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From: Rune Allnor on 10 Jul 2008 01:43
On 10 Jul, 04:28, Jerry Avins <j...(a)ieee.org> wrote:
> Rune Allnor wrote:
> > On 9 Jul, 16:09, "cpshah99" <cpsha...(a)rediffmail.com> wrote:
> >> Hi Rune
>
> >> As you said
>
> >>> Then use a different geometrical setup (2D only):
> >>> - Tx and Rx at different depths
> >>> - No movement; no velocities
> >>> 1) What is the horizontal distance between Tx and Rx?
> >>> 2) What is the vertical distance between Tx and Rx?
> >>> 3) What is the radial distance between Tx and Rx?
> >> So I selected depth and range of the ocean as 100 meters each.
>
> >> Now lets say TX is at depth of 80 meters from surface and RX is at depth
> >> of 20 meters from surface.
>
> >> So:
> >> 1. The horizontal distance between TX and RX is 100 meters
> >> 2. The vertical distance between TX and RX is 60 meters
> >> 3. The radial distance between TX and RX is 116.61 meters.
>
> >> Is this correct?
>
> > Correct. Now apply the same line of reasoning to the
> > path that is reflected at the surface.
>
> 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?

Are you asking me or commenting with further questions the OP?

In an attempt to answer while not giving the crux away: The
answer is straight-forward triangle geometry. The key is to
specify the correct triangle.

Rune
From: cpshah99 on 10 Jul 2008 04:54

>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
From: Jerry Avins on 10 Jul 2008 10:29
Rune Allnor wrote:

...

> Are you asking me or commenting with further questions the OP?
>
> In an attempt to answer while not giving the crux away: The
> answer is straight-forward triangle geometry. The key is to
> specify the correct triangle.

A second response to the OP. He seemed to be beyond help from hints. I
tried to go the Socratic question route.

Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
From: Fred Marshall on 10 Jul 2008 10:36

"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



From: Jerry Avins on 10 Jul 2008 10:36
cpshah99 wrote:
>> 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?

...

> Hi Jerry
>
> Thanks for your response.
>
> It is more like Llodys Mirror effect ( I just know what this term means),

*Not* Lloyd's mirror. Since the source is in the denser medium, there is
no destructive interference at glancing incidence.

> 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?

Evidently you can calculate the length of the hypotenuse for any length
of base. Now calculate the rate at which the hypotenuse shortens when
the base shortens at constant rate.

Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
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