Talk:A-law algorithm

how does the a-law algorithm look like? What's the difference to the mu-law algorithm? --Abdull 17:08, 4 Apr 2005 (UTC)

equation

This needs finishing:

A-law expansion is then given by the inverse equation:

$F^{-1}(y)={\begin{cases}{\operatorname {sgn}(y)|y|[1+\ln(A)]/A},&0\leq |y|<{1 \over 1+\ln(A)}\\\operatorname {sgn}(y)e^{|y|[1+ln(A)]-1)}/[A+A\ln(A)],&{1 \over 1+\ln(A)}\leq |y|\leq 1\end{cases}}$

It's not quite right. Check the external links and such. - Omegatron 15:12, May 24, 2005 (UTC)

retransformation wrong ?

I know these formulas are in many documents but I'm afraid the 2nd $F^{-1}$ is wrong.

I tried to build a VoIP (SIP) phone with A-Law audio codec. And I was very surprised because the decoded A-Law stream was very silent. I used exactly these formulas. And I think there is a mistake. Assume x = 1 then:

$y={\frac {1+\ln(A|x|)}{1+\ln(A)}}={\frac {1+\ln(A)}{1+\ln(A)}}=1$

But the retransformation gives

$x=F^{-1}(y)={\exp(|y|(1+\ln(A))-1) \over A(1+\ln(A))}={\exp((1+\ln(A))-1) \over A(1+\ln(A))}={A \over A(1+\ln(A))}=0.1827$

I think the A must be the denominator in both cases:

$x=F^{-1}(y)={\exp(|y|(1+\ln(A))-1) \over A}$

i.e.

$F^{-1}(1)={\exp(|1|(1+\ln(A))-1) \over A}={A \over A}=1$

Joerg Anders (GERMANY)