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.
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.
- 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
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.
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.)
XsⅇⅈΩTs=1Ts∑k=-∞∞XⅈΩ+2πkTs
Xs
Ω
Ts
1
Ts
k
∞
∞
X
Ω
2
k
Ts
xt=∑n=-∞∞xsnsinπTst−nTsπTst−nTs
x
t
n
xs
n
Ts
t
n
Ts
Ts
t
n
Ts