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)

1 equation
2 retransformation wrong ?
3 a or A
4 Legend of the graph uses wrong colours

[edit] equation

This needs finishing:

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

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

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

[edit] 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

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