Im not sure what is true, when I want to calculate it.

First, the fact is that classic homodyne is used only in receivers. (I'm not talking about quadrature homodyne.) But I can't understand how the spectrum behaves when a theoretical homodyne TX is transmitting.

Can someone answer me which one of my propositions are good: A or B?

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If I understand your question correctly, you are asking about the inputs and output of the frequency mixer in a direct-conversion transmitter.

Neither of your pictures is correct, but B is closer to correct. Here is the correct picture:

Here are some facts about frequency mixing that can be used to help see where you went wrong:

  • Until a filter is added, the output cannot have a partial copy of the input. Your input signal's spectrum has the shape |\___/|, so your output signal must be some sort of sum of copies of |\___/|. But both of your ideas don't: they contain |X| instead which is something different.

    Your option B does contain |\___/|, but it also contains /|___|\, which cannot be made up of complete copies of the input in any way.

  • Whenever the spectrum is inverted/reflected, it is always about some frequency that in some sense exists in the system — such as a carrier frequency, or zero. When you made the |X| shape, you instead reflected (part of) the input signal spectrum about the midpoint of the signal. That midpoint doesn't correspond to any signal we're presuming to exist exist, so the signal cannot be reflected about it.

  • $\begingroup$ thank You Kevin for your response. I thing that you are talking about Heterodyne mixing :) with IF I spoke with my teacher and he said, that pure homodyne is different and is nou useable beacuse of spectrums that are overlaping. Homodyne is only in receiver as I know. Heterodyne with IF is in transmitter. This is how I understand it for now, I may be wrong :) $\endgroup$ – Przemek Lewandowski May 1 '16 at 20:56

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