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Proof

Module by: Anders Gjendemsjø

Summary: Proof of Shannon's sampling theorem

Sampling theorem: In order to recover the signal xtxt from it's samples exactly, it is necessary to sample xtxt at a rate greater than twice it's highest frequency component.

Introduction

As mentioned earlier, sampling is the necessary fundament when we want to apply digital signal processing on analog signals.
Here we present the proof of the sampling theorem. The proof is divided in two. First we find an expression for the spectrum of the signal resulting from sampling the original signal xtxt. Next we show that the signal xtxt can be recovered from the samples. Often it is easier using the frequency domain when carrying out a proof, and this is also the case here.
    Key points in the proof
  • We find an equation for the spectrum of the sampled signal
  • We find a simple method to reconstruct the original signal
  • The sampled signal has a periodic spectrum...
  • ...and the period is 2πFs 2 π Fs

Proof part 1 - Spectral considerations

By sampling xtxt every TsTs second we obtain xsnxsn. The inverse fourier transform of this time discrete signal is
xsn=12π∫-ππXsⅇⅈωⅇⅈωndω xs n 1 2 π ω π Xs ω ω n (1)
For convenience we express the equation in terms of the real angular frequency ΩΩ using ω=ΩTs ω Ω Ts . We then obtain
xsn=Ts2π∫-πTsπTsXsⅇⅈΩTsⅇⅈΩTsndΩ xs n Ts 2 Ω π Ts π Ts Xs Ω Ts Ω Ts n (2)
The inverse fourier transform of a continuous signal is
xt=12π∫-∞∞XⅈΩⅇⅈΩtdΩ x t 1 2 Ω X Ω Ω t (3)
From this equation we find an expression for x nTs x n Ts
xnTs=12π∫-∞∞XⅈΩⅇⅈΩnTsdΩ x n Ts 1 2 Ω X Ω Ω n Ts (4)
To account for the difference in region of integration we split the integration in Equation 4 into subintervals of length 2πTs 2 π Ts and then take the sum over the resulting integrals to obtain the complete area.
xnTs=12π∑k=-∞∞∫2k-1πTs2k+1πTsXⅈΩⅇⅈΩnTsdΩ x n Ts 1 2 π k Ω 2 k 1 Ts 2 k 1 Ts X Ω Ω n Ts (5)
Then we change the integration variable, setting Ω=η+2πkTs Ω η 2 π k Ts
xnTs=12π∑k=-∞∞∫-πTsπTsXⅈη+2πkTsⅇⅈη+2πkTsnTsdη x n Ts 1 2 π k ∞ ∞ η Ts π Ts X η 2 π k Ts η 2 π k Ts n Ts (6)
We obtain the final form by observing that ⅇⅈ2πkn=1 2 π k n 1 , reinserting η=ΩηΩ and multiplying by TsTs Ts Ts
xnTs=Ts2π∫-πTsπTs∑k=-∞∞1TsXⅈΩ+2πkTsⅇⅈΩnTsdΩ x n Ts Ts 2 π Ω π Ts π Ts k ∞ ∞ 1 Ts X Ω 2 π k Ts Ω n Ts (7)
To make xsn=xnTs xs n x n Ts for all values of nn, the integrands in Equation 2 and Equation 7 have to agreee, that is
XsⅇⅈΩTs=1Ts∑k=-∞∞XⅈΩ+2πkTs Xs Ω Ts 1 Ts k ∞ ∞ X Ω 2 k Ts (8)
This is a central result. We see that the digital spectrum consists of a sum of shifted versions of the original, analog spectrum. Observe the periodicity!
We can also express this relation in terms of the digital angular frequency ω=ΩTs ω Ω Ts
Xsⅇⅈω=1Ts∑k=-∞∞Xⅈω+2πkTs Xs ω 1 Ts k ∞ ∞ X ω 2 π k Ts (9)
This concludes the first part of the proof. Now we want to find a reconstruction formula, so that we can recover xtxt from xsnxsn.

Proof part II - Signal reconstruction

For a bandlimited signal the inverse fourier transform is
xt=12π∫-πTsπTsXⅈΩⅇⅈΩtdΩ x t 1 2 Ω Ts Ts X Ω Ω t (10)
In the interval we are integrating we have: XsⅇⅈΩTs=XⅈΩTs Xs Ω Ts X Ω Ts . Substituting this relation into Equation 10 we get
xt=Ts2π∫-πTsπTsXsⅇⅈΩTsⅇⅈΩtdΩ x t Ts 2 Ω Ts Ts Xs Ω Ts Ω t (11)
Using the DTFT relation for XsⅇⅈΩTs Xs Ω Ts we have
xt=Ts2π∫-πTsπTs∑n=-∞∞xsnⅇ-ⅈΩnTsⅇⅈΩtdΩ x t Ts 2 Ω Ts Ts n xs n Ω n Ts Ω t (12)
Interchanging integration and summation (under the assumption of convergence) leads to
xt=Ts2π∑n=-∞∞xsn∫-πTsπTsⅇⅈΩt-nTsdΩ x t Ts 2 n xs n Ω Ts Ts Ω t n Ts (13)
Finally we perform the integration and arrive at the important reconstruction formula
xt=∑n=-∞∞xsnsinπTst-nTsπTst-nTs x t n xs n Ts t n Ts Ts t n Ts (14)
(Thanks to R.Loos for pointing out an error in the proof.)

Summary

spectrum sampled signal: XsⅇⅈΩTs=1Ts∑k=-∞∞XⅈΩ+2πkTs Xs Ω Ts 1 Ts k ∞ ∞ X Ω 2 k Ts
Reconstruction formula: xt=∑n=-∞∞xsnsinπTst-nTsπTst-nTs x t n xs n Ts t n Ts Ts t n Ts

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