1
\$\begingroup\$

I wonder if common electret microphones are frequency mixers. That is, say, when two signals pass through the diaphragm, will two new signals f1 + f2 and f1 - f2 be produced if the sound pressure is around 70-80dB-spl? If they are, how significant are the two signals - will the power levels be roughly the same as the input?

\$\endgroup\$
1
  • \$\begingroup\$ Just feed the mic into a very non-linear mic amplifier. \$\endgroup\$
    – user16324
    Feb 25, 2017 at 10:54

3 Answers 3

4
\$\begingroup\$

No, at that SPL, the microphone will be operating linearly, and not producing any significant harmonics or intermodulation products.

\$\endgroup\$
2
  • \$\begingroup\$ Right. So at around how much SPL is needed to produce them? \$\endgroup\$
    – John M.
    Feb 25, 2017 at 4:00
  • 1
    \$\begingroup\$ I don't know. Nobody tests them that way. What are you really trying to accomplish here, and why can't you do it by electronically processing the output of the microphone like everyone else? \$\endgroup\$
    – Dave Tweed
    Feb 25, 2017 at 5:19
2
\$\begingroup\$

will two new signals f1 + f2 and f1 - f2 be produced

only if the mic is non-linear. In reality, those devices are designed to be as linear as possible so the 'mixing' is quite limited.

it can become an issue when the mic is pushed outside of its envelope.

\$\endgroup\$
2
\$\begingroup\$

Why not? Almost every device can operate as a frequency mixer (albeit an inefficient one) when driven into its non-linear region. The more quadratic-type distortion a device has, the more efficient the mixing will be.

However, don't expect the mixing products to nowhere near to the power level of the inputs. An efficient diode-based mixer can have a conversion loss of -6dB. But a device not intended to be used as a mixer will be far far more lossy.

\$\endgroup\$

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.