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From: Fred Marshall on 10 Jul 2008 10:45

"Fred Marshall" <fmarshallx(a)remove_the_x.acm.org> wrote in message
news:ON6dnbsxEejWg-vVnZ2dnUVZ_jydnZ2d(a)centurytel.net...
>
> "cpshah99" <cpshah99(a)rediffmail.com> wrote in message
> news:k6mdnbNzRfbaU-jVnZ2dnUVZ_h_inZ2d(a)giganews.com...
>>
>>>My first response in this thread (intended to be a hint) was "geometry".
>>
>>>Consider a source and receiver 5 each meters below the surface and 25
>>>meters apart. The reflected source will be 5 meters *above* the surface.
>>
>>>How far is the phantom source from the receiver? If the source
>>>approaches to within 20 meters, how far will the phantom source be? 15
>>>meters? 10?
>>>
>>>Jerry
>>>--
>>>Engineering is the art of making what you want from things you can get.
>>>�����������������������������������������������������������������������
>>>
>>
>> Hi Jerry
>>
>> Thanks for your response.
>>
>> It is more like Llodys Mirror effect ( I just know what this term means),
>> in that we can say that the surface reflection is coming above the
>> surface.
>>
>> So if we keep Tx and Rx 25 meters apart and at depth of 5 meters each,
>> the
>> phantom source will be at 26.92 meters and if TX moves 5 meters towards
>> RX
>> then this phantom source will be at 22.36 meters.
>>
>> It is really simple geometry but I am still struggling as how to apply
>> this doppler expansion or compression to the signal?
>>
>> Thanks again
>>
>> Regards,
>>
>> Chintan
>
> Ah! I was wondering if/when you might ask that question.
>
> A lot depends on your implementation. As you already know, Doppler shift
> is about time stretching or compression. So, if you *really* want to do
> it, you might resample (interpolate) the waveform and redefine the time
> scale.
> Much depends on the context of the system architecture where you're
> wanting to do something like this. Of course, doing this will cause you
> to run out of signal at some point or to re-use signal at some point -
> just like in a video scan converter where you have to either skip a frame
> or repeat a frame because of physics.
>
> For example, if you're trying to develop an artificial non-moving target
> that seems to have motion then that's one thing.
> If you're trying to generate a Doppler-shifted echo then it's a similar
> issue.
> A lot depends on what you have control over and what is not in your
> control.
>
> What are you needing to do?
>
> Fred

Check this out and Search for REVGEN:

http://www.dsprelated.com/showmessage/27554/1.php

Fred


From: cpshah99 on 10 Jul 2008 11:18

>> Ah! I was wondering if/when you might ask that question.
>>
>> A lot depends on your implementation. As you already know, Doppler
shift
>> is about time stretching or compression. So, if you *really* want to
do
>> it, you might resample (interpolate) the waveform and redefine the time

>> scale.
>> Much depends on the context of the system architecture where you're
>> wanting to do something like this. Of course, doing this will cause
you
>> to run out of signal at some point or to re-use signal at some point -

>> just like in a video scan converter where you have to either skip a
frame
>> or repeat a frame because of physics.
>>
>> For example, if you're trying to develop an artificial non-moving
target
>> that seems to have motion then that's one thing.
>> If you're trying to generate a Doppler-shifted echo then it's a similar

>> issue.
>> A lot depends on what you have control over and what is not in your
>> control.
>>
>> What are you needing to do?
>>
>> Fred
>
>Check this out and Search for REVGEN:
>
>http://www.dsprelated.com/showmessage/27554/1.php
>
>Fred
>
>
>

%%%%%%

Hi Fred

Thanks again for your time and this link.

Now, I kept the motion of my TX in horizontal direction, i.e. line
connecting TX and RX. It seems that there is negligible effect on the
reflected path. For the geometry that i am using, the angle at which second
path goes is 11.30 deg. so V_r=v*cos(11.30), so if my v=1 then V_r=1.01,
which nearly same. (I think you have mentioned this point earlier). Is this
correct?

If this is correct than it means that both the direct and reflected path
will have same doppler, and same expansion or compression.

And regarding implementation, I have already posted one solution using
linear interpolator

http://www.dsprelated.com/showmessage/97986/1.php

But I dont know if it is correct. I am still using the same method.

Thanks again.

Chintan


From: Fred Marshall on 11 Jul 2008 13:17

"cpshah99" <cpshah99(a)rediffmail.com> wrote in message
news:f8qdnVIlucXUtevVnZ2dnUVZ_vninZ2d(a)giganews.com...
>
>>> Ah! I was wondering if/when you might ask that question.
>>>
>>> A lot depends on your implementation. As you already know, Doppler
> shift
>>> is about time stretching or compression. So, if you *really* want to
> do
>>> it, you might resample (interpolate) the waveform and redefine the time
>
>>> scale.
>>> Much depends on the context of the system architecture where you're
>>> wanting to do something like this. Of course, doing this will cause
> you
>>> to run out of signal at some point or to re-use signal at some point -
>
>>> just like in a video scan converter where you have to either skip a
> frame
>>> or repeat a frame because of physics.
>>>
>>> For example, if you're trying to develop an artificial non-moving
> target
>>> that seems to have motion then that's one thing.
>>> If you're trying to generate a Doppler-shifted echo then it's a similar
>
>>> issue.
>>> A lot depends on what you have control over and what is not in your
>>> control.
>>>
>>> What are you needing to do?
>>>
>>> Fred
>>
>>Check this out and Search for REVGEN:
>>
>>http://www.dsprelated.com/showmessage/27554/1.php
>>
>>Fred
>>
>>
>>
>
> %%%%%%
>
> Hi Fred
>
> Thanks again for your time and this link.
>
> Now, I kept the motion of my TX in horizontal direction, i.e. line
> connecting TX and RX. It seems that there is negligible effect on the
> reflected path. For the geometry that i am using, the angle at which
> second
> path goes is 11.30 deg. so V_r=v*cos(11.30), so if my v=1 then V_r=1.01,
> which nearly same. (I think you have mentioned this point earlier). Is
> this
> correct?
>
> If this is correct than it means that both the direct and reflected path
> will have same doppler, and same expansion or compression.
>

"Negligible" and "the same" aren't the same thing .... either things are the
same or they aren't strictly speaking. If they aren't then the differences
may be neglibible or not - and that is entirely up to you to decide in terms
of your application.

But, yes, unless you have very good frequency resolution then you may decide
that a 2% difference in the Doppler shift (not the absolute frequency) is
negligible beause it's undetectable or not important for you. Is the
indicated radial velocity 9.8 knots or 10 knots and who cares? (I get
cos(11.3) = .98)...

One needs to be careful about which percentages one is dealing with.
A 1% change in frequency due to motion amounts to 50fps in water or around
10fps in air.
A 2% reduction in Doppler shift depends on the maximum shift possible which
is dependent on the actual velocity converted to the radial velocity as
above.
So, if the object absolute velocity is 50fps then a 2% change would be
apparently 1fps and the Doppler shift would go from 1% of the transmit
frequency by 2% or 0.02% of the transmit frequency. Thus, detecting small
changes in apparent radial velocity can be a challenge.

Fred



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