"Ian Jackson" <IanJacksonRemoveThisBit@g3ohx.demon.co.uk> wrote in message
news:3s9WZMIfx2hGFwvm@g3ohx.demon.co.uk...
> In message <EXChi.23350$C96.1422@newssvr23.news.prodigy.net >, cledus
> <cledus@noemail.net> 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 Fc<20kHz. 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 Fc<20kHz 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.
> --
>

--
rb