Paolo, If the Q channel in your example has a phase shift of 90 degrees with respect to I, then the output is either 10.001 kHz (USB) or 9.999 kHz (LSB). If the phase shift is 0 degree, (I is same as Q), then the output is both on 10.001 kHz and 9.999 kHz (DSB).
Your idea is an interesting because it allows to make a WSPR transmitter with standard crystals. It would be interesting to see if this can be completely digitally built in a micro controller, i.e: A microcontroller clocked at 4 times the transmitting frequency, where the I and Q signals of a WSPR signal are mixed in software which is fed to and general purpose I/O port which contains the modulated WSPR signal. Possibly the sine wave versions of the I and Q signals must be sigma delta encoded to reduce spur. Goodluck with experiment, 73, Guido PE1NNZ On Tue, Mar 10, 2009 at 8:47 AM, Paolo Cravero <[email protected]> wrote: > I'm writing here to ask a SDR TX question, because I realized I've > completely forgotten the SDR basics I had read few years ago. > As I remember, if I "mix" an I+Q signal baseband (say, 1 kHz) with a > 4-phase carrier (say, 10'000 kHz) I obtain a single RF output at > 10'001 kHz (or 9'999 kHz if I invert I+Q). Correct? > If the above is correct, what if I feed the 4-phase switched-mixer > with I+I signal? Do I obtain a DSB output at both 9'999 and 10'001 kHz >
