| yes, there is a DC and that is what the code tries to remove. But I
would use a DC blocker instead.
Victor
On 25 Sep 2011, at 04:35, Partev Barr Sarkissian wrote:
> Victor,
>
> I think that makes sense. Correct me if I'm wrong, but when you
> mathemtically break down a signal into it's components, with a
> number of Low-Pass filter types, isn't there a DC component in
> there? That Split Side-Band paper you did (in JAES, September
> 2008 was it?), wasn't there a DC component in the Taylor Series
> Expansion you did?
>
>
> Regards, cheers,
> -Partev
>
>
> ================================================================
>
>
> --- Victor.Lazzarini@nuim.ie wrote:
>
> From: Victor.Lazzarini@nuim.ie
> To: csound@lists.bath.ac.uk
> Subject: Re: [Csnd] unexpected integ offset
> Date: Wed, 21 Sep 2011 20:12:18 +0100
>
> The integrtator is an IIR LP filter, so it will produce a signal with
> a certain amount of DC as it has a pole at 0Hz. The code assumes this
> to be about 0.5 and tries to remove it. I'd say a more robust
> approach is to use a DC blocker.
> Victor
> ----- Original Message -----
> From: ebmtranceboy
> Date: Wednesday, September 21, 2011 6:37 pm
> Subject: [Csnd] unexpected integ offset
> To: csound@lists.bath.ac.uk
>> Hi all,
>> Sorry in advance if this is not the right place to ask this kind of
>> question.
>> I'm trying to understand the following instrument and I'm
>> puzzled by the
>> "asaw = asawdc - .5" line
>>
>> ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
>> instr 1; band limited sawtooth wave
>> ; coded by Josep M Comajuncosas / jan98
>> ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
>>
>> iamp
>> = 10000; intended max. amplitude
>> kenv linen 1, .2,p3,.2; volume envelope
>> kfreq
>> = cpspch(p4)
>> ibound = sr/4;
>> set it to sr/2 for true BL square wave
>> apulse buzz 1, kfreq, ibound/kfreq, 1
>> asawdc integ apulse
>> asaw
>> = asawdc - .5
>> out asawdc*iamp*kenv
>> endin
>>
>> Indeed, if the fundamental period is 2pi and N the number of
> required
>> harmonics (ibound/kfreq in instr 1), buzz opcode provides a
>> signal (assumely
>> continuous)
>>
>> b(t) = (sum { cos(kt), k = 1..
> N}) / N
>>
>> so, by integration,
>>
>> saw(t) = integ [b(u)du, u = 0..t]
>> saw(t) = (sum { sin(kt) / k, k = 1..
> N}) / N
>>
>> which is precisely the reconstruction of the signal whose Fourier
>> coefficients are (N times smaller than) (1,1/2,1/3,...,1/N) that
>> is just the
>> asaw signal we're trying to build and I don't understand where
>> the 1/2
>> offset comes from.
>>
>> Is it a numerical (discrete signal) consequence ? If so, why
>> this 1/2
>> coefficient doesn't depend of N (setting ibound to sr/2 for
>> example still
>> requires the offset set to 1/2 to get the asaw expected).
>>
>> Help me with my maths please!
>>
>>
>>
>> --
>> View this message in context:
>> http://csound.1045644.n5.nabble.com/unexpected-integ-offset-
>> tp4827197p4827197.htmlSent from the Csound - General mailing
>> list archive at Nabble.com.
>>
>>
>> Send bugs reports to the Sourceforge bug tracker
>>
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>>
> Dr Victor Lazzarini, Senior Lecturer, Dept. of Music,
> National University of Ireland, Maynooth
>
>
>
>
> _____________________________________________________________
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Dr Victor Lazzarini
Senior Lecturer
Dept. of Music
NUI Maynooth Ireland
tel.: +353 1 708 3545
Victor dot Lazzarini AT nuim dot ie
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