# Theorem of the sampling of Nyquist-Shannon

theorem of the sampling of Nyquist- it is one of the more important theorems that it governs . It defines necessary the minimal tie to sampling of one it marks them in order to avoid distortions of the same one.

Given it marks them, with ended and famous, minimal of sampling of such it marks them must be at least the double quantity of the maximum frequency of the same one. Intuitivamente can be demonstrated in two ways as such frequency is the minim necessary not to introduce distortions in marks them:

• If we think next to the bandwidth of marks them, as an example 10 as to the maximum frequency of it marks them same; it turns out intuitivo to think that in order to sample adequately such member it is necessary to arrange at least 2 champions for every wave shape. Since the wave shapes are to maximum 10 mila the positive and negative members of the wave shape at least serve ventimila champions for being able to sample all.
• It marks them reconstructed, it is introduced, in frequency, as the repetition of many marks them of B band (the same one of marks them championship) centers frequencies to you K*F (where F it is the sampling frequency and K it is entire from a little infinite more infinite). If the distance between such marks them he was smaller of the double quantity of their band they would be overlapped generating interferences (in practical distorting marks them championship).

In truth the theorem would have to call Whittaker-Nyquist-Kotelnikov-Shannon, second the chronological order of who gradually demonstrated versions more generalized of the theorem.

For sampling of it marks them agrees the withdrawal of money of champions (samples) from it marks them analogic every $T_c$ second ($T_c$ it is in fact sampling step, while $f_c = \frac{1}{T_c}$ it is sampling frequency). The champions capture to you serve for a successive digitalizzazione of mark them, in such way to render it usufruibile from the computers and means of transmission.

## Enunciated

The theorem of the sampling, appeared for before the time in in an article of C. And. Shannon, it asserts that:

if:

• s(t) it is one it marks them to limited, that is its S(f) he is equal to 0 for |f|>BW (BandWidth, that is the band of marks them), that is the members of marks them with greater frequency of they are null;
• $f_c = \frac{1}{T_c} > 2BW$ , that is if the sampling frequency is greater of the double quantity of the band,

then:

• it marks them s(t) it is represented completely from the own champions;
• it marks them s(t) it can be reconstructed with one ideal having cut frequency $f_T$ such that $BW<f_T$.

FAMOUS: The formulation of which over sottintende that the inferior frequency of the band of marks them is zero. If instead draft of marks band-passing them, with various inferior frequency from zero, then the theorem goes reformulated in the following way. It is $BW = f_{H} - f_{L}$ the bandwidth of marks them, being $f_{H}$ and $f_{L}$ respective the advanced end and that inferior of the band. In order to avoid the superimposition of the phantom championship the sampling frequency $f_{c}$ it must satisfy the following diseguaglianze $\frac{2(f_{L} + BW)}{N + 1} < f_{c} < \frac{2f_{L}}{N}$, with $N<\frac{f_{L}}{W}$ entire. Every entire value del' $N$, included the zero, it supplies an allowed interval of sampling frequencies. These intervals allowed are separate to you from intervals of not acceptable frequencies, if the superimposition of the phantom is wanted to be avoided championship. Some of the values of $N$ they can give place to inferior frequencies of sampling to the minim of the band of marks them and this property has been variously called like undersampling, subsampling, sottocampionamento, subcampionamento, sub-Nyquist, super-Nyquist, harmonic sampling, etc. We consider the following example with $f_{H}=2100 kHz$ and $f_{L}=1550 kHz$. It is had $N<2,8$, that is $N=2, 1, 0$. For which the intervals allowed of sampling frequencies they are. With $N=2$: $1400 kHz<f_{c}<1550 kHz$, $N=1$: $2100 kHz<f_{c}<3100 kHz$, and finally with $N=0$: $4200 kHz<f_{c}$. To notice that it is mistaken to choose $f_c > 2BW$, that is $f_c > 1100 kHz$, in how much it is necessary to satisfy not one but two ties. For other details and applications to consult [ 1 ]

## Explanation

The explanation is simple for who knows. In fact, if it marks them s(t) like it has transformed of Fourier S(f), then it marks them championship with step $T_c$:

$s_c\left(t \right) = \sum_{k=-\infty}^\infty s\left(kT_c \right)\delta\left(t - kT_c \right)$

like it has transformed:

$S_c\left(f \right) = \frac{1}{T_c}\sum_{n=-\infty}^\infty S(f - nF_c)$.

This means that to sample in the time it is equivalent to periodicizzare in frequency. Therefore to say that the sampling step must be at least two times the band of marks means them that the repeated phantoms every $F_c$ they do not overlap between of they, avoiding therefore the phenomenon ofaliasing.

To this point, leaking it marks them with the described pass-low filter over, it captures it marks them with phantom centered in f = 0, than therefore, antitransformed, she supplies marks them s(t) of departure.

## Famous important

In the truth they do not exist marks them with closely limited band. For this reason it is introduced, to mount of the sampler, a filter low-pass filter (said filter anti-aliasing) that it eliminates the members of marks them so as to to respect the enunciated theorem over