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@majdak.com> Subject: Re: [PD] basic DSP stuff Cc: pd-list@iem.at Message-ID: 436FD99E.7000504@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
issue, much more philosophical than mathematical and thus outside my education focus.
br, Piotr
Hallo, Chuckk Hubbard hat gesagt: // Chuckk Hubbard wrote:
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.
It's not right, because: cos(x) != -cos(x). You probably intended to use cos(x) == cos(-x), but that's different thing.
Frank Barknecht _ ______footils.org_ __goto10.org__