Revision as of 08:04, 26 February 2007 by imported>Joseph Rushton Wakeling
Pink noise
In line with the issue a
spectrum satisfies the definition of equal power per octave, I put together a little proof of the converse. It's probably inelegant (my maths is rusty these days....). Anyway, here goes. Let
be the integral of the power spectrum. Then, since there is equal power per octave, we must have
where
is a constant.
If we write
, we get
, where
is constant. Since it's so, and
is also a constant, without loss of generality we can write,
![{\displaystyle {\frac {Z(\lambda f)}{Z(f)}}=\lambda K'}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ee130161fef8c216ec948fcbcc211c6dbd61193)
So,
. Now consider
, but also
, so
, so either
(boring, means the signal has no power) or
. We are left with,
![{\displaystyle Z(\lambda f)=\lambda Z(f)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3b5d6f496945eebd2ec39b1a489acd556e0cbfb5)
From which it follows that
is a linear function, so we can write
with c constant. Substituting into the previous equation, we get,
![{\displaystyle m\lambda f+c=\lambda (mf+c)=m\lambda f+\lambda c}](https://wikimedia.org/api/rest_v1/media/math/render/svg/acd50d3ba8e813ec513ce56d6e40f0f0c9e4bc4d)
So,
, and so
.
It follows that
, so
![{\displaystyle Y(f)=\log(mf)=\log(f)+\log(m)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/26873719683f1cce348fcf9d617212716c04c4fc)
So the differential, which gives us the power spectrum, is
.
Someone shout if there's a problem. Like I said, I'm rusty... :-) —Joseph Rushton Wakeling 06:38, 10 February 2007 (CST)
- Aaahhh. There is an error: one could use any power of
to give the value of the constant in the equation
. So you wind up with
, for arbitrary real
. The linear argument then won't work, but there must be some trick that brings out the more general solution giving us a power spectrum of
instead of straightforward
.