Classroom Exercise #2
Here’s the exercise that we did in class today.
I’ve added an extension to this exercise
below as well.
Problem:
Determine the Fourier transform of
v
(
t
) =
A
sin(2
πf
0
t
)
.
Solution #1:
The simplest approach is to express the sin function as a cos function.
Recall,
sin(
θ
) = cos(
θ
−
π/
2)
.
Thus,
v
(
t
) =
A
sin(2
πf
0
t
) =
A
cos(2
πf
0
t
−
π/
2)
.
Applying
cos(
θ
) =
1
2
(
e
jθ
+
e

jθ
)
we have
v
(
t
) =
A
2
e

jπ/
2
e
j
2
πf
0
t
+
A
2
e
jπ/
2
e
j
2
π
(

f
0
)
t
.
(1)
Apply equation (13) in Chap. 1 yields the desired Fourier transform of
v
(
t
)
:
V
(
f
) =
A
2
e

jπ/
2
δ
(
f
−
f
0
) +
A
2
e
jπ/
2
δ
(
f
+
f
0
)
.
An alternative approach is to express the given sin function as a sum of complex expo
nentials directly, i.e., we apply the identity
sin(
θ
) =
1
2
j
(
e
jθ
−
e

jθ
)
.
This is described in the second solution below.
Solution #2:
First, we write
v
(
t
)
as
v
(
t
) =
A
2
j
e
j
2
πf
0
t
−
A
2
j
e
j
2
π
(

f
0
)
t
=
A
2
1
j
e
j
2
πf
0
t
+
A
2
−
1
j
e
j
2
π
(

f
0
)
t
.
Now, since
j
=
√
−
1
, we have that
1
j
=
j
j
2
=
−
j
=
e

jπ/
2
since
e
j
(
θ
−
π/
2
)
= cos(
θ
−
π/
2) +
j
sin(
θ
−
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 Spring '11
 phoon
 Fourier Series, Sin, Cos, Complex number

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