[PD] Re: PD-list Digest, Vol 8, Issue 34
badmuthahubbard at gmail.com
Tue Nov 8 02:19:23 CET 2005
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
= 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)
-(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.
"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 at majdak.com>
> Subject: Re: [PD] basic DSP stuff
> Cc: pd-list at iem.at
> Message-ID: <436FD99E.7000504 at 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
> 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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