Metinks y'all are all messed up on electromagnetics and everythang
else.If y'all cats ever listen to me, y'all might learn something.But, I
wouldn't bet a plugged nicklel on any of y'alls.Easy Rider movie is on
Radio tb right now.Don't bother me!
cuhulin

"Hein ten Horn" wrote in message
...
Ron Baker, Pluralitas! wrote:
Hein ten Horn wrote:
A vibrating element here (such as a cubic micrometre
of matter) experiences different changing forces. Yet
the element cannot follow all of them at the same time.

It does. Not identically but it does follow all of them.

Impossible. Remember, we're talking about sound. Mechanical
forces only. Suppose you're driving, just going round the corner.
From the outside a fistful of forces is working on your body,
downwards, upwards, sidewards. It is absolutely impossible
that your body's centre of gravity is following different forces in
different directions at the same time. Only the resulting force is
changing your movement (according to Newton's second law).

As a matter of fact the resulting force (the resultant) is
fully determining the change of the velocity (vector) of
the element.
The resulting force on our element is changing at the
frequency of 222 Hz, so the matter is vibrating at the
one and only 222 Hz.

Your idea of frequency is informal and leaves out
essential aspects of how physical systems work.

Nonsense. Mechanical oscillations are fully determined by
forces acting on the vibrating mass. Both mass and resulting force
determine the frequency. It's just a matter of applying the laws of
physics.

Let me call you an idiot now and get that out of the way.
You're an idiot.
You don't know the laws of physics or how to
apply them.

How do you determine "the frequency"?
Show me the math.

What is "the frequency" of
cos(2pi 200 t) + cos(2pi 210 t) + cos(2pi 1200 t) + cos(2pi 1207 t)



Question

Is our auditory system in some way acting like a spectrum analyser?
(Is it able to distinguish the composing frequencies from a vibration?)

Ron?
Somebody else?

Thanks

gr, Hein

Already, this thread has gone on about half as long as when somebody in
the pipebombnews (now, pipelinenews.org, or whatever it is) thread a few
years ago said somethings about, Beavers.I sure did contribute my share
of opinionated opinions in them Beaver threads.
cuhulin

"Hein ten Horn" wrote in message
...
Ron Baker, Pluralitas! wrote:
Hein ten Horn wrote:
Ron Baker, Pluralitas! wrote:

How do you arrive at a "beat"?

Not by train, neither by UFO.
Sorry. English, German and French are only 'second'
languages to me.
Are you after the occurrence of a beat?

Another way to phrase the question would have been:
Given a waveform x(t) representing the sound wave
in the air how do you decide whether there is a
beat in it?

Oh, nice question. Well, usually (in my case) the functions
are quite simple (like the ones we're here discussing) so that
I see the beat in a picture of a rough plot in my mind.

Sorry, but your mental images are hardly linear
or authoritative on the laws of physics.


Then: a beat appears at constructive interference, thus
when the cosine function becomes maximal (+1 or -1).
Or are you after the beat frequency (6 Hz)?
Then: the cosine function has two maxima per period
(one being positive, one negative) and with three
periodes a second it makes six beats/second.

Hint: Any such assessment is nonlinear.

Mathematical terms like linear, logarithmic, etc. are familiar
to me, but the guys here use linear and nonlinear in another
sense. Something to do with harmonics or so? Anyway,
that's why the hint isn't working here.

You have a system decribed by a function f( ) with
input x(t) and output y(t).
y(t) = f( x(t) )
It is linear if
a * f( x(t) ) = f( a*x(t) )
and
f( x1(t) ) + f( x2(t) ) = f( x1(t) + x2(t) )

A linear amplifier is
y(t) = f( x(t) ) = K * x(t)
A ("double balanced") mixer is
y(t) = f( x1(t), x2(t) ) = x1(t) * x2(t)
(which is not linear.)


(And kudos to you that you can do the math.)

Simplifying the math:
x = cos(a) * cos(b) = 0.5 * (cos[a+b] + cos[a-b])
(Where a = 2 * pi * f_1 * t and b = same but f_2.)
All three of the above are equivalent. There is no difference.
You get x if you add two sine waves or if you
multiply two (different) sine waves.

??
sin(a) + sin(b) sin(a) * sin(b)
In case you mistyped cosine:
sin(a) + sin(b) cos(a) * cos(b)

It would have been more proper of me to say
"sinusoid" rather than "sine wave". I called
cos() a "sine wave". If you look at cos(2pi f1 t)
on an oscilloscope it looks the same as sin(2pi f t).
In that case there is essentially no difference.
Yes, there are cases where it makes a difference.
But at the beginning of an analysis it is rather
arbitrary and the math is less cluttered with
cos().

Got the (co)sin-stuff. But the unequallities are still there.
It's easy to understand: the left-hand term is sooner or later
greater then one, the right-hand term not (in both unequalities).
As a consequence we've two different x's.

You left out the "0.5".
cos(a) * cos(b) = 0.5 * (cos(a+b) + cos(a-b))



So which is it really? Hint: If all you have is x then
you can't tell how it was generated.

Yep.

What you do with it afterwards can make a
difference.

Sure (but nature doesn't mind).

Referring to the physical system, what's now your point?
That system contains elements vibrating at 25003 Hz.
There's no math in it.

Whoops. You'll need math to understand it.

I would say we need the math to work with it, to get our
things done. Understanding nature is not self-evident the
same thing.

Nature works by laws. We express the laws
in math. Nature is complicated and it may be
difficult to simplify the system one is looking
at or to put together of mathematical model
that is representative of the system but it
is often possible. It is regularly possible to
have a mathematical model of a system that
is accurate to better than 5 decimal places.

I would really appreciate it if you would take
the time to read my UTC 9:57 reply to JK once again,
but then with a more open mind. Thanks.

Hmm. You're definitely not going to like part of it.
(You seem so much more reasonable here.)


gr, Hein

y'alls waves are full of crap!

www.devilfinder.com Mississippi Roads
cuhulin
.................................................. .........
I knowwwwwww a placeeee wherrrre the Riverrrrrrr is
windingggggg,,,,,,,,,, oh, those Roadssssss,,,,,,, Mississippi
Roads,,,,,,,,,, they are windinnnng from the Deltaaaaaaa,,,,,, all the
wayyyy tooooo the Coast,,,,,,,,,
.................................................. .........

Or Guitar Amps.Its been over a year and a month and a half already.But,
I will get blamed for no reason at all.
cuhulin

On Sat, 14 Jul 2007 23:43:55 +0200, "Hein ten Horn"
wrote:

Ron Baker, Pluralitas! wrote:
Hein ten Horn wrote:
Ron Baker, Pluralitas! wrote:

How do you arrive at a "beat"?

Not by train, neither by UFO.
Sorry. English, German and French are only 'second'
languages to me.
Are you after the occurrence of a beat?

Another way to phrase the question would have been:
Given a waveform x(t) representing the sound wave
in the air how do you decide whether there is a
beat in it?

Oh, nice question. Well, usually (in my case) the functions
are quite simple (like the ones we're here discussing) so that
I see the beat in a picture of a rough plot in my mind.

---
And what does it look like, then?
---

Then: a beat appears at constructive interference, thus
when the cosine function becomes maximal (+1 or -1).
Or are you after the beat frequency (6 Hz)?
Then: the cosine function has two maxima per period
(one being positive, one negative) and with three
periodes a second it makes six beats/second.

Hint: Any such assessment is nonlinear.

Mathematical terms like linear, logarithmic, etc. are familiar
to me, but the guys here use linear and nonlinear in another
sense.

---
Where is "here"?

I'm writing from sci.electronics.basics and, classically, a device
with a linear response will provide an output signal change over its
linear dynamic range which varies as a function of an input signal
amplitude change and some system constants and is described by:


Y = mx+b


Where Y is the output of the system, and is the distance traversed
by the output signal along the ordinate of a Cartesian plot,

m is a constant describing the slope (gain) of the system,

x is the input to the system, is the distance traversed by
the input signal along the abscissa of a Cartesian plot, and

b is the DC offset of the output, plotted on the ordinate.

In the context of this thread, then, if a couple of AC signals are
injected into a linear system, which adds them, what will emerge
from the output will be an AC signal which will be the instantaneous
arithmetic sum of the amplitudes of both signals, as time goes by.

As nature would have it, if the system was perfectly linear, the
spectrum of the output would contain only the lines occupied by the
two inputs.

Kinda like if we listened to some perfectly recorded and played back
music...

If the system is non-linear, however, what will appear on the output
will be the AC signals input to the system as well as some new
companions.

Those companions will be new, real frequencies which will be located
spectrally at the sum of the frequencies of the two AC signals and
also at their difference.
---

Something to do with harmonics or so? Anyway,
that's why the hint isn't working here.

---
Harmonics _and_ heterodynes.

If the hint isn't working then you must confess ignorance, yes?



--
JF

On Jul 14, 6:31 am, John Fields wrote:
On Thu, 05 Jul 2007 07:32:20 -0700, Keith Dysart

Since your modulator version has a DC offset applied to
the 1e5 signal, some of the 1e6 signal is present in the
output, so your spectrum has components at .9e6, 1e6 and
1.1e6.

---
Yes, of course, and 1e5 as well.

There is no 1e5 if the modulator is a perfect multiplier. A
practical multiplier will leak a small amount of 1e5.

Don't be fooled by the apparent 1e5 in the FFT from your
simulation. This is an artifact. Run the simulation with
a maximum step size of 0.03e-9 and it will completely
disappear. (Well, -165 dB).

To generate the same signal with the summing version you
need to add in some 1e6 along with the .9e6 and 1.1e6.

---
That wouldn't be the same signal since .9e6 and 1.1e6 wouldn't have
been created by heterodyning and wouldn't be sidebands at all.

It does not matter how the .9e6, 1.0e6 and 1.1e6 are put into
the resulting signal. One can multiply 1e6 by 1e5 with a DC
offset, or one can add .9e6, 1.0e6 and 1.1e6. The resulting
signal is identical.

This can be seen from the mathematical expression
0.5 * (cos(a+b) + cos(a-b)) + cos(a)
= (1 + cos(b)) * cos(a)

Note that cos(b) is not prsent in the spectrum, only a,
a+b and a-b are there. And a will go away if the DC offset
is removed.

The results will be identical and the results of summing
will be quite detectable using an envelope detector just
as they would be from the modulator version.

---
The results would certainly _not_ be identical, since the 0.9e6

To clearly see the equivalency, in the summing version of the
circuit, add in the 1.0e6 signal as well. The resulting
signal will be identical to the one from the multiplier
version.

(You can improve the fidelity of the resulting summed version
by eliminating the op-amp. Just use three resistors. The op-amp
messes up the signal quite a bit.)

If you have access to Excel, you might try the spreadsheet
available here (http://keith.dysart.googlepages.com/radio5).
It plots the results of adding and multiplying, and lets
you play with the frequencies and phases.

....Keith

Ron Baker, Pluralitas! wrote:
Hein ten Horn wrote:
Ron Baker, Pluralitas! wrote:
Hein ten Horn wrote:

As a matter of fact the resulting force (the resultant) is
fully determining the change of the velocity (vector) of
the element.
The resulting force on our element is changing at the
frequency of 222 Hz, so the matter is vibrating at the
one and only 222 Hz.

Your idea of frequency is informal and leaves out
essential aspects of how physical systems work.

Nonsense. Mechanical oscillations are fully determined by
forces acting on the vibrating mass. Both mass and resulting force
determine the frequency. It's just a matter of applying the laws of
physics.

You don't know the laws of physics or how to apply them.

I'm not understood. So, back to basics.
Take a simple harmonic oscillation of a mass m, then
x(t) = A*sin(2*pi*f*t)
v(t) = d(x(t))/dt = 2*pi*f*A*cos(2*pi*f*t)
a(t) = d(v(t))/dt = -(2*pi*f)^2*A*sin(2*pi*f*t)
hence
a(t) = -(2*pi*f)^2*x(t)
and, applying Newton's second law,
Fres(t) = -m*(2*pi*f)^2*x(t)
or
f = ( -Fres(t) / m / x(t) )^0.5 / (2pi).

So my statements above, in which we have
a relatively slow varying amplitude (4 Hz),
are fundamentally spoken valid.
Calling someone an idiot is a weak scientific argument.
Hard words break no bones, yet deflate creditability.

gr, Hein

Back to basics,,,, Turn your Radio on and listen to whatever you can
pick up on there.I was in an Irish computer newsgroup a short while ago
and an Irish guy in Ireland said his wife was reading my stuff and she
likes me.
cuhulin