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:

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
WIRE -768 -320 -912 -320
WIRE -1312 -304 -1344 -304
WIRE -1200 -304 -1200 -336
WIRE -1200 -304 -1232 -304
WIRE -1200 -288 -1200 -304
WIRE -1344 -256 -1344 -304
WIRE -912 -256 -912 -320
WIRE -544 -240 -544 -256
WIRE -464 -240 -544 -240
WIRE -544 -224 -544 -240
WIRE -1552 -144 -1552 -416
WIRE -1440 -144 -1440 -272
WIRE -1440 -144 -1552 -144
WIRE -1344 -144 -1344 -176
WIRE -1344 -144 -1440 -144
WIRE -1200 -144 -1200 -208
WIRE -1200 -144 -1344 -144
WIRE -1552 -128 -1552 -144
WIRE -912 -128 -912 -176
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
WINDOW 0 -48 76 Left 0
WINDOW 3 -52 34 Left 0
SYMATTR InstName Rs4
SYMATTR Value 1000
SYMBOL res -1216 -416 R90
WINDOW 0 -28 61 VBottom 0
WINDOW 3 -30 62 VTop 0
SYMATTR InstName Rs2
SYMATTR Value 1000
SYMBOL res -1216 -528 R90
WINDOW 0 -32 59 VBottom 0
WINDOW 3 -30 62 VTop 0
SYMATTR InstName Rs1
SYMATTR Value 1000
SYMBOL voltage -1440 -368 R0
WINDOW 3 -210 108 Left 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
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
SYMBOL voltage -912 -272 R0
WINDOW 3 14 106 Left 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName Vp1
SYMATTR Value SINE(1 1 100)
SYMBOL res -560 -240 R0
SYMATTR InstName Rp2
SYMATTR Value 1000
SYMBOL res -560 -352 R0
SYMATTR InstName Rp1
SYMATTR Value 3000
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