No, regardless of the type of detector, both receivers will produce the 100

Hz. Difference signal.

Don

Looks like you didn't carefully read between the lines in Jeroen's

posting.

Clearly, he had in mind a product detector driven by a

1MHz-50Hz LO. On the other hand, he better be careful about the

phases of the signals. If the product detector is driven by a LO of

sin(2*pi*999950*t), and if the 1MHz is sin(2*pi*1e6*t), the difference

is a cosine: A*cos(2*pi*50*t). If the 999.99kHz signal is

sin(2*pi*999900*t) the difference is the same cosine term, so the

total output is 2*A*cos(2*pi*50*t). But if the 999.99kHz signal is

the opposite polarity, -sin(2*pi*999900*t), then the outputs from the

two cancel and you get zero. I suppose most receivers that use

product detectors either have some carrier to lock their LO to, or

just receive the signal as a single sideband and suppress the other--

or just detect the two sidebands independently.

Although mixing in the output amplifiers is tough to avoid if you just

blindly combine the signals, you can use a circuit to keep the

signals out of the alternate amplifiers. One such circuit is a

Wilkinson combiner. Problem: it wastes half the power. It could be

done with filters, but Qu would have to be very high (incredibly

high?) to avoid significant power loss and get good isolation, given

such close frequency spacing.

Cheers,

Tom