I'm curious...can you test what happens if you make it a one-sided test? That
would eliminate local minimums.
if((smspec[j]>smspec[j-1]) || (smspec[j]>smspec[j+1]) && (npk<200))
{
Also...what's the distribution of the snr values for the successful vs
unsuccessful decodes? Maybe there's a better value to be had there.
73
Mike W9MDB
-----Original Message-----
From: Steven Franke [mailto:s.j.fra...@icloud.com]
Sent: Sunday, June 28, 2015 9:47 AM
To: WSJT software development
Subject: Re: [wsjt-devel] wspr two-pass decoder status and .c2 file problem
Joe,
Frequency-sorting has now been committed.
It dawned on me that it is no longer really appropriate to insist that a
decoding candidate correspond to a local maximum in the average spectrum. I
tried this simple change:
$ diff wsprd_exp.c ~/Builds/wsjtx/lib/wsprd/wsprd_exp.c
848,850c848,849
< for(j=1; j<410; j=j+2) {
< // if((smspec[j]>smspec[j-1]) && (smspec[j]>smspec[j+1]) &&
(npk<200)) {
< if((smspec[j]>min_snr) && (npk<200)) {
---
> for(j=1; j<410; j++) {
> if((smspec[j]>smspec[j-1]) && (smspec[j]>smspec[j+1]) &&
> (npk<200)) {
On my 20m wav files it gives us another 10% over the current wsprd_exp with
subtraction and two-passes, or a whopping 47% improvement over the current
default wsprd:
Default wsprd: 1315 in 163s
wsprd_exp: 1758 in 270s
no peak requirement: 1934 in 1487s
With this change, a case that yields 16 decodes from the current wsprd_exp now
yields 19 decodes from one .wav file!
It is dog-slow as it stands. But I’m sure that there are things that we can do
to speed it up significantly. This might be a good application for openmpi, no?
Steve k9an
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