Graphics Reference
In-Depth Information
y
=
b
(
t
)
v
y
= sin(2
p
1.2
t
)
1
0
]
1
]
]
]
]
2
1.5
1
0.5
0
0.5
1
1.5
2
t
1
3
0
2
1
]
1
]
2
1
v
0.5
0
0
t
0
2
−
0.5
Figure 18.39: To compute the Fourier transform of the box b, for each frequency
ω
,we
multiply
sin(
2
πω
x
)
by b and compute the area beneath the resultant function, with positive
area (above the t
ω
-plane) shown in green and negative area in red. Top: The computation
for
ω
=
1.2
. Bottom: Computations for several values of
ω
. For each one, we plot, at the
right, the total area computed. This gives a function of the frequency
ω
, shown as a smooth
magenta curve; the result is evidently
ω →
5
p
y
cos(2
x
)
−
1
−
0.5
0
0.5
sinc
(
ω
)
.
Figure 18.40: x
→
cos(
2
π
x
)
.
The function name is often pronounced “sink.” Despite being described by cases,
the function is smooth and infinitely differentiable; its Taylor series is just the
series for
sin(
1
ˆ
y
5
f
(
v
)
π
x
)
divided by
π
x
:
0.5
x
)
2
3
!
x
)
4
5
!
−...
(
π
+
(
π
sinc
(
x
)=
1
−
.
(18.63)
0
Consider the function
f
(
x
)=cos(
2
−
2
0
2
x
)
(see Figure 18.40) on the interval
H
.
Direct evaluation of the integral shows that
π
1
)=
2
, and
FT
(
f
)(
k
)=
0 for all other
k
(see Figure 18.41). Thus,
f
(
x
)=
2
e
1
(
x
)+
2
e
−
1
(
x
)
,
which is also obvious from the definition of
e
k
(
x
)
.
(
f
)(
1
)=
2
and
F
F
(
f
)(
−
Figure 18.41: The Fourier trans-
form
for
Figure
18.40,
k
→
F(
f
)(
k
)
.