# Appendix D: Exercises

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Series Geophysical References Series Digital Imaging and Deconvolution: The ABCs of Seismic Exploration and Processing Enders A. Robinson and Sven Treitel 4 http://dx.doi.org/10.1190/1.9781560801610 9781560801481 SEG Online Store
For the want of a nail the shoe was lost,
For the want of a shoe the horse was lost,
For the want of a horse the rider was lost,
For the want of a rider the battle was lost,
For the want of a battle the kingdom was lost,
And all for the want of a horseshoe nail.

—Benjamin Franklin


1. Fill in the table:

Sampling interval ${\displaystyle \Delta t}$ Sampling rate ${\displaystyle f_{s}}$ Nyquist frequency ${\displaystyle f_{n}}$
0.001 s
0.002 s
0.003 s
0.004 s
0.005 s
0.006 s
0.007 s
0.008 s

2. For ${\displaystyle \Delta t{\ =\ 0.004s}}$, fill in the table:

Actual frequency f Aliased frequency ${\displaystyle f_{a}}$ resulting from sampling
0
100
200
300
400
500

3. Show that Euler’s equation ${\displaystyle e^{i\theta }{\rm {=\ cos\ }}\theta {\ +\ }i{\rm {\ sin\ }}\theta }$ for ${\displaystyle \theta {\ =\ }\pi }$ gives ${\displaystyle e^{i\pi }{\ =\ }-{\rm {1}}}$. This equation is considered to be one of the most beautiful equations in mathematics because it relates the four special numbers ${\displaystyle e,\;i,\;\pi ,\;-1}$ to each other.

4. Discuss why the alias phenomenon represents a frequency ambiguity resulting from the sampling process.

5. Suppose there are fewer than two samples per cycle of an input sinusoidal signal of a given frequency. Show that the same sample values would occur for a sinusoidal signal of a still lower frequency. Hence, that signal looks the same as the original signal.

6. Half of the sampling frequency is called the folding or Nyquist frequency ${\displaystyle f_{n}}$. Show that the frequency ${\displaystyle {f_{n}}+\;\delta }$ appears as the smaller apparent frequency ${\displaystyle {f_{n}}-\;\delta }$. These two frequencies are aliases of each other.

7. To avoid aliasing, show why frequencies above the Nyquist frequency must be removed by an anti-alias filter before sampling.

8. Discuss why aliasing is an inherent property of all sampling systems. For example, aliasing applies to (1) sampling done at discrete time intervals, as with digital seismic recording, and (2) sampling done at discrete spatial intervals by separate elements of geophone and source-point arrays.

9. Suppose that ${\displaystyle f\;=\;1.3\;{f_{n}}}$. Show that f is indistinguishable from the lower (alias) frequency ${\displaystyle {f_{a}}\;=\;0.7\;{f_{n}}}$.

10. Describe what happens when a signal is sampled with fewer than two points per cycle. [Hint: The effect is somewhat similar to the effect observed if a wheel rotating at a high rate of speed is illuminated stroboscopically at a different rate. The wheel appears to be rotating at a speed different from its true rate.]

11. Illustrate Exercise 10 for a few specific cases, using a 0.004-s sampling interval. [Hint: Consider a 250-Hz sine wave signal sampled 250 times per second. A constant value (anywhere between -1 and +1) will result. Thus, a sampled 250-Hz signal has the appearance of a zero-frequency signal. The analogous situation for the rotating wheel analogy is that the stroboscopic rate causes the wheel to appear to be standing still.]

12. In Exercise 11, has the sampling process caused a 250-Hz signal to appear to be a zero-frequency signal? [Hint: Now consider an original 150-Hz sine-wave signal. The signal resulting from sampling at a 4-ms interval has the appearance of a 100-Hz signal, because 150 – 250 = –100 Hz, which is the same as 100 Hz. In other words, a signal above 125 Hz takes on an alias as a result of sampling. The critical frequency (125 Hz in this case) is called the Nyquist frequency, or the half-sampling frequency.]

13. At a 4-ms sampling interval, what happens to sinusoidal signals above 250 Hz in frequency? This case corresponds to the situation in the stroboscope analogy in which the wheel appears to turn counter to its true direction of rotation. Frequencies between 250 and 375 Hz appear as aliased frequencies between –125 Hz and 0 Hz.

14. In terms of frequency expressed as cycles per sampling interval, show that the Nyquist frequency always turns out to be 0.5. The corresponding Nyquist angular frequency is ${\displaystyle {\rm {2}}\pi \left({\rm {0.5}}\right){\ =\ }\pi }$. For example, for a sampling interval of 0.004 s, the Nyquist frequency in Hz is given as (0.5 cycles/time-unit)/(0.004 s/time-unit) = (125 cycles/s).