Constant coefficient linear systems describe LTI
systems and form the beginnings of the study of state-space
systems. In general, an nn-th
order system such as this can be described by the following
differential equation, where uu and yy
represent the input and output variables, respectively:
dndtnyt+α
n
-
1
dn-1dtn-1yt+ … +α1ddtyt+α0yt=βmdmdtmut+
β
m
-
1
dm-1dtm-1ut+ … +β1ddtut+β0ut
t
n
y
t
α
n
-
1
t
n1
y
t
…
α1
t
1
y
t
α0
y
t
βm
t
m
u
t
β
m
-
1
t
m1
u
t
…
β1
t
1
u
t
β0
u
t
(1)
This is a linear differential equation with real-valued constant coefficients
αiαi
and
βjβj.
We can represent this equation with a much more compact
notation by thinking of the differential operator as a
variable ss,
where the degree of the derivative corresponds to the power of
ss. We will define
qsqs
and
ps p s
as nn-th and mm-th order polynomials in ss.
qs=sn+
α
n
-
1
sn-1+…+α1s+α0
q
s
s
n
α
n
-
1
s
n1
…
α1
s
α0
(2)
ps=βmsm+
β
m
-
1
sm-1+…+β1s+β0
p
s
βm
s
m
β
m
-
1
s
m1
…
β1
s
β0
(3)
If we go ahead and say that
qsq s
and
psp s
will take the differential operator as arguments, we can rewrite Equation 1 as
qⅆⅆtyt=pⅆⅆtut
q
ⅆ
ⅆ
t
y
t
p
ⅆ
ⅆ
t
u
t
(4)
Looking at differential equations in terms of polynomials such as
qsq s
and
psp s
right away reminds us the Laplace transform. Using the Laplace transform can often simplify the calculations involved in system analysis.
By using Laplace domain tools to examine the RLC circuit seen before, the differential equation describing the behavior of this system is easy to deduce.
YsUs=Z2sZ1s+Z2s
Y
s
U
s
Z2
s
Z1
s
Z2
s
(5)
where
Z1s=1+s2
Z1
s
1
s
2
(6)
and
Z2s=112+s3
Z2
s
1
1
2
s
3
(7)
This yields
Ys=6s2+72s+9Us
Y
s
6
s
2
7
2
s
9
U
s
(8)
By multiplying both sides by the denominator of the fraction and taking the inverse Laplace transform, the final differential equation describing the system is determined:
d2dt2yt+72ddtyt+9yt=6ut
t
2
y
t
72
t
1
y
t
9
y
t
6
u
t
(9)
The results from this example encourage us to apply Laplace techniques to earlier equations in this section. By taking the Laplace transform of Equation 4, we find that the transfer function of this system is simply the ratio of the two polynomials
psp s
and
qsq s.
By saying above that the Laplace transform of
ddtyt
t1
y
t
is
sYs
s
Y
s
we are assuming that the system has zero initial conditions . Taking a more general approach, a factor to characterize the initial conditions must be included:
ℒddtyt=sYs-y0-
ℒ
t
y
t
s
Y
s
y
0-
(10)
Note:
0-0-
here means the instant immediately preceding time
t=0t0;
it represents the time right before we start paying attention to the system. Also, some readers may be worried that by using
y0-
y
0-
in this equation, we are mixing frequency domain and time domain variables here. However, as
y0-
y
0-
is simply a constant, there is no time domain component involved.
Let's look at an example to see how these non-zero initial conditions come into play. A first order system is described by the following differential equation:
ddtyt+α0yt=ut
t
1
y
t
α0
y
t
u
t
(11)
When we take the Laplace transform of the derivative of
ytyt,
we must remember to include a term that represents the initial values of the system output.
sYs-y0-+α0Ys=Us
s
Y
s
y
0-
α0
Y
s
U
s
(12)
By combining the
YsYs
terms we get
s+α0Ys-y0-=Us
s
α0
Y
s
y
0-
U
s
(13)
If we say that
qs=s+α0
qs
s
α0
and
ps=1
ps
1
,
and define
rs
rs
as
y0-
y
0-
,
we can rearrange terms to get an expression relating
Ys
Y
s
to
Us
U
s
that takes the initial conditions into account:
Ys=psqsUs+rsqs
Y
s
p
s
q
s
U
s
r
s
q
s
(14)
What we have here is the Laplace domain solution to a differential equation describing a dynamical system. There are two terms in this solution: one that relies on the input and one that does not. These parts correspond to the particular and homogeneous solutions, respectively. Taking the inverse Laplace transform, we can write (14) as:
yt=ypartt+yhomot
y
t
ypart
t
yhomo
t
(15)
Here,
ypartt
ypart
t
corresponds to
psqsUs
p
s
q
s
U
s
and
yhomot
yhomo
t
corresponds to
rsqs
r
s
q
s
.
This makes a lot of sense. The particular solution (forced response) depends on a combination of
qs
q
s
,
which describes how the system behaves independently, and
psUs
p
s
U
s
,
which describes how the system reacts to the input
UU.
The homogeneous solution (natural response) depends on a combination of
qs
q
s
and
rs
r
s
,
the latter of which contains information about the initial conditions of the system.
Let's say that we know the homogeneous solution,
yt
y
t
,
to a differential equation describing a system.
yt=ⅇt
y
t
t
(16)
Goal: Using this solution, we want to try and figure out the system's
qⅆⅆt
q
ⅆ
ⅆ
t
function given zero initial conditions.
Solution:
From above, we know that for a homogeneous solution
yt=rⅆⅆtqⅆⅆt
y
t
r
ⅆ
ⅆ
t
q
ⅆ
ⅆ
t
(17)
We can clear the denominator by moving the
qⅆⅆt
q
ⅆ
ⅆ
t
to the left-hand side. And since we have zero initial conditions,
rⅆⅆt
r
ⅆ
ⅆ
t
goes to 0:
qⅆⅆtyt=0
q
ⅆ
ⅆ
t
y
t
0
(18)
The solution can quickly be determined by inspection because we know that the derivative of
ⅇt
t
is
ⅇt
t
.
Therefore a solution of
qⅆⅆt=ⅆⅆt-1
q
ⅆ
ⅆ
t
ⅆ
ⅆ
t
1
would work. However, a more systematic approach will be necessary for more difficult situations. We will investigate this approach here.
Again, we will do our work in the Laplace domain. By equating the Laplace transform of our homogeneous solution with the ratio of
rs
r
s
and
qs
q
s
as discussed above, we have:
Ys=1s-1=rsqs
Y
s
1
s
1
r
s
q
s
(19)
Directly, we can see the solution for
qs
q
s
:
by simply setting the denominators equal to each other,
qs=s-1
q
s
s
1
.
This, of course, is the Laplace transform of the solution of
qⅆⅆt
q
ⅆ
ⅆ
t
that we found by inspection above.
Now that we have the basics down, we'll look at a more complicated example. We are given
yt=Aⅇat+Btⅇbt
y
t
A
a
t
B
t
b
t
(20)
Goal: We would like to find the differential equation whose homogeneous solution is
yt
y
t
.
Solution:
Again, we take the Laplace transform of
yt
y
t
,
and then combine the two resultant fractions into one ratio of polynomials:
Ys=As-a+Bs-b2=As-b2+Bs-as-as-b2=rsqs
Y
s
A
s
a
B
s
b
2
A
s
b
2
B
s
a
s
a
s
b
2
r
s
q
s
(21)
Next, we equate the denominators of the last two fractions to find
qs
q
s
:
qs=s-as-b2=s3-2b+as2+b2+2abs-ab2
q
s
s
a
s
b
2
s
3
2
b
a
s
2
b
2
2
a
b
s
a
b
2
(22)
Recalling the start of this module, multiplying
qs
q
s
by
Ys
Y
s
and taking the inverse Laplace transform will yield the differential equation whose homogeneous solution is
yt
y
t
:
d3dt3yt-2b+ad2dt2yt+b2+2abddtyt-ab2yt=0
t
3
y
t
2
b
a
t
2
y
t
b
2
2
a
b
t
1
y
t
a
b
2
y
t
0
(23)
