1
\$\begingroup\$

I try to make a system that generates high frequency tones ( 12-17kHz ) only.

Use some cheap tweeters like : https://www.banggood.com/100W-Piezo-Horn-Speaker-Tweeter-30KHZ-Piezoelectric-Head-Driver-Loudspeaker-Treble-p-1418120.html?cur_warehouse=CN

Is the size and material of a speaker enclosure important to get the highest possible sound pressure out of a tweeter?

How far can high frequencies ( > 12 kHz ) reach outdoors?

Why low frequencies travel more far outdoors?

Thx!

\$\endgroup\$
3
  • \$\begingroup\$ I’m voting to close this question because this is not about electronic design \$\endgroup\$
    – Eugene Sh.
    Commented Aug 28, 2020 at 20:21
  • \$\begingroup\$ Steve Gibson's Quiet Canine is not all that effective if that's what you had in mind but the horn amplifies the gain while reducing beamwidth making it somewhat more directional. But a long Helholtz tube would be even more effective if that was your goal. His BOM cites Clone of Motorola/CTS KSN-1 grc.com/tqc/TQC_v2.2.2.pdf \$\endgroup\$ Commented Aug 28, 2020 at 23:53
  • \$\begingroup\$ the Military have a xx kilowatt HF sound generator on a truck to clear riots , fortunately not seen in use. \$\endgroup\$ Commented Aug 29, 2020 at 0:00

2 Answers 2

1
\$\begingroup\$

You aim to work at sound wavelengths about one inch. High efficiency cabinet would need a horn and matching structure "pressure chamber" at the throat side of the horn. The pressure chamber should be so small that no standing wave could fit into it. The dimensions (=far too big) of your speaker make this impossible, it is already its own cabinet.

I guess you are going to produce intolerable sound and you want to make the effective distance as long as possible. In that case you can try to use the speaker as a radiator in front of a big parabolic reflector. That's like the radiator horn in old style microwave radar antennas. Narrow beam projects the soundwave to much smaller area than the speaker alone would do. In the middle of the beam you get the same effect as the radiation power was increased.

The apparent power increasing factor (with properly designed and built reflector) that you can expect is (A/B)^2 where A is the diameter of your reflector and B is the diameter of your speaker. This assumes that the beam of your speaker is already so narrow that nearly all of the acoustic power from the speaker meets the reflector. The formula is taken from a book which presents coarse formulas for radar antennas.

About the attenuation: Air isn't frictionless, it has viscosity. Stoke's law gives a rule for attenuation vs distance in free air. https://en.wikipedia.org/wiki/Stokes%27s_law_of_sound_attenuation It states that energy loss due the viscosity grows as the frequency increases.

How long your sound will reach? Impossible to say even if I knew what's the acoustic output power, how it's directed and what level is considered as "vanished". I guess you operate near the ground so ground and all kind of obstacles cause scattering and absorbtion which can only be measured "on site". Wind and temperature variations make their own effects. If you are lucky they can increase the effective distance.

\$\endgroup\$
7
  • \$\begingroup\$ Thank you for your answer! The parabolic reflector is probably the best option because the beam can be very narrow indeed. Is the material of the parabola important? The distance between the tweeter and the center is that important? A is the size at the end of the parabolic reflector? \$\endgroup\$
    – user261847
    Commented Aug 30, 2020 at 12:53
  • \$\begingroup\$ Material: it must be hard and heavy. Soft, light and flexible absorb sound. You should learn in math what is the parabola and the revolution surface paraboloid which is used in parabolic reflectors. The placement of the speaker = the focal point, fine tune by measuring the distant sound level at the operating frequency. A=the diameter of the edge circle of the parabolic reflector assuming it's round. Red this en.wikipedia.org/wiki/Parabolic_reflector if you do not know the meaning of the focal point. \$\endgroup\$
    – user136077
    Commented Aug 30, 2020 at 14:23
  • \$\begingroup\$ Ok very interesting, I will try to find a parabolic enclosure. So the audio frequency used is not affecting the dimensions of the parabool neither the distance between speaker and parabool? Could I use fro example a parabolic wifi antenne and just replace the Wifi emitter by a speaker? \$\endgroup\$
    – user261847
    Commented Sep 1, 2020 at 11:32
  • \$\begingroup\$ Parabolic wifi antennas are not known where I live because our local regulations prohibit increasing beam strength with drectivity. A parabolic satellite antenna reflector is probably a good idea. The optimal placement isn't frequency independent. Some fine tuning is needed due the properties of the speaker. Its effective center surely depends on frequency. Prepare to make some measurements with a mic which catches the sound at some distance. You need a n echoless chamber but a lawn very likely is good enough at 12 kHz. \$\endgroup\$
    – user136077
    Commented Sep 1, 2020 at 11:55
  • 1
    \$\begingroup\$ Ok parabolic satellite antenna is a great idea and cheap. I will try that. Thanks a lot. \$\endgroup\$
    – user261847
    Commented Sep 1, 2020 at 12:01
0
\$\begingroup\$

Those tweeters are sealed at the back and have an integral horn a tuned cabinet is not needed

the horn appears to be designed to be mounted in a flat surface (like in the front of a cabinet) but you could mount them in flat panel with no back and get much the same result.

How far can high frequencies ( > 12 kHz ) reach outdoors?

how do you measure?

Why low frequencies travel more far outdoors?

diffusion. also air is not an ideal gas.

\$\endgroup\$
3
  • \$\begingroup\$ Thanks for your answer. What is the effect of mounting the tweeter in a panel ? Why would it sound louder? Is it because sound at the back of the tweeter will interfere and lower the sound at the front? \$\endgroup\$
    – user261847
    Commented Aug 30, 2020 at 12:55
  • \$\begingroup\$ it would reduce the amount of sound that goes to the back of the speaker, reflecting it forwards instead. \$\endgroup\$ Commented Aug 30, 2020 at 22:31
  • \$\begingroup\$ ok thanks I will test it. \$\endgroup\$
    – user261847
    Commented Sep 1, 2020 at 11:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.