# Talk:1-f noise

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 Definition:  Electromagnetic noise proportional to f-n, where f = frequency and n={0,1,2). [d] [e]
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## Status?

Wasn't sure whether this is "developed" or "developing". --Larry Sanger 08:52, 9 March 2007 (CST)

Hi Larry — I'm afraid I've been on hiatus for a while due to other pressures. I will come back to it but if someone else wants to write an article and wants to start from scratch, that's fine. Regarding the Wikipedia source, that was all mine (I think there's maybe one line that someone else added, which will be deleted anyway). —Joseph Rushton Wakeling 12:55, 12 March 2007 (CDT)

## 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 power 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)

Aaahhh. There is an error: one could use any power of ${\displaystyle \lambda }$ to give the value of the constant in the equation ${\displaystyle Z(\lambda f)/Z(f)=K}$. So you wind up with ${\displaystyle Z(\lambda f)=\lambda ^{n}Z(f)}$, for arbitrary real ${\displaystyle n}$. 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 ${\displaystyle {\frac {\textrm {constant}}{f}}}$ instead of straightforward ${\displaystyle 1/f}$.