−A]−1[x(0)+Bu]=(S+1)(S+3)1⎣⎡S+4−31S⎦⎤⎣⎡−3S2S+4⎦⎤
=(S+1)(S+3)1⎣⎡S−3S2−12S+2S+49+2S+4⎦⎤
=⎣⎡S(S+1)(S+3)−3S2−10S+4(S+1)(S+3)2S+13⎦⎤
x1=S4/3+S+1−11/2+S+37/6=34−211e−t+67e−3t
x2=S+111/2+S+3−7/2=211e−t−27e−3t
Another solution
x=ϕ(t)x(0)+∫0tϕ(t−τ)Bu(τ)dτ
ϕ=21⎣⎡3e−t−e−3t−3e−t+3e−3te−t−e−3t−e−t+3e−3t⎦⎤
ϕ(t)x(0)=⎣⎡−3.5e−t+0.5e−3t3.5e−t−1.5e−3t⎦⎤
∫0tϕ(t−τ)Bu(τ)dτ=21∫0t⎣⎡3e−t−e−3t−3e−t+3e−3te−t−e−3t−e−t+3e−3t⎦⎤⎣⎡04⎦⎤=2∫0t⎣⎡e−t+τ−e−3t+3τ−e−t+τ+3e−3t+3τ⎦⎤
=2⎣⎡e−t+τ−31e−3t+3τ−e−t+τ+e−3t+3τ⎦⎤0t
x1(t)=−3.5e−t+0.5e−3t+2−32−2e−t+32e−3t=34−5.5e−t+67e−3t
x2(t)=3.5e−t−1.5e−3t−2+2+2e−t−2e−3t=5.5e−t−3.5e−3t
Another solution
x˙=⎣⎡0−31−4⎦⎤⎣⎡x1x2⎦⎤+⎣⎡04⎦⎤u
x1(0)=−3,x2(0)=2,u=1
Sx1−x1(0)=x2
Sx2−x2(0)=−3x1−4x2+s4
S(Sx1+3)−2=−3x1−4(Sx1+3)+s4
(S2+4S+3)x1=−3S+2−12+S4
x1=S(S+1)(S+3)−3S2−10S+4
x1=S4/3+S+1−11/2+S+37/6=34−211e−t+67e−3t
x2=x1˙=211e−t−27e−3t