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I'm trying to build an AM transmitter to remotely listen to sounds within a fairly small distance from the microphone (<30m). One thing I wanted to try was to use the microphone jack on my laptop to pick up the signal. The idea is that I could pick up all signals (my laptop can record up to 192,000 samples per second) and have the transmitter transmit at a frequency such as 48khz. I could then demodulate the signal in software. I figure this could reduce costs and it should be fairly easy to pick up RF frequencies through the microphone jack. The main issue is that the transmission frequency would be very low.

My questions are:

  1. Is it practical to transmit audio at such a low frequency?
  2. What can I expect in terms of signal quality?
  3. What can I expect in terms of signal range at this frequency (the transmission power would probably be around 100mw)?
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    \$\begingroup\$ The microphone input on your computer is not likely to pick up much above 20,000 Hz. \$\endgroup\$ – Peter Bennett Jan 23 '17 at 4:47
  • \$\begingroup\$ @PeterBennett As I said it has a 192khz sample rate. It can pick up up to a 96khz frequency. \$\endgroup\$ – LiquidFeline Jan 23 '17 at 5:11
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    \$\begingroup\$ Sample rate is one thing. Analog bandwidth is another, separate thing. \$\endgroup\$ – mkeith Jan 23 '17 at 5:31
  • \$\begingroup\$ @mkeith intended input is one thing, susceptibility to other inputs is... another thing. Of course there is attenuation and aliasing. But aliasing can also be an advantage, mixing a modulated signal down to baseband. Going a step further by adding an intentional external mixer, a sound card can make a very high performance narrowband radio receiver - but at the frequency proposed the architecture of the sampler may recover something usable for you. It's interesting to try sweeping a piezo source above audio while watching an "audio" spectrum of the data. \$\endgroup\$ – Chris Stratton Jan 23 '17 at 6:35
  • \$\begingroup\$ Yes, the aliasing can be an advantage for radio. Some high-speed ADC's have analog bandwidth to multiples of the sampling frequency specifically to take advantage of this. But, like Peter said, I doubt the analog path on a microphone has bandwidth to 96kHz. Let me know if I am wrong. \$\endgroup\$ – mkeith Jan 23 '17 at 7:43
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No, it is not practical to transmit audio nor any other information at such low frequencies. The passive antenna would have to be much bigger than the very distance you want to transmit. When I am saying bigger I am talking about a dipole hundreds of meters long.

An alternative would be to use an active antenna but without proper experience in RF circuits I think this would complicate your design a lot.

For such an application a better approach is to use IR modulation. Or use an RF transmitter on frequencies much simpler to deal with. Like an FM microphone, for which plenty of commercial solutions exist.

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  • \$\begingroup\$ Out of curiosity, what is the reason why I'd need such a long antenna? And what kind of frequency would I need to use if I had an antenna no more than a meter long? \$\endgroup\$ – LiquidFeline Jan 23 '17 at 5:17
  • \$\begingroup\$ @LiquidFeline See: qsl.net/k/kd4sai/antencal.html \$\endgroup\$ – jonk Jan 23 '17 at 5:34
  • \$\begingroup\$ A shorter antenna is less efficient and has to generate higher voltage, eg. 0.000025% and 27 thousand volts for 100mW into a 1m antenna at 48kHz. people.physics.anu.edu.au/~dxt103/calculators/Rrad.php \$\endgroup\$ – Bruce Abbott Jan 23 '17 at 6:01
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Perhaps FM modulate a tone, which drives a piezo tweeter above the audio range. Use your laptop microphone to capture the acoustic energy.

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  • \$\begingroup\$ Possibly, but you'll get a fairly limited bandwidth between what you can argue as audio cutoff and where you'll first start to run into issues with typical audio inputs - forget about FM, even with single sideband you'd struggle to fit telephone quality audio. Also, while smartphone MEMS microphones often work well above audio, many regular computers and laptops have cheap electrets that have poor response at those frequencies. \$\endgroup\$ – Chris Stratton Feb 1 '17 at 8:06

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