On Sun, 15 Jul 2007 04:14:33 -0700, Keith Dysart
wrote:

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

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

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

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

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

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

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

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

---
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.

He

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TEXT -64 344 Left 0 !.tran .02


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JF