Speex dev - Jan 2007 - questions concerning adaptation rate

Hello Jean-Marc,
 
When you count the adaptation step size in case where the filter has already had
minimal adaptation, you use the following expression:
 
r = (0.7*r + 0.3*15*RER*e)/e*(power[i] + 10) = (0.7*leak_estimate*Yf[i] +
0.3*15*RER*(Rf[i] + 1))/((Rf[i] + 1)*(power[i] + 10)). 
 
Why do we need this weighted sum and the component 0.3*15*RER*(Rf[i] + 1)? Why
use the correlation-based RER if we can do the same in the frequency domain? If
we removed this component, the expression would be exactly like in the method
described in "On adjusting the learning rate in frequency domain echo
cancellation with double-talk". Or am I missing something?

Also, why do we use an increasing adaptation rate as the filter gets closer to
adaptation but isn't yet adapted? Wouldn't it make sense to decrease it?

Thank you,
Julia

> When you count the adaptation step size in case where the filter has
already had
> minimal adaptation, you use the following expression:
>  
> r = (0.7*r + 0.3*15*RER*e)/e*(power[i] + 10) = (0.7*leak_estimate*Yf[i] +
> 0.3*15*RER*(Rf[i] + 1))/((Rf[i] + 1)*(power[i] + 10)). 
Actually, the QCONST16(.3,15) doesn't mean (.3*15), but the value .3 in
Q15 format (fixed-point).
> Why do we need this weighted sum and the component 0.3*15*RER*(Rf[i] + 1)? 
First, the reason for multiplying by (Rf[i]+1) is because I'll be
dividing by it in the next line (and I didn't want to do divisions for
both terms).
> Why
> use the correlation-based RER if we can do the same in the frequency
domain? If
> we removed this component, the expression would be exactly like in the
method
> described in "On adjusting the learning rate in frequency domain echo
> cancellation with double-talk". Or am I missing something?
The reason for using the RER is a bit of a kludge, that is I have found
that it helps, but I cannot explain it for now, except by saying that it
forces the learning rate to be more similar across frequency.
> Also, why do we use an increasing adaptation rate as the filter gets closer
to
> adaptation but isn't yet adapted? Wouldn't it make sense to
decrease it?
That's not the case. What makes you say think that?

	Jean-Marc

Thank you for answering the previous question!

The expression we use in this case is 0,25*Sxx/See. Won't See decrease as
the filter gets closer to adaptation?