[PD] basic DSP stuff

Chuckk Hubbard badmuthahubbard at gmail.com
Tue Nov 8 02:30:13 CET 2005

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

= 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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