Digital Signal Processing Reference
In-Depth Information
Therefore, the partial fraction expansion for
X
(
s
)isgivenby
2
s
+
1
1
s
+
2
3
(
s
+
2)
2
2
(
s
+
2)
3
.
X
(
s
)
=
−
+
+
(D.25)
(ii) Assuming that the inverse Laplace transform
x
(
t
) is right-sided, we use
Table 6.1 to determine the inverse Laplace transform
x
(
t
)ofthe
X
(
s
):
−
t
−
2
t
−
2
t
+
t
2
e
−
2
t
)
u
(
t
)
x
(
t
)
=
(2e
−
e
+
3
t
e
−
t
+
(
t
2
+
3
t
−
1)e
−
2
t
]
u
(
t
)
.
=
[2e
(D.26)
D.2 Continuous-ti
me Fourier transform
The partial fraction expansion method, described above, may also be applied
to decompose the CTFT functions to a summation of simpler terms. Consider
the following rational function for CTFT:
=
b
m
( j
ω
)
m
+
b
m
−
1
( j
ω
)
m
−
1
++
b
1
( j
ω
)
+
b
0
a
n
( j
ω
)
n
+
a
n
−
1
( j
ω
)
n
−
1
++
a
1
( j
ω
)
+
a
0
X
(
ω
)
=
N
(
ω
)
D
(
ω
)
,
(D.27)
where the numerator
N
(
ω
) is a polynomial of degree
m
and the denominator
D
(
ω
) is a polynomial of degree
n
.If
m
≥
n
, we can divide
N
(
ω
)by
D
(
ω
) and
express
X
(
ω
) as follows:
m
−
n
N
1
(
ω
)
D
(
ω
)
X
1
(
ω
)
α
ℓ
( j
ω
)
−ℓ
+
X
(
ω
)
=
.
(D.28)
ℓ=
0
The procedure for decomposing
X
1
(
ω
) in simpler terms remains the same as
that discussed for the Laplace transform, except that the expansion is now made
with respect to (j
ω
). For example, if the denominator polynomial
D
(
ω
) has
n
first-order, non-repeated roots,
p
1
,
p
2
,...,
p
n
, such that
N
1
(
ω
)
D
(
ω
)
N
1
(
ω
)
( j
ω −
p
1
)( j
ω −
p
2
)
( j
ω −
p
n
)
,
X
1
(
ω
)
=
=
(D.29)
the function
X
1
(
ω
) may be decomposed as follows:
N
1
(
ω
)
D
(
ω
)
k
1
j
ω −
p
1
k
2
j
ω −
p
2
k
n
j
ω −
p
n
,
=
+
++
(D.30)
where the partial fraction coefficients
k
r
are calculated using the Heaviside
formula:
( j
ω −
p
r
)
N
1
(
ω
)
D
(
ω
)
k
r
=
.
(D.31)
j
ω=
p
r
Using the CTFT pair
1
a
+
j
ω
,
CTFT
←→
e
−
at
u
(
t
)
Search WWH ::
Custom Search