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The Shannon-Whitaker Sampling Theorem

Module by: Michael A Lexa, Ronald DeVore. E-mail the authors

The classical theory behind the encoding analog signals into bit streams and decoding bit streams back into signals, rests on a famous sampling theorem which is typically refereed to as the Shannon-Whitaker Sampling Theorem. In this course, this sampling theory will serve as a benchmark to which we shall compare the new theory of compressed sensing.

To introduce the Shannon-Whitaker theory, we first define the class of bandlimited signals. A bandlimited signal is a signal whose Fourier transform only has finite support. We shall denote this class as B A B A and define it in the following way:

B A :={fL 2 (R):f ^(ω)=0,|ω|Aπ}.B A :={fL 2 (R):f ^(ω)=0,|ω|Aπ}.
(1)
Here, the Fourier transform of ff is defined by
f ^(ω):=1 2π R f(t)e -iωt dt. f ^(ω):=1 2π R f(t)e -iωt dt.
(2)
This formula holds for any fL 1 fL 1 and extends easily to fL 2 fL 2 via limits. The inversion of the Fourier transform is given by
f(t):=1 2π R f ^(ω)e iωt dω.f(t):=1 2π R f ^(ω)e iωt dω.
(3)

Theorem 1: Shannon-Whitaker Sampling Theorem

If fB A fB A , then ff can be uniquely determined by the uniformly spaced samples fn Afn A and in fact, is given by

f(t)= nZ fn A sinc (π( At -n)),f(t)= nZ fn A sinc (π( At -n)),
(4)
where sinc (t)=sint t sinc (t)=sint t.

Proof

It is enough to consider A=1A=1, since all other cases can be reduced to this through a simple change of variables. Because fB A=1 fB A=1 , the Fourier inversion formula takes the form

f(t)=1 2π -π π f ^(ω)e iωt dω.f(t)=1 2π -π π f ^(ω)e iωt dω.
(5)
Define F(ω)F(ω) as the 2π2π periodization of f ^f ^,
F(ω):= nZ f ^(ω-2nπ).F(ω):= nZ f ^(ω-2nπ).
(6)
Because F(ω)F(ω) is periodic, it admits a Fourier series representation
F(ω)= nZ c n e -inω ,F(ω)= nZ c n e -inω ,
(7)
where the Fourier coefficients c n c n given by
c n =1 2π -π π F(ω)e inω dω=1 2π -π π f ^(ω)e inω dω.c n =1 2π -π π F(ω)e inω dω=1 2π -π π f ^(ω)e inω dω.
(8)
By comparing (Equation 8) with (Equation 5), we conclude that
c n =1 2πf(n).c n =1 2πf(n).
(9)
Therefore by plugging (Equation 9) back into (Equation 6), we have that
F(ω)=1 2π nZ f(n)e -inω .F(ω)=1 2π nZ f(n)e -inω .
(10)
Now, because
f ^(ω)=F(ω)χ [-π,π] =1 2π nZ f(n)e -inω χ [-π,π] ,f ^(ω)=F(ω)χ [-π,π] =1 2π nZ f(n)e -inω χ [-π,π] ,
(11)
and because of the facts that
F(χ [-π,π] )=1 2π sinc (πω)andF(g(t-n))=e -inω F(g(t)),F(χ [-π,π] )=1 2π sinc (πω)andF(g(t-n))=e -inω F(g(t)),
(12)
we conclude
f(t)= nZ f(n) sinc (π(t-n)).f(t)= nZ f(n) sinc (π(t-n)).
(13)

Comments:

  1. (Good news) The set { sinc (π(t-n))} nZ { sinc (π(t-n))} nZ is an orthogonal system and therefore, has the property that the L 2 L 2 norm of the function and its Fourier coefficients are related by,
    f L 2 2 =2π nZ |f(n)| 2 f L 2 2 =2π nZ |f(n)| 2
    (14)
  2. (Bad news) The representation of ff in terms of sinc functions is not a stable representation, i.e.
    nZ | sinc (π(t-n))| nZ 1 |t-n|+1divergences nZ | sinc (π(t-n))| nZ 1 |t-n|+1divergences
    (15)

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