In order for a linear constant-coefficient difference equation
to be useful in analyzing a LTI system, we must be able to
find the systems output based upon a known input,
xn
x
n
, and a set of initial conditions. Two common
methods exist for solving a LCCDE: the direct
method and the indirect method, the later
being based on the z-transform. Below we will briefly discuss
the formulas for solving a LCCDE using each of these methods.
The final solution to the output based on the direct method
is the sum of two parts, expressed in the following
equation:
yn=
y
h
n+
y
p
n
y
n
y
h
n
y
p
n
(11)
The first part,
y
h
n
y
h
n
, is referred to as the
homogeneous
solution and the second part,
y
h
n
y
h
n
, is referred to as
particular
solution. The following method is very similar to
that used to solve many differential equations, so if you
have taken a differential calculus course or used
differential equations before then this should seem very
familiar.
We begin by assuming that the input is zero,
xn=0
x
n
0
.
Now we simply need to solve the homogeneous difference
equation:
∑
k
=0N
a
k
yn−k=0
k
0
N
a
k
y
n
k
0
(12)
In order to solve this, we will make the assumption that
the solution is in the form of an exponential. We will
use lambda,
λλ, to
represent our exponential terms. We now have to solve the
following equation:
∑
k
=0N
a
k
λn−k=0
k
0
N
a
k
λ
n
k
0
(13)
We can expand this equation out and factor out all of the
lambda terms. This will give us a large polynomial in
parenthesis, which is referred to as the
characteristic polynomial. The roots of this
polynomial will be the key to solving the homogeneous
equation. If there are all distinct roots, then the
general solution to the equation will be as follows:
y
h
n=
C
1
λ
1
n+
C
2
λ
2
n+…+
C
N
λ
N
n
y
h
n
C
1
λ
1
n
C
2
λ
2
n
…
C
N
λ
N
n
(14)
However, if the characteristic equation contains multiple
roots then the above general solution will be slightly
different. Below we have the modified version for an
equation where
λ
1
λ
1
has
KK multiple
roots:
y
h
n=
C
1
λ
1
n+
C
1
n
λ
1
n+
C
1
n2
λ
1
n+…+
C
1
nK−1
λ
1
n+
C
2
λ
2
n+…+
C
N
λ
N
n
y
h
n
C
1
λ
1
n
C
1
n
λ
1
n
C
1
n
2
λ
1
n
…
C
1
n
K
1
λ
1
n
C
2
λ
2
n
…
C
N
λ
N
n
(15)
The particular solution,
y
p
n
y
p
n
, will be any solution that will solve the
general difference equation:
∑
k
=0N
a
k
y
p
n−k=∑
k
=0M
b
k
xn−k
k
0
N
a
k
y
p
n
k
k
0
M
b
k
x
n
k
(16)
In order to solve, our guess for the solution to
y
p
n
y
p
n
will take on the form of the input,
xn
x
n
. After guessing at a solution to the above
equation involving the particular solution, one only
needs to plug the solution into the difference equation
and solve it out.
The indirect method utilizes the relationship between the
difference equation and z-transform, discussed earlier, to find a
solution. The basic idea is to convert the difference
equation into a z-transform, as described above, to get the
resulting output,
Yz
Y
z
.
Then by inverse transforming this and using partial-fraction
expansion, we can arrive at the solution.
Z
y
(
n
+
1
)
-
y
(
n
)
=
z
Y
(
z
)
-
y
(
0
)
Z
y
(
n
+
1
)
-
y
(
n
)
=
z
Y
(
z
)
-
y
(
0
)
(17)
This can be interatively extended to an arbitrary order derivative as in Equation Equation 18.
Z
-
∑
m
=
0
N
-
1
y
(
n
-
m
)
=
z
n
Y
(
z
)
-
∑
m
=
0
N
-
1
z
n
-
m
-
1
y
(
m
)
(
0
)
Z
-
∑
m
=
0
N
-
1
y
(
n
-
m
)
=
z
n
Y
(
z
)
-
∑
m
=
0
N
-
1
z
n
-
m
-
1
y
(
m
)
(
0
)
(18)
Now, the Laplace transform of each side of the differential equation can be taken
Z
∑
k
=
0
N
a
k
y
(
n
-
m
+
1
)
-
∑
m
=
0
N
-
1
y
(
n
-
m
)
y
(
n
)
=
Z
x
(
n
)
Z
∑
k
=
0
N
a
k
y
(
n
-
m
+
1
)
-
∑
m
=
0
N
-
1
y
(
n
-
m
)
y
(
n
)
=
Z
x
(
n
)
(19)
which by linearity results in
∑
k
=
0
N
a
k
Z
y
(
n
-
m
+
1
)
-
∑
m
=
0
N
-
1
y
(
n
-
m
)
y
(
n
)
=
Z
x
(
n
)
∑
k
=
0
N
a
k
Z
y
(
n
-
m
+
1
)
-
∑
m
=
0
N
-
1
y
(
n
-
m
)
y
(
n
)
=
Z
x
(
n
)
(20)
and by differentiation properties in
∑
k
=
0
N
a
k
z
k
Z
y
(
n
)
-
∑
m
=
0
N
-
1
z
k
-
m
-
1
y
(
m
)
(
0
)
=
Z
x
(
n
)
.
∑
k
=
0
N
a
k
z
k
Z
y
(
n
)
-
∑
m
=
0
N
-
1
z
k
-
m
-
1
y
(
m
)
(
0
)
=
Z
x
(
n
)
.
(21)
Rearranging terms to isolate the Laplace transform of the output,
Z
y
(
n
)
=
Z
x
(
n
)
+
∑
k
=
0
N
∑
m
=
0
k
-
1
a
k
z
k
-
m
-
1
y
(
m
)
(
0
)
∑
k
=
0
N
a
k
z
k
.
Z
y
(
n
)
=
Z
x
(
n
)
+
∑
k
=
0
N
∑
m
=
0
k
-
1
a
k
z
k
-
m
-
1
y
(
m
)
(
0
)
∑
k
=
0
N
a
k
z
k
.
(22)
Thus, it is found that
Y
(
z
)
=
X
(
z
)
+
∑
k
=
0
N
∑
m
=
0
k
-
1
a
k
z
k
-
m
-
1
y
(
m
)
(
0
)
∑
k
=
0
N
a
k
z
k
.
Y
(
z
)
=
X
(
z
)
+
∑
k
=
0
N
∑
m
=
0
k
-
1
a
k
z
k
-
m
-
1
y
(
m
)
(
0
)
∑
k
=
0
N
a
k
z
k
.
(23)
In order to find the output, it only remains to find the Laplace transform X(z)X(z) of the input, substitute the initial conditions, and compute the inverse Z-transform of the result. Partial fraction expansions are often required for this last step. This may sound daunting while looking at Equation 23, but it is often easy in practice, especially for low order difference equations. Equation 23 can also be used to determine the transfer function and frequency response.
As an example, consider the difference equation
y
[
n
-
2
]
+
4
y
[
n
-
1
]
+
3
y
[
n
]
=
cos
(
n
)
y
[
n
-
2
]
+
4
y
[
n
-
1
]
+
3
y
[
n
]
=
cos
(
n
)
(24)
with the initial conditions y'(0)=1y'(0)=1 and y(0)=0y(0)=0
Using the method described above, the Z transform of the solution y[n]y[n] is given by
Y
[
z
]
=
z
[
z
2
+
1
]
[
z
+
1
]
[
z
+
3
]
+
1
[
z
+
1
]
[
z
+
3
]
.
Y
[
z
]
=
z
[
z
2
+
1
]
[
z
+
1
]
[
z
+
3
]
+
1
[
z
+
1
]
[
z
+
3
]
.
(25)
Performing a partial fraction decomposition, this also equals
Y
[
z
]
=
.
25
1
z
+
1
-
.
35
1
z
+
3
+
.
1
z
z
2
+
1
+
.
2
1
z
2
+
1
.
Y
[
z
]
=
.
25
1
z
+
1
-
.
35
1
z
+
3
+
.
1
z
z
2
+
1
+
.
2
1
z
2
+
1
.
(26)
Computing the inverse Laplace transform,
y
(
n
)
=
(
.
25
z
-
n
-
.
35
z
-
3
n
+
.
1
cos
(
n
)
+
.
2
sin
(
n
)
)
u
(
n
)
.
y
(
n
)
=
(
.
25
z
-
n
-
.
35
z
-
3
n
+
.
1
cos
(
n
)
+
.
2
sin
(
n
)
)
u
(
n
)
.
(27)
One can check that this satisfies that this satisfies both the differential equation and the initial conditions.
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