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aliasing condition for radon transform?

Started by Unknown April 10, 2007
I am talking about geophysical data here, where you have a record
consisting of a few hundred traces, so the data has two dimensions
time and distance (or t-x in short).
If an event has a dip or slowness less than one sample per trace,
it is not aliased.  If it has greater dip, it would alias, unless
the temporal frequency content is limited.

Now radon transform can be used to remove certain types of noise
from this data.  This transforms data from t-x to tau-p domain
and back again. I looked at the documentation for a radon
transform program, and in the section about aliasing, it gave
two rather complicated equations about how many p traces you need
to avoid aliasing.  When I entered the parameters of this data,
I got number of p traces being twice the number of x traces in the
raw data.  It happened that the data had dips of up to 2 samples
per trace.  Coincidence I thought.  Then I recalculated for data
that has dips of 1 sample per trace, and end up with number of
required p traces equal to number of x traces.

So I thought, why can't the aliasing limit be expressed more
simply as:
p traces = (x traces) * (maximum dip in samples per trace) ?

efffemm@f-m.fm wrote:
> I am talking about geophysical data here, where you have a record > consisting of a few hundred traces, so the data has two dimensions > time and distance (or t-x in short). > If an event has a dip or slowness less than one sample per trace, > it is not aliased. If it has greater dip, it would alias, unless > the temporal frequency content is limited. > > Now radon transform can be used to remove certain types of noise > from this data. This transforms data from t-x to tau-p domain > and back again. I looked at the documentation for a radon > transform program, and in the section about aliasing, it gave > two rather complicated equations about how many p traces you need > to avoid aliasing. When I entered the parameters of this data, > I got number of p traces being twice the number of x traces in the > raw data. It happened that the data had dips of up to 2 samples > per trace. Coincidence I thought. Then I recalculated for data > that has dips of 1 sample per trace, and end up with number of > required p traces equal to number of x traces. > > So I thought, why can't the aliasing limit be expressed more > simply as: > p traces = (x traces) * (maximum dip in samples per trace) ?
Before I start thinking hard, define "dip" and "trace" unambiguously. What is a sample that it can be a measure of "dip"? Jerry -- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
efffemm@f-m.fm wrote:

> I am talking about geophysical data here, where you have a record > consisting of a few hundred traces, so the data has two dimensions > time and distance (or t-x in short). > If an event has a dip or slowness less than one sample per trace, > it is not aliased. If it has greater dip, it would alias, unless > the temporal frequency content is limited.
I usually hear about Radon transforms for tomography, with two spatial dimensions. I will guess that your t-x data can be processed similarly to data with two spatial dimensions. I don't know the aliasing rules for that, though. -- glen
On 11 Apr, 03:24, efff...@f-m.fm wrote:
> I am talking about geophysical data here, where you have a record > consisting of a few hundred traces, so the data has two dimensions > time and distance (or t-x in short). > If an event has a dip or slowness less than one sample per trace, > it is not aliased. If it has greater dip, it would alias, unless > the temporal frequency content is limited. > > Now radon transform can be used to remove certain types of noise > from this data. This transforms data from t-x to tau-p domain > and back again. I looked at the documentation for a radon > transform program,
What radon transform program was this? Was it intended for geophyscs or for general image processing?
> and in the section about aliasing, it gave > two rather complicated equations about how many p traces you need > to avoid aliasing. When I entered the parameters of this data, > I got number of p traces being twice the number of x traces in the > raw data. It happened that the data had dips of up to 2 samples > per trace. Coincidence I thought. Then I recalculated for data > that has dips of 1 sample per trace, and end up with number of > required p traces equal to number of x traces. > > So I thought, why can't the aliasing limit be expressed more > simply as: > p traces = (x traces) * (maximum dip in samples per trace) ?
The one explanation I can think of, is that geophysical data are constrained in a way that does not generally hold for images. If you use one-sided NMO images you know that the waves in a geophysics image will behave corresponding to outwards-only wave fronts. In general image processing one needst to account for what you as a geophycisist might call "inwards propagating waves." Rune
> > Before I start thinking hard, define "dip" and "trace" unambiguously. > What is a sample that it can be a measure of "dip"? >
Consider a signal propagating across with a certain velocity. It has a linear alignment across successive traces, so appears to dip. A horizontal event is considered to be zero dip, as is has the same time on each trace. A dipping event advances a fixed time on each trace. Example: time sampling interval is 4 ms and spatial sampling interval is 12 m. A signal propagating at 1500 m/s advances 8 ms each trace, so you can say the dip is 2 samples per trace.
efffemm@f-m.fm wrote:
>> Before I start thinking hard, define "dip" and "trace" unambiguously. >> What is a sample that it can be a measure of "dip"? >>
I still don't get it.
> Consider a signal propagating across with a certain velocity. > It has a linear alignment across successive traces, so appears > to dip.
Propagating across what? What dips? Why? What does the dip look like ("appears to dip").
> A horizontal event is considered to be zero dip, as is has > the same time on each trace. A dipping event advances a fixed > time on each trace. Example: time sampling interval is 4 ms > and spatial sampling interval is 12 m. A signal propagating > at 1500 m/s advances 8 ms each trace, so you can say the > dip is 2 samples per trace.
I'm glad you aren't teaching a course I have to pass. Will you keep trying to explain what it's all about, or do you give up? Jerry -- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯