Talk:A-law algorithm

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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)

a or A

Is it a-law or A-law and what (if anything) does the a/A stand for? --Abdull 16:54, 11 August 2006 (UTC)[reply]

Legend of the graph uses wrong colours

The green straight line corresponds to no companding... 192.54.144.226 08:11, 22 August 2006 (UTC)Denis[reply]

Whoops - You are right - It's fixed now Ozhiker 08:35, 22 August 2006 (UTC)[reply]

Interoperability

The article states: "A-law is used for an international connection if at least one country uses it." Wouldn't both ends of the connection need to support the algorithm to send signals using it? Or am i missing something? Foobaz·o< 17:35, 17 May 2007 (UTC)[reply]