Pd-list November 2005

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143 participants
298 discussions

Re: PD-list Digest, Vol 8, Issue 34
by Chuckk Hubbard 08 Nov '05

08 Nov '05

Oops, responded without changing the subject: No sweat. I guess I could elaborate on the specific fudging I see: When multiplying Z1 and Z2, you take r1*r2*(cos x + i sin x)(cos y + i sin y) = r1r2((cos x cos y) + i(cos x sin y) + i(cos y sin x) + i*i*(sin x sin y)) = r1r2((cos x cos y) - (sin x sin y) + i((cos x sin y) + cos y sin x))) (there's where I see the fudging, in making the sines negative) = r1r2(cos (x+y) + i sin (x+y)) Now I'm seeing something. Ah ha. (cos x cos y) = (-cos x * cos y) = -(cos x cos y) and -(cos x cos y) + (sin x sin y) = -(cos(x+y)) = cos (x+y) So the negative is a red herring since cosine is an even function? (I use x and y not as polar terms, but because I have no theta key) Crazy. I wonder if that's right. I wasn't thinking philosophically, but turning an imaginary number into a real number seems questionable when the only reason for using an imaginary number is to keep it separated from its real counterpart. The fact that i^2 = -1 has no bearing on a + ib alone. I'm not very educated- I'm a composition major at a jazz school- but I've spent some time with a couple books on the subject. -Chuckk -- "It is not when truth is dirty, but when it is shallow, that the lover of knowledge is reluctant to step into its waters." -Friedrich Nietzsche, "Thus Spoke Zarathustra" > Date: Mon, 07 Nov 2005 23:47:58 +0100 > From: Piotr Majdak <piotr(a)majdak.com> > Subject: Re: [PD] basic DSP stuff > Cc: pd-list(a)iem.at > Message-ID: <436FD99E.7000504(a)majdak.com> > Content-Type: text/plain; charset=ISO-8859-1; format=flowed > > Chuckk Hubbard wrote: > > Not that I don't appreciate the snide commentary, but this is why I'm > > asking. > > If I offended you, please forgive me. You asked a simple question and I > tried to answer, as simple as possible (I don't know your level of > education). > > Back to your question, you asked: > > > So why, when you multiply Z1 and Z2, do i*sin(a) and i*sin(b) > multiply to -sin(a)sin(b)? > > You see, you wrote an "i" there. If you define "i" by "sqrt(-1)" (I > admit I was implying that), then my answer: > > i*sin(a) * i*sin(b) = -1 * sin(a)*sin(b) > > is correct. And, as you see, "i" is there :-) > > But, if you wanted to discuss the fact, that "i" has no physical meaning > - in this case I misunderstood your question. This is complete another > issue, much more philosophical than mathematical and thus outside my > education focus. > > br, Piotr > > > >

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Re: PD-list Digest, Vol 8, Issue 34
by Chuckk Hubbard 08 Nov '05

08 Nov '05

No sweat. I guess I could elaborate on the specific fudging I see: When multiplying Z1 and Z2, you take r1*r2*(cos x + i sin x)(cos y + i sin y) = r1r2((cos x cos y) + i(cos x sin y) + i(cos y sin x) + i*i*(sin x sin y)) = r1r2((cos x cos y) - (sin x sin y) + i((cos x sin y) + cos y sin x))) (there's where I see the fudging, in making the sines negative) = r1r2(cos (x+y) + i sin (x+y)) Now I'm seeing something. Ah ha. (cos x cos y) = (-cos x * cos y) = -(cos x cos y) and -(cos x cos y) + (sin x sin y) = -(cos(x+y)) = cos (x+y) So the negative is a red herring since cosine is an even function? (I use x and y not as polar terms, but because I have no theta key) Crazy. I wonder if that's right. I wasn't thinking philosophically, but turning an imaginary number into a real number seems questionable when the only reason for using an imaginary number is to keep it separated from its real counterpart. The fact that i^2 = -1 has no bearing on a + ib alone. I'm not very educated- I'm a composition major at a jazz school- but I've spent some time with a couple books on the subject. -Chuckk -- "It is not when truth is dirty, but when it is shallow, that the lover of knowledge is reluctant to step into its waters." -Friedrich Nietzsche, "Thus Spoke Zarathustra" Date: Mon, 07 Nov 2005 23:47:58 +0100 > From: Piotr Majdak <piotr(a)majdak.com> > Subject: Re: [PD] basic DSP stuff > Cc: pd-list(a)iem.at > Message-ID: <436FD99E.7000504(a)majdak.com> > Content-Type: text/plain; charset=ISO-8859-1; format=flowed > > Chuckk Hubbard wrote: > > Not that I don't appreciate the snide commentary, but this is why I'm > > asking. > > If I offended you, please forgive me. You asked a simple question and I > tried to answer, as simple as possible (I don't know your level of > education). > > Back to your question, you asked: > > > So why, when you multiply Z1 and Z2, do i*sin(a) and i*sin(b) > multiply to -sin(a)sin(b)? > > You see, you wrote an "i" there. If you define "i" by "sqrt(-1)" (I > admit I was implying that), then my answer: > > i*sin(a) * i*sin(b) = -1 * sin(a)*sin(b) > > is correct. And, as you see, "i" is there :-) > > But, if you wanted to discuss the fact, that "i" has no physical meaning > - in this case I misunderstood your question. This is complete another > issue, much more philosophical than mathematical and thus outside my > education focus. > > br, Piotr > > > >

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biquad~ yet again
by david golightly 08 Nov '05

08 Nov '05

Hey list, I just did a search in the list-archive and, yes, I realize that there was a dicussion on biquad~ back in April, but the thread referred to a link that is now broken. Suffice to say, it seems the programming of this external could be a bit more user-friendly, targeted toward the people who are likely to actually USE pd, not people with advanced degrees in math. (If you fall into both categories, more power to ya). If I could just get some help here, I'd like: f(x, a) = (ff1, ff2, ff3, fb1, fb2) where these are the biquad coefficients for x = notch frequency and a = attenuation While I understand how it's useful to have tools for building filters in any DSP software, it's too bad this isn't also wrapped up as a more straightforward notch filter. Or maybe I'm just being a wuss here. I'm sure there's someone out there who could implement this as an external. Thanks, David - ---=---^--- -^- -=- - - ^------ - -^---^- -- --^^ --=== homepage: http://www.davidgolightly.net

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Re: [PD] basic DSP stuff
by David Powers 08 Nov '05

08 Nov '05

I think you are taking the "imaginary" part too seriously... In terms of material reality, "-1" and "1" are equally imaginary as "i" - their "reality" is dependent on the system of mathematics where they are allowed to combine with other mathematical entities following particular rules. "i" is not a REAL number but it IS a number... and "REAL" in this sense, refers to a specific mathematical definition of the set of real numbers, and not the ontological status in the material world. In other words, I wouldn't worry too much about whether it exists if the math is correct (ie. i becomes -1 when squared) ~David ---------- Original Message ------------- Subject: Re: [PD] basic DSP stuff Date: Mon, 7 Nov 2005 16:48:23 -0500 From: Chuckk Hubbard <badmuthahubbard(a)gmail.com> To: pd-list(a)iem.at Not that I don't appreciate the snide commentary, but this is why I'm asking. You can't forget the sin*sin part. i stands for "imaginary". (slowly: i is "imaginary"). Turning the product of sines into the negative product of sines is imaginary. There is no number that, squared, equals -1. There is no number that, squared, equals -1. It's not really there. Date: Mon, 07 Nov 2005 20:58:36 +0100 > From: Piotr Majdak <piotr(a)majdak.com> > Subject: Re: [PD] basic DSP stuff > Cc: Pure Data List <pd-list(a)iem.at> > Message-ID: <436FB1EC.80908(a)majdak.com> > Content-Type: text/plain; charset=us-ascii; format=flowed > > Chuckk Hubbard wrote: > > i*sin(a) and i*sin(b) multiply to -sin(a)sin(b)? > > let's forget the sin*sin and look at the i*i part only: since i = > sqrt(-1) by definition, i*i is -1 (slowly: i*i= sqrt(-1)*sqrt(-1) = > sqrt(-1 * -1) = sqrt((-1)^2) = -1) > > Thus, i*sin(a) * i*sin(b) = -1 * sin(a)*sin(b) > > br, piotr > > > > ------------------------------ > -- ~David Powers

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Pd tutorials!
by Hans-Christoph Steiner 07 Nov '05

07 Nov '05

So I just added some more sketches for the Pd intro tutorials. They are attached. If anyone feels like trying to capture core concepts of Pd in clever little patches, please post them here so they can be gathered into a coherent tutorial .hc ________________________________________________________________________ ____ "I have the audacity to believe that peoples everywhere can have three meals a day for their bodies, education and culture for their minds, and dignity, equality and freedom for their spirits." - Martin Luther King, Jr.

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guitarplug crashes PD
by Benjamin Brown 07 Nov '05

07 Nov '05

I bought a usb audio plug made by MacMice called the guitar plug. It is a 48k 16bit device and when I have it pluged in and I load PD , Pd crashes at startup. I can get Pd to stay open if in startup I specify the bitrate, however this still doesnt not allow any input from the audio interface. Any help would be nice, running panther 10.3.9 on dual g4 1gigram Ben Brown

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gui problem in pd-0.39-1test1
by Ales Zemene 07 Nov '05

07 Nov '05

hi, objects like pyext from py are created, but invisible. they work, you can wire them, but dont see. dunno if it is onother version than 0.39-1test1 the same, but THIS is now on download pd.iem.at [Greetings~] Ales Zemene --

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.pdrc in osx
by Ed Kelly 07 Nov '05

07 Nov '05

Hi all, I've developed an application for PD and I want to be able to package the whole thing as an installer, so that non-PD people (like my boss) can try out the patch straight out of the box. The only problem I have is that the .pdrc file seems not to have any effect on pd 0.39 in osx, i.e. the libraries are not loaded at startup. Is there another way to include the information on which libraries to load etc as a file, so that paths are configured by default? We all have os 10.3.9 here, so Hans' installer doesn't work, and I have custom externals (which I will release to cvs soon ;-) Best, Ed -> -> --> ---> -----> --------> -------------> r3search + praktik EK5perimenz ___________________________________________________________ To help you stay safe and secure online, we've developed the all new Yahoo! Security Centre. http://uk.security.yahoo.com

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Re: .pdrc on osx
by Ed Kelly 07 Nov '05

07 Nov '05

Apologies, I found out how to do it by searching the list! Whoops, Ed -> -> --> ---> -----> --------> -------------> r3search + praktik EK5perimenz ___________________________________________________________ How much free photo storage do you get? Store your holiday snaps for FREE with Yahoo! Photos http://uk.photos.yahoo.com

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RME - Hemmerfall Multiface Digiface Linux
by zack 07 Nov '05

07 Nov '05

Quick question to the well-seasoned: Can the Hammerfall Multiface / Digiface run multichannel under Linux without JACK? If so, what's needed: ALSA? OSS? Which Linux (debian, ubunto etc.?) thanks, Zack

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Pd-list November 2005