Proof of Shannon's sampling theorem
In order to recover the signal
$x(t)$ from it's samples exactly, it is necessary to sample
$x(t)$ 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 signalprocessing 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 fromsampling the original signal
$x(t)$ .
Next we show that the signal
$x(t)$ 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
$2\times \pi {F}_{s}$
Proof part 1  spectral considerations
By sampling
$x(t)$ every
${T}_{s}$ second we obtain
${x}_{s}(n)$ .
The inverse fourier transform of this
time discrete signal is
${x}_{s}(n)=\frac{1}{2\pi}\int_{\pi}^{\pi} {X}_{s}(e^{i\omega})e^{i\omega n}\,d \omega $
For convenience we express the equation in terms of the real angular
frequency
$\Omega $ using
$\omega =\Omega {T}_{s}$ .We then obtain
${x}_{s}(n)=\frac{{T}_{s}}{2\pi}\int_{\frac{\pi}{{T}_{s}}}^{\frac{\pi}{{T}_{s}}} {X}_{s}(e^{i\Omega {T}_{s}})e^{i\Omega {T}_{s}n}\,d \Omega $
The inverse fourier transform of a continuous signal is
$x(t)=\frac{1}{2\pi}\int_{()} \,d \Omega $∞
∞
X
Ω
Ω
t From this equation we find an expression for
$x(n{T}_{s})()$
$x(n{T}_{s})=\frac{1}{2\pi}\int_{()} \,d \Omega $∞
∞
X
Ω
Ω
n
T
s To account for the difference in region of integration we split the integration in
into subintervals of length
$\frac{2\pi}{{T}_{s}}$ and then take the sum over the resulting integrals to obtain the complete area.
$x(n{T}_{s})=\frac{1}{2\pi}\sum_{k=()} $∞
∞
Ω
2
k
1
T
s
2
k
1
T
s
X
Ω
Ω
n
T
s Then we change the integration variable, setting
$\Omega =\eta +\frac{2\times \pi k}{{T}_{s}}$
$x(n{T}_{s})=\frac{1}{2\pi}\sum_{k=\infty}^{\infty} \int_{\frac{\pi}{{T}_{s}}}^{\frac{\pi}{{T}_{s}}} X(i(\eta +\frac{2\times \pi k}{{T}_{s}})())e^{i(\eta +\frac{2\times \pi k}{{T}_{s}})()n{T}_{s}}\,d \eta $
We obtain the final form by observing that
$e^{i\times 2\times \pi kn}=1$ ,
reinserting
$\eta =\Omega $ and multiplying by
$\frac{{T}_{s}}{{T}_{s}}$
$x(n{T}_{s})=\frac{{T}_{s}}{2\pi}\int_{\frac{\pi}{{T}_{s}}}^{\frac{\pi}{{T}_{s}}} \sum_{k=\infty}^{\infty} \frac{1}{{T}_{s}}Xi(\Omega +\frac{2\times \pi k}{{T}_{s}})()e^{i\Omega n{T}_{s}}\,d \Omega $
To make
${x}_{s}(n)=x(n{T}_{s})$ for all values of
$n$ , the integrands in
and
have to agreee, that is
${X}_{s}(e^{i\Omega {T}_{s}})=\frac{1}{{T}_{s}}\sum_{k=\infty}^{\infty} Xi(\Omega +\frac{2\pi k}{{T}_{s}})()$
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
$\omega =\Omega {T}_{s}$
${X}_{s}(e^{i\omega})=\frac{1}{{T}_{s}}\sum_{k=\infty}^{\infty} Xi\frac{\omega +2\times \pi k}{{T}_{s}}$
This concludes the first part of the proof. Now we want to find a reconstruction formula, so
that we can recover
$x(t)$ from
${x}_{s}(n)$ .
Proof part ii  signal reconstruction
For a
bandlimited signal the inverse fourier transform is
$x(t)=\frac{1}{2\pi}\int_{\frac{\pi}{{T}_{s}}}^{\frac{\pi}{{T}_{s}}} X(i\Omega )e^{i\Omega t}\,d \Omega $
In the interval we are integrating we have:
${X}_{s}(e^{i\Omega {T}_{s}})=\frac{X(i\Omega )}{{T}_{s}}$ . Substituting this relation into
we get
$x(t)=\frac{{T}_{s}}{2\pi}\int_{\frac{\pi}{{T}_{s}}}^{\frac{\pi}{{T}_{s}}} {X}_{s}(e^{i\Omega {T}_{s}})e^{i\Omega t}\,d \Omega $
Using the
DTFT relation for
${X}_{s}(e^{i\Omega {T}_{s}})$ we have
$x(t)=\frac{{T}_{s}}{2\pi}\int_{\frac{\pi}{{T}_{s}}}^{\frac{\pi}{{T}_{s}}} \sum_{n=()} \,d \Omega $∞
∞
x
s
n
Ω
n
T
s
Ω
t Interchanging integration and summation (under the assumption of convergence) leads to
$x(t)=\frac{{T}_{s}}{2\pi}\sum_{n=()} $∞
∞
x
s
n
Ω
T
s
T
s
Ω
t
n
T
s Finally we perform the integration and arrive at the important reconstruction formula
$x(t)=\sum_{n=()} $∞
∞
x
s
n
T
s
t
n
T
s
T
s
t
n
T
s (Thanks to R.Loos for pointing out an error in the proof.)
Summary
${X}_{s}(e^{i\Omega {T}_{s}})=\frac{1}{{T}_{s}}\sum_{k=\infty}^{\infty} Xi(\Omega +\frac{2\pi k}{{T}_{s}})()$
$x(t)=\sum_{n=()} $∞
∞
x
s
n
T
s
t
n
T
s
T
s
t
n
T
s
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