The theorem of Nyquist-Shannon , named according to Harry Nyquist and Claude Shannon, states that the Fréquence of sampling of a signal must be equal or higher than the double of the maximum frequency contained in this signal, in order to convert this signal of an analogical form with a numerical form . This theorem is at the base of the numerical conversion of the signals.

The best illustration of the application of this theorem is the determination of the sampling rate of a audio CD, which is of 44,1  K Hz. Indeed, the human ear can collect the sounds until 16  Khz, sometimes until 20  Khz. It is thus appropriate, during conversion, to sample the audio signal with at least 40  Khz. 44,1  Khz is the standardized value by industry.

Elementary considerations

If one wants to use a signal sampled, it is necessary to be sure that this one contains all the information of the analogical signal of origin. It is often convenient to consider this one as a sum of Sinusoïde S (cf spectral Analyze). However it is intuitively obvious that a loss of information occurs if the step of sampling is too large by comparison with the periods in question, the sampling rate being too weak compared to the frequencies considered.

That is to say a sinusoidal signal of amplitude has and frequency F:

$x\; (T)\; =\; has\; \backslash \; cos\; (2\; \backslash \; pi\; F\; T)\; \backslash ,$

By sampling it with a step T 1/T one is a frequency obtains the continuation of numerical values

$x\_n\; =\; has\; \backslash \; cos\; (2\; \backslash \; pi\; N\; F\; T)\; \backslash ,$

Now let us consider the signal of amplitude B and frequency 1/T - F:

$\backslash \; textstyle\; there\; (T)\; =\; B\; \backslash \; cos\; \backslash \; left\; (2\; \backslash \; pi\; \backslash \; left\; (\backslash \; frac1T\; -\; F\; \backslash \; right)\; T\; \backslash \; right)$

Once sampled at the same frequency, it becomes

$\backslash \; textstyle\; y\_n\; =\; B\; \backslash \; cos\; \backslash \; left\; (2\; \backslash \; pi\; N\; \backslash \; left\; (\backslash \; frac1T\; -\; F\; \backslash \; right)\; T\; \backslash \; right)\; =\; B\; \backslash \; cos\; \backslash \; left\; (2\; \backslash \; pi\; N\; \backslash \; left\; (1\; -\; F\; T\; \backslash \; right)\; \backslash \; right)\; \backslash ,$

The elementary Trigonométrie leads to

$y\_n\; =\; B\; \backslash \; cos\; (2\; \backslash \; pi\; N\; F\; T)\; \backslash ,$

Thus, in the sum x_n + y_n, it is impossible to distinguish what belongs to the signal of frequency F and that of frequency 1/T - F. This result led to the effect of Crenellation , fold of spectrum or aliasing , which indicates that one takes a sinusoid for another ( alias ).

If more the high frequency of a signal is f_M, the frequency 1/T - f_M should not belong to the spectrum of the signal, which leads to the inequality:

$\backslash \; frac\; 1T\; >\; 2\; f\_M$

So that a signal is not disturbed by sampling, the sampling rate must be higher than the double of the more high frequency contained in the signal. This limiting frequency is called the frequency of Nyquist.

Precise details

One can interpret the preceding result by considering a transitory signal X (T), therefore provided with a Transformée of Fourier X (F).

Let us consider the function obtained by multiplying signal X (T) by a Peigne of Dirac, nap of deltas of intensity T distant of T.

$x^*\; (T)\; =\; X\; (T)\; \backslash \; delta\_T\; (T)\; \backslash ,$

Taking into account the fundamental property of the comb of Dirac, the transform of Fourier of x* (T) is the approximation of the transform of X (T) obtained by the method of the rectangles:

$X^*\; (F)\; =\; T\; \backslash \; sum\_\; \{n=-\; \backslash \; infty\}\; ^\; \{+\; \backslash \; infty\}\; X\; (N\; T)\; e^\; \{-\; I\; N2\; \backslash \; pi\; F\; T\}$

By using the development in Fourier series of the comb, this transform is also calculated in the form

$X^*\; (F)\; =\; \backslash \; int\_\; \{-\; \backslash \; infty\}\; ^\; \{+\; \backslash \; infty\}\; \backslash \; sum\_\; \{n=-\; \backslash \; infty\}\; ^\; \{+\; \backslash \; infty\}\; e^\; \{I\; 2\; \backslash \; pi\; N\; t/T\}\; X\; (T)\; e^\; \{-\; I\; 2\; \backslash \; pi\; F\; \backslash \; mathrm\; dt\}\; \backslash \; mathrm\; dt$

By gathering the exponential ones and by exchanging the operators, one obtains:

$X^*\; (F)\; =\; \backslash \; sum\_\; \{n=-\; \backslash \; infty\}\; ^\; \{+\; \backslash \; infty\}\; X\; (F\; -\; n/T)$

the bringing together of the two results shows that the calculation of the transform of a signal sampled with the step T by the method of the rectangles gives the sum of the true transform and all relocated this one with a step equal to the sampling rate 1/T.

All the useful information is contained in the interval.

If the frequencies present in the signal do not overflow of this interval, i.e. if the sampling rate is higher than the double of the more high frequency, one obtains the true transform. In the contrary case, the relocated close ones come to be superimposed. This phenomenon is called " covering of the specter"

Because of symmetry, all occurs as if the true spectrum were folded up (energy associated with the frequencies higher than half of the sampling rate is transferred in lower part from this frequency). If one wants to avoid the Franglais one in general uses the folding up term preferably with Aliasing.

These results apply without modification to a signal to finished variance.

Formulate of Shannon

Since the transform X* (F) of the correctly sampled signal contains, in the interval, the transform of the signal of origin X (T), one can reconstitute this one by calculating the opposite transform, the Intégration being limited with this interval.

One obtains thus

$x\; (T)\; =\; \backslash \; sum\_\; \{n=-\; \backslash \; infty\}\; ^\; \{+\; \backslash \; infty\}\; X\; (N\; T)\; \backslash \; frac\; \{\backslash \; sin\; \backslash \; left\; (\backslash \; frac\; \backslash \; pi\; T\; (T\; -\; NT)\; \backslash \; right)\}\{\backslash \; frac\; \backslash \; pi\; T\; (T\; -\; NT)\}$

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