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In-Depth Information
2, then
f
can be recovered from
f
, that is,
•f
F
(
f
)(
ω
)=
0for
|ω|≥
1
/
f
is invertible once the domain is restricted to the functions
strictly band-limited at 1
the map
f
→
/
2.
2
, there are multiple functions whose samples are
given by
g
, so in general, it's impossible to reconstruct an arbitrary
L
2
function from its samples.
• For any function
g
∈
L
2
(
H
)
, sampled at a
collection of
n
evenly spaced points in the interval; in that case, if the function is
band-limited at any frequency below
n
The corresponding statements hold for a function
f
∈
/
2, then it's reconstructible from its sam-
ples. (Indeed, that's the essence of the sampling theorem.)
We'll actually show exactly
how
to reconstruct a band-limited function from
its samples, in that rather than just showing that the sampling process is invertible,
we'll explicitly describe an inverse. We'll also describe some approximations to
the inverse that are easily computable, that is, functions with the property that if
f
is band-limited, then sampling
f
, followed by the approximate-reconstruction
operation, will yield a function very close to
f
.
We'll start with the easy part: showing there are multiple functions with the
same samples. Recall that for a continuous function, “sampling at
x
” simply means
“evaluating the function at
x
.” So to show that there are two different functions
with the same samples at integer points, consider
f
1
(
x
)=
0 and
f
2
(
x
)=sin(
x
)
.
At every integer point, these two have exactly the same value (namely 0). Thus,
if you're given the samples of one of the two functions, you cannot possibly tell
which one it was. Of course,
f
2
isn't in
L
2
(
R
)
, because it doesn't approach zero as
x
π
→±∞
, but this quibble is easily resolved:
Inline Exercise 18.8:
Show that if
f
is any function in
L
2
, then
x
f
(
x
)
f
2
(
x
)
is a function whose samples match those of
f
1
, so the sampling operation is a
many
-to-one map from
L
2
(
R
)
→
2
(
Z
)
.
→
1
Note, however, that the function
f
2
has frequency
2
, so it's
just
above the
Nyquist limit: We expect it to produce aliasing.
For the corresponding situation on the interval
H
=(
2
,
2
]
, a similar example
suffices. If we consider the
n
equally spaced sample points
x
j
=
1
−
2
+
n
,
j
=
1
−
0, 1,
...
,
n
−
1, then the function
g
1
(
x
)=
0 and the function
g
2
(
x
)=sin(
π
n
(
x
+
1
2
))
both take on the value 0 at every
x
j
.
We say that the function
g
2
is an
alias
of the function
g
1
, because from the
point of view of their samples, they appear to be the same. If we had a way to
construct a function on the whole interval from a set of samples, then the samples
of
g
1
and
g
2
, being identical, would produce identical results.
Now let's turn to the more difficult part: showing that
if a function is appro-
priately band-limited, it can be reconstructed from its samples.
As we saw earlier, if we have a function
f
L
2
(
R
)
and multiply it by a “comb-
like” function in
L
2
, the Fourier transform of the result starts to look periodic,
consisting of multiple copies of the transform of
f
. As we take better and better
approximations of the comb, the Fourier transform becomes closer and closer to
the convolution of
f
with a comb. The limit of the comb approximations doesn't
exist, of course, but the collection of samples of
f does
exist, and is a function
f
in
∈
2
. In the frequency domain, the Fourier transforms of the convolutions of
f
with
more and more comblike approximations of the comb get closer and closer to a
