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RE: Dumb Question
- Subject: RE: [amsat-bb] Dumb Question
- From: "hdefelice" <hdefelice@xxxxxxxxxxxxxx>
- Date: Wed, 19 Jan 2000 12:48:49 -0500
I found that the best way to visualize SSB operation is to think in terms
of generating or receiving a signal using a balanced mixer. In this case, a
signal is received and "downconverted" to audio or baseband by eventually
mixing it with a local oscillator. Selecting upper or lower sideband just
determines the sense or polarity of the demodulated signal. In general, a
LSB signal is produced when the modulating information subtracts from the
carrier frequency and USB is produced when the modulating information adds
to the carrier frequency. In order to properly demodulate a SSB signal, the
demodulator must reproduce the modulating information in the same sense. If
the modulation information caused the signal to increase in frequency, then
when that signal is demodulated, the output of the demod must also increase
in frequency. If the opposite sideband were selected at the demod
(receiver), then an increase in frequency at the modulator (transmitter)
would cause a decrease of frequency in the demodulator (receiver). When
using SSB to transmit and receive digital information a "1" or "0" is
determined by the change of frequency or phase of a carrier. If the
transmitter is set to USB, let's assume that a "1" causes an increase in
frequency and a "0" causes a decrease in frequency. On the receiving end,
if the receiver is also set to USB, an increase in frequency will cause a
"1" to be output and a decrease in frequency will output a "0". If the same
signal is received using LSB, an increase in frequency from the transmitter
will cause a DERCREASE in frequency at the receiver demod and therefore
output a "0" instead of a one. Therefore selecting USB or LSB allows you to
decide whether you want your data NORMAL or INVERTED. This is very
convenient because it is very easy for a modem designer to add an odd
number of flip-flops to a circuit or an extra transistor driver that may
invert the data somewhere along the path.
This is a condensed explanation of a subject not so easy to condense. I'm
sure it will spur comments. I hope I haven't confused you even more.
Howie
AB2S
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