xx and uu
are functions of time, and the notation
ddtx
t1
x
implies
ddtxt
t1
x
t
.
Given the state equations and output equations
ddtxt=Ax+Bu
t1
x
t
A
x
B
u
(1)
y=Cx+Du
y
C
x
D
u
(2)
we would like to be able to find solutions for xx and
yy in terms of the initial state of the system and the system's
input. To find such solutions, we begin with an examination of
a scalar (n=1n1, m=1m1)
state equation:
ddtxt=ax+bu
t1
x
t
a
x
b
u
(3)
If we looked at a special case of this equation, one
where the input uu was 00,
we'd have
ddtxt=ax
t1
x
t
a
x
.
We've seen this many times before; to solve this, we need a
function whose derivative is directly proportional to itself.
This function is the exponential function. Therefore, in
solving the more general case presented by the state equation, we expect that the exponential function will also come into play.
Starting with the state equation, we can collect like terms, multiply through by
ⅇ-at
a
t
,
and rewrite the left-hand side of the derivative equation in terms of the
derivative. (We take this last step after noticing that the
left-hand side of the derivation equation looks like the derivative product rule has already been applied to it.)
ⅇ-atddtxt-aⅇ-atxt=bⅇ-atut
a
t
t1
x
t
a
a
t
x
t
b
a
t
u
t
(4)
ddtⅇ-atxt=ⅇ-atbut
t1
a
t
x
t
a
t
b
u
t
(5)
Since we are searching for
xtxt
instead of its derivative, we will integrate both sides from
t0t0
to tt.
∫t0tddtⅇ-atxtdt=∫t0tⅇ-atbutdt
t
t0
t
t1
a
t
x
t
t
t0
t
a
t
b
u
t
(6)
In the left-hand side of this equation, the integral and the derivative counteract each other and we are left with the difference of the function
ⅇ-atxt
a
t
x
t
evaluated at the upper and lower integration limits. To avoid
confusion, the variable of integration will be changed from tt
(which is now a constant limit in the integral) to ττ.
ⅇ-atxt-ⅇ-at0xt0=∫t0tⅇ-aτbuτdτ
a
t
x
t
a
t0
x
t0
τ
t0
t
a
τ
b
u
τ
(7)
We now move the
xt0xt0
term to the other side and divide through by
ⅇ-atat.
This leaves us with a solution for the state variable
xtxt
in the scalar case:
xt=ⅇat-t0xt0+∫t0tⅇat-τbuτdτ
x
t
a
t
t0
x
t0
τ
t0
t
a
t
τ
b
u
τ
(8)
What happens if we let
t0t0
go to -∞∞?
The first term on the right-hand side will go to zero since
x-∞=0
x
0
.
Then, if we say that
ht=ⅇatb
ht
a
t
b
,
the second term can be rewritten as
∫-∞tht-τuτdτ
τ
t
h
t
τ
u
τ
(9)
This is the convolution equation h*uhu.
For the scalar case, the solution to the output equation
ytyt
has the same basic form as the solution to the state equation:
yt=cⅇat-t0xt0+∫t0tcⅇat-τbuτdτ+ⅆut
y
t
c
a
t
t0
x
t0
τ
t0
t
c
a
t
τ
b
u
τ
ⅆ
u
t
(10)
Again, we can see the convolution in the second term.
The general matrix forms of the solutions for the
state and output equations follow the same pattern. The only
differences are that we use the matrix exponential instead of
the scalar exponential, and that we use the matrices
AA,
BB,
CC, and
DD
instead of the scalars
aa,
bb,
cc, and
dd.
xt=ⅇAtx0+∫0tⅇAt-τBuτdτ
x
t
A
t
x
0
τ
0
t
A
t
τ
B
u
τ
(11)
yt=CⅇAtx0+∫0tCⅇAt-τBuτdτ+Dut
y
t
C
A
t
x
0
τ
0
t
C
A
t
τ
B
u
τ
D
u
t
(12)
The convolution term is easy to see in the solution for
xtxt.
However, it takes a little regrouping to find it in the solution for
ytyt.
If we pull the DD
matrix into the integral (by multiplying it by the impulse function,
δδ,
we once again have the integral of a function of
t-τtτ
being multiplied by the input (convolution).
yt=CⅇAtx0+∫0tCⅇAt-τB+Dδt-τuτdτ
y
t
C
A
t
x
0
τ
0
t
C
A
t
τ
B
D
δ
t
τ
u
τ
(13)
