674 Chapter 9 / Control Systems Analysis in State SpaceSubstituting 0 for l 1 and–2forl 2 in this last equation, we obtainExpanding the determinant, we obtainorAn alternative approach is to use Equation (9–48). We first determine a 0 (t)anda 1 (t)fromSincel 1 =0andl 2 =–2,the last two equations becomeSolving for a 0 (t)anda 1 (t)givesThen can be written asLinear Independence of Vectors. The vectors x 1 ,x 2 ,p,xnare said to be linearly
independent if
wherec 1 ,c 2 ,p,cnare constants, implies that
Conversely, the vectors x 1 ,x 2 ,p,xnare said to be linearly dependent if and only if xican
be expressed as a linear combination of xj(j=1, 2,p,n; jZi),or
xi= a
nj= 1
jZicj xj
c 1 =c 2 =p=cn= 0
c 1 x 1 +c 2 x 2 +p+cn xn= 0
eAt=a 0 (t) I+a 1 (t) A=I+1
2
A 1 - e-2tB A=B
1
0
1
2 A^1 - e- 2tB
e-2t
R
eAta 0 (t)=1, a 1 (t)=
1
2
A 1 - e-2tBa 0 (t)- 2 a 1 (t)=e-2ta 0 (t)= 1a 0 (t)+a 1 (t)l 2 =el^2 ta 0 (t)+a 1 (t)l 1 =el^1 t= B
1
0
1
2 A^1 - e- 2tB
e-2t
R
=
1
2
bB
0
0
1
- 2
R+ B
2
0
0
2
R - B
0
0
1
- 2
Re-2tr
eAt=^12 AA+ 2 I-Ae-2tB- 2eAt+A+ 2 I-Ae-2t= 0
3
1
1
I
0
- 2
A
1
e-2t
eAt3 = 0
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