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Old July 4th 07, 03:52 PM posted to sci.electronics.basics,rec.radio.shortwave,rec.radio.amateur.antenna,alt.cellular.cingular,alt.internet.wireless
Ron Baker, Pluralitas![_2_] Ron Baker,    Pluralitas![_2_] is offline
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First recorded activity by RadioBanter: May 2007
Posts: 92
Default AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency


"Keith Dysart" wrote in message
ps.com...
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.

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.

And the amplitudes of the the sum and difference
frequencies need to be one half of the amplitude of
the frequencies being multiplied.

...Keith


You win.

When I conceived the problem I was thinking
cosines actually. In which case there are no
phase shifts to worry about in the result.

I also forgot the half amplitude factor.

While it might not be obvious, the two cases I
described are basically identical. And this
situation occurs in real life, i.e. in radio signals,
oceanography, and guitar tuning.

It follows from what is taught in high school
geometry.

cos(a) * cos(b) = 0.5 * (cos[a+b] + cos[a-b])

Basically: multiplying two sine waves is
the same as adding the (half amplitude)
sum and difference frequencies.

(For sines it is
sin(a) * sin(b) = 0.5 * (cos[a-b]-cos[a+b])
= 0.5 * (sin[a-b+90degrees] - sin[a+b+90degrees])
= 0.5 * (sin[a-b+90degrees] + sin[a+b-90degrees])
)

--
rb