Some beginner questions concerning YM2612
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Re: Some beginner questions concerning YM2612
Most of the info is found in this post and this thread in general : viewtopic.php?p=5716#p5716
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Re: Some beginner questions concerning YM2612
THANK YOU !
I knew this thread, but since it's very technical (and, as I said before, I have absolutely no knowledge in music nor FM synthesis - and generally speaking very few when it comes to hardware), I progressed slowly in reading it and hadn't come to this post. I have to digest all the informations, but it seems to answer most of my questions about Envelope (envelop? enveloppe ?) Generator !
I knew this thread, but since it's very technical (and, as I said before, I have absolutely no knowledge in music nor FM synthesis - and generally speaking very few when it comes to hardware), I progressed slowly in reading it and hadn't come to this post. I have to digest all the informations, but it seems to answer most of my questions about Envelope (envelop? enveloppe ?) Generator !
Re: Some beginner questions concerning YM2612
Something I can't find in the above thread (but it may have overlooked it) :
In the Envelop Generator, how is the attenuation initialized in the attack phase ?
is it 1023 ? if so it means that if TL isn't 0 dB (maximal volume), then operator will remain silent for some time ?
or is it initialized at TL ?
In the Envelop Generator, how is the attenuation initialized in the attack phase ?
is it 1023 ? if so it means that if TL isn't 0 dB (maximal volume), then operator will remain silent for some time ?
or is it initialized at TL ?
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Re: Some beginner questions concerning YM2612
There is no initialization, the attack begins wherever last state left off. Only way to get maximum attenuation is to wait out the envelope, using fastest release to reduce the time needed. Only on chip reset the internal attenuation is at max attenuation, once first note begins all bets are off.
Last edited by TmEE co.(TM) on Wed May 01, 2019 11:58 am, edited 1 time in total.
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Re: Some beginner questions concerning YM2612
Stupid me. Of course it makes sense ! Thanks
Re: Some beginner questions concerning YM2612
Another question : when I translate decibel coordinates to linear ones (whatever they are), I did :
attenuation_in_dB = 10*log(I_max/I)
where I is my linear amplitude and I_max the maximum amplitude (since I use 16 bits integer, I_max = 32767), which led to :
I = I_max*10^(-attenuation_in_dB/10) = I_max*10^(-attenuation*9.6/1024)
where attenuation is the attenuation computed by the envelope generator.
Am I right ? Because if so, it means that any value of attenuation >= 482 results in an amplitude of 0...
attenuation_in_dB = 10*log(I_max/I)
where I is my linear amplitude and I_max the maximum amplitude (since I use 16 bits integer, I_max = 32767), which led to :
I = I_max*10^(-attenuation_in_dB/10) = I_max*10^(-attenuation*9.6/1024)
where attenuation is the attenuation computed by the envelope generator.
Am I right ? Because if so, it means that any value of attenuation >= 482 results in an amplitude of 0...
Re: Some beginner questions concerning YM2612
Isn't decibels-to-linear something like pow(2,dB/6)? (in the case of attenuation the decibels would be negative) That returns a multiplier (in floating point), mind you, but should give you a clue of where to go.
I may or may not be misunderstanding your question.
I may or may not be misunderstanding your question.
Sik is pronounced as "seek", not as "sick".
Re: Some beginner questions concerning YM2612
Your formula is an approximation of mine (with a division by 3 rather than 6, unless I missed something, like efficace value or something like that, which is quite possible).
10^(db/10) = (2^1/log(2))^(db/10) = 2^(db/(10log(2))
and 10log(2) = 3.0103...
Unless the true formula is 10^(db/20) ?
10^(db/10) = (2^1/log(2))^(db/10) = 2^(db/(10log(2))
and 10log(2) = 3.0103...
Unless the true formula is 10^(db/20) ?
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Re: Some beginner questions concerning YM2612
This page seems pretty good to me.
Re: Some beginner questions concerning YM2612
So, according to this page, it seems that in audio we use the dB for tensions (why ?), which is different (why ?).
So it's indeed 10^(dB/20), which indeed equals roughly 2^(dB/6).
Thanks
So it's indeed 10^(dB/20), which indeed equals roughly 2^(dB/6).
Thanks