[ ultrasound rangfinder, differential phase shift method ] - Page 1
Forum section: electronics / projects

author: texane
author_rank: Regular Contributor
timestamp: August 13, 2010, 04:52:37 pm
content: Hi,I am investigating the use of differential phase shift method to measure the distance between an ultrasound transducerand 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 adistance on [0; 2pi[, but I don t understand how, with 2 frequencies, we extend the concept above [0; 2pi[...Regards, thanks for helping,Fabien.

author: Unknown
author_rank: Unknown
timestamp: August 13, 2010, 07:28:34 pm
content: 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.

author: DJPhil
author_rank: Frequent Contributor
timestamp: August 13, 2010, 11:39:05 pm
content: Quote from: alm on August 13, 2010, 07:28:34 pmThat'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 founda patent that might help. I kept getting swamped with hits forDPSK, which uses phase shift to encode binary data.Another possible avenue of research is weather radar.In the 80s they started playingwith 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.

author: texane
author_rank: Regular Contributor
timestamp: August 14, 2010, 09:07:51 am
content: Hi,Thanks for helping. I am to read the paper you mentioned. It isquite funny, I was looking for this paper but did not find anywhereto download itI will reply as soon as I am done, but I think I can be more precisenow 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 waveperiods have elapsed. the an ambiguity arises, and we cannot relyon 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 limitationas above. Here is why:If f1 and f2 2 freqs, p1 and p2 the corresponding phase shift measuredon 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) < 2pino?For instance, let f1 = 40khz and f2 = 38khz. f1 - f2 = 2khz.If the phase shift difference has to be in [0, 2pi[ (right?), themeasurable 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 thefreq (in the example -> 2khz from 40khz and 38khz) thus increasing"a bit" the measurable distance.... but our phase shift is always in therange [0,2pi[, and we cant measure a distance above this range.Where am I wrong?Thanks very much for helping,Fabien.

author: Unknown
author_rank: Unknown
timestamp: August 14, 2010, 10:37:24 am
content: Quote from: texane on August 14, 2010, 09:07:51 amMy 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.

author: texane
author_rank: Regular Contributor
timestamp: August 14, 2010, 10:56:56 am
content: Thanks for replying. I will read the paper in the hours to come, justa few thing to do before, but knowing I am not completely wrong willhelp...Regards,Fabien.