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Old July 2nd 07, 03:19 AM posted to sci.electronics.basics,rec.radio.shortwave,rec.radio.amateur.antenna,alt.cellular.cingular,alt.internet.wireless
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Default AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-lowcarrier frequency

Ian Jackson wrote:
In message , cledus
writes
Radium wrote:
Hi:
Please don't be annoyed/offended by my question as I decreased the
modulation frequency to where it would actually be realistic.
I have a very weird question about electromagnetic radiation,
carriers, and modulators.





No offense but please respond with reasonable answers & keep out the
jokes, off-topic nonsense, taunts, insults, and trivializations. I am
really interested in this.
Thanks,
Radium



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
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.

[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,
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.

Ian.



Ian,

I believe your analysis is correct. But if you expect to build a
receiver that uses a filter centered at 1 MHz with a BW of 20+ MHz to
recover a DSB AM signal, I don't believe that the DBM approach will
accomplish this. With your approach, you could filter out the sidebands
by centering a filter around 10 MHz (the baseband freq). This could be
used to recover the baseband 10 MHz signal. But the OP asked about AM
of a carrier at very low frequencies. Good explanation of what happens
when using a DBM, though.

Regards,
-C


 
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