# Doppler effect detection

Started by October 6, 2019
Hi all,

is it possible to know, without any prior,
if an incoming signal has a Doppler effect?

That's it, to know if the source is moving
to (or away from) the detector without any
knowledge of the frequency.

I can imagine that if the source is not
moving directly towards (or away) the
detector, but with an angle, the Doppler
effect changes hence it is detectable.

bye,

--

piergiorgio

Piergiorgio Sartor  <piergiorgio.sartor.this.should.not.be.used@nexgo.REMOVETHIS.de> wrote:

>Hi all,
>
>is it possible to know, without any prior,
>if an incoming signal has a Doppler effect?
>
>That's it, to know if the source is moving
>to (or away from) the detector without any
>knowledge of the frequency.
>
>I can imagine that if the source is not
>moving directly towards (or away) the
>detector, but with an angle, the Doppler
>effect changes hence it is detectable.
>

Not as you have phrased it.  Even if the relative
motion is at an angle, the component of motion
along the line connecting the source and the receiver
determines the Doppler shift, and is still constant.

One needs to know something about the sgnal.

Steve

On 06/10/2019 22.25, Steve Pope wrote:
> Piergiorgio Sartor  <piergiorgio.sartor.this.should.not.be.used@nexgo.REMOVETHIS.de> wrote:
>
>> Hi all,
>>
>> is it possible to know, without any prior,
>> if an incoming signal has a Doppler effect?
>>
>> That's it, to know if the source is moving
>> to (or away from) the detector without any
>> knowledge of the frequency.
>>
>> I can imagine that if the source is not
>> moving directly towards (or away) the
>> detector, but with an angle, the Doppler
>> effect changes hence it is detectable.
>>
>> How about a direct motion?
>
> Not as you have phrased it.  Even if the relative
> motion is at an angle, the component of motion
> along the line connecting the source and the receiver
> determines the Doppler shift, and is still constant.

Why it is constant?

If the motion does not go towards to (or away
from) the receiver the relative speed is not
constant, hence the Doppler should also not be.

Let me try to portrait it:

------ source motion ------->
a\          b|         c/
\          |         /
\         |        /

When on the left (a), there is relative
fast motion, like on the right (c).
When exactly above (b), in that moment, there
is no motion, or minimal.
At least in the horizontal direction (more or
less, clearly)

There should be a trigonometric (sin(a) or
cos(a)) relationship, I guess.

Am I missing something?

> One needs to know something about the sgnal.

Let's assume the signal is (more or less) white
noise, could this help?

bye,

--

piergiorgio

Piergiorgio Sartor  <piergiorgio.sartor.this.should.not.be.used@nexgo.REMOVETHIS.de> wrote:

>On 06/10/2019 22.25, Steve Pope wrote:

>> Not as you have phrased it.  Even if the relative
>> motion is at an angle, the component of motion
>> along the line connecting the source and the receiver
>> determines the Doppler shift, and is still constant.

>Why it is constant?

>If the motion does not go towards to (or away
>from) the receiver the relative speed is not
>constant, hence the Doppler should also not be.
>
>Let me try to portrait it:
>
>------ source motion ------->
>a\          b|         c/
>   \          |         /
>    \         |        /
>
>When on the left (a), there is relative
>fast motion, like on the right (c).
>When exactly above (b), in that moment, there
>is no motion, or minimal.
>At least in the horizontal direction (more or
>less, clearly)
>
>There should be a trigonometric (sin(a) or
>cos(a)) relationship, I guess.

time though it will look constant.

If you know something such as, the source is known to be moving at
a constant rate for a sufficiently long interval of time,
such that you may infer something about the likely geometry,
then yes.

Steve

On 10/6/2019 16:39, Piergiorgio Sartor wrote:
> On 06/10/2019 22.25, Steve Pope wrote:
>> Piergiorgio Sartor
>> <piergiorgio.sartor.this.should.not.be.used@nexgo.REMOVETHIS.de> wrote:
>>
>>> Hi all,
>>>
>>> is it possible to know, without any prior,
>>> if an incoming signal has a Doppler effect?
>>>
>>> That's it, to know if the source is moving
>>> to (or away from) the detector without any
>>> knowledge of the frequency.
>>>
>>> I can imagine that if the source is not
>>> moving directly towards (or away) the
>>> detector, but with an angle, the Doppler
>>> effect changes hence it is detectable.
>>>
>>> How about a direct motion?
>>
>> Not as you have phrased it.&#2013266080; Even if the relative
>> motion is at an angle, the component of motion
>> along the line connecting the source and the receiver
>> determines the Doppler shift, and is still constant.
>
> Why it is constant?
>
> If the motion does not go towards to (or away
> from) the receiver the relative speed is not
> constant, hence the Doppler should also not be.
>
> Let me try to portrait it:
>
> ------ source motion ------->
> a\&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080; b|&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080; c/
>  &#2013266080; \&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080; |&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080; /
>  &#2013266080;&#2013266080; \&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080; |&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080; /
>
> When on the left (a), there is relative
> fast motion, like on the right (c).
> When exactly above (b), in that moment, there
> is no motion, or minimal.
> At least in the horizontal direction (more or
> less, clearly)
>
> There should be a trigonometric (sin(a) or
> cos(a)) relationship, I guess.
>
> Am I missing something?
>
>> One needs to know something about the sgnal.
>
> Let's assume the signal is (more or less) white
> noise, could this help?
What does frequency shifted white noise look like?  If I had to guess,
I'd say white noise, so that is probably the hardest case.  Start with a
sinusoid.  The frequency change will be fairly obvious.
Best wishes,
--Phil
>
> bye,
>


On 07/10/2019 15.59, Phil Martel wrote:
> On 10/6/2019 16:39, Piergiorgio Sartor wrote:
>> On 06/10/2019 22.25, Steve Pope wrote:
>>> Piergiorgio Sartor
>>> <piergiorgio.sartor.this.should.not.be.used@nexgo.REMOVETHIS.de> wrote:
>>>
>>>> Hi all,
>>>>
>>>> is it possible to know, without any prior,
>>>> if an incoming signal has a Doppler effect?
>>>>
>>>> That's it, to know if the source is moving
>>>> to (or away from) the detector without any
>>>> knowledge of the frequency.
>>>>
>>>> I can imagine that if the source is not
>>>> moving directly towards (or away) the
>>>> detector, but with an angle, the Doppler
>>>> effect changes hence it is detectable.
>>>>
>>>> How about a direct motion?
>>>
>>> Not as you have phrased it.&#2013266080; Even if the relative
>>> motion is at an angle, the component of motion
>>> along the line connecting the source and the receiver
>>> determines the Doppler shift, and is still constant.
>>
>> Why it is constant?
>>
>> If the motion does not go towards to (or away
>> from) the receiver the relative speed is not
>> constant, hence the Doppler should also not be.
>>
>> Let me try to portrait it:
>>
>> ------ source motion ------->
>> a\&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080; b|&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080; c/
>> &#2013266080;&#2013266080; \&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080; |&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080; /
>> &#2013266080;&#2013266080;&#2013266080; \&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080; |&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080; /
>>
>> When on the left (a), there is relative
>> fast motion, like on the right (c).
>> When exactly above (b), in that moment, there
>> is no motion, or minimal.
>> At least in the horizontal direction (more or
>> less, clearly)
>>
>> There should be a trigonometric (sin(a) or
>> cos(a)) relationship, I guess.
>>
>> Am I missing something?
>>
>>> One needs to know something about the sgnal.
>>
>> Let's assume the signal is (more or less) white
>> noise, could this help?
> What does frequency shifted white noise look like?&#2013266080; If I had to guess,
> I'd say white noise, so that is probably the hardest case.&#2013266080; Start with a
> sinusoid.&#2013266080; The frequency change will be fairly obvious.

I do not think the white noise will lead
to white noise.
If the shift is upward, there will be
no low frequencies, i.e. not white anymore.

On the other hand, the white noise is
"quasi white", since it is anyway low
pass, hence a shift downward could be
still detectable. It depends.

In any case, I do not have control over
the source, I can only speculate on it.
One "speculation" is that it could be
(quasi) white noise or similar.

bye,

--

piergiorgio

On 07/10/2019 00.03, Steve Pope wrote:
> Piergiorgio Sartor  <piergiorgio.sartor.this.should.not.be.used@nexgo.REMOVETHIS.de> wrote:
>
>> On 06/10/2019 22.25, Steve Pope wrote:
>
>>> Not as you have phrased it.  Even if the relative
>>> motion is at an angle, the component of motion
>>> along the line connecting the source and the receiver
>>> determines the Doppler shift, and is still constant.
>
>> Why it is constant?
>
>> If the motion does not go towards to (or away
>> from) the receiver the relative speed is not
>> constant, hence the Doppler should also not be.
>>
>> Let me try to portrait it:
>>
>> ------ source motion ------->
>> a\          b|         c/
>>    \          |         /
>>     \         |        /
>>
>> When on the left (a), there is relative
>> fast motion, like on the right (c).
>> When exactly above (b), in that moment, there
>> is no motion, or minimal.
>> At least in the horizontal direction (more or
>> less, clearly)
>>
>> There should be a trigonometric (sin(a) or
>> cos(a)) relationship, I guess.
>
> time though it will look constant.

Yes, of course. Here we are in the
sound wave domain and the range of
~+/-100m or similar.

> If you know something such as, the source is known to be moving at
> a constant rate for a sufficiently long interval of time,
> such that you may infer something about the likely geometry,
> then yes.

This could be an assumption, even if
maybe not really with 100% guarantee...

Of course, an other assumption is that
the source, independently from the
direction, does not have constant speed.

So, even in this case it would be interesting
to know if this helps.

bye,

--

piergiorgio

On 10/7/2019 12:08, Piergiorgio Sartor wrote:
> On 07/10/2019 15.59, Phil Martel wrote:
>> On 10/6/2019 16:39, Piergiorgio Sartor wrote:
>>> On 06/10/2019 22.25, Steve Pope wrote:
>>>> Piergiorgio Sartor
>>>> <piergiorgio.sartor.this.should.not.be.used@nexgo.REMOVETHIS.de> wrote:
>>>>
>>>>> Hi all,
>>>>>
>>>>> is it possible to know, without any prior,
>>>>> if an incoming signal has a Doppler effect?
>>>>>
>>>>> That's it, to know if the source is moving
>>>>> to (or away from) the detector without any
>>>>> knowledge of the frequency.
>>>>>
>>>>> I can imagine that if the source is not
>>>>> moving directly towards (or away) the
>>>>> detector, but with an angle, the Doppler
>>>>> effect changes hence it is detectable.
>>>>>
>>>>> How about a direct motion?
>>>>
>>>> Not as you have phrased it.&#2013266080; Even if the relative
>>>> motion is at an angle, the component of motion
>>>> along the line connecting the source and the receiver
>>>> determines the Doppler shift, and is still constant.
>>>
>>> Why it is constant?
>>>
>>> If the motion does not go towards to (or away
>>> from) the receiver the relative speed is not
>>> constant, hence the Doppler should also not be.
>>>
>>> Let me try to portrait it:
>>>
>>> ------ source motion ------->
>>> a\&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080; b|&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080; c/
>>> &#2013266080;&#2013266080; \&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080; |&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080; /
>>> &#2013266080;&#2013266080;&#2013266080; \&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080; |&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080;&#2013266080; /
>>>
>>> When on the left (a), there is relative
>>> fast motion, like on the right (c).
>>> When exactly above (b), in that moment, there
>>> is no motion, or minimal.
>>> At least in the horizontal direction (more or
>>> less, clearly)
>>>
>>> There should be a trigonometric (sin(a) or
>>> cos(a)) relationship, I guess.
>>>
>>> Am I missing something?
>>>
>>>> One needs to know something about the sgnal.
>>>
>>> Let's assume the signal is (more or less) white
>>> noise, could this help?
>> What does frequency shifted white noise look like?&#2013266080; If I had to guess,
>> I'd say white noise, so that is probably the hardest case.&#2013266080; Start with
>> a sinusoid.&#2013266080; The frequency change will be fairly obvious.
>
> I do not think the white noise will lead
> to white noise.
> If the shift is upward, there will be
> no low frequencies, i.e. not white anymore.
I disagree with the "no low frequencies".  The shift is dependent on
frequency.
https://en.wikipedia.org/wiki/Doppler_effect

f=\left(1+{\frac  {\Delta v}{c}}\right)f_{0}

Given that white noise has energy down to 0 Hz, the energy will be
shifted up, so there will be some energy at any arbitrary low frequency.
That said, I'm not certain if the energy distribution will still be
flat. Perhaps someone more familiar with this can comment, both on
whether measuring Doppler shift on white noise is possible and whether
it would be feasible in the real world.
>
> On the other hand, the white noise is
> "quasi white", since it is anyway low
> pass, hence a shift downward could be
> still detectable. It depends.
>
> In any case, I do not have control over
> the source, I can only speculate on it.
> One "speculation" is that it could be
> (quasi) white noise or similar.
>
> bye,
>


On 07/10/2019 22.16, Phil Martel wrote:
[...]
> Given that white noise has energy down to 0 Hz, the energy will be
> shifted up, so there will be some energy at any arbitrary low frequency.
> That said, I'm not certain if the energy distribution will still be
> flat. Perhaps someone more familiar with this can comment, both on

That's is my conjecture, it will not
be white, I just simplified.

> whether measuring Doppler shift on white noise is possible and whether
> it would be feasible in the real world.

Yep, that's one question!

bye,

--

piergiorgio

Piergiorgio Sartor  <piergiorgio.sartor.this.should.not.be.used@nexgo.REMOVETHIS.de> wrote:

>On 07/10/2019 22.16, Phil Martel wrote:

>> Given that white noise has energy down to 0 Hz, the energy will be
>> shifted up, so there will be some energy at any arbitrary low frequency.
>> That said, I'm not certain if the energy distribution will still be
>> flat. Perhaps someone more familiar with this can comment, both on

>That's is my conjecture, it will not
>be white, I just simplified.

>> whether measuring Doppler shift on white noise is possible and whether
>> it would be feasible in the real world.

>Yep, that's one question!

Depends on the white noise.  Not all white noise is created equal.
Seems to me something like the "random telegraph process", which is
white but not Gaussian, could have its Doppler shift measured.

Steve