[ ultrasound rangfinder, differential phase shift method ]

[texane]

Hi,

I am investigating the use of differential phase shift method to measure the distance between an ultrasound transducer
and an immobile object. I have some trouble to understand the formula that gives you the distance given:
. 2 phase p1 and p2, 2 freq f1 and f2
. the propagation speed of the wave C (~343m/s in my case)
the distance is then: d = (p2 - p1) / (4PI * (f2 - f1))

Could someone points me what leads to this formula? Documents, websites... I understand how a phase shift relates to a
distance on [0; 2pi[, but I don t understand how, with 2 frequencies, we extend the concept above [0; 2pi[...

Regards, thanks for helping,

Fabien.

[alm]

Not sure where the extra factor 2 comes from, This paper (page 90) appears to explain the principle, and might help you understanding the other formula.

[DJPhil]

Not sure where the extra factor 2 comes from, This paper (page 90) appears to explain the principle, and might help you understanding the other formula.

That's slick, they're using the change in the speed of sound through air to measure temperature. This paper looks like the best explanation of the formula I can find in twenty minutes. Good catch Alm!

I think I've seen this technique used in RF for distance measurement. I found a patent that might help. I kept getting swamped with hits for DPSK, which uses phase shift to encode binary data.

Another possible avenue of research is weather radar. In the 80s they started playing with differential phase measurements of radar signals with two different polarizations and found it useful for gathering more data about rainfall than where the edge of the fall was.

Hope that helps, I hope it's not all old news.

[texane]

Hi,

Thanks for helping. I am to read the paper you mentioned. It is
quite funny, I was looking for this paper but did not find anywhere
to download it

I will reply as soon as I am done, but I think I can be more precise
now concerning what I don t understand.

Here my reasoning, and what I don t understand:
. we can measure a distance by using a signal phase shift
-> ok
. but if the phase shift is >= 2pi, we don t know how many wave
periods have elapsed. the an ambiguity arises, and we cannot rely
on phase shift anymore
-> ok
. thus a second freq is used.  this leads to a second phase shift.
-> I think I understand, but my opinion is we have the same limitation
as above. Here is why:

If f1 and f2 2 freqs, p1 and p2 the corresponding phase shift measured
on the receiver side, the distance formula is given by:
d = (p2 - p1) * c / 2pi (f2 - f1)

But even here, it must be that:
0 < (p2 - p1) < 2pi
no?

For instance, let f1 = 40khz and f2 = 38khz. f1 - f2 = 2khz.
If the phase shift difference has to be in [0, 2pi[ (right?), the
measurable distance is:
340 * 1 / 2000 = 17 cms no?
(340 sound speed in the air in cms/s, 2000 the freq in hertz)

If I am right, the only advantage of the method is to reduce the
freq (in the example -> 2khz from 40khz and 38khz) thus increasing
"a bit" the measurable distance.... but our phase shift is always in the
range [0,2pi[, and we cant measure a distance above this range.
Where am I wrong?

Thanks very much for helping,

Fabien.

[alm]

. thus a second freq is used.  this leads to a second phase shift.
-> I think I understand, but my opinion is we have the same limitation
as above. Here is why:

If f1 and f2 2 freqs, p1 and p2 the corresponding phase shift measured
on the receiver side, the distance formula is given by:
d = (p2 - p1) * c / 2pi (f2 - f1)

But even here, it must be that:
0 < (p2 - p1) < 2pi
no?

For instance, let f1 = 40khz and f2 = 38khz. f1 - f2 = 2khz.
If the phase shift difference has to be in [0, 2pi[ (right?), the
measurable distance is:
340 * 1 / 2000 = 17 cms no?
(340 sound speed in the air in cms/s, 2000 the freq in hertz)

If I am right, the only advantage of the method is to reduce the
freq (in the example -> 2khz from 40khz and 38khz) thus increasing
"a bit" the measurable distance.... but our phase shift is always in the
range [0,2pi[, and we cant measure a distance above this range.

My knowledge of this is limited to what I read in the above paper (Tsai, 2005), but I believe you are correct (although there might be an extra factor two in there according to your original source). You are basically simulating a low-frequency signal (fig. 2 from Tsai, 2005 is the (p1-p2) as function of time/distance), Tr >> 1/f1 or 1/f2, without exceeding the bandwidth of the ultrasonic transducers. But you can trade off resolution for range by changing (f1-f2). With a 40kHz and 39kHz signal, you'd have twice the range, even more with a 40kHz and 39.9kHz signal.

[texane]

Thanks for replying. I will read the paper in the hours to come, just
a few thing to do before, but knowing I am not completely wrong will
help...

Regards,

Fabien.