1
1.1
Solution to Linear Time-Invariant Systems
Scalar equation
Homogeneous equation
dx
= ax,
dt
Separation of variables
x(0) = x0
1
dx = a dt
x
Integrating both sides
¯x(t) Z
¯
=
ln x¯¯
x(0)
Solution
ln x(t) − ln x(0) = ln
MAE 280A
x(t)
x(0)
1
dx =
x
x(t)
= at
x(0)
1
Z
⇒
t
0
¯t
¯
adτ = aτ ¯¯
0
x(t) = eat x(0) = eat x0
Maurı́cio de Oliveira
1.2
Scalar system
LTI scalar system
ẋ(t) = ax(t) + bu(t),
y(t) = cx(t) + du(t).
(1)
Useful properties
d at
e = aeat ,
dt
¤
d £ −at
e x(t) = e−at ẋ(t) − ae−at x(t).
dt
Multiply (1) by e−at on both sides
e−at ẋ(t) − ae−at x(t) = e−at bu(t)
Integrate on both sides
¯t Z
¯
−at
e x(t)¯¯ =
0
Solution
e
−at
x(t) − x(0) =
Z
t
0
⇒
¤
d £ −at
e x(t) = e−at bu(t)
dt
¤
d £ −aτ
e x(τ ) dτ =
dτ
Z
t
e−aτ bu(τ )dτ
0
t
e−aτ bu(τ )dτ
0
⇒
at
x(t) = e x(0) +
Z
t
ea(t−τ ) bu(τ )dτ
0
Substitute into the output equation
Z t
at
y(t) = ce x(0) +
cea(t−τ ) bu(τ )dτ + du(t)
0
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1.3
The matrix exponential
Taylor series (convergent for all x finite!)
∞
X
1
1
1 i
e =
x = 1 + x + x2 + · · · + xn + · · ·
i!
2!
n!
i=0
x
Generalization for matrices
∞
X
1 i
X.
e =
i!
i=0
X
Properties:
1) e0 = I,
2) eA+B = eA eB , when AB = BA,
3) [eX ]−1 = e−X .
Proof:
1) By definition.
2) Use the definition to compute
µ
¶µ
¶
1 2 1 3
1 2 1 3
A B
e e = I + A + A + A + ···
I + B + B + B + ···
2!
3!
2!
3!
¢
1¡ 2
A + 2AB + B 2
= I + (A + B) +
2!
¢
1¡ 3
+
A + 3A2 B + 3AB 2 + B 3 + · · ·
3!
and compare with
1
1
(A + B)2 + (A + B)3 + · · ·
2!
3!
¢
1¡ 2
= I + (A + B) +
A + AB + BA + B 2
2!
eA+B = I + (A + B) +
+
¢
1¡ 3
A + BA2 + ABA + B 2 A + A2 B + BAB + AB 2 + B 3 + · · ·
3!
3) X commutes with X. Then from 1) and 2)
I = eX−X = eX e−X
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3
⇒
[eX ]−1 = e−X .
Maurı́cio de Oliveira
1.4
Time dependent matrix exponential
We are interested in
e
At
Properties:
∞
X
1 ii
At.
=
i!
i=0
4) eA(t1 +t2 ) = eAt1 eAt2 ,
d At
dt e
= AeAt = eAt A.
£
¤
6) dtd e−At x(t) = e−At ẋ(t) − e−At Ax(t).
5)
Proof:
4) At1 and At2 commute.
5) Use definition
d
d At
e =
dt
dt
=
Ã
∞
X
i=0
Ã
!
∞
X
1 ii
At ,
i!
i=0
1
Ai ti−1 ,
(i − 1)!
!
∞
X
1
ti−1 Ai−1 = AeAt ,
(i − 1)!
à ∞i=0
!
X
1
=
ti−1 Ai−1 A = eAt A.
(i − 1)!
i=0
=A
6) Chain rule
¤
d £ −At
d
d £ −At ¤
x(t),
e x(t) = e−At x(t) +
e
dt
dt
dt
= e−At ẋ(t) − e−At Ax(t).
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Maurı́cio de Oliveira
1.5
Complete solution
LTI System
ẋ(t) = Ax(t) + Bu(t),
y(t) = Cx(t) + Du(t),
Multiply by e−At on both sides
e−At ẋ(t) − e−At Ax(t) = e−At Bu(t)
Integrate on both sides
¯t Z
¯
e−At x(t)¯¯ =
0
Solution
e
−At
x(t) − x(0) =
Z
t
0
⇒
¤
d £ −At
e x(t) = e−At Bu(t)
dt
¤
d £ −Aτ
e x(τ ) dτ =
dτ
Z
t
e−Aτ Bu(τ )dτ
0
t
e−Aτ Bu(τ )dτ
0
At
x(t) = e x(0) +
⇒
Z
t
eA(t−τ ) Bu(τ )dτ
0
Substitute on the output equation
At
y(t) = Ce x(0) +
MAE 280A
Z
t
CeA(t−τ ) Bu(τ )dτ + Du(t)
0
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Maurı́cio de Oliveira
1.6
Impulse response (SIMO)
Recall that
δ(t − τ ) = δ(τ − t)
Therefore
Z t
Z t
Z t
A(t−τ )
A(t−τ )
Dδ(τ − t)u(τ )dτ,
Ce
Bu(τ )dτ +
Ce
Bu(τ )dτ + Du(t) =
0
0
0
Z t
£ A(t−τ )
¤
Ce
B + Dδ(t − τ ) u(τ )dτ,
=
|
{z
}
0
h(t − τ )
where
h(t) = CeAt B + Dδ(t)
is the impulse response of the system.
System response:
y(t) =
At
+
Ce x(0)
| {z }
Initial conditions response
Z
|0
t
Repeat the above for each input for MIMO systems.
MAE 280A
6
h(t − τ )u(τ )dτ .
{z
}
Input response
Maurı́cio de Oliveira
1.7
Application: discretization
LTI continuous-time system
ẋ(t) = Ax(t) + Bu(t),
y(t) = Cx(t) + Du(t),
If u(t) comes from a computer
u(t) = u(k) := u(kh),
∀ kh ≤ t < (k + 1)h
h : update period
The solution x(t) of the LTI continuous-time system at t = kh is
Z kh
eA(kh−τ ) Bu(τ )dτ
x(k) := x(kh) = eAkh x(0) +
0
and at t = (k + 1)h
A(k+1)h
x(k + 1) := x[(k + 1)h] = e
·
Z
= eAh eAkh x(0) +
x(0) +
kh
0
Z
(k+1)h
0
eA[(k+1)h−τ ] Bu(τ )dτ,
¸
eA(kh−τ ) Bu(τ )dτ
Z (k+1)h
+
eA[(k+1)h−τ ] Bu(τ )dτ,
kh
= eAh x(k) +
Z
(k+1)h
eA[(k+1)h−τ ] Bu(τ )dτ.
kh
Since u(t) = u(k),
∀ kh ≤ t < (k + 1)h
Z (k+1)h
x(k + 1) = eAh x(k) +
eA[(k+1)h−τ ] dτ B u(k).
| kh
{z
}
|{z}
F
G
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Maurı́cio de Oliveira
Note that
G=
Z
(k+1)h
eA[(k+1)h−τ ] dτ B
kh
and for σ := (k + 1)h − τ
G=−
Z
0
e
Aσ
dσB =
h
Z
h
eAσ dσB
0
Discretized LTI system
x(k + 1) = F x(k) + Gu(k),
y(k) = Cx(k) + Du(k).
where
F = eAh ,
Z h
eAσ dσB.
G=
0
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Maurı́cio de Oliveira
2
2.1
How to compute the matrix exponential?
Eigenvalues and Eigenvectors
Definition: λ ∈ C is an eigenvalue of A ∈ Rn×n (or ∈ Cn×n ) if there exists
v ∈ Cn such that
Av = λv,
v 6= 0.
If so, v is an eigenvector associated with λ.
From linear algebra
Av = λv
(λI − A)v = 0.
⇔
and v 6= 0 if and only if
d(λ) = det(λI − A) = 0.
Conclusion:
Eigenvalues of A are the roots of the characteristic polynomial of A.
⇒ A has n eigenvalues.
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Maurı́cio de Oliveira
2.2
Diagonalizable matrix
Definition: Matrix A ∈ Rn×n (or ∈ Cn×n ) is said to be diagonalizable if it is
similar to a diagonal matrix, i.e., if there exists a nonsingular matrix T such
that T −1 AT is diagonal.
Theorem: Matrix A ∈ Rn×n (or ∈ Cn×n ) is diagonalizable if and only if it
has n linearly independent eigenvectors.
From linear algebra
? The vectors v1 , . . . , vn ∈ Cn are linearly dependent if there exists
α1 , . . . , αn ∈ C not all null such that
n
X
αi vi = 0.
i=1
Otherwise, they are linearly independent.
? If v1 , . . . , vn ∈ Cn are linearly independent then the matrix
£
¤
T = v1 · · · v n
(2)
is nonsingular. Proof: there exists no

such that T x = 0.

α1
x =  ...  6= 0
αn
Proof of the Theorem:
Let v1 , . . . , vn be n linearly independent eigenvectors of A and build T as
in (2). Therefore
£
¤
£
¤
Av1 · · · Avn = AT = T Λ = λ1 v1 · · · λn vn ,
where
As T is nonsingular

Λ=
λ1
...
0
0
λn

.
T −1 AT = Λ.
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Maurı́cio de Oliveira
2.3
Matrix function
Let f (x) : C → C be a continuous scalar-valued function.
Definition: If X is a diagonalizable matrix, i.e., X = T ΛT −1 , then


f (λ1 )
0
...
 T −1 .
f (X) = T 
0
f (λn )
If f has a power series expansion
f (x) =
∞
X
α i xi
i=0
then

f (X) = T 
=T
=
=
f (λ1 )
Ã
∞
X
i=0
∞
X
0
∞
X
0
...
f (λn )
!

 T −1 ,
αi Λi T −1 ,
i=0
¢
¡
αi T Λi T −1 ,
αi X i .
i=0
Compare the above with the formula previously used for eX !
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Maurı́cio de Oliveira