The question is simple, the answer I'm not sure...

How could I pick up an analog audio signal, divide it's frequency by 2 (one octave lower) and send it back on it's way?

I could do this by sampling the input signal and generating an output signal that is half of it, but the generated signal would be a digital signal, not an analog, and when we speak of audio, tone means a lot, and the less differences there is between the input and output signal (besides the fact that the output is an octave lower, obviously...), the better...

Is this possible in anyway?

There can be digital circuitry, e.g. for sampling, but the least, the better.

  • \$\begingroup\$ there are fixed frequency divider ICs but I don't know if they can deal with the complexity of audio signals \$\endgroup\$
    – KyranF
    Nov 2, 2014 at 19:34
  • 1
    \$\begingroup\$ Set the turntable to 16 2/3 rpm? \$\endgroup\$ Nov 3, 2014 at 1:39
  • \$\begingroup\$ lool thats the point, but its a guitar, not a turntable xD \$\endgroup\$ Nov 3, 2014 at 21:48
  • \$\begingroup\$ If the signal is coming from a guitar, play it a octave lower instead. \$\endgroup\$ Dec 2, 2014 at 21:47
  • \$\begingroup\$ @OlinLathrop unless i don't have an octave lower... that's what bass guitars are for... and when there's no bass guitar, guitar players use effect pedals, and that's what i want to build... if you have anything else that "useful" to add, feel free :) \$\endgroup\$ Dec 2, 2014 at 22:07

4 Answers 4


You could use a technique similar to digital processing, but without converting the signal to and from digital codes. For example you could use bucket-brigades clocked in and out at different frequencies, or a tape recorder with two heads and transport mechanisms (real time recording, half speed playback).

However if you want to halve the frequency of an arbitrarily long and complex audio waveform then you have a problem - whether using analog or digital processing. The signal is coming in twice as fast as it is being sent out, so in order to exactly preserve the original waveform you have to continuously store the input. Eventually you must run out of storage space, then you will have to 'catch up' to real time and lose a chunk of the signal.

A sufficiently powerful digital system could simply include massive amounts of storage, or it could apply Fourier transforms to break the signal up into its component frequencies, halve each one and then recombine them. The resulting waveform might not be identical to the original, but it should sound virtually the same.

If you just want to change the frequency of a repetitive waveform (eg. single note from a musical instrument) then you only need enough space to store a single cycle, or perhaps the duration of one note. You then have to accurately detect the end of the waveform so that it can be repeated seamlessly, and decide what to do about its envelope (eg. do you let the note play out at half real time, or force a faster decay?).


We are in a digital world and clearly this can be done digitally so all that remains is for you to decide how many bits of resolution the analogue and digital conversion processes need. 24 bits is plenty for me.

However, if you have the belief that any digital processing is unacceptable there is no useful analogue answer I could give.

  • \$\begingroup\$ There are times when no complex digital circuitry is advantageous (normally due to certification process) but that doesn't mean you shouldn't take the hit - weigh up engineering time and the end complexity \$\endgroup\$
    – user16222
    Dec 2, 2014 at 22:15

DSP. ADC, FFT, resample, IFFT, DAC. You can translate frequencies in analog (add or subtract) but you cannot rescale them. The only way to do that is in the frequency domain. The only easy way to get to frequency domain with a reasonable amount of frequency resolution is with a digital signal processor of some sort. You may be able to do all of this in one chip if you can get a DSP with built-in ADC and DAC.


If it's just one note, use a single transistor running as an amplifier with a tuned input circuit for the halved frequency.

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    \$\begingroup\$ Welcome to EE.SE. I'm a guitarist, an electronics enthusiast and an engineer and have no idea what this means. I think you need to supply a lot more detail on your idea. \$\endgroup\$
    – Transistor
    Oct 5, 2017 at 21:51

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