# Talk:1-f noise

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Pink noise

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

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

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

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

${\displaystyle Z(\lambda f)=\lambda Z(f)}$

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

${\displaystyle m\lambda f+c=\lambda (mf+c)=m\lambda f+\lambda c}$

So, ${\displaystyle c=\lambda c}$, and so ${\displaystyle c=0}$.

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

${\displaystyle Y(f)=\log(mf)=\log(f)+\log(m)}$

So the differential, which gives us the power spectrum, is ${\displaystyle 1/f}$.

Someone shout if there's a problem. Like I said, I'm rusty... :-) —Joseph Rushton Wakeling 06:38, 10 February 2007 (CST)