On Mon, 15 Oct 2018, Ralph Mowery wrote:
In article ,
says...
Consider a 2 tone signal at the 9MHz USB IF, comprising 900Hz and
1300Hz tones.
The components will be 9.0009 and 9.0013
Subtract the VFO at 5.5MHz:
9.0009 - 5.5 = 3.50009
9.0013 - 5.5 = 3.50013
Nothing has been inverted. The 80m signal is still upper sideband.
GB3BERNIE
Ralph is posting from rec.radio.amateur.antenna and google groups
strips the crosspost - without a repeater, he's not going to answer you.
Try it the other way around and use a ssb generated at 5 mhz and the vfo
at 9 mhz. It is difficult for me to remember which was used for the vfo
and ssb generator.
I think the origins are with a 5MHz IF. This has come up before, the same
explanation given, yet if I wasn't sick and did the figuring, I think it's
that the 9Mhz one wasn't it, but a 5MHz IF does do the inversion.
But I can't remember what rig had a 5MHz IF. THey existed, but the ones I
can think of came later. So maybe it was a phasing rig, but which did
conversion rather than generate the SSB signal on the signal frequency.
The Central Electronics 10 transmitter maybe, reinforced by their later
20, but I havent' checked.
There was a popular rig in QST that worked out the figures so the low IF
was converted up to an intermediate frequency with one crystal, one caused
no inversion, but if you multiplied the crystal frequency by three, it was
high side and inverted the sideband. But since it did both sidebands, it
wouldn't have been a standard for LSB below a certain frequency, There
were early ssb rigs that didn't have selectable sideband, they just picked
conversion frequency properly so below 10MHz, it was LSB, and above was
USB. Since nobody used the opposite sideband, no need for a switch.
Michael