Digital Signal Processing Reference
In-Depth Information
TABLE 2.2 Form of Input Function and Forced Response
Input or Forcing Function
Particular Function or Forced Response
A(α)
n
,α
B(α)
n
1
=
c
1
(i
=
1
,
2
,...)
A(α)
n
,α
B
1
n
]
(α)
n
2
=
c
i
[
B
0
+
3
A
cos
(ω
0
n
+
θ)
B
cos
(ω
0
n
+
φ)
i
=
0
A
i
n
i
α
n
i
=
0
B
i
n
i
α
n
4
equation, we compute the coefficient
B
or the coefficients
B
i
. Next we apply
the given initial conditions on the sum of the complementary function and the
particular function, in which there are
n
unknown constants of the complemen-
tary function. When we obtain these constants that satisfy the initial conditions,
and substitute them, the solution for the total output is complete. Example 2.12
illustrates the classical method of solving a difference equation.
Example 2.12
Solve the linear difference equation given below, using the classical method:
2
(
0
.
1
)
n
y(n)
−
0
.
5
y(n
−
1
)
+
0
.
06
y(n
−
2
)
=
(2.49)
y(
−
1
)
=
1 d
y(
−
2
)
=
0
(2.50)
The characteristic polynomial is
z
2
0
.
2
)
,which
has the characteristic roots at
z
1
=
0
.
3and
z
2
=
0
.
2. Since these are simple
zeros, the complementary function
y
c
(n)
−
0
.
5
z
+
0
.
06
=
(z
−
0
.
3
)(z
−
A
2
(
0
.
2
)
n
. Since the input
x(n)
is given as 2
(
0
.
1
)
n
, we choose from Table 2.2, the particular function
y
p
to
be of the form
y
p
(n)
A
1
(
0
.
3
)
n
=
+
B(
0
.
1
)
n
. Thus we substitute
y
p
(n
B(
0
.
1
)
n
−
1
and
=
−
1
)
=
B(
0
.
1
)
n
−
2
and get the following:
y
p
(n
−
2
)
=
B(
0
.
1
)
n
−
0
.
5
B(
0
.
1
)
n
−
1
+
0
.
06
B(
0
.
1
)
n
−
2
=
2
(
0
.
1
)
n
B(
0
.
1
)
n
0
.
5
(
10
)B(
0
.
1
)
n
0
.
06
(
100
)B(
0
.
1
)
n
2
(
0
.
1
)
n
−
+
=
[1
.
0
−
5
+
6]
B(
0
.
1
)
n
=
2
(
0
.
1
)
n
(
0
.
1
)
n
. So the total solu-
Therefore
B
=
1 and the particular function
y
p
(n)
=
tion is
y(n)
=
y
c
(n)
+
y
p
(n)
A
1
(
0
.
3
)
n
A
2
(
0
.
2
)
n
(
0
.
1
)
n
=
+
+
When we apply the initial conditions on this total response, we get
y(
−
1
)
=
3
.
3333
A
1
+
5
A
2
+
10
=
1
y(
−
2
)
=
11
.
111
A
1
+
25
A
2
+
100
=
0
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