On Jul 3, 4:19 pm, John Fields wrote:
On Tue, 03 Jul 2007 12:05:52 -0700, Keith Dysart





wrote:
On Jul 3, 2:07 pm, Keith Dysart wrote:
On Jul 3, 12:50 pm, John Fields wrote:

On Mon, 2 Jul 2007 23:03:36 -0700, "Ron Baker, Pluralitas!"

wrote:

"John Smith I" wrote in message
...
Radium wrote:

snip

Suppose you have a 1 MHz sine wave whose amplitude
is multiplied by a 0.1 MHz sine wave.
What would it look like on an oscilloscope?

snip

What would it look like on a spectrum analyzer?

| |
| | | |
--------+--------------------+-------+------+----
100kHz 0.9MHz 1MHz 1.1MHz

Then suppose you have a 1.1 MHz sine wave added
to a 0.9 MHz sine wave.
What would that look like on an oscilloscope?

snip

Tricky!!!

It looks like AM but it isn't, it's just the phases sliding past
each other slowly and algebraically adding which creates the
illusion.

What would that look like on a spectrum analyzer?

| |
| |
-----------------------------+--------------+----
0.9MHz 1.1MHz

--
JF

But if you remove the half volt bias you put on the
100 kHz signal in the multiplier version, the results
look exactly like the summed version, so I suggest
that results are the same when a 4 quadrant multiplier
is used.

And since the original request was for a "1 MHz sine
wave whose amplitude is multiplied by a 0.1 MHz sine
wave" I think a 4 quadrant multiplier is in order.

...Keith-

Ooops. I misspoke. They are not quite the same.

---
That's right. They can't possibly be because the first instance
_was_ multiplication and the second instance addition.

Quite counter intuitive, I agree, but none-the-less true.
To convince myself, I once created an Excel spreadsheet
to demonstrate the fact.

It along with some other discussion and plots are available
here http://keith.dysart.googlepages.com/radio5

The spectrum is the same, but if you want to get exactly
the same result, the lower frequency needs a 90 degree
offset and the upper frequency needs a -90 degree offset.

---
That makes no sense since the frequencies are different and,
consequently, the phase difference between the signals will be
constantly changing.

To get exactly the same results, if, at time t0, the phases
for the signals being multiplied together are 0, then at
time t0, the initial phases for the signals being added
must be 90 and -90.

....Keith