Capitolo 0. INTRODUCTION 2.1
Example. Compute the exponential of matrix ejα: j =
0 1
−1 0
→ jωt =
0 ωt
−ωt 0
→ ejωt =
cos ωt sin ωt
−sin ωt cos ωt
The characteristic polynomial of matrix jωt is:
∆jωt(λ) = det(λI − jωt) =
λ −ωt ωt λ
= (λ + ωt2) = (λ − jωt)(λ + jωt) The eigenvector v1 corresponding to the eigenvalue λ1 = jωt is:
(λ1I − jωt)v1 = 0 →
jωt −ωt ωt jωt
1 j
= 0 → v1 = 1 j
The eigenvector v2 corresponding to eigenvalue λ2 = −jωt is the complex conjugate of the eigenvector v1:
v2 = v1∗ =
1
−j
The transformation matrix T which transforms matrix jωt in the Jordan ca- nonical form is:
T = v1 v2 = 1 1 j −j
→ T−1 = 1 2j
j 1 j −1
So, the exponential of matrix ejωt can be expressed as follows:
ejωt = Te
jωt 0 0 −jωt
T−1 = 1 1 j −j
ejωt 0 0 e−jωt
j 1 j −1
1 2j
=
ejωt+e−jωt 2
ejωt−e−jωt 2j
−ejωt−e−jωt
2j
ejωt+e−jωt 2
=
cos ωt sin ωt
−sin ωt cos ωt
Zanasi Roberto - System Theory. A.A. 2015/2016
Capitolo 2. CANONICAL FORMS 2.2
Example. Given the following dynamic system:
˙x(t) =
0 1 0 1 0 0 0 0 1
x(t) y(t) = 1 1 1 x(t)
→
( ˙x(t) = Ax(t) y(t) = Cx(t) compute the output free evolution y(t) of the system.
Solution. The output free evolution can be expressed in the following way:
y(t) = CeAtx0 = CTeAtT−1x0 x = Tx
where T is transformation matrix which brings matrix A in the Jordan cano- nical form A. The characteristic polynomial A is
det(λI − A) = (λ − 1)2(λ + 1)
The eigenvalues of matrix A are: λ1 = 1 with molteplicity 2 and λ2 = −1. The eigenvectors corresponding to λ1 = 1 can be determined solving the following linear system:
(λ1I − A)v1,2 = o →
1 −1 0
−1 1 0
0 0 0
v1,2 = 0 In this case there exist two linearly independent eigenvectors:
v1 =
1 1 0
, v2 =
0 0 1
The eigenvector corresponding to λ2 = −1 can be determined in a similar way:
(λ2I − A)v3 = o, ↔
−1 −1 0
−1 −1 0 0 0 −2
v3 = 0 → v3 =
1
−1 0
Zanasi Roberto - System Theory. A.A. 2015/2016
Capitolo 2. CANONICAL FORMS 2.3
Since the 3 eigenvectors are linearly independent, matrix A is diagonalizable.
The transformation matrix T has the following form:
T = v1 v2 v3 =
1 0 1 1 0 −1 0 1 0
, T−1 =
0.5 0.5 0
0 0 1
0.5 −0.5 0
The transformed matrices A and C are:
A = T−1AT =
1 0 0 0 1 0 0 0 −1
, C = CT = 2 1 0 The output free evolution can be expressed as follows:
y(t) = CTeAtT−1x0
= 1 1 1
1 0 1 1 0 −1 0 1 0
et 0 0 0 et 0 0 0 e−t
0.5 0.5 0
0 0 1
0.5 −0.5 0
x10
x20
x30
= et[x10 + x20 + x30] In Matlab:
-- Matlab commands --- syms t x1 x2 x3
A=[0 1 0; 1 0 0; 0 0 1];
C=[1 1 1];
x0=[x1; x2; x3];
yt=simplify(C*expm(A*t)*x0)
-- Matlab output --- yt =
exp(t)*(x1 + x2 + x3)
---
Zanasi Roberto - System Theory. A.A. 2015/2016
Capitolo 2. CANONICAL FORMS 2.4
Example. Compute the free evolution x(t) of the following dynamic system:
˙x(t) =
0 0 1 0 1 1
−1 0 0
x(t) Solution. The characteristic polynomial of matrix A is:
det(sI − A) =
s 0 −1
0 s − 1 −1
1 0 s
= (s − 1)(s2 + 1) = (s − 1)(s + j)(s − j) The three eigenvalues of matrix A are: s1,2 = ±j and s3 = 1. The complex eigenvector v1 associated to the eigenvalue s1 = j can be computed as follows:
(s1I − A)v1 = o ↔
j 0 −1
0 j − 1 −1
1 0 j
v1 = o → v1 =
2 1 − j
2j
The eigenvector v3 associated to the eigenvalue s3 = 1 is the following:
(s3I − A)v1 = o ↔
1 0 −1 0 0 −1 1 0 1
v1 = o → v3 =
0 1 0
Let us consider the following state space transformation x = Tx:
T =
2 0 0 1 −1 1 0 2 0
, A = T−1AT =
0 1 0
−1 0 0 0 0 1
The free evolution of the system is:
x(t) = eAtx0 = TeAtT−1x0
=
2 0 0 1 −1 1 0 2 0
cos t sin t 0
−sin t cos t 0
0 0 et
0.5 0 0 0 0 0.5
−0.5 1 0.5
x10
x20
x30
=
x10cos t + x30sin t
0.5x10(cos t + sin t − et) + x20et + 0.5x30(sin t − cos t + et) x30cos t − x10sin t
Zanasi Roberto - System Theory. A.A. 2015/2016
