2
\$\begingroup\$

I'm looking for a quick-and-dirty way of converting something close to a sine wave to a narrow pulse wave. Please see the image: a low bass note played through a broken transformer. I'm trying to re-create the effect, preferably using discrete components.

narrow pulse waveform

\$\endgroup\$
8
  • 1
    \$\begingroup\$ try 'Low pass filter + zero crossing comparator'. \$\endgroup\$
    – nidhin
    Nov 19, 2014 at 8:12
  • \$\begingroup\$ Yeah I was thinking of some sort of a zero crossing detector. But I'm not good at designing circuits and I haven't really found a simple low voltage one (preferably with transistor(s)). \$\endgroup\$
    – Jari
    Nov 19, 2014 at 8:31
  • \$\begingroup\$ It is very important to perform lowpass filtering (integration) prior to zero-crossing detection. \$\endgroup\$
    – LvW
    Nov 19, 2014 at 10:27
  • \$\begingroup\$ OK thank you! I found this circuit, a good one to start with? sound.westhost.com/appnotes/an005-f1.gif \$\endgroup\$
    – Jari
    Nov 19, 2014 at 12:06
  • \$\begingroup\$ You show an output waveform and clearly there are pulses but what is the stuff in-between - is it noise that you don't want or part of the input signal? Are the pulses coincident with a certain part of the input waveform i.e. zero cross of input or possibly the peak voltage? Are there in fact many pulses per cycle? I cannot tell because I can't see the input waveform. \$\endgroup\$
    – Andy aka
    Nov 19, 2014 at 13:14

1 Answer 1

-4
\$\begingroup\$

There is a way to convert a cosine and sine waves into a pulse. I do this in RF.

First of all you have to have both real and imaginary waves. We start with just a real wave and then perform a complex multiply. After some digital filtering you get both real and imaginary waves. For my project there are about 2 or 3 cycles at some frequency.

From trigonometry, cosine^2 + sine^2 = 1;

Your perform this equation on each sample.

This will give you a pulse.

\$\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.