```Jerry Avins wrote:
> Fred Marshall wrote:
>> "Jerry Avins" <jya@ieee.org> wrote in message
>> news:zJSdnRT7h4OhE_vanZ2dnUVZ_rSrnZ2d@rcn.net...
>>> Fred Marshall wrote:
>>>
>>>   ...
>>>
>>>> But the Doppler effect doesn't modify F(w) in that manner.
>>>> Take time compression as an example (positive Doppler / closing range).
>>>> f(t) goes to f[(1+k)t] where k=v/V and v is the closing velocity and
>>>> V is the speed in the medium.
>>> That applies to a moving source, does it not? A closing observer
>>> yields f[t/(1-k)]. It hardly matters when k is small enough.
>>>
>>> Jerry
>>> --
>>
>> Jerry,
>>
>> I'm confused by your comment.  Maybe I'm not thinking straight and I'm
>> surely not putting much time into it (thinking) right now....

Fred,

I won't repeat the derivation here, but here's a consistent convention
for the signs: take the positive direction of motion as the direction of
sound propagation in the space between source and observer. An observer
with a positive velocity hears a lower pitch. A source with a positive
velocity generates a higher pitch. There is an obvious difference
between the two cases. When the source is stationary with respect to the
medium, the wavelength in the medium is everywhere c/f. (c is the
velocity of sound and f is the source frequency.) When the source moves,
the wavelength depends on the position within the medium; it differs
from place to place. Call the source velocity v_s, the observer velocity
v_o and f' the frequency at the observer.

f'    c - v_o
Then  -- = ---------.
f     c - v_s

When the velocities of source and observer are equal, there is no
Doppler shift. When both are small, the approximation
1/(c - v) = (c + v)
holds quite well. That can lead one to overlook the different behaviors.

Jerry
--
Engineering is the art of making what you want from things you can get.
&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
```
```Fred Marshall wrote:
> "Jerry Avins" <jya@ieee.org> wrote in message
> news:zJSdnRT7h4OhE_vanZ2dnUVZ_rSrnZ2d@rcn.net...
>> Fred Marshall wrote:
>>
>>   ...
>>
>>> But the Doppler effect doesn't modify F(w) in that manner.
>>> Take time compression as an example (positive Doppler / closing range).
>>> f(t) goes to f[(1+k)t] where k=v/V and v is the closing velocity and V is
>>> the speed in the medium.
>> That applies to a moving source, does it not? A closing observer yields
>> f[t/(1-k)]. It hardly matters when k is small enough.
>>
>> Jerry
>> --
>
> Jerry,
>
> I'm confused by your comment.  Maybe I'm not thinking straight and I'm
> surely not putting much time into it (thinking) right now....
>
> Whether the source or the reflector move is immaterial.  It's the radial
> velocity (or rate of change of range or distance) that matters.  In the
> formulation above the "closing velocity" could be negative.  But, the
> formula should have been f[t/(1+k)] where k=v/V and v is the closing
> velocity and V is the speed in the medium.  The function of t has to shrink
> on the time scale.  Did I get it right this time?

velocity of propagation through the medium is constant relative to the
medium, The shift due to motion of the source is different from the
shift due to motion of the observer. In the case of light, Einstein told
us that the velocity of propagation is independent of motion; but that's
not true with sound. There's an attempt at explanation at
http://www.devmaster.net/articles/openal-tutorials/lesson7.php , but the
equations aren't difficult to work out.  The upshot is (arbitrarily
taking closing velocity as positive) is deltaF = F*(Vc + Vs)/(Vc - Vo)
where Vs is velocity of sound, Vs is closing velocity of observer, Vs is
closing velocity of source.* All veli=ocities are relative to the air. A

> It seems that your argument could lead to "the Doppler shift is
> unnoticeable" but I'm sure that's not where you were going.  Generally it
> does matter.  Now, it should be noted that some things matter more or less
> if you're talking about radar vs. sonar because of the huge difference
> between the speed of sound and the speed of light.  For example, in radar
> it's common to determine range rate by measuring time difference between
> returns.  In sonar there's generally not enough time to do that and the
> range rate is almost always determined by the Doppler shift based on a
> single return/echo.

Think about sonar in a fast-moving stream.

Jerry
_____________________________________
* When of motion takes its sign from a fixed direction (e.g., positive
to the right) numerator and denominator have the same sign.
--
Engineering is the art of making what you want from things you can get.
&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
```
```"Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message
news:AOWdnfmJbJnujfXanZ2dnUVZ_sytnZ2d@centurytel.net...
>
> "Jerry Avins" <jya@ieee.org> wrote in message
> news:zJSdnRT7h4OhE_vanZ2dnUVZ_rSrnZ2d@rcn.net...
>> Fred Marshall wrote:

> Whether the source or the reflector move is immaterial.  It's the radial
> velocity (or rate of change of range or distance) that matters.  In the
> formulation above the "closing velocity" could be negative.  But, the
> formula should have been f[t/(1+k)] where k=v/V and v is the closing
> velocity and V is the speed in the medium.  The function of t has to
> shrink on the time scale.  Did I get it right this time?
>

I should probably expand on this comment:

If the platform is moving then there will be "own Doppler" generated in the
returns and its spectrum can be complicated (spread) by beamwidths and look
angle.  In sonar this is important because volume reverberation is centered
at this shifted frequency (relative to onboard time/frequency references re
the transmitted waveform) and "zero Doppler" echoes are at this same
frequency.

All this means is that it's sometimes useful to separate out the platform
radial velocity from any radial velocity of the reflectors - and that's
usually pretty easy to do in sonar.  The volume reverberation can be a
reference for this purpose.

Fred

```
```"Jerry Avins" <jya@ieee.org> wrote in message
news:zJSdnRT7h4OhE_vanZ2dnUVZ_rSrnZ2d@rcn.net...
> Fred Marshall wrote:
>
>   ...
>
>> But the Doppler effect doesn't modify F(w) in that manner.
>> Take time compression as an example (positive Doppler / closing range).
>> f(t) goes to f[(1+k)t] where k=v/V and v is the closing velocity and V is
>> the speed in the medium.
>
> That applies to a moving source, does it not? A closing observer yields
> f[t/(1-k)]. It hardly matters when k is small enough.
>
> Jerry
> --

Jerry,

I'm confused by your comment.  Maybe I'm not thinking straight and I'm
surely not putting much time into it (thinking) right now....

Whether the source or the reflector move is immaterial.  It's the radial
velocity (or rate of change of range or distance) that matters.  In the
formulation above the "closing velocity" could be negative.  But, the
formula should have been f[t/(1+k)] where k=v/V and v is the closing
velocity and V is the speed in the medium.  The function of t has to shrink
on the time scale.  Did I get it right this time?

It seems that your argument could lead to "the Doppler shift is
unnoticeable" but I'm sure that's not where you were going.  Generally it
does matter.  Now, it should be noted that some things matter more or less
if you're talking about radar vs. sonar because of the huge difference
between the speed of sound and the speed of light.  For example, in radar
it's common to determine range rate by measuring time difference between
returns.  In sonar there's generally not enough time to do that and the
range rate is almost always determined by the Doppler shift based on a
single return/echo.

Fred

```
```Fred Marshall wrote:

...

> But the Doppler effect doesn't modify F(w) in that manner.
> Take time compression as an example (positive Doppler / closing range).
> f(t) goes to f[(1+k)t] where k=v/V and v is the closing velocity and V is
> the speed in the medium.

That applies to a moving source, does it not? A closing observer yields
f[t/(1-k)]. It hardly matters when k is small enough.

Jerry
--
Engineering is the art of making what you want from things you can get.
&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
```
```<mrabcx@gmail.com> wrote in message
> Hi,
>
> I have a simple question regarding radars and Doppler shifts.
> If the radars sends out a pulse p(t) then lets assume one
> moving targets reflects it, so that the receiving echo can
> be written as
> r(t)=a*p(t-d)*exp^(j*v),
> where a is the reflecting coefficient, d is some delay and
> exp^(j*v) coresponds to a Doppler shift due to moving target.
> I assume that * denotes standard multiplication.
>
> I think the model I have set up for r(t) should be quite reasable.
> Now what if I take a Fourier transform of it, can I then
> write it like this:
> R(w)=a*P(w)*exp(-jd) # F(exp^(j*v))
> where # denotes a convolution in Fourier domain and
> F(exp^(j*v)) is the Fourier transform of exp^(j*v) ?
>
> The reason I ask is that I have not seen this kind of notation
> being used in books but multiplication in time-domain corresponds
> to a convolution in Fourier so I hope writing it out this should
> be fine, or have I missed out something ?
> Or are there reasons why this notation should not be used or
> other simplified expressions are rather preferred ?
>
> Thanks for any help!

A true Doppler effect is really a compression or expansion of time and not a
modulation.  So, the relative bandwidth has to be somewhat low in order to
model it as a shift in frequency.

The Frequency Shifting Property says:

F[f(t)*e^jw0t] =  F(w-w0)

But the Doppler effect doesn't modify F(w) in that manner.
Take time compression as an example (positive Doppler / closing range).
f(t) goes to f[(1+k)t] where k=v/V and v is the closing velocity and V is
the speed in the medium.

What I think you want to apply is the Scaling Property:

f(at) = (1/|a|)*F(w/a)

Compression in the time domain is equivalent to expansion in the frequency
domain.
{Lathi; Signals, Systems and Communication)

Fred

```
```On Dec 14, 7:54 am, "mra...@gmail.com" <mra...@gmail.com> wrote:
> Hi,
>
> I have a simple question regarding radars and Doppler shifts.
> If the radars sends out a pulse p(t) then lets assume one
> moving targets reflects it, so that the receiving echo can
> be written as
> r(t)=a*p(t-d)*exp^(j*v),
> where a is the reflecting coefficient, d is some delay and
> exp^(j*v) coresponds to a Doppler shift due to moving target.
> I assume that * denotes standard multiplication.
>
> I think the model I have set up for r(t) should be quite reasable.
> Now what if I take a Fourier transform of it, can I then
> write it like this:
> R(w)=a*P(w)*exp(-jd) # F(exp^(j*v))
> where # denotes a convolution in Fourier domain and
> F(exp^(j*v)) is the Fourier transform of exp^(j*v) ?
>
> The reason I ask is that I have not seen this kind of notation
> being used in books but multiplication in time-domain corresponds
> to a convolution in Fourier so I hope writing it out this should
> be fine, or have I missed out something ?
> Or are there reasons why this notation should not be used or
> other simplified expressions are rather preferred ?
>
> Thanks for any help!

The multiplication by a complex exponential in the time domain
corresponds to a frequency shift in the frequency domain, so instead
of the convolution (which is correct) it is simpler to see (f - f0) or
(w-w0) where w0 is the amount of the frequency shift - this is a much
simpler notation than the convolution. If you simplify the convolution
you should get the same result.

Cheers,
David

```
```Hi,

I have a simple question regarding radars and Doppler shifts.
If the radars sends out a pulse p(t) then lets assume one
moving targets reflects it, so that the receiving echo can
be written as
r(t)=a*p(t-d)*exp^(j*v),
where a is the reflecting coefficient, d is some delay and
exp^(j*v) coresponds to a Doppler shift due to moving target.
I assume that * denotes standard multiplication.

I think the model I have set up for r(t) should be quite reasable.
Now what if I take a Fourier transform of it, can I then
write it like this:
R(w)=a*P(w)*exp(-jd) # F(exp^(j*v))
where # denotes a convolution in Fourier domain and
F(exp^(j*v)) is the Fourier transform of exp^(j*v) ?

The reason I ask is that I have not seen this kind of notation
being used in books but multiplication in time-domain corresponds
to a convolution in Fourier so I hope writing it out this should
be fine, or have I missed out something ?
Or are there reasons why this notation should not be used or
other simplified expressions are rather preferred ?

Thanks for any help!

```