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Proof

Module by: Anders Gjendemsjø. E-mail the author

Summary: Proof of Shannon's sampling theorem

Sampling theorem:

In order to recover the signal x⁢txt from it's samples exactly, it is necessary to sample x⁢txt 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 x⁢txt. Next we show that the signal x⁢txt 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 2×π⁢Fs 2 π Fs

Proof part 1 - Spectral considerations

By sampling x⁢txt every TsTs second we obtain xs⁢nxsn. The inverse fourier transform of this time discrete signal is

xs⁢n=12⁢π⁢∫−ππXs⁢ei⁢ω⁢ei⁢ω⁢ndω 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=Ω⁢Ts ω Ω Ts . We then obtain
xs⁢n=Ts2⁢π⁢∫−πTsπTsXs⁢ei⁢Ω⁢Ts⁢ei⁢Ω⁢Ts⁢ndΩ 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
x⁢t=12⁢π⁢∫−∞∞X⁢i⁢Ω⁢ei⁢Ω⁢tdΩ x t 1 2 Ω X Ω Ω t (3)
From this equation we find an expression for x (n⁢Ts) x n Ts
x⁢n⁢Ts=12⁢π⁢∫−∞∞X⁢i⁢Ω⁢ei⁢Ω⁢n⁢Tsd Ω 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 2⁢πTs 2 π Ts and then take the sum over the resulting integrals to obtain the complete area.
x⁢n⁢Ts=12⁢π⁢∑k=−∞∞∫(2⁢k−1)⁢πTs(2⁢k+1)⁢πTsX⁢i⁢Ω⁢ei⁢Ω⁢n⁢TsdΩ 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+2×π⁢kTs Ω η 2 π k Ts
x⁢n⁢Ts=12⁢π⁢∑k=−∞∞∫−πTsπTsX⁢i⁢(η+2×π⁢kTs)⁢ei⁢(η+2×π⁢kTs)⁢n⁢Tsdη 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 ei⁢2×π⁢k⁢n=1 2 π k n 1 , reinserting η=ΩηΩ and multiplying by TsTs Ts Ts
x⁢n⁢Ts=Ts2⁢π⁢∫−πTsπTs∑k=−∞∞1Ts⁢X⁢(i⁢(Ω+2×π⁢kTs))⁢ei⁢Ω⁢n⁢TsdΩ 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 xs⁢n=x⁢n⁢Ts xs n x n Ts for all values of nn, the integrands in Equation 2 and Equation 7 have to agreee, that is
Xs⁢ei⁢Ω⁢Ts=1Ts⁢∑k=−∞∞X⁢(i⁢(Ω+2⁢π⁢kTs)) 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=Ω⁢Ts ω Ω Ts

Xs⁢ei⁢ω=1Ts⁢∑k=−∞∞X⁢(i⁢ω+2×π⁢kTs) 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 x⁢txt from xs⁢nxsn.

Proof part II - Signal reconstruction

For a bandlimited signal the inverse fourier transform is

x⁢t=12⁢π⁢∫−πTsπTsX⁢i⁢Ω⁢ei⁢Ω⁢tdΩ x t 1 2 Ω Ts Ts X Ω Ω t (10)
In the interval we are integrating we have: Xs⁢ei⁢Ω⁢Ts=X⁢i⁢ΩTs Xs Ω Ts X Ω Ts . Substituting this relation into Equation 10 we get
x⁢t=Ts2⁢π⁢∫−πTsπTsXs⁢ei⁢Ω⁢Ts⁢ei⁢Ω⁢tdΩ x t Ts 2 Ω Ts Ts Xs Ω Ts Ω t (11)
Using the DTFT relation for Xs⁢ei⁢Ω⁢Ts Xs Ω Ts we have
x⁢t=Ts2⁢π⁢∫−πTsπTs∑n=−∞∞xs⁢n⁢e−(i⁢Ω⁢n⁢Ts)⁢ei⁢Ω⁢tdΩ 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
x⁢t=Ts2⁢π⁢∑n=−∞∞xs⁢n⁢∫−πTsπTsei⁢Ω⁢(t−n⁢Ts)dΩ 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
x⁢t=∑n=−∞∞xs⁢n⁢sinπTs⁢(t−n⁢Ts)πTs⁢(t−n⁢Ts) 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:

Xs⁢ei⁢Ω⁢Ts=1Ts⁢∑k=−∞∞X⁢(i⁢(Ω+2⁢π⁢kTs)) Xs Ω Ts 1 Ts k ∞ ∞ X Ω 2 k Ts

Reconstruction formula:

x⁢t=∑n=−∞∞xs⁢n⁢sinπTs⁢(t−n⁢Ts)πTs⁢(t−n⁢Ts) x t n xs n Ts t n Ts Ts t n Ts

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