Talk:1-f noise

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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z(f) = mf + c} with c constant. Substituting into the previous equation, we get,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m\lambda f + c = \lambda (mf + c) = m\lambda f + \lambda c}

So, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c = \lambda c} , and so Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c = 0} .

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Y(f) = \log(mf) = \log(f) + \log(m)}

So the differential, which gives us the power spectrum, is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda} to give the value of the constant in the equation Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z(\lambda f)/Z(f) = K} . So you wind up with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z(\lambda f) = \lambda^{n}Z(f)} , for arbitrary real Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\textrm{constant}}{f}} instead of straightforward Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1/f} .