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***************** Begin Lecture Notes on State Variables (1) *****************
Dr. Shoane
General Solution of the State Equations
A solution to the state equation, Eq can be found in any one of the texts on LinearControl Theory listed in the References. The solution is expressed in terms of an n n matrix ( t), called the transition matrix of the system.
+=t
duBtxttx0
)()()0()()(
The transition matrix depends solely on the system matrix A. One method for finding
( t) uses a definition based on an infinite series.
Note that
...!3!2
1!
32
0
++++==
=
xxx
n
xe
n
n
X(6.27a)
We define the state transition matix as
Atet =)(
(6.27b)
Hence,
...!3
)(
!2
)(
!
)()(
32
0
++++===
=
tAtAtAI
k
tAet
k
k
At
(6.27c)or
...)(!3
1)(
!2
1)()( 32 ++++= tAtAtAIt
It can further be shown that the terms of A with power n can be written as a sum ofterms up to power n-1, where n is the linear dimension of the A square matrix. Thus, thehighest term that needs to be written is n-1. Hence, we can write
1
110)(...)()()(
+++==n
n
tAAtAtItet (6.28a)
************************************************************************
Example A - Initial Condition Only (Zero Input)
The solution to this unforced network is given by
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)0(xex At= (6.28b)
Now we would like to find the state transition matrix. Since A is a 2x2 (i.e., n = 2)
matrix, we have
AIAAeAfk
k
k
n
k
k
k
tA
1
1
0 0
1
0
)( +====
=
= (6.28c)
The solution to the characteristic equation 0= IA is
As an illustration of how the transition matrix is used to solve the linear state
equations, suppose the system matrix for an autonomous system ( 0u = ) is
0 1
2 3A
=
0-3-2-
1=
=
IA (6.28d)
0232 =++ (6.28e)
Hence 1 = -1 and 2 = -2.
The Cayley-Hamilton theorem states that every matrix A is a zero of its own
characteristic equation. That is, Eq (6.28c) is also satisfied for the eigenvalues 1 and 2. Therefore, we have
1
110)(...)()()(
+++==n
n
tAAtAtItet (6.28f)
2102
1101 an d
+=+=
te
te (6.28g)
or
10102a n d
2 ==
te
te (6.28h)
Subtracting the two equations in (6.28h) gives
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te
te
2
1
= 6.28i)
Substitute Eq (6.28i) into the first equation in Eq (6.28h) gives
te
te
22
0
=(6.28j)
Substitute Eqs (6.28h) and (6.28i) into Eq (6.28c) gives
0
0
0
233-
222-
2
22
22
10
tetetete
tete
tete
tete
AI +
+
+=+ (6.28k)
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Therefore
22
222
2
2
2
=
+
+
tetetete
tete
tete( t ) (6.28l)
Another way to obtain the state transition matrix is via the transformed domain.
It can be shown that in the transform domain, the general solution is given by
)()()0()()( sBUsxssX += (6.28m)
where )(s is called the resolvent matrix, which is also the Laplace transform of the
state transition matrix. It can be shown that the resolvent matrix is given by
1)()(
= AsIs (6.28n)
So that for our example (Example A),
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1
1
3s2
1-
3-2-
10
s0
0)()(
1
+== =
ssAs Is
(6.28o)
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++++
+++++
=++
+
=++
+
=
21( s
s
2 )1 )( s
2-
)2)1(
1)2) (1(
3
23
s2-13
2)3(
s2-13
2
ssss
s
ss
s
ss
s
(6.28p)
Using partial fraction expansion, we get
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++
+++
+
++
++
+
+=
2s
2
1s
1-
2
2
1
2
2
1-
1s
1
2
1
1
2
ss
sss(6.28q)
The inverse Laplace transform of the resolvent matrix, )(s , is the state transitionmatrix. Hence,
22222
2
22
=
++
tetetete
tetetete( t ) (6.28r)
which is the same as that obtained above in Eq. (6.28k).
Then we can solve for x(t), with u(t)=0 (zero input).
+=t
duBtxttx0
)()()0()()( (6.27 again)
Example B - Input Only (Zero Initial Conditions)
2x 1x
u1
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The equations for the model are given by
21
222
1211
xxy
uaxx
uxaxx
+=
+=
++=
(6.28A)
.11a10
01
0
1H e ,
B,
- aA (6.28B)
The characteristic equation can be obtained from
0--0
1=
=
a
aIA (6.28C)
or
y
u2
-a -a
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0222=++ aa (6.28D)
two.oftymultipliciwith,2
442 22
2,1a
aaa=
=
(6.28E)
1
110)(...)()()(
+++==n
n
tAAtAtItet (6.28f again)
)(10
+==
tef (6.28F)
and because of the multiplicity of two, we take the derivative of f( ) with respect to
)(1
=
=t
ted
df
(6.28G)
Substituting = -a into the above equation give
atte=1
(6.28H)
Substituting (6.28H) back to (6.28F), with = -a, gives
a ta ta ta t
a t eea t ee
+== 00
(6.28I)
Hence, the state transition matrix is given by
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+
+
+=+==
a
aa
a ta t
a ta t
ta
tt ea
t eae
t eaeAIAft
00
0)()(
10
=
a
aa t
e
t ee
0
(6.28J)
The response of a system, with zero initial conditions and no direct connection between
input and output (i.e., D = 0), is given by the following convolution integral
dutHtyt
)()()(0
= (6.28K)
where H(t) is the impulse response matrix is given by
BtCtH )()( = (6.28L)
or
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=
= +
aa
a t
a ta t
ee
e
t eetH )(11001
0
11)(
(6.28M)
For inputs u1 = u(t) and u2 = )(tuet
, the response is given by
[ ]
[ ]
)()1(
112
)1(1
)()1)(()(
)(
)()1)(()()()(
22
0
))1()(
0
)()(
0
tueaaa
teaa
a
dueetue
due
uetedxtty
att
taatta
ttata
t
+
+
+=
++=
+==
(6.28N)
-----------------------------------------------------------------------------------------------------------
-
Now consider the transformed domain. The solution with zero initial condition isgiven by (see Eq. 6.28m)
)()()( sBUssX = (6.28O)
where the resolvent matrix given by
1)()(
= AsIs (6.28P)
Hence, we have
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1
1
as0
1-
a-0
1a
s0
0)()(
1
+
+== = ass
As Is(6.28Q)
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+
++=++
+
+
=++
+
+
=
)(
1
0
(1
(1
(
as01
(
as01
2
a
asas
a )a )s
as
a )a ) (s
as
(6.28R)
Taking the Inverse Laplace transform gives
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=
a
aa t
e
t ee
0
( t )(6.28T)
which is the same as that obtained using the Cayley-Hamilton approach.
But we also want to obtain the output Y(s), which is given by
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[ ] [ ]
+++
+++
+=
++
+++
+
=
+
+
++
=
==
a )) ( s( s
a )( ssa )s ( s
a )) ( s( s
a( ss
a )s ( s
s
s
a )( s
a )( s
a )( s
B U ( s )sCH ( s ) U ( sY ( s )
1
1
)1(
11
1
1
)1(
11
11
1
1
1
10
11
11
)(
2
22
(6.28U)
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+
++
+
+
+
+
+
+=
asa
sa
as
a
as
a
s
a
asa
sa 1
1
11
1)1(
1
)(
)1(1
1
)1(111
Y(s)
obtainweexpansion,fractionpartialApplying
2
2
2
(6.28V)
(6.28X))()1(
1
1
2
)1(
1)(
givestransformLaplaceInverse
(6.28W))(
1)1(
11)1(
11
21
1)1(s
1a1Y(s)
)(
1
)1(
11
)1(
1)1(2
1
1
)1(s
1
a
1Y(s)
)(
1
)1(
11
)1(
1
)1(
11
1
1
)1(
1
)1(
1
s
1
a
1Y(s)
)1(
1)1()1(
ds
dwhere
22
222
222
222
2
21
tueaaa
te
a
a
aty
asaasaaasaa
asaasaa
aa
sa
a
asaasaaasaa
ass
att
asas
++
+=
+ + ++ +=
+
+
++
+
=
+
++
++
+
+
=
=+=+
which is the same as the solution above (Eq. 6.28N) using the Cayley-Hamilton theorem.
-----------------------------------------------------------------------------------------------------------
-
Summary of Equations
Time Domain
+=t
dutHxtCty0
)()()0()()( (6.28Y)
Zero Input Zero State
where BtCtH )()( = Impulse Response Function
and1
110)(...)()()(
+++== nn
tAAtAtItet
State Transition Matrix
Transformed Domain
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MatrixResolvent()(and
MatrixFunctionTransferewher
StateZeroInputZero
)0()(
1sI-A)s
(s)BCH(s)
H(s)U(s)xsCY(s)
-=
=
+=
(6.28Z)
***************** End Lecture Notes on State Variables (1) *****************
