On Jul 15, 3:12 pm, John Fields 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 we 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

On Sun, 15 Jul 2007 14:57:17 -0700, Keith Dysart
wrote:

On Jul 15, 3:12 pm, John Fields 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 we 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

On Jul 16, 11:31 am, John Fields
wrote:
On Sun, 15 Jul 2007 14:57:17 -0700, Keith Dysart
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 we 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

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TEXT -1592 -560 Left 0 !.tran 0 .02 0 .3e-7

In article .com,
Keith Dysart wrote:

On Jul 16, 11:31 am, John Fields
wrote:
On Sun, 15 Jul 2007 14:57:17 -0700, Keith Dysart
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 wrote:
In article .com,
Keith Dysart 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 more, so that the sum and
product versions can be directly compared).

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.

I am not convinced. Can you explain? AM modulation with a single
frequency produces a single sum and difference for the sidebands
while FM has an infinite number of frequencies in the sidebands.
This does not seem like a simple phase difference.

....Keith

On Mon, 16 Jul 2007 16:00:29 -0700, Keith Dysart
wrote:

On Jul 16, 11:31 am, John Fields
wrote:
On Sun, 15 Jul 2007 14:57:17 -0700, Keith Dysart
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 we 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.

---
LOL... Except for scaling, Mr. President, the mirror on the Hubble
would have worked the first time out.
---

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).

---
That's easy enough to make happen, but that's not what this is
about; it's about the out signal from a 3 input adder being
identical to the output signal from a mixer.

It seems, you want to try to prove that multiplication is tha same
as addition, with:

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TEXT -1592 -560 Left 0 !.tran 0 .02 0 .3e-7

---
But, using the adder, if you change the amplitude and/or the
frequency of any input what happens to the output? The spectral
line corresponding to the changed input will change, but the lines
corresponding to the other inputs will not.

Use the multiplier and what happens if you change either the carrier
or the LO? The output lines will _all_ change even though only one
input has been changed.

On top of that, even if the output spectrum of the adder is made to
look like the output spectrum of the mixer, the phase relationship
between the outputs of the two will be different unless
extraordinary care is taken to make them the same and, even then, if
information is impressed on the adder's center frequency it will not
convey to the "sidebands".


--
JF