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
18.13.3 An Example on an Interval
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 ) .
 
 
 
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