>>
>>From: "Neil Gould"
>>Subject: Re: What is "Realtime Convolution and Modulation Synthesis"?
>>Date: Sunday, July 12, 2009 7:20 AM
>>
>>Laurence Payne wrote:
>>> >>>
>>>> The definition of convolution is set in stone, and that is what I
>>>> provided. Yamaha's adspeak version does not explain it.
>>>
>>> Yamaha may be using the term inaccurately. But it's their inaccurate
>>> meaning that we're chasing here!
>>>
>>The thing about definitions is that a general term may be correctly >>applied
>>in different fields to describe very different phenomena. Those whose >>usage
>>of a term is limited to a single field may come to think that a particular
>>application sets that definition in stone, but context is relevant! >>
>>[snip]
>>
>>Neil,
>>Yamaha uses two phrases , "Realtime Convolution" and "Modulation", and you
>>seem to have an irresistable urge to merge the two. It is common for both
>>techniques to appear in a synthesizer product, and be used at the same >>time.
>>
>>1. FM Synthesis makes the funky sound. This is further operated on by "2", >>below.
>>
>>2. Convolution, in this form, (F*G)(t)= Sum( f(tau)*g(t-tau) from
>>tau=(-infinity) to tau=t) , is used to make the instrument sound "real." >>The
>>above mathematical formula is the basis of ambience synthesis, which >>includes both:
>> a. The effect of the instrument body.
>> b. The effect of an artificial venue, if required.
>>
>>Bob Morein
>>(310) 237-6511
>>
>
> What convolution actually is (rather than what it is used for in this
> application) is a mathematical technique for performing operations in
> a perpendicular axis or domain. In audio what this normally means is
> performing a frequency domain multiplication (as you would for a
> filter, say) in the time domain. The effect is the same, but it can
> be done in near real time, rather than having to gather up a heap of
> samples, perform an FFT, do the multiplication and then an inverse FFT
> to bring it back to time domain.
>
> d
----------------------------------------------------------------------- This is true, though I would disagree with the "actually is". It depends on what your definition of "is", is. You've stated an identity, and one identity does not take precedence over another except for those damn voices in my head. It is true that the Fourier transform of the product of two functions is equal to the product of the Fourier transform of two functions, times a constant.
What any of that means is a mystery to me, however.
Bob Morein (215) 646-4894