# Difference between revisions of "Extension of the sampling theorem"

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 9 295 - 366 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem

Explain why sampling at 4 ms is sufficient to reproduce exactly a signal whose spectrum is within the range ${\displaystyle 0}$ to ${\displaystyle 125\ {\hbox{Hz}}=f_{N}}$, but not if the bandwidth is shifted upward?

### Background

To convert an analog (continuous) signal to the digital form, values of the signal are measured at a fixed interval ${\displaystyle \Delta }$, called the sampling interval. The result is a series of numbers representing the amplitudes and polarities of the signal at the times ${\displaystyle t=n\Delta }$, where ${\displaystyle n}$ is an integer (often we omit ${\displaystyle \Delta }$ and specify the time by giving ${\displaystyle n}$ only). We write ${\displaystyle g_{t}}$ for the digital function corresponding to ${\displaystyle g(t)}$.

Sampling is equivalent to multiplying the analog signal by a comb, a function consisting of an infinite series of unit impulses spaced at a fixed interval ${\displaystyle \Delta }$. The equation of comb(t) is

 {\displaystyle {\begin{aligned}\mathrm {comb} (t)=\sum \limits _{r=-\infty }^{+\infty }\delta (t+r\Delta ),\\{\mbox{so}}\qquad \qquad g_{t}=g(t)\sum \limits _{r=-\infty }^{+\infty }\delta (t+r\Delta ).\end{aligned}}} (9.4a)

The transform of comb(t) is another comb with impulses in the frequency domain at intervals ${\displaystyle 1/\Delta }$:

 {\displaystyle {\begin{aligned}\mathrm {comb} (t)\leftrightarrow k\,\mathrm {comb} (f),\end{aligned}}} (9.4b)

where ${\displaystyle k=2\pi /\Delta }$ [see Sheriff and Geldart, 1995, equation (15.155)].

Figure 9.4a.  Sampling and recovering a waveform. The left column is time domain, the right is frequency domain.

The Nyquist frequency, ${\displaystyle f_{N}}$, is half the frequency of sampling ${\displaystyle 1/\Delta }$, that is,

 {\displaystyle {\begin{aligned}f_{N}=1/2\Delta .\end{aligned}}} (9.4c)

While a signal that does not contain frequencies higher than ${\displaystyle f_{N}}$ will be recorded accurately, a frequency higher than ${\displaystyle f_{N}}$ by the amount ${\displaystyle \Delta f}$ (i.e., a frequency that is not sampled at least twice per cycle) will produce an alias frequency equal to ${\displaystyle (f_{N}-\Delta f)}$ (Sheriff and Geldart, 1995, section 9.2.2c).

The convolution theorem, equation (9.3f), has a converse [see Sheriff and Geldart, 1995, equation (15.146)]:

 {\displaystyle {\begin{aligned}g(t)h(t)\leftrightarrow {\frac {1}{2\pi }}G(\omega )*H(\omega ).\end{aligned}}} (9.4d)

### Solution

We examine first the case where the signal frequency spectrum extends from ${\displaystyle 0}$ to 125 Hz, next consider modifications when the spectrum is shifted upward 125 Hz ${\displaystyle (=f_{N}}$ for 4-ms sampling), and finally, the case where the signal is shifted upward by an amount ${\displaystyle f\neq 125n}$.

Part (i) of Figure 9.4a shows the signal and its frequency spectrum, the latter being confined to the range ${\displaystyle -125\ {\hbox{to}}+125}$ Hz [the negative frequencies arise when we use Euler’s equations (Sheriff and Geldart, 1995, problem 15.12a) to transform equation (9.1a) into (9.1c). Part (ii) shows comb(t), with elements 4 ms apart, and its transform, a comb with elements spaced ${\displaystyle 1/\Delta =250}$ Hz apart. Digitizing the signal is equivalent to multiplying it by comb(t) [part (iii)], which is equivalent to convolving the spectrum ${\displaystyle G(f)}$ with comb(f) [see equation (9.4b)]; this results in a repetition of the spectrum for each impulse in comb(f) [see equation (9.3j)]. To eliminate the repeated spectra, we multiply by a boxcar [part (iv)], thus getting back the original spectrum [part (v)]. Multiplication in the frequency domain is equivalent to convolution in the time domain, so we convolve with a sinc function in part (V) and recover the original time-domain signal.

Figure 9.4b.  Repeat of Figure 9.4a for a 125–250 Hz signal.
Figure 9.4c.  Repeat of Figure 9.4a for spectrum overlapping the Nyquist frequency.

If the signal is shifted upward by ${\displaystyle 125n}$, where ${\displaystyle n}$ is an integer, then the sampling interval is larger than 1/2 cycle for all frequencies and the resulting spectrum is that of the alias function rather than of the signal, as illustrated in Figure 9.4b. Because we no longer have the signal spectrum after sampling, we cannot recover the signal.

If the signal is shifted upward so that it overlaps the Nyquist frequency (Figure 9.4c), the sampling causes the frequencies above ${\displaystyle f_{\rm {N}}}$ to overlap and distort the signal spectrum. Now we cannot separate the signal spectrum from the distorted spectrum and hence cannot recover the signal.