EL 625 Lecture 4 1
EL 625 Lecture 4
Solution of the dynamic state equations
x(t) = A(t)x(t) + B(t)u(t)
y(t) = C(t)x(t) + D(t)u(t) (1)
Homogeneous equation (Unforced system):
x = A(t)x (2)
Consider the matrix differential equation, Q(t) = A(t)Q(t). If
this can be solved, the solution to the homogeneous equation is
x(t) = Q(t)Q−1(t0)x(t0) (3)
x(t) = Q(t)Q−1(t0)x(t0)
= A(t)Q(t)Q−1(t0)x(t0)
= A(t)x(t) (4)
Also, evaluating the left side of (3) at t = t0, we have
Q(t0)Q−1(t0)x(t0) = x(t0) (5)
So, the solution satisfies the initial condition.
EL 625 Lecture 4 2
‘Transition matrix’, φ(t, t0) =4 Q(t)Q−1(t0) (6)
x(t) = φ(t, t0)x(t0) (7)
The transition matrix characterizes the ‘flow’ of the differential e-
quation.
Properties of the transition matrix:
1. x(t0) = φ(t0, t0)x(t0)
φ(t0, t0) = I (8)
2.
φ(t2, t0)x(t0) = x(t2)
= φ(t2, t1)x(t1)
= φ(t2, t1)φ(t1, t0)x(t0)
φ(t2, t1)φ(t1, t0) = φ(t2, t0) (9)
3.
x(t2) = φ(t2, t1)x(t1) = φ(t2, t1)φ(t1, t2)x(t2)
EL 625 Lecture 4 3
φ(t2, t1)φ(t1, t2) = I (10)
φ(t1, t2) = φ−1(t2, t1) (11)
Given a transition matrix, φ(t, t0), A(t) can be evaluated as follows.
x(t) = φ(t, t0)x(t0)
Also, x(t) = A(t)x(t)
= A(t)φ(t, t0)x(t0)
φ(t, t0) = A(t)φ(t, t0) (12)
φ(t, t0)|t0=t = A(t) (13)
We can also determine an expression for φ−1(t, t0).
φ(t, t0)φ−1(t, t0) = I (14)
ddtφ(t, t0)
φ−1(t, t0) + φ(t, t0)
ddtφ−1(t, t0)
= 0 (15)
ddtφ(t, t0)
φ−1(t, t0) = −φ(t, t0)
ddtφ−1(t, t0)
ddtφ−1(t, t0) = −φ−1(t, t0)
ddtφ(t, t0)
φ−1(t, t0)
= −φ−1(t, t0)A(t)φ(t, t0)φ−1(t, t0)
= −φ−1(t, t0)A(t) (16)
EL 625 Lecture 4 4
Solution of the forced system equations:
Assume:
x(t) = φ(t, t0)f(t) (17)
=⇒
x(t) = φ(t, t0)f(t) + φ(t, t0)f(t)
= A(t)φ(t, t0)f(t) + φ(t, t0)f(t)
= A(t)x(t) + φ(t, t0)f(t)
=⇒ φ(t, t0)f(t) = B(t)u(t) (18)
Thus,
f(t) = f(t0) +∫ t
t0φ(t0, λ)B(λ)u(λ)dλ (19)
x(t) = φ(t, t0)f(t0) +∫ t
t0φ(t, t0)φ(t0, λ)B(λ)u(λ)dλ (20)
x(t) = φ(t, t0)x(t0) +∫ t
t0φ(t, λ)B(λ)u(λ)dλ (21)
EL 625 Lecture 4 5
y(t) = C(t)φ(t, t0)x(t0)︸ ︷︷ ︸
Zero-input response
+∫ t
t0C(t)φ(t, λ)B(λ)u(λ)dλ +D(t)u(t)
︸ ︷︷ ︸
Zero-state response (22)
Using
u(t) =∫ ∞−∞ u(λ)δ(t− λ)dλ (23)
and∫ t
t0C(t)φ(t, λ)B(λ)u(λ)dλ =
∫ ∞t0C(t)φ(t, λ)B(λ)u(λ)1(t− λ)dλ
(24)
With t0 = −∞ and x(t0) = 0,∫ ∞t0C(t)φ(t, λ)B(λ)u(λ)1(t− λ)dλ =
∫ ∞−∞C(t)φ(t, λ)B(λ)u(λ)1(t− λ)dλ
(25)
Hence,
Zero-state
response =∫ ∞−∞[C(t)φ(t, λ)B(λ)1(t− λ) + D(t)δ(t− λ)]u(λ)dλ
=∫ ∞−∞H(t, λ)u(λ)dλ (26)
where,
EL 625 Lecture 4 6
H(t, λ) = C(t)φ(t, λ)B(λ)1(t− λ) + D(t)δ(t− λ)
For fixed systems, A,B,C and D are constant matrices
H(t, λ) = H(t− λ) (27)
H(t) = Cφ(t)B1(t) + Dδ(t) (28)
A different approach to the derivation of the forced response
x(t) = A(t)x(t) + B(t)u(t)
φ(t0, t)x(t) = φ(t0, t)A(t)x(t) + φ(t0, t)B(t)u(t)
φ(t0, t)x(t)− φ(t0, t)A(t)x(t) = φ(t0, t)B(t)u(t)
φ(t0, t)x(t) + φ(t0, t)x(t) = φ(t0, t)B(t)u(t) (29)ddt
(φ(t0, t)x(t)) = φ(t0, t)B(t)u(t) (30)
φ(t0, τ )x(t)|tτ=t0 =∫ t
t0φ(t0, τ )B(τ )u(τ )dτ (31)
φ(t0, t)x(t)− x(t0) =∫ t
t0φ(t0, τ )B(τ )u(τ )dτ (32)
φ(t0, t)x(t) = x(t0) +∫ t
t0φ(t0, τ )B(τ )u(τ )dτ(33)
x(t) = φ(t, t0)x(t0)
+φ(t, t0)∫ t
t0φ(t0, τ )B(τ )u(τ )dτ(34)
EL 625 Lecture 4 7
=⇒ x(t) = φ(t, t0)x(t0) +∫ t
t0φ(t, τ )B(τ )u(τ )dτ
(35)
Computing the transition matrix
The physical meaning of φ(t, t0):
x1(t)
x2(t)...
xn(t)
=
φ11(t, t0) φ12(t, t0) . . . φ1n(t, t0)
φ21(t, t0) φ22(t, t0) . . . φ2n(t, t0)... . . . . . . ...
φn1(t, t0) φn2(t, t0) . . . φnn(t, t0)
x1(t0)
x2(t0)...
xn(t0)
(36)
=⇒ xi(t) =n∑
j=1φij(t, t0)xj(t0) (37)
If xk(t0) = 1 with xj(t0) = 0 for all j 6= k,
xi(t) = φik(t, t0) (38)
φij(t, t0)→ response observed at the output of the ith integrator
at time t when a unit initial condition is placed on the jth
integrator at t = t0 and all other integrators have zero initial
conditions (with all inputs being zero).(39)
Example:
x1 = x2
x2 = − 1t + 1
x2 (40)
EL 625 Lecture 4 8
x1(t)
x2(t)
=
φ11(t, t0) φ12(t, t0)
φ21(t, t0) φ22(t, t0)
x1(t0)
x2(t0)
(41)
1. With x1(t0) = 1 and x2(t0) = 0,
x1 = 0
x2 = 0
=⇒ x1(t) = 1
x2(t) = 0
=⇒ φ11(t, t0) = 1
φ21(t, t0) = 0
2. With x1(t0) = 0 and x2(t0) = 1,
x2 = − 1t+1x2 with x2(t0) = 1
Solving,
x2(t) =t0 + 1t + 1
x2(t0) =t0 + 1t + 1
=⇒ φ22(t, t0) =t0 + 1t + 1
x1(t) = x2(t) =t0 + 1t + 1
EL 625 Lecture 4 9
x1(t) =∫ t
t0
t0 + 1τ + 1
dτ
= (t0 + 1) ln
t + 1t0 + 1
=⇒ φ21(t, t0) = (t0 + 1) ln
t + 1t0 + 1
φ(t, t0) =
1 (t0 + 1) ln(
t+1t0+1
)
0 t0+1t+1
Disadvantage of this method: We must integrate the unforced sys-
tem equations to calculate the transition matrix. For fixed systems,
simpler solutions which do not require this integration are available.
Fixed systems: A is a constant matrix.
First order case:
x(t) = ax(t)
=⇒ x(t) = ea(t−t0)x(t0)
=⇒ φ(t, t0) = ea(t−t0) (42)
Is the general solution φ(t, t0) = eA(t−t0)? Yes
eAt =4 I + At + A2t2
2!+ . . . =
∞∑
i=0Ait
i
i!(43)
EL 625 Lecture 4 10
deAt
dt= A[I + At + A2t
2
2!+ . . .] = AeAt (44)
x(t) = eA(t−t0)x(t0) (45)
Evaluating the left hand side at t = t0
x(t)|t = t0 = eA0x(t0) = x(t0) (46)
x(t) = AeA(t−t0)x(t0) = Ax(t) (47)
Thus,
φ(t, t0) = eA(t−t0) (48)
Properties of eAt:
eA0 = I + A0 + A202
2!+ . . . = I (49)
(eAt)−1 = e−At (50)ddteAt = AeAt (51)
Solution of the forced system equations
x(t) = eAtx(0) +∫ t
t0eA(t−τ)Bu(τ )dτ (52)
Impulse response:
h(t, τ ) = h(t− τ ) = CeA(t−τ)1(t− τ ) +Dδ(t− τ ) (53)
h(t) = CeAtB1(t) + Dδ(t) (54)
EL 625 Lecture 4 11
Calculating eAt:
Classical approach for finding the solutions of the unforced system:
Guess an exponential solution. . .
x(t) = eλtξ (ξ 6= 0) (55)
EL 625 Lecture 4 12
x(t) = λeλtξ
A(t)x(t) = λeλtξ
=⇒ Aeλtξ = λeλtξ
λξ = Aξ (56)
This is the ‘eigenvalue’ problem.
λ −→ eigenvalue
ξ −→ eigenvector (ξ 6= 0)
(λI − A)ξ = 0
det[λI − A] = 0 (57)
‘characteristic equation’ of the matrix A
p(λ) =4 det[λI − A] : an nth degree polynomial in λ, called the
characteristic polynomial.
EL 625 Lecture 4 13
Characteristic equation, p(λ) = 0 has n solutions,
the eigenvalues,λ1, λ2, . . . , λn −→ not necessarily distinct
For each eigenvalue, associated eigenvector equation:
[λiI − A]ξi = 0 (58)
has a nontrivial solution, ξi
Useful theorems from matrix theory:
1. Eigen vectors corresponding to distinct eigen values are linearly
independent.
Reminder: Vectors v1, v2, . . . vn are linearly independent if ∑ni=1 αivi =
0 =⇒ αi = 0∀i = 1, 2, ...n. In the figure below, v1 and v2 are
linearly independent. But, v1 and v3 are linearly dependent.
-
6
0
x
y�������
��
��
��
�������������
v1
v3
v2
EL 625 Lecture 4 14
2. An n × n matrix can be diagonalized if it has n distinct eigen-
values.
x(t) = k1eλ1tξ1 + k2eλ2tξ2 + . . . + kneλntξn (59)
x(0) = k1ξ1 + k2ξ2 + . . . + knξn (60)
= [ξ1, ξ2, . . . , ξn]
k1
k2
...
kn
(61)
Let T = [ξ1, ξ2, . . . , ξn] and k =
k1
k2
...
kn
(62)
Thus,
x(0) = Tk = x0
k = T−1x0 (63)
EL 625 Lecture 4 15
x(t) = [eλ1tξ1, eλ2tξ2, . . . , eλntξn]
k1
k2
...
kn
(64)
= [ξ1, ξ2, . . . , ξn]
eλ1t 0 . . . 0
0 eλ2t . . . 0... ... . . . ...
0 0 . . . eλnt
k1
k2
...
kn
(65)
= T
eλ1t 0 . . . 0
0 eλ2t . . . 0... ... . . . ...
0 0 . . . eλnt
T−1x0 (66)
= TeΛtT−1x0 (67)
where
Λ =
λ1 0 . . . 0
0 λ2 . . . 0... ... . . . ...
0 0 . . . λn
(68)
EL 625 Lecture 4 16
eΛt =
eλ1t 0 . . . 0
0 eλ2t . . . 0... ... . . . ...
0 0 . . . eλnt
(69)
Λi =
λi1 0 . . . 0
0 λi2 . . . 0... ... . . . ...
0 0 . . . λin
(70)
deΛt
dt=
λ1eλ1t 0 . . . 0
0 λ2eλ2t . . . 0... ... . . . ...
0 0 . . . λneλnt
= ΛeΛt (71)
x(t) = TeΛtT−1x0 (72)
x(t) = TΛeΛtT−1x0
= TΛT−1TeΛtT−1x0
= TΛT−1x(t)
= Ax(t) (73)
EL 625 Lecture 4 17
A = TΛT−1 (74)
T−1AT = Λ (75)
Thus, A is diagonalized by the similarity transformation, T .
If f (λ) is any function which can be expanded in a power series,
f (λ) =∞∑
i=0aiλi (76)
Define: f(A) = ∑∞i=0 aiA
i Thus,
f (A) =∞∑
i=0ai(TΛT−1)i (77)
=∞∑
i=0aiTΛiT−1 (78)
=∞∑
i=0aiT
λi1 0 . . . 0
0 λi2 . . . 0... ... . . . ...
0 0 . . . λin
T−1 (79)
= T
∑∞i=0 aiλ
i1 0 . . . 0
0 ∑∞i=0 aiλ
i2 . . . 0
... ... . . . ...
0 0 . . . ∑∞i=0 aiλ
in
︸ ︷︷ ︸
=4 f(Λ)
T−1 (80)
EL 625 Lecture 4 18
f(A) = Tf (Λ)T−1 (81)
Example:
A =
0 1
−2 3
The characteristic equation is
p(λ) = det[λI − A] = 0
Thus,
λ −1
2 λ− 3= 0
=⇒ λ2 − 3λ + 2 = 0
The eigenvalues are:
λ1 = 1
λ2 = 2 (82)
(λ1I − A)ξ1 = 0
1 −1
2 −2
ξ11
ξ12
=
0
0
=⇒ ξ11 = ξ12
EL 625 Lecture 4 19
Thus, the first eigenvector is
ξ1 =
1
1
(83)
(λ2I − A)ξ2 = 0
2 −1
2 −1
ξ21
ξ22
=
0
0
=⇒ ξ22 = 2ξ21
Thus, the second eigenvector is
ξ2 =
1
2
(84)
T = [ξ1, ξ2]
=
1 1
1 2
(85)
T−1 =
2 −1
−1 1
(86)
T−1AT =
1 0
0 2
= Λ (87)
EL 625 Lecture 4 20
eAt = TeΛtT−1 (88)
=
1 1
1 −2
et 0
0 e2t
2 −1
−1 1
(89)
=
2et − e2t e2t − et
2et − 2e2t 2e2t − et
(90)
But, this requires finding the eigenvectors. . . Can we avoid this?
Yes
f (Λ) =n∑
k=0f (λk)Ek (91)
Ei is a matrix with a 1 in the i, i position and zeros everywhere else.ith column
↓
Ei =
0 . . . . . . . . . . . . . . . 0... . . . ... ... ... ... ...
0 . . . 0 . . . . . . . . . 0
0 . . . 0 1 0 . . . 0
0 . . . . . . 0 0 . . . 0... ... ... ... ... ... ...
0 . . . . . . . . . . . . . . . 0
←− ith row (92)
EL 625 Lecture 4 21
f (A) = Tf (Λ)T−1
=n∑
k=0f(λk)TEkT−1
=n∑
k=0f(λk)Zk0 (93)
where
Zk0 = TEkT−1 (94)
The matrices Zk0 are independent of the function f . Once
these matrices are evaluated, they can be used to find any function,
f(A).
Can we find Zk0 without finding T ? Yes
Method of Trial Functions: Choose a number of ‘trial func-
tions’ for which f(A) is easy to calculate and use these to calculate
Zk0.
Example:
A =
0 1
−2 3
The eigenvalues of this matrix are λ1 = 1 and λ2 = 2.
Using the method of trial functions, the calculation of the eigen-
EL 625 Lecture 4 22
vectors can be avoided.
f (A) = f(1)Z10 + f (2)Z20 (95)
Choose as the first trial function,
f1(λ) = (λ− 2) (96)
f1(A) = A− 2I
=
−2 1
−2 1
= f1(1)Z10 + f1(2)Z20
= −Z10
=⇒ Z10 =
2 −1
2 −1
(97)
Choose as the second trial function,
f2(λ) = (λ− 1) (98)
f2(A) = A− I
=
−1 1
−2 2
= f2(1)Z10 + f2(2)Z20
= Z20
EL 625 Lecture 4 23
=⇒ Z20 =
−1 1
−2 2
(99)
Thus, for any function, f(λ),
f (A) = f(1)
2 −1
2 −1
+ f(2)
−1 1
−2 2
(100)
With f (λ) = eλt,
eAt = et
2 −1
2 −1
+ e2t
−1 1
−2 2
(101)
=
2et − e2t e2t − et
2et − 2e2t 2e2t − et
(102)
In general, if A has n distinct eigenvalues, λ1, λ2, . . . , λn, choose
as the kth trial function,
fk(λ) =n∏
i = 1
i 6= k
(λ− λi) (103)
fk(λj) =
0 for j 6= kn∏
i = 1i 6= k
(λk − λi) for j = k (104)
Zk0 =fk(A)fk(λk)
(105)
EL 625 Lecture 4 24
(106)
=
n∏
i = 1i 6= k
(A− λiI)
n∏
i = 1i 6= k
(λk − λi)(107)
(108)
f (A) =n∑
k=0f (λk)
n∏
i = 1i 6= k
(A− λiI)
n∏
i = 1i 6= k
(λk − λi)
(109)
If the matrix, A has repeated eigenvalues, i.e. the characteristic
equation has repeated roots,
p(λ) = (λ− λ1)n1(λ− λ2)n2 · · · (λ− λs)ns (110)
ni is the algebraic multiplicity of the eigenvalue λi. In this case,
f (A) =s∑
i=1
ni−1∑
j=0
djf(λ)dλj
λ=λi
Zij(111)
Example:
A =
1 1
0 1
(112)
λI − A =
λ− 1 −1
0 λ− 1
p(λ) = det(λ) = λ2 − 2λ + 1 = 0 (113)
EL 625 Lecture 4 25
Thus, the matrix has a repeated eigen value at 1 i.e. λ1 = 1, λ2 = 1.
f (A) = f(1)Z10 + f ′(1)Z11 (114)
Choosing
f1(λ) = (λ− 1) (115)
f1(A) = A− I
=
0 1
0 0
= f1(1)Z10 + f ′1(λ)|λ=1Z11
= Z11
=⇒ Z11 =
0 1
0 0
(116)
Choosing,
f2(λ) = λ (117)
f2(A) = A
= f2(1)Z10 + f ′1(λ)|λ=1Z11
= Z10 + Z11
=⇒ Z10 = A− Z11
=
1 1
0 1
−
0 1
0 0
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