# Effect of signal/noise ratio on event picking

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Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 8 253 - 294 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 8.16a

Take a wavelet that has amplitudes at successive 4-ms intervals of ${\displaystyle 0,\ldots ,0,8,7,-8,-6,0,4,2,0,\ldots ,0}$ (with 10 zeros at each end) and add random noise (use Table 6.22a) in the range from ${\displaystyle +10}$ to ${\displaystyle -10}$, giving ${\displaystyle {\mbox{S/N}}\approx 1.0}$. Do this five times for different noise values and plot the results to illustrate how coherence helps in detecting the wavelet.

### Background

Variable-area recording is usually achieved by blacking in the peaks of ordinary wiggly-trace recording.

In sign-bit recording the only data recorded are the algebraic signs of the signal at each sampling instant. Band-pass signals have roughly equal probability of being either positive or negative. The superposition of signal on random noise biases this probability in favor of the signal and this bias increases with signal strength. As more traces are added, the stacking tends to reduce the random noise so that the sum is more likely to be that of the signal. By stacking large numbers of traces, a record is obtained that is comparable to a normal record.

Table 8.16a. Sequences of signal plus random noise; ${\displaystyle {\mbox{S/N}}\approx 1.0.}$
S ${\displaystyle {\rm {N}}_{1}}$ ${\displaystyle \sum _{1}}$ ${\displaystyle {\rm {N}}_{2}}$ ${\displaystyle \sum _{2}}$ ${\displaystyle {\rm {N}}_{3}}$ ${\displaystyle \sum _{3}}$ ${\displaystyle {\rm {N}}_{4}}$ ${\displaystyle \sum _{4}}$ ${\displaystyle {\rm {N}}_{5}}$ ${\displaystyle \sum _{5}}$ C D
0 2 2 0 0 3 3 5 5 3 3 13 5
0 0 0 1 1 1 1 3 3 8 8 13 3
0 8 8 –3 –3 5 5 –10 –10 –3 –3 –3 –1
0 9 9 8 8 –2 –2 4 4 –2 –2 17 1
0 –7 –7 0 0 –2 –2 0 0 –4 –4 –13 –3
0 –1 –1 –7 –7 –5 –5 –8 –8 –8 –8 –29 –5
0 3 3 9 9 –9 –9 4 4 5 5 12 3
0 0 0 –5 –5 –2 –2 1 1 –10 –10 –16 –1
0 10 10 10 10 –8 –8 –1 –1 6 6 17 1
0 –7 –7 –8 –8 2 2 –7 –7 0 0 –20 –3
8 –9 –1 –1 7 1 9 –5 3 –6 2 20 3
7 5 12 –2 5 6 13 0 7 –1 6 43 5
–8 2 –6 –7 –15 8 0 3 –5 –2 –10 –36 –3
–6 –1 –7 –2 –8 –5 –11 9 3 –2 –8 –31 –3
0 –7 –7 –1 –1 –6 –6 1 1 –9 –9 –22 –3
4 –10 –6 –3 1 3 7 7 11 10 14 27 3
2 –9 –7 4 6 –8 –6 3 5 2 4 2 1
0 –2 –2 8 8 6 6 7 7 –10 –10 9 1
0 –2 –2 –10 –10 5 5 0 0 7 7 0 –1
0 1 1 6 6 5 5 –2 –2 3 3 13 3
0 –1 –1 –6 –6 –3 –3 –1 –1 –7 –18 –5
0 –5 –5 0 0 1 1 –7 –7 0 0 –11 –1
0 –4 –4 –6 –6 –8 –8 –5 –5 –4 –4 –27 –5
0 –3 –3 0 0 6 6 6 6 9 9 18 3
0 3 3 5 5 2 2 –8 –8 8 8 10 3
0 9 9 9 9 8 8 1 1 3 3 30 5
0 4 4 –7 –7 –4 –4 4 4 2 2 –1 –1
 S = signal; ${\displaystyle {\rm {N}}_{i}=i^{\rm {th}}}$ set of random noises, ${\displaystyle i=1,2,3,4,5}$; ${\displaystyle \Sigma _{i}=\Sigma ({\rm {S}}+{\rm {N}}_{i})}$,

${\displaystyle {\rm {C}}=\Sigma _{i}}$ for part (c), D = sign-bit sum for part (d)

Figure 8.16a.  Vartiable-area display of signal plus random noise; ${\displaystyle {\mbox{S/N}}\approx 1.0.}$

### Solution

Although we have few data, we calculated S/N for each of the five sets (rather than use an average value). Using absolute values (since S/N does not depend upon the polarity) we get for the averages ${\displaystyle {\rm {S}}=5}$, ${\displaystyle {\rm {S/N}}_{1}=0.8}$, ${\displaystyle {\rm {S/N}}_{2}=1.7}$, ${\displaystyle {\rm {S/N}}_{3}=1.0}$, ${\displaystyle {\rm {S/N}}_{4}=1.2}$, ${\displaystyle {\rm {S/N}}_{5}=1.0}$. The average of these five, 1.1, is close to 1.

Table 8.16a lists five random noise sequences and ${\displaystyle {\hbox{S}}+{\hbox{N}}}$ for each noise sequence. The sums are then plotted in Figure 8.16a with positive values shaded in. The first peak and trough of the signal are coherent enough on the five traces that they probably would be picked, but the second signal peak is lost in the background noise.

Figure 8.16b.  Variable-area display of signal plus random noise; ${\displaystyle {\rm {S/N}}=0.5}$
Table 8.16b. Sequences of signal plus random noise; ${\displaystyle {\rm {S/N}}\approx 0.5.}$
S ${\displaystyle {\rm {N}}_{1}}$ ${\displaystyle \sum _{1}}$ ${\displaystyle {\rm {N}}_{2}}$ ${\displaystyle \sum _{2}}$ ${\displaystyle {\rm {N}}_{3}}$ ${\displaystyle \sum _{3}}$ ${\displaystyle {\rm {N}}_{4}}$ ${\displaystyle \sum _{4}}$ ${\displaystyle {\rm {N}}_{5}}$ ${\displaystyle \sum _{5}}$ C D
0 2 2 0 0 3 3 5 5 3 3 13 5
0 0 0 1 1 1 1 3 3 8 8 13 3
0 8 8 –3 –3 5 5 –10 –10 –3 –3 –3 –1
0 9 9 8 8 –2 –2 4 4 –2 –2 17 1
0 –7 –7 0 0 –2 –2 0 0 –4 –4 –13 –3
0 –1 –1 –7 –7 –5 –5 –8 –8 –8 –8 –29 –5
0 3 3 9 9 –9 –9 4 4 5 5 12 3
0 0 0 –5 –5 –2 –2 1 1 –10 –10 –16 –1
0 10 10 10 10 –8 –8 –1 –1 6 6 17 1
0 –7 –7 –8 –8 2 2 –7 –7 0 0 –20 –3
8 –9 –1 –1 7 1 9 –5 3 –6 2 20 3
7 5 12 –2 5 6 13 0 7 –1 6 43 5
–8 2 –6 –7 –15 8 0 3 –5 –2 –10 –36 –3
–6 –1 –7 –2 –8 –5 –11 9 3 –2 –8 –31 –3
0 –7 –7 –1 –1 –6 –6 1 1 –9 –9 –22 –3
4 –10 –6 –3 1 3 7 7 11 10 14 27 3
2 –9 –7 4 6 –8 –6 3 5 2 4 2 1
0 –2 –2 8 8 6 6 7 7 –10 –10 9 1
0 –2 –2 –10 –10 5 5 0 0 7 7 0 –1
0 1 1 6 6 5 5 –2 –2 3 3 13 3
0 –1 –1 –6 –6 –3 –3 –1 –1 –7 –18 –5
0 –5 –5 0 0 1 1 –7 –7 0 0 –11 –1
0 –4 –4 –6 –6 –8 –8 –5 –5 –4 –4 –27 –5
0 –3 –3 0 0 6 6 6 6 9 9 18 3
0 3 3 5 5 2 2 –8 –8 8 8 10 3
0 9 9 9 9 8 8 1 1 3 3 30 5
0 4 4 –7 –7 –4 –4 4 4 2 2 –1 –1
 S = signal; ${\displaystyle {\rm {N}}_{i}=i^{\rm {th}}}$ set of random noises; ${\displaystyle i=1,2,3,4,5}$; ${\displaystyle \Sigma _{i}=\Sigma ({\rm {S}}+{\rm {N}}_{i})}$; ${\displaystyle {\rm {C}}=\Sigma _{i}}$ for part (c);

D = sign-bit sum for part (d).

Figure 8.16c.  Display of sums of five samples of signal plus random noise. Top, signal S; center, ${\displaystyle S/N\approx 0.8}$; bottom, ${\displaystyle {\rm {S/N}}\approx 0.4}$.

## Problem 8.16b

Repeat for noise ranging from ${\displaystyle \pm 20}$, giving ${\displaystyle {\rm {S/N}}\approx 0.5.}$

### Solution

The data are shown in Table 8.16b and the results graphed in Figure 8.16b. The signal appears to be completely lost at this S/N. The value ${\displaystyle {\rm {S/N}}\approx 1}$ is roughly the limit of our ability to extract a signal visually from a noisy trace.

## Problem 8.16c

Sum the five waveforms in parts (a) and (b) to show how stacking enhances the signal.

### Solution

The columns headed C in Tables 8.16a and 8.16b are the sums of the five values of ${\displaystyle \Sigma _{i}}$; these are plotted in Figure 8.16c. The signal is clearly evident in the curve for part (a) where ${\displaystyle {\rm {S/N}}\approx 1}$, but not in the curve for part (b) where ${\displaystyle {\rm {S/N}}\approx 0.5}$. However, even with ${\displaystyle {\rm {S/N}}\approx 1}$ false signals can be seen (such as a negative wavelet at the right-hand end).

Figure 8.16d.  Display of sign-bit sums of five samples of signal plus random noise. Top, signal; center, ${\displaystyle {\hbox{S/N}}\approx 0.8}$; bottom, ${\displaystyle {\hbox{S/N}}\approx 0.4}$.

## Problem 8.16d

Replace the elements in the wavelets in parts (a) and (b) with ${\displaystyle +1}$ or ${\displaystyle -1}$ as the value is positive or negative (sign-bit expression) and repeat part (c).

### Solution

We have replaced each value of the sums of S + N in Tables 8.16a and 8.16b with +1 or –1 as the value was positive or negative (zero values are alternately made + and –) and summed the five values in the columns headed D. The results, plotted in Figure 8.16d, are similar to Figure 8.16c, because the same sequences are used. The signal is becoming evident although noise can lead to false picks. More than five sequences need to be summed to make the signal stand out clearly.