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

[John Fields]

---
Yes, of course.

The cilia in the cochlea are different lengths and, consequently,
"tuned" to different frequencies to which they respond by undulating
and sending electrical signals to the brain when the nerves to which
they're connected fire. See:

http://en.wikipedia.org/wiki/Organ_of_Corti

 

[Hein ten Horn]

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.

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.

Yep.

Sure (but nature doesn't mind).

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

 

[Ron Baker, Pluralitas!]

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).


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)

 

[Ron Baker, Pluralitas!]

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.

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))

Sure (but nature doesn't mind).


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.)

 

[John Fields]

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?
---

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.

 

[J. Mc Laughlin]

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


---
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.

 

[Keith Dysart]

On Thu, 05 Jul 2007 07:32:20 -0700, Keith Dysart

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).

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.

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

 

[Hein ten Horn]

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

 

[Ron Baker, Pluralitas!]

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".

 

[John Fields]

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).


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.


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:

Version 4
SHEET 1 980 680
WIRE 160 -48 -80 -48
WIRE 272 -48 240 -48
WIRE 160 64 32 64
WIRE 272 64 272 -48
WIRE 272 64 240 64
WIRE 320 64 272 64
WIRE 528 64 400 64
WIRE 448 112 352 112
WIRE 352 144 352 112
WIRE 160 160 128 160
WIRE 272 160 272 64
WIRE 272 160 240 160
WIRE 320 160 272 160
WIRE 528 176 528 64
WIRE 528 176 384 176
WIRE 320 192 272 192
WIRE -80 208 -80 -48
WIRE 32 208 32 64
WIRE 128 208 128 160
WIRE 352 224 352 208
WIRE 448 224 448 112
WIRE -80 320 -80 288
WIRE 32 320 32 288
WIRE 32 320 -80 320
WIRE 128 320 128 288
WIRE 128 320 32 320
WIRE 272 320 272 192
WIRE 272 320 128 320
WIRE 352 320 352 304
WIRE 352 320 272 320
WIRE 448 320 448 304
WIRE 448 320 352 320
WIRE -80 368 -80 320
FLAG -80 368 0
SYMBOL voltage -80 192 R0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR Value SINE(0 1 900)
SYMATTR InstName V1
SYMBOL voltage 128 192 R0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR Value SINE(0 1 1100)
SYMATTR InstName V2
SYMBOL res 256 144 R90
WINDOW 0 -26 57 VBottom 0
WINDOW 3 -25 58 VTop 0
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

 

[Hein ten Horn]

And what does it look like, then?

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

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.

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

 

[Hein ten Horn]

Only for a single sinusoid.


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.

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. ).

gr, Hein

 

[Keith Dysart]

---
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.

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

 

[Ron Baker, Pluralitas!]

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)


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

 

[John Fields]

John Fields wrote:

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.
--

In this thread.


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


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.
---

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.

 

[John Fields]

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.
---

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.

 

[Jim Kelley]

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).

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.

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

 

[Jim Kelley]

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

 

[Jim Kelley]

Hi Craig -

Yes, it's pretty consistent alright. It gets even more consistent as
$f1 moves closer to $f2. But you'll note that it gets much less
consistent the further the two frequencies are apart. Which is what
I've been saying. You can also see the periodicity I described.

jk

 

[Keith Dysart]

---
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.

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>-----------+

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.

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
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WIRE -1200 -512 -1232 -512
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TEXT -1592 -560 Left 0 !.tran 0 .02 0 .3e-7