Page 16-47
∫
∞
⋅
⋅
⋅
⋅
=
=
0
)
sin(
)
(
2
)
(
)}
(
{
dt
t
t
f
F
t
f
ω
π
ω
s
F
Inverse sine transform
∫
∞
−
⋅
⋅
⋅
=
=
0
1
)
sin(
)
(
)
(
)}
(
{
dt
t
F
t
f
F
s
ω
ω
ω
F
Fourier cosine transform
∫
∞
⋅
⋅
⋅
⋅
=
=
0
)
cos(
)
(
2
)
(
)}
(
{
dt
t
t
f
F
t
f
ω
π
ω
c
F
Inverse cosine transform
∫
∞
−
⋅
⋅
⋅
=
=
0
1
)
cos(
)
(
)
(
)}
(
{
dt
t
F
t
f
F
c
ω
ω
ω
F
Fourier transform (proper)
∫
∞
∞
−
−
⋅
⋅
⋅
=
=
dt
e
t
f
F
t
f
t
i
ω
π
ω
)
(
2
1
)
(
)}
(
{
F
Inverse Fourier transform (proper)
∫
∞
∞
−
−
−
⋅
⋅
=
=
dt
e
F
t
f
F
t
i
ω
ω
ω
)
(
)
(
)}
(
{
1
F
Example 1 – Determine the Fourier transform of the function f(t) = exp(- t), for t
>0, and f(t) = 0, for t<0.
The continuous spectrum, F(
ω
), is calculated with the integral:
∫
∫
+
−
∞
∞
→
+
−
=
ε
ω
ε
ω
π
π
0
)
1
(
0
)
1
(
2
1
lim
2
1
dt
e
dt
e
t
i
t
i
.
1
1
2
1
1
)
)
1
(
exp(
1
2
1
lim
ω
π
ω
ε
ω
π
ε
i
i
i
+
⋅
=
+
+
−
−
=
∞
→
This result can be rationalized by multiplying numerator and denominator by
the conjugate of the denominator, namely, 1-i
ω
. The result is now:
