AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency

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.

>> 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.

>>> (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.

>>> 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. 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.

gr, Hein

"Hein ten Horn" <tenhornRemovE@ThiSraketnet.nl> wrote in message
news:469927cb$0$336$e4fe514c@news.xs4all.nl...
> 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
>
>

"Hein ten Horn" <tenhornRemovE@ThiSraketnet.nl> wrote in message
news:469943cd$0$322$e4fe514c@news.xs4all.nl...
> 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
>
>

On Sat, 14 Jul 2007 23:43:55 +0200, "Hein ten Horn"
<tenhornRemovE@ThiSraketnet.nl> 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

"John Fields" <jfields@austininstruments.com> wrote in message
news:dnii93top5no78hmv9013gutn3jafcra82@4ax.com...
> On Sat, 14 Jul 2007 23:43:55 +0200, "Hein ten Horn"
> <tenhornRemovE@ThiSraketnet.nl> 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 <jfie...@austininstruments.com> 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 d.

> >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

"Hein ten Horn" <tenhornRemovE@ThiSraketnet.nl> wrote in message
news:469a27e2$0$333$e4fe514c@news.xs4all.nl...
> 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)

Only for a single sinusoid.

> 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).

Only for a single sinusoid.
What if x(t) = sin(2pi f1 t) + sin(2pi f2 t)

>
> 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.

Yes.
And so is "Nonsense." And so is your idea of
"the frequency".

> Hard words break no bones, yet deflate creditability.
>
> gr, Hein
>
>

On Sun, 15 Jul 2007 04:14:33 -0700, Keith Dysart
<Keith.Dysart@gmail.com> wrote:

>On Jul 14, 6:31 am, John Fields <jfie...@austininstruments.com> 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 d.
>
>> >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.

---
No, it isn't, since in the additive mode any modulation impressed on
the carrier (1.0e6) will not affect the .9e6 and 1.1e6 in any way
since they're unrelated.
---

>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.

---
It will _look_ identical, but it won't be because there will be
nothing locking the three frequencies together. Moreover, as I
stated earlier, any amplitude changes (modulation) impressed on the
1.0e6 signal won't cause the 0.9e6 and 1.1e6 signals to change in
any way.
---

>(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.)

---
Actually the resistors "mess up the signal" more than the opamp does
since the signals aren't really adding in the resistors. That is,
if f1 is at 1V, and f2 is at 1V, and f3 is also at 1V, the output
of the resistor network won't be at 3V, it'll be at 1V. By using
the opamp as a current-to-voltage converter, all the input signals
_will_ be added properly since the inverting input will be at
virtual ground and will sink all the current supplied by the
resistors, making sure the sources don't interact.

Here:

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SYMATTR InstName R2
SYMATTR Value 1000
SYMBOL Opamps\\UniversalOpamp 352 176 R0
SYMATTR InstName U1
SYMBOL res 416 48 R90
WINDOW 0 -36 60 VBottom 0
WINDOW 3 -30 62 VTop 0
SYMATTR InstName R3
SYMATTR Value 1000
SYMBOL voltage 448 208 R0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR Value 12
SYMATTR InstName V3
SYMBOL voltage 352 320 R180
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR Value 12
SYMATTR InstName V4
SYMBOL res 256 48 R90
WINDOW 0 -28 61 VBottom 0
WINDOW 3 -30 62 VTop 0
SYMATTR InstName R6
SYMATTR Value 1000
SYMBOL res 256 -64 R90
WINDOW 0 -32 59 VBottom 0
WINDOW 3 -30 62 VTop 0
SYMATTR InstName R7
SYMATTR Value 1000
SYMBOL voltage 32 192 R0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR Value SINE(0 1 1000)
SYMATTR InstName V5
TEXT -64 344 Left 0 !.tran .02


--
JF

John Fields wrote:
> 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?

Roughly like the ones in your Excel(lent) plots.

>>>> 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"?

In this thread.

> I'm writing from sci.electronics.basics

Subscribing to that group would be a good
thing to do, I suspect.

> 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.

In general: that sum times a constant factor.
Perhaps the factor being one is usually tacitly assumed.

> 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.

From physics (and my good old radio hobby)
I'm familiar with the phenomenon. The meanwhile
cleared using of the word non-linear in a narrower
sense made me sometimes too careful, I guess.

>> 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?

The continuous thread was clear to me.

Thanks.

gr, Hein

Ron Baker, Pluralitas! wrote:
> "Hein ten Horn" wrote:
>> 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)
>
> Only for a single sinusoid.
>
>> 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).
>
> Only for a single sinusoid.
> What if x(t) = sin(2pi f1 t) + sin(2pi f2 t)

In the following passage I wrote "a relatively
slow varying amplitude", which relates to the
4 Hz beat in the case under discussion (f1 =
220 Hz and f2 = 224 Hz) where your
expression evaluates to
x(t) = 2 * cos(2pi 2 t) * sin(2pi 222 t),
indicating the matter is vibrating at 222 Hz.

>> 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.
>
> Yes.
> And so is "Nonsense." And so is your idea of
> "the frequency".

Note the piquant difference: nonsense points
to content and we're not discussing idiots
(despite a passing by of some very strange
postings. ).

>> Hard words break no bones, yet deflate creditability.

gr, Hein

On Jul 15, 3:12 pm, John Fields <jfie...@austininstruments.com> wrote:
> On Sun, 15 Jul 2007 04:14:33 -0700, Keith Dysart
> >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.
>
> ---
> No, it isn't, since in the additive mode any modulation impressed on
> the carrier (1.0e6) will not affect the .9e6 and 1.1e6 in any way
> since they're unrelated.
> ---

I thought the experiment being discussed was one where the
modulation was 1e5, the carrier 1e6 and the resulting
spectrum .9e6, 1e6 and 1.1e6.

Read my comments in that context, or just ignore them if
that context is not of interst.

> >(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.)
>
> ---
> Actually the resistors "mess up the signal" more than the opamp does
> since the signals aren't really adding in the resistors.

I did not write clearly enough. The three resistors I had
in mind were: one to each voltage source and one to ground.

To get there from your latest schematic, discard the op-amp
and tie the right end of R3 to ground.

To get an AM signal that can be decoded with an envelope
detector, V5 needs to have an amplitude of at least 2 volts.

....Keith

"Hein ten Horn" <tenhornRemovE@ThiSraketnet.nl> wrote in message
news:469a9790$1$339$e4fe514c@news.xs4all.nl...
> Ron Baker, Pluralitas! wrote:
>> "Hein ten Horn" wrote:
>>> 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)
>>
>> Only for a single sinusoid.
>>
>>> 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).
>>
>> Only for a single sinusoid.
>> What if x(t) = sin(2pi f1 t) + sin(2pi f2 t)
>
> In the following passage I wrote "a relatively
> slow varying amplitude", which relates to the
> 4 Hz beat in the case under discussion (f1 =
> 220 Hz and f2 = 224 Hz) where your
> expression evaluates to
> x(t) = 2 * cos(2pi 2 t) * sin(2pi 222 t),
> indicating the matter is vibrating at 222 Hz.

So where did you apply the laws of physics?
You said, "It's just a matter of applying the laws of
physics." Then you did that for the single sine case. Where
is your physics calculation for the two sine case?
Where is the expression for 'f' as in your first
example? Put x(t) = 2 * cos(2pi 2 t) * sin(2pi 222 t)
in your calculations and tell me what you get
for 'f'.

And how do you get 222 Hz out of
cos(2pi 2 t) * sin(2pi 222 t)
Why don't you say it is 2 Hz? What is your
law of physics here? Always pick the bigger
number? Always pick the frequency of the
second term? Always pick the frequency of
the sine?
What is "the frequency" of
cos(2pi 410 t) * cos(2pi 400 t)


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


>
>>> So my statements above, in which we have
>>> a relatively slow varying amplitude (4 Hz),

How do you determine amplitude?
What's the math (or physics) to derive
amplitude?

>>> are fundamentally spoken valid.
>>> Calling someone an idiot is a weak scientific argument.
>>
>> Yes.
>> And so is "Nonsense." And so is your idea of
>> "the frequency".
>
> Note the piquant difference: nonsense points
> to content and we're not discussing idiots
> (despite a passing by of some very strange
> postings. ).
>
>>> Hard words break no bones, yet deflate creditability.
>
> gr, Hein
>
>

On Sun, 15 Jul 2007 23:49:04 +0200, "Hein ten Horn"
<tenhornRemovE@ThiSraketnet.nl> wrote:

>John Fields wrote:

>> And what does it look like, then?
>
>Roughly like the ones in your Excel(lent) plots.

---
I've posted nothing like that, so if you have graphics which support
your position I'm sure we'd all be happy to see them.
--

>>> Mathematical terms like linear, logarithmic, etc. are familiar
>>> to me, but the guys here use linear and nonlinear in another
>>> sense.
>>
>> Where is "here"?
>
>In this thread.
>
>> I'm writing from sci.electronics.basics
>
>Subscribing to that group would be a good
>thing to do, I suspect.
>
>> 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.
>
>In general: that sum times a constant factor.
>Perhaps the factor being one is usually tacitly assumed.

---
That's not right.

The output of the system will be the input signal multiplied by the
gain of the system, with the offset added to that product.
---

>> 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.
>
>From physics (and my good old radio hobby)
>I'm familiar with the phenomenon. The meanwhile
>cleared using of the word non-linear in a narrower
>sense made me sometimes too careful, I guess.

---
OK, I guess...
---

>>> 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?
>
>The continuous thread was clear to me.
>
>Thanks.

---
:-)


--
JF

On Sun, 15 Jul 2007 14:57:17 -0700, Keith Dysart
<Keith.Dysart@gmail.com> wrote:

>On Jul 15, 3:12 pm, John Fields <jfie...@austininstruments.com> wrote:
>> On Sun, 15 Jul 2007 04:14:33 -0700, Keith Dysart
>> >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.
>>
>> ---
>> No, it isn't, since in the additive mode any modulation impressed on
>> the carrier (1.0e6) will not affect the .9e6 and 1.1e6 in any way
>> since they're unrelated.
>> ---
>
>I thought the experiment being discussed was one where the
>modulation was 1e5, the carrier 1e6 and the resulting
>spectrum .9e6, 1e6 and 1.1e6.

---
That was my understanding, and is why I was surprised when you made
the claim, above:

"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."

which I interpret to mean that three unrelated signals occupying
those spectral positions were identical to three signals occupying
the same spectral locations, but which were created by heterodyning.

Are you now saying that wasn't your claim?
---

>Read my comments in that context, or just ignore them if
>that context is not of interst.

---
What I'd prefer to do is point out that if your comments were based
on the concept that the signals obtained by mixing are identical to
those obtained by adding, then the concept is flawed.
---

>> >(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.)
>>
>> ---
>> Actually the resistors "mess up the signal" more than the opamp does
>> since the signals aren't really adding in the resistors.
>
>I did not write clearly enough. The three resistors I had
>in mind were: one to each voltage source and one to ground.
>
>To get there from your latest schematic, discard the op-amp
>and tie the right end of R3 to ground.

That really doesn't change anything, since no real addition will be
occurring. Consider:


f1>---[1000R]--+-->E2
|
f2>---[1000R]--+
|
f3>---[1000R]--+
|
[1000R]
|
GND>-----------+

Assume that f1, f2, and f3 are 2VPP signals and that we have sampled
the signal at E2 at the instant when they're all at their positive
peak.

Since the resistors are essentially in parallel, the circuit can be
simplified to:


E1
|
[333R] R1
|
+---->E2
|
[1000R] R2
|
GND

a simple voltage divider, and E2 can be found via:

E1 * R2 1V * 1000R
E2 = --------- = -------------- = 0.75V
R1 + R2 333R + 1000R

Note that 0.75V is not equal to 1V + 1V + 1V.
---

>To get an AM signal that can be decoded with an envelope
>detector, V5 needs to have an amplitude of at least 2 volts.

---
Ever heard of galena? Or selenium? Or a precision rectifier?


--
JF

Hein ten Horn wrote:

> That's a misunderstanding.
> 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.
> 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.

Under the stated conditions there is no sine wave oscillating at 222
Hz. The wave has a complex shape and contains spectral components at
two distinct frequencies (neither of which is 222Hz).

>>It might be correct to say that matter is vibrating at an
>>average, or effective frequency of 222 Hz.
>
>
> No, it is correct. A particle cannot follow two different
> harmonic oscillations (220 Hz and 224 Hz) at the same
> time.

The particle also does not average the two frequencies. The waveform
which results from the sum of two pure sine waves is not a pure sine
wave, and therefore cannot be accurately described at any single
frequency.

>>Obviously. It's a very simple matter to verify this by experiment.
>
>
> Indeed, it is. But watch out for misinterpretations of
> the measuring results! For example, if a spectrum
> analyzer, being fed with the 222 Hz signal, shows
> that the signal can be composed from a 220 Hz and
> a 224 Hz signal, then that won't mean the matter is
> actually vibrating at those frequencies.

:-) Matter would move in the same way the sound pressure wave does,
the amplitude of which is easily plotted versus time using
Mathematica, Mathcad, Sigma Plot, and even Excel. I think you should
still give that a try.

jk

John Fields wrote:

> On Fri, 06 Jul 2007 19:04:00 -0000, Jim Kelley <jwkelley@uci.edu>
> wrote:

> In your example, with 300Hz and 400Hz as the carriers, the sidebands
> would be located at:
>
> f3 = f1 + f2 = 300Hz + 400Hz = 700Hz
>
> and
>
> f4 = f2 - f1 = 400Hz - 300Hz = 100Hz
>
>
> both of which are clearly within the range of frequencies to which
> the human ear responds.

Indeed. We would hear f3 and f4 if they were in fact there.

> Your use of the term "beat frequency" is confusing since it's
> usually used to describe the products of heterodyning, not the
> audible warble caused by the vector addition of signals close to
> unison.

The term is commonly used in describing the results of interference in
time, as well as for mixing.

> Since the response of the ear is non-linear in amplitude it has no
> choice _but_ to be a mixer and create sidebands.

Perhaps you're confusing log(sin(a)+sin(b)) with
log(sin(a))+log(sin(b)).

If you don't mind me asking, where did you get this notion about the
ears creating sidebands?

jk

On Jul 16, 11:31 am, John Fields <jfie...@austininstruments.com>
wrote:
> On Sun, 15 Jul 2007 14:57:17 -0700, Keith Dysart
> <Keith.Dys...@gmail.com> wrote:
> >I thought the experiment being discussed was one where the
> >modulation was 1e5, the carrier 1e6 and the resulting
> >spectrum .9e6, 1e6 and 1.1e6.
>
> ---
> That was my understanding, and is why I was surprised when you made
> the claim, above:
>
> "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."
>
> which I interpret to mean that three unrelated signals occupying
> those spectral positions were identical to three signals occupying
> the same spectral locations, but which were created by heterodyning.
>
> Are you now saying that wasn't your claim?
> ---

No, that was indeed the claim. As a demonstration, I've
attached a variant of your original LTspice simulation.
Plot Vprod and Vsum. They are on top of each other.
Plot the FFT for each. They are indistinguishable.

> >Read my comments in that context, or just ignore them if
> >that context is not of interst.
>
> ---
> What I'd prefer to do is point out that if your comments were based
> on the concept that the signals obtained by mixing are identical to
> those obtained by adding, then the concept is flawed.

See the simulation results.

> >I did not write clearly enough. The three resistors I had
> >in mind were: one to each voltage source and one to ground.
>
> >To get there from your latest schematic, discard the op-amp
> >and tie the right end of R3 to ground.
>
> That really doesn't change anything, since no real addition will be
> occurring. Consider:
>
> f1>---[1000R]--+-->E2
> |
> f2>---[1000R]--+
> |
> f3>---[1000R]--+
> |
> [1000R]
> |
> GND>-----------+
<snip>
>
> Note that 0.75V is not equal to 1V + 1V + 1V.

E2 = (V1+V2+V3)/4 -- a scaled sum

Except for scaling, the result is the sum of the inputs.

> >To get an AM signal that can be decoded with an envelope
> >detector, V5 needs to have an amplitude of at least 2 volts.
>
> ---
> Ever heard of galena? Or selenium? Or a precision rectifier?

Oh, yes. And cat whiskers too.

But that was not my point. Because the carrier level was not
high enough, the envelope was no longer a replica of the signal
so an envelope detector would not be able to recover the signal
(no matter how sensitive it was).

....Keith

Version 4
SHEET 1 980 680
WIRE -1312 -512 -1552 -512
WIRE -1200 -512 -1232 -512
WIRE -1552 -496 -1552 -512
WIRE -1312 -400 -1440 -400
WIRE -1200 -400 -1200 -512
WIRE -1200 -400 -1232 -400
WIRE -768 -384 -976 -384
WIRE -976 -368 -976 -384
WIRE -1440 -352 -1440 -400
WIRE -544 -352 -624 -352
WIRE -1200 -336 -1200 -400
WIRE -1136 -336 -1200 -336
WIRE -544 -336 -544 -352
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WIRE -544 -224 -544 -240
WIRE -1552 -144 -1552 -416
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WIRE -1200 -144 -1344 -144
WIRE -1552 -128 -1552 -144
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WIRE -544 -128 -544 -144
FLAG -1552 -128 0
FLAG -1136 -336 Vsum
FLAG -976 -368 0
FLAG -912 -128 0
FLAG -544 -128 0
FLAG -464 -240 Vprod
SYMBOL voltage -1552 -512 R0
WINDOW 3 -216 102 Left 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR Value SINE(0 .5 900 0 0 90)
SYMATTR InstName Vs1
SYMBOL voltage -1344 -272 R0
WINDOW 3 -228 104 Left 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR Value SINE(0 .5 1100 0 0 -90)
SYMATTR InstName Vs3
SYMBOL res -1216 -320 R90
WINDOW 0 -26 57 VBottom 0
WINDOW 3 -25 58 VTop 0
SYMATTR InstName Rs3
SYMATTR Value 1000
SYMBOL res -1184 -192 R180
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SYMATTR InstName Rs4
SYMATTR Value 1000
SYMBOL res -1216 -416 R90
WINDOW 0 -28 61 VBottom 0
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SYMATTR InstName Rs2
SYMATTR Value 1000
SYMBOL res -1216 -528 R90
WINDOW 0 -32 59 VBottom 0
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SYMATTR Value 1000
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WINDOW 3 -210 108 Left 0
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SYMATTR Value SINE(0 1 1000 0 0 0)
SYMATTR InstName Vs2
SYMBOL SpecialFunctions\\modulate -768 -384 R0
WINDOW 3 -66 -80 Left 0
SYMATTR InstName A1
SYMATTR Value space=1000 mark=1000
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In article <1184626829.678755.104800@q75g2000hsh.googlegroups .com>,
Keith Dysart <Keith.Dysart@gmail.com> wrote:

> On Jul 16, 11:31 am, John Fields <jfie...@austininstruments.com>
> wrote:
> > On Sun, 15 Jul 2007 14:57:17 -0700, Keith Dysart
> > <Keith.Dys...@gmail.com> wrote:
> > >I thought the experiment being discussed was one where the
> > >modulation was 1e5, the carrier 1e6 and the resulting
> > >spectrum .9e6, 1e6 and 1.1e6.
> >
> > ---
> > That was my understanding, and is why I was surprised when you made
> > the claim, above:
> >
> > "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."
> >
> > which I interpret to mean that three unrelated signals occupying
> > those spectral positions were identical to three signals occupying
> > the same spectral locations, but which were created by heterodyning.
> >
> > Are you now saying that wasn't your claim?
> > ---
>
> No, that was indeed the claim. As a demonstration, I've
> attached a variant of your original LTspice simulation.
> Plot Vprod and Vsum. They are on top of each other.
> Plot the FFT for each. They are indistinguishable.

-- lots o' snipping goin' on --

OK. I haven't been (had the patience to keep on) following this
discussion, so I apologize if this is totally inappropriate, but

If the statements above refer to creating that set of signals by using a
bunch of signal generators, or alternately by using some sort of actual
"modulation", the answer is, there is a very significant difference.

In the case where the set is created by modulating the "carrier" with
the low frequency, there is a very specific phase relationship between
the signals which would be essentially impossible to achieve if the
signals were to be generated independently. In fact, the only difference
between AM and FM/PM is that the phase relationship between the carrier
and the sideband set differs by 90 degrees between the two.

Isaac

On Jul 16, 11:30 pm, isw <i...@witzend.com> wrote:
> In article <1184626829.678755.104...@q75g2000hsh.googlegroups .com>,
> Keith Dysart <Keith.Dys...@gmail.com> wrote:
> > No, that was indeed the claim. As a demonstration, I've
> > attached a variant of your original LTspice simulation.
> > Plot Vprod and Vsum. They are on top of each other.
> > Plot the FFT for each. They are indistinguishable.
>
> -- lots o' snipping goin' on --
>
> OK. I haven't been (had the patience to keep on) following this
> discussion, so I apologize if this is totally inappropriate, but
>
> If the statements above refer to creating that set of signals by using a
> bunch of signal generators, or alternately by using some sort of actual
> "modulation", the answer is, there is a very significant difference.
>
> In the case where the set is created by modulating the "carrier" with
> the low frequency, there is a very specific phase relationship between
> the signals which would be essentially impossible to achieve if the
> signals were to be generated independently.

All true. The simulation offered previously achieves the
required phase relationship (and