I am planning to make a sonar application using smart-phone speaker and microphone, and it is given that speaker's power output is 3 watts and sound frequency is 20 kHz. I wanted to understand how will the quality of detected signal decrease with increased distance and at what distance limit the received signal would be too weak for the readings. I made some research and my assumption according to it is that reliable maximum range would be between 10-25 meters, yet I am unsure. But it is still unclear for me how the range is being calculated and how does it relate to the properties(size and shape mainly) of the objects that are going to be detected. And I wanted to know if knowledge of power output and frequency is sufficient to answer the above question? I would appreciate if you can bring some clarification to the topic or maybe if you can share some reference material for better understanding. Thank you very much.

  • \$\begingroup\$ In dry air at 20 °C (68 °F), the speed of sound is 343.2 metres per second (varies with temperature and humidity). With a target at 25m the pulse covers twice that - you do the maths for timing the return pulse to the microphone. It is unlikely that the speaker will be outputting full power at 20kHz which is the limit of human hearing (unless you're a child). The magnitude of the return signal will depend upon the size, material and shape of the target. \$\endgroup\$ Jul 1, 2016 at 18:46
  • \$\begingroup\$ Thanks for your comment. I actually know these details, what you wrote is the working principle of sonar. I understand it, but I don't understand why it is stated 25 meters, not 50. How do people calculate effective distances, and how can I calculate it for my specifications? \$\endgroup\$
    – UserRR
    Jul 1, 2016 at 19:19
  • \$\begingroup\$ As with radar, the 4th power law applies. \$\endgroup\$
    – user16324
    Jul 1, 2016 at 21:41

1 Answer 1


how the range is being calculated and how does it relate to the properties(size and shape mainly) of the objects that are going to be detected

I'm going to make an answer based on what I think you should concentrate on.

Output power increases range for sure. If 3 watts is being outputted to the speaker maybe, due to speaker defficiencies only 10% of 3 watts is actually converted to sound energy. If we stick with 300 mW output power (rather like you would make calculations in radio wave transmission) then that power can be roughly said to emit in all directions in a hemisphere.

You can then talk sensibly about watts per square metre and, at say 10m the area of a sphere is \$4\pi r^2\$ = 1257 sq metres. Half that area is infiltrated by the speaker output power so power density at 10m is 300 mW / 628 = 0.48 mW/sq m.

So, if a 1 sq metre object at 10m totally reflects all the incident power then it regenerates 0.48 mW. This is also subject to the same rule as before i.e. all that power is distributed as watts per sq metre on a hemi sphere and, at 10m distance the power per sq metre is 0.76 uW / sq metre.

The microphone receiving that incident power might only be 1 cm square so the power it actually receives is 10,000 times lower i.e. 76 pico watts. If the microphone is 10% efficient at converting that to electrical power then you should see an electrical signal of 7.6 pico watts.

Converting that power to voltage calls for a leap of faith in terms of impedance presented to the microphone and sensitivity of the JFET inside the capsule but maybe you can assume that power is amplified a 1000 times by the JFET and dumped across a 2k2 resistor (pretty standard circuit but could be as high as 10 kohm).

So 7.6 pico watts becomes 7.6 nano watts due to the JFET inside the capsule and this power produces a voltage across the 2k2 resistor. So the output voltage = \$\sqrt{P \times R}\$ = 4.1 mV RMS. A pretty good signal really.

Of course, this assumes 100% reflection of power and is 100% unlikely - at best/typical you might see 1% reflected power so that 4.1 mV RMS is probably going to be more like 0.13 mV RMS.

Bottom line is that the numbers are guesswork given that you don't know how much output power can be transmitted in a pulse and that the microphone size and internals are guesswork too. Efficiencies of both speaker and microphone may be way off the mark too.

Just to try and put things in perspective, a signal from a pretty average electret capsule of a few mV RMS is typical of 92 dB SPL.

  • \$\begingroup\$ Then it seems that the trick is to look for minimum requirements of the receiver signal, and calculate max. distance that could provide that minimum(reverse calculation). Thanks a lot for your help. \$\endgroup\$
    – UserRR
    Jul 1, 2016 at 20:39

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