[sdiy] why 10V

[Don Tillman]

   > Date: Fri, 19 Sep 2003 12:29:50 +0200
   > From: jhaible at debitel.net
   > 
   > >    > Next question: If you run a +/-5V audio signal thru a chain of N
   > >    > unity-gain all pass filters, what will be your worst case maximum
   > >    > level at the end of the chain ?
   > > 
   > > A Hilbert transform is probably the worst case scenario.
   > > 
   > > The Hilbert tranform is a 90-degree phase shift that's constant over
   > > frequency.  If you apply a square wave to a Hilbert tranform, the
   > > result is a waveform with huge spikes.
   > > 
   > > So if the all pass filter aproximates a constant 90 degree phase
   > > shift, the output peaks could be maybe ten times the input square wave
   > > voltage.
   > 
   > Thanks for adressing this, Don!
   > 10 times is more than I expected. How did you calculate this?

I guessed!  :-)  Hey, it was late at night.

You weren't specific about the all pass filter, so I considered the
case where the all pass filter shifts phase by a constant 90 degrees.
As such this isn't realizable, but perhaps you'd want to come
arbitrariliy close to that.

Apply a 1 volt zero-to-peak square wave.  The harmonic content is:

  square = 1.414 (sinx - (1/3)sin3x + (1/5)sin5x - (1/7)sin7x ...)

Shifting the phase 90 deegrees:

  shifted square = 1.414 (cosx - (1/3)cos3x + (1/5)cos5x - (1/7)cos7x ...)

For cosines the sign of the coefficient doesn't matter, and so the
peaks of this waveform add up to:

  peaks = 1.414 (1 + 1/3 + 1/5 + 1/7...) 

For 100 harmonics that's 4.15 V.
For 1000 harmonics it's 5.78 V.
For 10000 harmonics it's 7.5 V.

If the realization of the Hibert transform was sloppy, one way or
another, you could have larger peaks.

Sure, it's a silly case.  What I really meant to say is that it
completely depends on the all pass filter, and it's possible to come
up with some crazy all pass filters.

  -- Don

-- 
Don Tillman
Palo Alto, California
don at till.com
http://www.till.com