PHYSICS 218
SOLUTION TO HW 4
Created: September 18, 2004
8:05am
Last updated: September 26, 2004
11:24am
1. (a) Recall that integration by parts is
b
a
u
d
v
=
u v
b
a

b
a
v
d
u .
So for this problem, it is of our interest to let
u
=
F
(
t
) and d
v
= e

iωt
d
t
. Starting with
G
(
ω
),
we integrate by parts to get
G
(
ω
) =
∞
∞
F
(
t
) e

iωt
d
t
=
F
(
t
) e

iωt

i ω
∞
∞

∞
∞
d
F
d
t
e

iωt

i ω
d
t
=
1
i ω
∞
∞
f
(
t
) e

iωt
d
t
=
g
(
ω
)
i ω
,
where the first term vanishes due to boundary conditions lim
t
→±∞
F
(
t
) = 0.
(b) Here, we let
u
= e

iωt
d
t
and d
v
=
f
(
t
) =
d
F
d
t
to get
g
(
ω
) =
∞
∞
d
F
d
t
e

iωt
d
t
=
F
(
t
) e

iωt
∞
∞

∞
∞
F
(
t
) (

i ω
) e

iωt
d
t
=
i ω
∞
∞
F
(
t
) e

iωt
d
t
=
i ω G
(
ω
)
.
2. (a) The function
f
(
t
) =
0
for

t

> τ ,
h
for

τ < t <
0
,

h
for
0
< t < τ ,
(1)
is sketched in figure (1).
(b) For piecewise continuous function like
f
(
t
), we can separate the domain of integration into
several regions. So the Fourier transform of
f
(
t
) is
g
(
ω
) =
∞
∞
f
(
t
) e

iωt
d
t
=

τ
∞
+
0

τ
+
τ
0
+
∞
τ
f
(
t
) e

iωt
d
t
=
0

τ
h
e

iωt
d
t
+
τ
0
(

h
) e

iωt
d
t
=
h
e

iωt

i ω
0

τ
+

h
e

iωt

iω
τ
0
=
h
i ω
(
e
iωτ
+ e

iωτ

2
)
=
2
h
i ω
cos(
ω τ
)

1
=
4
h i
ω
sin
2
ω τ
2
.
(c) The graph of
g
(
ω
) is shown in figure (2).
3. (a) Note that the function
f
(
t
) actually has four regions. To be more precise, we can write it as
f
(
t
) =
0
for

t

> τ ,
b
(
1 +
t/τ
)
for

τ
t
0
,
b
(
1

t/τ
)
for
0
t
τ .
(2)
The sketch of it is shown in figure (3).
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
PHYSICS 218
SOLUTION TO HW 4
0

τ
τ
t
f
(
t
)

h
h
Figure 1:
Sketch of
f
(
t
).
1.5
1
0.5
0
0.5
1
1.5
6
4
2
0
2
4
6
g
(
ω
)
ω
(in units of
π
)
Figure 2:
Graph of
g
(
w
) with
h
= 1. Note that we have left out the imaginary number
i
.
(b) Similar to problem (2), the Fourier transform of
f
(
t
) is
g
(
ω
) =
0

τ
b
(
1 +
t/τ
)
e

iωt
d
t
+
τ
0
b
(
1

t/τ
)
e

iωt
d
t
=
b
τ

τ
e

iωt
d
t
+
b
τ
0

τ
t
e

iωt
d
t

b
τ
τ
0
t
e

iωt
d
t .
(3)
The first integral is easily evaluated to be
τ

τ
e

iωt
d
t
=
2
ω
sin(
ωτ
).
The second and third
integral can be evaluated by employing the trick of partial differentiation under the integral
2
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '04
 POLLACK
 Physics, Magnetism, group velocity, e−iωt dt

Click to edit the document details