On Fri, 2006-06-16 at 13:22 +0200, Alex Polite wrote: Where n is a fixed number I suppose? Otherwise it won't work as a LADSPA. -- Lars Luthman - please encrypt any email sent to me if possible PGP key: http://pgp.mit.edu:11371/pks/lookup?op=get&search=0x04C77E2E Fingerprint: FCA7 C790 19B9 322D EB7A E1B3 4371 4650 04C7 7E2E

Hi Alex, Alex Polite a écrit : If I have correctly understood the problem, here is how it could be implemented in Faust : envelop = abs : max ~ -(0.00001); loudest(x,y) = x*(envelop(x)>envelop(y)) + y*(envelop(x) <= envelop(y)); process = loudest; The first line compute a very simple envelop from an input signal : y(t) = max( abs(x(t)), y(t-1) - 0.000001 ) The second line computes the loudest of two signals by comparing their envelops. The last line defines the 'process' we want to compute (the equivalent of main in C). Here is the core of the code generated by the Faust compiler : for (int i=0; i<count; i++) { float fTemp0 = input1[i]; float fRec0Temp = max((fRec0 - 1.000000e-05f), fabsf(fTemp0)); float fTemp1 = input0[i]; float fRec1Temp = max((fRec1 - 1.000000e-05f), fabsf(fTemp1)); output0[i] = ((fTemp1 * (fRec1Temp > fRec0Temp)) + (fTemp0 * (fRec1Temp <= fRec0Temp))); fRec1 = fRec1Temp; fRec0 = fRec0Temp; } If you need more than two input signals you can cascade several loudest modules. For example : process = loudest : loudest; will take the loudest of 3 signals And : process = loudest : loudest : loudest; will take the loudest of 4 signals. Instead of cascading 'loudest' manually so to speak, you could also write : process = seq(i,7,loudest); This will take the loudest of 8 signals by cascading 'loudest' 7 times. Obviously this is not the most efficient way as it computes 2*7 envelops instead of the 8 needed. Here is a more efficient solution that computes the right number of envelops: envelop = abs : max ~ -(0.00001); loudest(ex,x,y) = max(ex, envelop(y)), x*(ex>envelop(y)) + y*(ex <= envelop(y)); chain(n) = first : seq(i,n-1,loudest) : ! ; first(x) = envelop(x), x; process = chain (8); As you can see, loudest now takes 3 arguments and produces 2 signals. The first argument is the envelop of x and the first result is the combined envelop of x and y. Hoping this will help... Yann