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Old July 1st 07, 10:44 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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Default AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency


"Ian Jackson" wrote in message
...
In message , cledus
writes


snip


The fundamental answer is no, it is not possible to generate AM where the
baseband signal is a pure 20 kHz sinewave and Fc20kHz. The reason is
that the modulated waveform consists of the sum of a sinewave at Fc, a
sinewave at Fc+20kHz, and a sinewave at Fc-20kHz. If Fc20kHz then one of
the components becomes a "negative" frequency. So the carrier must be
greater than the baseband signal to prevent this.

I'm afraid that this is not correct. The 'laws of physics' don't suddenly
stop working if the carrier is lower than the modulating frequency.
However, there's no need to get into complicated mathematics to illustrate
this. Here is a simple example:

(a) If you modulate a 10MHz carrier with a 1MHz signal, you will produce
two new signals (the sidebands) at the difference frequency of 10 minus 1
= 9MHz, and the sum frequency of 10 plus 1 = 11MHz. So you have the
original carrier at 10MHz, and sideband signals at 9 and 11MHz (with a
balanced modulator - no carrier - only 9 and 11MHz).

(b) If you modulate a 1MHz carrier with a 10MHz signal, you will produce
two new signals (the sidebands) at the difference frequency of 1 minus 10
= minus 9MHz, and the sum frequency of 1 plus 10 = 11MHz. The implication
of the negative 'minus 9' MHz signal is that the phase of the 9MHz signal
is inverted, ie 180 degrees out-of-phase from 9MHz


Actually there would be no phase flip.
cos(-a) = cos(a)

produced in (a). So you have the original carrier at 1MHz, and sidebands
at 9 and 11MHz (again, with a balanced modulator - no carrier - only 9 and
11MHz).

The waveforms of the full composite AM signals of (a) and (b) will look
quite different. The carriers are at different frequencies, and the phase
of the 9MHz signal is inverted. However, with a double-balanced modulator,
you will only have the 9 and 11MHz signal so, surprisingly, the resulting
signals of (a) and (b) will look the same.


A double-balanced mixer is a multiplier. A * B = B * A


[Note that, in practice, many double-balanced modulators/mixers put loads
of unwanted signals - mainly due the effects of harmonic mixing. However,
the basic 'laws of physics' still apply.]

Finally, although I have spoken with great authority, when I get a chance
I WILL be doing at test with a tobacco-tin double-balanced mixer,


What's a tobacco-tin double-balanced mixer?

a couple of signal generators and a spectrum analyser - just to make sure
that I'm not talking rubbish. In the meantime, I'm sure that some will
correct me if I'm wrong.


You did pretty good.


Ian.
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rb