Will an electric megaphone amplify "ultrasound" and still send it in a narrow cone ? If So, how will such a device distort sound at frequencies higher than the human ear can detect ?
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1$\begingroup$ Very much depends on the device. Most are for audible purposes, and will focus on that job. $\endgroup$– JMLCarterCommented Mar 5, 2017 at 4:01
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1 Answer
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Ditto for @JMLCarter comment.
To elaborate on this, Wikipedia refers for some data below, while other sourced from technical literature.
The average reflex or reentrant horn speaker has a typical frequency range of 450Hz - 6KHz.
Also called a folded horn, the reflex type has several horns facing each other to force the sound-path into a Z pattern.
This simply shortens the overall body length to make megaphone more portable, while the horn remains long.
The mouth (front flare) matches the impedance of vocal chords/throat to the air.
Both the amplifier circuit and horn driver (a small but powerful speaker) would be optimized (wide audio band-pass filtration) for vocal range frequency.
Ultrasound is above human hearing, way above human voice frequency (U/S starts at around 15KHz-16KHz).
The horn itself is way too large to be resonant at such high frequency, so there will be attenuation.
Plus, some ultrasound generating circuits produce a square wave signal. A 16-40 KHz square wave alone would introduce harmonics.
Taking all above negatives into account, the answer is no, not feasible.
I don't know what U/S circuitry you have, and for what purpose you want to broadcast/amplify.
However, you could build or buy an amp to operate at HF audio, then use a shorter smaller resonant horn to "channel" the sound.
It is unclear whether a horn profile follows resonance formulas for audio. Speed of sound (symbol v) at sea-level 25~C is 341.376 meters per second.
Standard Loudspeaker enclosures usually use Half-Wave or Quarter-Wave dimensions to reinforce a frequency range, depending on ports.
To convert to mm, v*1000, so v=341376 mm/s. f=Hz. WL in mm. (WL symbol lowercase Greek lambda). As Lambda is unicode font I'll use WL for wavelength.
FWL=full wave length (WL x 1); HWL=half wave length (WL/2); QWL=quarterwave length (WL/4).
f=v/WL. WL=v/f
Here are some calculated figures
FWL (Lf horn)=341376/450=758.6133 mm. HWL (Lf horn)=379.3067mm. QWL=(Lf horn)=189.65335 mm.
FWL (Hf horn)=341376/6000=56.896 mm. HWL (Hf horn)=28.448 mm. QWL=(Hf horn)=14.224 mm.
FWL (Lf horn)=341376/16000=21.336 mm. HWL (Lf horn)=10.668 mm. QWL=(Lf horn)= 5.334 mm.
