Radius of the Fresnel zone is given by $$Rf=(v/2)(t_0/f_\mathrm{dom})^{1/2}$$ where $v$: velocity of layer

$t_0$: two way travel time

$f_\mathrm{dom}$ :dominant frequency in the spectrum

This shows that high frequencies give better resolution than lower frequencies and resolution deteriorates with depth and increasing velocities.

However I found some text in "Seismic Data Analysis- Yilmaz" which is contradictory. "Just having low or high frequencies does not improve temporal resolution. Both low and high frequencies are needed to increase temporal resolution. This is demonstrated further in Figure 1.1-30."

enter image description here

Could someone explain what the author is trying to say here or am I making a mistake in interpretation somewhere ?

up vote 8 down vote accepted

In seismic data analysis we have to distinguish between vertical (temporal) and horizontal resolution.

The (First) Fresnelzone is linked to the vertical resolution and defines the area in which the collected energy will still stack constructively.

The temporal resolution however, defines what the wave "sees". The events can still be resolved if their thickness is within $\frac{\lambda}{4}$. However, in some cases it can go down to $\frac{\lambda}{32}$. This is a matter of detectability of events.

Seismic Resolution

Don't forget that we're dealing with superpositions of different frequencies so basically low frequencies will give use the necesseray energy content of general trends, while high frequencies enable us to make a distinction between geologic features.

This file from the university of Oslo gives a nice overview.

(The picture is from a blog post I've written on knowing your wavelength it's CC-BY-SA Source1 Source2

  • 2
    Nice answer and especially with the Tardis! – user889 Oct 24 '14 at 12:16

The radius of the Fresnel zone formula gives resolution as a function of the dominant frequency in the spectrum. And indeed, the higher the dominant frequency, the higher the resolution.

In addition to what Way of the Geophysicist explained, I'd add specifically on Yilmaz that waht he is talking about is the importance of bandwidth. The plot you attached shows that with progressively increasing bandwidth, you are able to resolve better and better the reflectors in the reflectivity model (top panel) and eventually, with a bandwidth upwards of 10-50 Hz, even the reflectors separated by only 12 ms (third from the left in each panel).

To understand this I find it helpful to also look at Figure 1.1.29 [note 1 ], which shows a series of band-limited responses of the same reflectivity model as in Figure 1.1.30. Figure 1.1.29 demonstrates that if you increase the dominant frequency (the center of the narrow band range) only, you do not get improved resolution; you need also to increase the bandwidth as in Figure 1.1.30.

Figure 1.1.23 will also help understanding: it shows a series of zero-phase wavelets in which, with progressively increasing bandwidth, the wavelets are more compressed in time (hence higher temporal resolution). Notice that the amplitude (size) of the side lobes is decreasing as well with increasing bandwidth.

1 There's a google book preview and the three figures I refer to are included. Scroll to the book TOC from here and click on link to Ch1.

  • where are these figures? – user889 Dec 17 '14 at 20:13
  • Hi Sabre Tooth. They are copyrighted from a book by the Society of Exploration Geophysicists, so I am not going to include them. But the question is specific about one of the figures, and I assume user3365247 has he/she the book since it was included. For the rest of readers, there's a google book preview and the three figures I refer to are included. Scroll to TOC and click on link to Ch1: books.google.ca/… – MyCarta Dec 17 '14 at 21:35
  • 1
    The point of stack exchange is that the answers are useful for anyone reading this - perhaps provide a link to the book – user889 Dec 17 '14 at 21:37
  • +1 for the reminder, added link in the comments, and to the answer. – MyCarta Dec 17 '14 at 21:44

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