Talk:1-f noise

Pink noise

In line with the issue a $1/f$ 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 $Y(f)$ be the integral of the power spectrum. Then, since there is equal power per octave, we must have $Y(\lambda f)-Y(f)=k$ where $k$ is a constant.

If we write $Z(f)=\exp[Y(f)]$, we get $Z(\lambda f)/Z(f)=K$, where $K=\exp(k)$ is constant. Since it's so, and $\lambda$ is also a constant, without loss of generality we can write,

${\frac {Z(\lambda f)}{Z(f)}}=\lambda K'$

So, $Z(\lambda f)=\lambda K'Z(f)$. Now consider $Z(\lambda ^{2}f)=\lambda K'Z(\lambda f)=\lambda ^{2}K'^{2}Z(f)$, but also $Z(\lambda ^{2}f)=\lambda ^{2}K'Z(f)$, so $K'^{2}=K'$, so either $K'=0$ (boring, means the signal has no power) or $K'=1$. We are left with,

$Z(\lambda f)=\lambda Z(f)$

From which it follows that $Z$ is a linear function, so we can write $Z(f)=mf+c$ with c constant. Substituting into the previous equation, we get,

$m\lambda f+c=\lambda (mf+c)=m\lambda f+\lambda c$

So, $c=\lambda c$, and so $c=0$.

It follows that $\exp[Y(f)]=mf$, so

$Y(f)=\log(mf)=\log(f)+\log(m)$

So the differential, which gives us the power spectrum, is $1/f$.

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 $\lambda$ to give the value of the constant in the equation $Z(\lambda f)/Z(f)=K$. So you wind up with $Z(\lambda f)=\lambda ^{n}Z(f)$, for arbitrary real $n$. 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 ${\frac {\textrm {constant}}{f}}$ instead of straightforward $1/f$.