<p><strong>Because those frequencies are not present in the seismic data.</strong></p>

<p>Here's an example with some typical numbers:</p>

<ul>
<li>Seismic sample interval: 4 ms</li>
<li>Therefore <a href="https://en.wikipedia.org/wiki/Nyquist_frequency">Nyquist</a> is 0.5 $\times$ 1/0.004 = 125 Hz</li>
<li>Usually there is a filter on the recorder at 0.8 $\times$ Nyquist = 100 Hz</li>
<li>So the seismic does not contain frequencies beyond 100 Hz</li>
</ul>

<p>The maximum frequency $f_\text{max}$ of the data determines its resolving power. There are <a href="http://www.agilegeoscience.com/blog/2011/7/8/tuning-geology.html">various ways</a> to think about and calculate this; here's one formula (Kallweit &amp; Wood, 1982):</p>

<p>$$\tau_\text{min} = \frac{1}{1.4\,f_\text{max}}$$</p>

<p>So in our example, the thinnest resolvable bed is about 1 / (1.4 $\times$ 100) = 7.1 ms, and it will be thicker than that if there is a lot of noise or the bandwidth is not optimal (the earth attenuates high frequencies more than low ones, so this is an absolute minimum). </p>

<p>If we assume a P-wave velocity of 2500 m/s, then 7.1 ms two-way time is about 2500 * (7.1/2) = 8.9 m (55 ft). Ordinary wireline logs have a resolution of 0.15 m or 6 in, so there's a substantial discrepancy. I call this <a href="http://www.agilegeoscience.com/blog/2011/1/5/the-integration-gap.html">'the integration gap'</a>.</p>

<p>Because the seismic contains no information about geology thinner than this 8.9 m (in this example), not only is it pointless trying to invert for it, but it would be dangerous: anything we thought we'd resolved would be erroneous. This is why it's necessary to filter the logs back to the resolution of the seismic.</p>

<p><strong>Reference</strong></p>

<p>Kallweit, R &amp; L Wood (1982). The limits of resolution of zero-phase wavelets. <em>Geophysics</em> <strong>47</strong> (7), p 1035–1046, <a href="http://dx.doi.org/10.1190/1.1441367">DOI:10.1190/1.1441367</a>.</p>

CC BY-SA 3.0
2015-04-16T12:28:20.443Z
2015-04-16T12:28:20.443Z
Matt Hall
<p>Different post-stack inversion methods such as model based, sparse spike, colored and recursive inversion methods use high cut frequency impedance logs (low frequency) and mid-frequency seismic trace in order to generate an inverted impedance model. Then why are high frequency components are not taken into account during inversion?</p>

CC BY-SA 3.0
2015-04-16T09:29:00.607Z
Do you mean high frequency components from the well log or high frequency components from the seismic are not taken into account?  If you meant why are well log higher frequencies not taken into account, that is because wireline logs contain higher frequencies which are not measured by (bandlimited) seismic. You can only invert seismic for the bandwidth that has been measured. Caveat there is when performing an absolute inversion; low-frequencies are incorporated into the inversion through a low-frequency model.
stevej
<geophysics><seismology><inversion>
Why do most seismic inversion methods ignore high frequencies?
Kunal Rathod
# Why do most seismic inversion methods ignore high frequencies?

(Asked by: username_0 on 2015-04-16)

Different post-stack inversion methods such as model based, sparse spike, colored and recursive inversion methods use high cut frequency impedance logs (low frequency) and mid-frequency seismic trace in order to generate an inverted impedance model. Then why are high frequency components are not taken into account during inversion?

**Tags:** geophysics, seismology, inversion

> **Comments:**
> **username_1**: Do you mean high frequency components from the well log or high frequency components from the seismic are not taken into account?  If you meant why are well log higher frequencies not taken into account, that is because wireline logs contain higher frequencies which are not measured by (bandlimited) seismic. You can only invert seismic for the bandwidth that has been measured. Caveat there is when performing an absolute inversion; low-frequencies are incorporated into the inversion through a low-frequency model.  

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## 1 answers

### Answer by username_2 on 2015-04-16. Score: 10 | Accepted answer

**Because those frequencies are not present in the seismic data.**

Here's an example with some typical numbers:

* Seismic sample interval: 4 ms
* Therefore [Nyquist](https://en.wikipedia.org/wiki/Nyquist_frequency) is 0.5 $\times$ 1/0.004 = 125 Hz
* Usually there is a filter on the recorder at 0.8 $\times$ Nyquist = 100 Hz
* So the seismic does not contain frequencies beyond 100 Hz

The maximum frequency $f\_\text{max}$ of the data determines its resolving power. There are [various ways](http://www.agilegeoscience.com/blog/2011/7/8/tuning-geology.html) to think about and calculate this; here's one formula (Kallweit & Wood, 1982):

$$\tau\_\text{min} = \frac{1}{1.4\,f\_\text{max}}$$

So in our example, the thinnest resolvable bed is about 1 / (1.4 $\times$ 100) = 7.1 ms, and it will be thicker than that if there is a lot of noise or the bandwidth is not optimal (the earth attenuates high frequencies more than low ones, so this is an absolute minimum).

If we assume a P-wave velocity of 2500 m/s, then 7.1 ms two-way time is about 2500 \* (7.1/2) = 8.9 m (55 ft). Ordinary wireline logs have a resolution of 0.15 m or 6 in, so there's a substantial discrepancy. I call this ['the integration gap'](http://www.agilegeoscience.com/blog/2011/1/5/the-integration-gap.html).

Because the seismic contains no information about geology thinner than this 8.9 m (in this example), not only is it pointless trying to invert for it, but it would be dangerous: anything we thought we'd resolved would be erroneous. This is why it's necessary to filter the logs back to the resolution of the seismic.

**Reference**

Kallweit, R & L Wood (1982). The limits of resolution of zero-phase wavelets. *Geophysics* **47** (7), p 1035–1046, [DOI:10.1190/1.1441367](http://dx.doi.org/10.1190/1.1441367).


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