On Wed, 26 Jan 2011 15:50:25 +0100, Michał Baszyński napisał:
> W dniu 2011-01-26 15:22, Verox pisze:
>
>> Potrzebuje z sygnału wejściowego (AFSK, 1200 + 2400 Hz) wyciągnąć
>> te prążki celem poddania dalszej detekcji. Wiem, że można to zrobić przy
>> pomocy FFT, ale może są jakieś łatwiejsze do implementacji algorytmy.
>> Bardzo proszę grupowych matematyków o nakierowanie :-)
>
> nie jestem matematykiem, ale popatrz na Goerztla

Dziękuję pięknie, wziąłem algo z wikipedii:

http://en.wikipedia.org/wiki/Goertzel_algorithm

I działa :-)

wyniki symulacji dla f_probkowania=16 kHz

ponośna 2200 Hz:

freq = 500.000000 Hz pwr = 3630.037598 lvl=0
freq = 600.000000 Hz pwr = 6390.092285 lvl=0
freq = 700.000000 Hz pwr = 6390.092285 lvl=0
freq = 800.000000 Hz pwr = 6390.092285 lvl=0
freq = 900.000000 Hz pwr = 10942.559570 lvl=0
freq = 950.000000 Hz pwr = 10942.559570 lvl=0
freq = 1000.000000 Hz pwr = 10942.559570 lvl=0
freq = 1100.000000 Hz pwr = 21924.453125 lvl=1
freq = 1180.000000 Hz pwr = 21924.453125 lvl=1
freq = 1190.000000 Hz pwr = 21924.453125 lvl=1
freq = 1200.000000 Hz pwr = 21924.453125 lvl=1
freq = 1210.000000 Hz pwr = 21924.453125 lvl=1
freq = 1220.000000 Hz pwr = 21924.453125 lvl=1
freq = 1230.000000 Hz pwr = 21924.453125 lvl=1
freq = 1250.000000 Hz pwr = 21924.453125 lvl=1
freq = 1300.000000 Hz pwr = 21924.453125 lvl=1
freq = 1500.000000 Hz pwr = 44501.601562 lvl=2
freq = 2000.000000 Hz pwr = 412674.156250 lvl=20
freq = 2100.000000 Hz pwr = 15191353.000000 lvl=247
freq = 2150.000000 Hz pwr = 15191353.000000 lvl=247
freq = 2180.000000 Hz pwr = 15191353.000000 lvl=247
freq = 2190.000000 Hz pwr = 15191353.000000 lvl=247
freq = 2200.000000 Hz pwr = 15191353.000000 lvl=247
freq = 2210.000000 Hz pwr = 15191353.000000 lvl=247
freq = 2220.000000 Hz pwr = 15191353.000000 lvl=247
freq = 2250.000000 Hz pwr = 15191353.000000 lvl=247
freq = 2300.000000 Hz pwr = 1229633.875000 lvl=61
freq = 2400.000000 Hz pwr = 1229633.875000 lvl=61
freq = 3000.000000 Hz pwr = 69920.218750 lvl=3
freq = 3500.000000 Hz pwr = 34523.902344 lvl=1
freq = 4000.000000 Hz pwr = 21944.585938 lvl=1

podnośna 1200 Hz

freq = 500.000000 Hz pwr = 419.161530 lvl=0
freq = 600.000000 Hz pwr = 1563.212891 lvl=0
freq = 700.000000 Hz pwr = 1563.212891 lvl=0
freq = 800.000000 Hz pwr = 1563.212891 lvl=0
freq = 900.000000 Hz pwr = 8458.195312 lvl=0
freq = 950.000000 Hz pwr = 8458.195312 lvl=0
freq = 1000.000000 Hz pwr = 8458.195312 lvl=0
freq = 1100.000000 Hz pwr = 17934266.000000 lvl=128
freq = 1180.000000 Hz pwr = 17934266.000000 lvl=128
freq = 1190.000000 Hz pwr = 17934266.000000 lvl=128
freq = 1200.000000 Hz pwr = 17934266.000000 lvl=128
freq = 1210.000000 Hz pwr = 17934266.000000 lvl=128
freq = 1220.000000 Hz pwr = 17934266.000000 lvl=128
freq = 1230.000000 Hz pwr = 17934266.000000 lvl=128
freq = 1250.000000 Hz pwr = 17934266.000000 lvl=128
freq = 1300.000000 Hz pwr = 17934266.000000 lvl=128
freq = 1500.000000 Hz pwr = 14059.884766 lvl=0
freq = 2000.000000 Hz pwr = 2022.837646 lvl=0
freq = 2100.000000 Hz pwr = 1150.462524 lvl=0
freq = 2150.000000 Hz pwr = 1150.462524 lvl=0
freq = 2180.000000 Hz pwr = 1150.462524 lvl=0
freq = 2190.000000 Hz pwr = 1150.462524 lvl=0
freq = 2200.000000 Hz pwr = 1150.462524 lvl=0
freq = 2210.000000 Hz pwr = 1150.462524 lvl=0
freq = 2220.000000 Hz pwr = 1150.462524 lvl=0
freq = 2250.000000 Hz pwr = 1150.462524 lvl=0
freq = 2300.000000 Hz pwr = 806.450989 lvl=0
freq = 2400.000000 Hz pwr = 806.450989 lvl=0
freq = 3000.000000 Hz pwr = 394.281921 lvl=0
freq = 3500.000000 Hz pwr = 356.505646 lvl=0
freq = 4000.000000 Hz pwr = 229.909637 lvl=0

Dziwi mnie tylko to, że przy jednakowej amplitudzie sygnału wejściowego
(bo jest wyliczony :-), algorytm daje wyniki mocno różniące się na wyjściu. W moim
zastosowaniu to nie przeszkadza - ot taka ciekawostka :-)

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