Re: randomized white noise = white noise?

I'll try to answer two posts at once here...

Note that when a rolloff slope changes from 3dB/octave to 6dB/octave,
it's a gradual thing. You can draw lines on a Bode plot that are
straight and intersect at a sharp corner, but in practice, the
transition is gradual. So the difference in response between just the
3dB/octave and the added 6dB/octave in the top octave will be
relatively small. In fact, if you get Scilab and run the simple code I
suggested, you can see a plot of the response--easy to plot the
response of your filter, or the difference between it and "true pink."
And what you will see with your original coefficients is that they
cause the high end to drop off faster than "pink" anyway, by just about
what you are suggesting with that "6dB slope in the top octave". So
you may have that already. In fact, in Scilab (or Matlab) it's easy to
make most of what I posted be a routine that will plot the response or
whatever, and you just change the coefficients and run the routine to
see the effect. Even with a slow processor, it should do the calcs in
a second or less, especially if you used fewer points logarithmically
spaced, but even if you do that one point per Hz hack I posted, it's
still fast.

OK, have a look at the relationships in your filter to see why you get
large values pretty often. Notice that b2, for example, remembers over
half its previous value, and adds MORE than one times the input to
itself. So let's say you have three white samples in a row that are in
the range 0.5 to 1.0 (assuming that white runs from -1 to +1). Note
that:

b2(k+1) = 0.57*b2(k) + 1.05*white(k)
= 0.57* ( 0.57*b2(k-1) + 1.05*white(k-1)) + 1.05*white(k)
=
0.57*(0.57*(0.57*b2(k-2)+1.05*white(k-2))+1.05*white(k-1))+1.05*white(k)
= 0.185*b(k-2)+0.34*white(k-2)+0.60*white(k-1)+1.05*white(k)

If the average value of white = 0 (no DC bias), then on average b(k-2)
will be zero. A string of three "whites" in a row at 0.5 or above will
then, on average, result in an output from just THAT equation that's
greater than 1.

The way around this is to scale "white" down, and don't let it cover
the whole range from -1 to +1. How much do you scale it? The absolute
worst case would be if white stayed at +1 (or -1) for a long time. For
your filter (which technically is three one-pole resonators), you can
easily get this value: first you find the value for each of the "b"
values, then add them plus 0.1848 for the current "white" value that's
in the output. How do you get the value for each "b"? Simple: it
will reach steady state when the contribution from the "old" b decays
by exactly the amount of the new input. So for b0, white*0.099 =
(1-0.99765)*b0(steady-state). So b0(steady-state)=42.12*white. That
is, if the input is locked at 1, b0 will eventually get to 42.12. So
an absolutely safe scaling would be to not let "white" get above
1/42.12. Now in practice, the probability of "white" staying that
close to full scale in one direction for long enough to do that is so
tiny (and in fact would be zero for any practical pseudorandom number
generator) that you can safely make the scaling larger than that. For
b2, the safe scaling (by a similar calculation) would be 0.408.

Note that since you are dealing with NOISE here, if you can gracefully
handle the overflow, letting it happen once every few thousand samples
is likely not going to even be noticable. I suspect that if you scale
the white input by 0.25, you will be fine. Try it and see how that
works. Your filter probably does a very decent job of turning the
uniform amplitude distribution into a Gaussian amplitude distribution,
so the mere fact that you see about 3 or 4 out of 10 that's larger than
the range of the WHITE input tells me that the standard deviation is
quite close to full scale. Then the scaling I suggested, 0.25, should
get you to about one in 10,000 that overflows. Note that you do not
have to SHIFT your white; you can just duplicate the top bit into the
next two bits (for a signed value), and it will still be white. The
bits in a uniform-amplitude-distribution random number should be all
uncorrelated, and it doesn't matter which bits you drop.

Cheers,
Tom

.

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