FPGA Central

Hello People

In underwater communications, assume there is one direct path and on
surface reflected path. And the receiver is moving at 0.1 m/s (eithe
towards or away from transmitter).

My question is: do both the paths have same doppler effect?

Your guidance will be greatly appreciated.

Regards,

Chintan

"cpshah99" <[email protected]> wrote in message
news:[email protected]..
> Hello People
>
> In underwater communications, assume there is one direct path and one
> surface reflected path. And the receiver is moving at 0.1 m/s (either
> towards or away from transmitter).
>
> My question is: do both the paths have same doppler effect?
>
> Your guidance will be greatly appreciated.
>
> Regards,
>
> Chintan

The surface acts like a mirror so the reflection looks like it's coming from
a source that's above the surface. Motion would be the same but the
geometry is not - whether it's negligible or not is something you can
decide.

If the source and the receiver are at the same depth then the radial
velocity (which is what's important for Doppler) is the same as the
horizontal velocity - which I assume is what you mean as "moving toward or
moving away".

The reflected source would have the same horizontal velocity but not the
same radial velocity. Can you see why?

Fred

Fred Marshall wrote:

...

> The reflected source would have the same horizontal velocity but not the
> same radial velocity. Can you see why?
Sure. Geometry. :-)

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

>
>"cpshah99" <[email protected]> wrote in message
>news:[email protected]..
>> Hello People
>>
>> In underwater communications, assume there is one direct path and one
>> surface reflected path. And the receiver is moving at 0.1 m/s (either
>> towards or away from transmitter).
>>
>> My question is: do both the paths have same doppler effect?
>>
>> Your guidance will be greatly appreciated.
>>
>> Regards,
>>
>> Chintan
>
>The surface acts like a mirror so the reflection looks like it's comin
from
>a source that's above the surface. Motion would be the same but the
>geometry is not - whether it's negligible or not is something you can
>decide.
>
>If the source and the receiver are at the same depth then the radial
>velocity (which is what's important for Doppler) is the same as the
>horizontal velocity - which I assume is what you mean as "moving towar
or
>moving away".
>
>The reflected source would have the same horizontal velocity but not th

>same radial velocity. Can you see why?
>
>Fred
>
>
>
%%%

HI Fred

Thanks very much for replying.

I am sorry but I am not getting your point. Just assume that there is n
surface reflection. Also the TX and RX are at same depth. TX is steady an
RX is moving. So the signal will be expnaded by the velocity. Is thi
correct?

And as you are saying, 'The reflected source would have the sam
horizontal velocity but not the same radial velocity.' I am not gettin
this point.

I have found all the angles and lenths of signal but I have no idea how t
use them.

Thanks again.

Chintan

cpshah99 wrote:
> Hello People
>
> In underwater communications, assume there is one direct path and one
> surface reflected path. And the receiver is moving at 0.1 m/s (either
> towards or away from transmitter).
>
> My question is: do both the paths have same doppler effect?
>
> Your guidance will be greatly appreciated.
>
> Regards,
>
> Chintan

No they don't, because the reflected path-length is changing at a
different rate from the direct path-length.

Regards,
John

"cpshah99" <[email protected]> wrote in message
news:[email protected]..
> >
>>"cpshah99" <[email protected]> wrote in message
>>news:[email protected]..
>>> Hello People
>>>
>>> In underwater communications, assume there is one direct path and one
>>> surface reflected path. And the receiver is moving at 0.1 m/s (either
>>> towards or away from transmitter).
>>>
>>> My question is: do both the paths have same doppler effect?
>>>
>>> Your guidance will be greatly appreciated.
>>>
>>> Regards,
>>>
>>> Chintan
>>
>>The surface acts like a mirror so the reflection looks like it's coming
> from
>>a source that's above the surface. Motion would be the same but the
>>geometry is not - whether it's negligible or not is something you can
>>decide.
>>
>>If the source and the receiver are at the same depth then the radial
>>velocity (which is what's important for Doppler) is the same as the
>>horizontal velocity - which I assume is what you mean as "moving toward
> or
>>moving away".
>>
>>The reflected source would have the same horizontal velocity but not the
>
>>same radial velocity. Can you see why?
>>
>>Fred
>>
>>
>>
> %%%
>
> HI Fred
>
> Thanks very much for replying.
>
> I am sorry but I am not getting your point. Just assume that there is no
> surface reflection. Also the TX and RX are at same depth. TX is steady and
> RX is moving. So the signal will be expnaded by the velocity. Is this
> correct?
>
> And as you are saying, 'The reflected source would have the same
> horizontal velocity but not the same radial velocity.' I am not getting
> this point.
>
> I have found all the angles and lenths of signal but I have no idea how to
> use them.
>
> Thanks again.
>
> Chintan

Presumably you have the expression for Doppler shift as a function of speed
of sound and radial velocity.

If we ignore the reflected path for a moment then:

- if the depths of source and receiver are the same then the radial
velocity is the same as the horizontal velocity toward or away from. This
resolves two ways:
.. If the horizontal velocity V is 100% radial then you know the radial
velocity.
.. If the horizontal velocity V is at a right angle (90 degrees) from
the line connecting the source and receiver, then the radial velocity is
zero.
.. If the horizontal velocity V is A degrees from the line connecting
the source and receiver, then the radial velocity is V*cos(a) and the
tangential velocity is V*sin(A), contributing zero Doppler as above.

So, it's simpler to deal with the 100% case isn't it? Let's stick with that
for now in the discussion below:

If the source and the receiver are at different depths and the movement is
purely a constant horizontal velocity V then we are motivated to calculate
the vertical angle B between the source and the receiver. The radial
velocity is V*cos(B) and is ever-changing as the distance between the source
and receiver changes. In the long-distance limit, the angle goes to zero
and the radial velocity is V. In the short-distance limit, the source and
receiver are at the same x,y coordinate and the radial velocity is zero per
V*cos(B).

The reflected path situation is the same as the above where the depths are
different. In this special case, the vertical "distance" between the source
and the receiver is 2 times the depth of the observed object.

Fred

On 8 Jul, 22:21, "cpshah99" <cpsha...@rediffmail.com> wrote:
> >"cpshah99" <cpsha...@rediffmail.com> wrote in message
> >news:[email protected]..
> >> Hello People
>
> >> In underwater communications, assume there is one direct path and one
> >> surface reflected path. And the receiver is moving at 0.1 m/s (either
> >> towards or away from transmitter).
>
> >> My question is: do both the paths have same doppler effect?

.....
> Just assume that there is no
> surface reflection.

Underwater acoustics? Famous last words...

> Also the TX and RX are at same depth. TX is steady and
> RX is moving. So the signal will be expnaded by the velocity. Is this
> correct?

If you mean 'non-zero horizontal velocity might cause a Doppler
shift'
then yes, this is correct.

> And as you are saying, 'The reflected source would have the same
> horizontal velocity but not the same radial velocity.' I am not getting
> this point.

Then re-introduce the surface reflection. You need it to understand
Fred's statement.

> I have found all the angles and lenths of signal but I have no idea how to
> use them.

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?

- Let the Transmitter move in some direction.

1) What is the horizontal velocity of Tx relative to Rx?
2) What is the vertical velocity of Tx relative to Rx?
3) What is the radial velocity of Tx relative to Rx?

- Go back and read Fred's post once more.

Rune

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 dept
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?

Hi

Fred, thanks you very much. Now again I am keeping both TX and RX at sam
depth, 20 meters from surface.

Starting with a simple case, both are steady: So no doppler.

Now, TX is steady and RX is moving in straight line with velocity 0.
m/s.
So as the angle is 0, there for radial velocity is equal to horizonta
velocity, i.e. v_r=v_h=0.1 m/s;

So my doppler shift fd=v/c*f;

Is this correct?

Then I will move forward.

Thanks again, Fred and Rune, for spending ur valueable time.

Regards,

Chintan

On 9 Jul, 16:09, "cpshah99" <cpsha...@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.

Rune

Rune Allnor wrote:
> On 9 Jul, 16:09, "cpshah99" <cpsha...@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?

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

On 10 Jul, 04:28, Jerry Avins <j...@ieee.org> wrote:
> Rune Allnor wrote:
> > On 9 Jul, 16:09, "cpshah99" <cpsha...@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

>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.
>ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï ¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿ ½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï ¿½ï¿½ï¿½ï¿½ï¿½
>

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 th
surface.

So if we keep Tx and Rx 25 meters apart and at depth of 5 meters each, th
phantom source will be at 26.92 meters and if TX moves 5 meters towards R
then this phantom source will be at 22.36 meters.

It is really simple geometry but I am still struggling as how to appl
this doppler expansion or compression to the signal?

Thanks again

Regards,

Chintan

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.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

"cpshah99" <[email protected]> wrote in message
news:[email protected]..
>
>>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.
>>���������������� ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿ ½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï ¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ �����
>>
>
> 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

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.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

"Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message
news:[email protected]..
>
> "cpshah99" <[email protected]> wrote in message
> news:[email protected]..
>>
>>>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.
>>>ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿ ½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï ¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿ ½ï¿½ï¿½ï¿½ï¿½ï¿½
>>>
>>
>> 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

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

>> scale.
>> Much depends on the context of the system architecture where you're
>> wanting to do something like this. Of course, doing this will caus
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
frame
>> or repeat a frame because of physics.
>>
>> For example, if you're trying to develop an artificial non-movin
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 simila

>> 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. lin
connecting TX and RX. It seems that there is negligible effect on th
reflected path. For the geometry that i am using, the angle at which secon
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 thi
correct?

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

And regarding implementation, I have already posted one solution usin
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

"cpshah99" <[email protected]> wrote in message
news:[email protected] ...
>
>>> 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