Ken Kundert
|
This is always a confusing issue because the graphics used to plot the transfer functions is not powerful enough to visually convey the fact that the inputs and output of the transfer functions are at different frequencies. If you plot versus the output frequency, then you don’t see that the input frequencies of the various transfer functions differ, and if you plot versus the input frequency, then you don’t see that the output frequencies differ. The situation is made worse because the input and output frequencies are specified indirectly. When I get confused, I always carefully identify the input and output frequencies; that always clears things up.
For PXF analysis, when the sweeptype=relative, as it is in this case, the output frequency (shared with all transfer functions) is fout = relharmnum*ffund + f where f is the sweep variable (in this case it is the relative frequency offset). In your example, f ranges between 1kHz and 10MHz and relharmnum=1, and so fout ranges from 6.1MHz to 16MHz. The input frequency for a particular transfer function is fin = sideband * ffund + fout
In your question, you ask about two particular transfer functions. The first is where f=8MHz and sideband=-1. In this case fout = 1*6MHz + 8MHz = 14MHz and fin = -1*6MHz + 14MHz = 8MHz. In the second, f=2MHz and sideband=0. In this case fout = 1*6MHz + 2MHz = 8MHz and fin = 0*6MHz + 8MHz = 8MHz.
The input frequencies for the transfer functions are the same, but the output frequencies differ. This is why one should not expect the transfer functions to be the same.
|