The Designer's Guide Community Forum

The Designer's Guide Community Forum
https://designers-guide.org/forum/YaBB.pl
Simulators >> RF Simulators >> PXF Gain
https://designers-guide.org/forum/YaBB.pl?num=1036181769

Message started by emad on Nov 1^st, 2002, 12:16pm

Title: PXF Gain
Post by emad on Nov 1^st, 2002, 12:16pm

Ever tried to simulate a circuit using PXF and made the range frequency larger than the harmonic beat? Well I did. I have a circuit with a fundamental frequency of 6 MHz and the frequency of interest around the fundamental is 1kHz-10 MHz. "Why the strange range" one would say! I have to because this fundamental frequency is a subharmoic of the main fundamental of interest.

The result of the simulation is some PXF gains from various sidebands but to my surprise, in the overlaping zone between sidebands (for example 6-10 MHz) the PXF gains from sideband 0 and sideband 1 at the same offset are not identical. Why is that?

Title: Re: PXF Gain
Post by Ken Kundert on Nov 1^st, 2002, 11:53pm

Emad,
I'm not sure I understand the specifics of your question, but I'll try my best to answer it. Please let me know if I have misunderstood your question.

Here is what I understood you to say. You have a fundamental frequency of 6MHz, and the output frequency range for the PXF analysis is 1kHz - 10MHz relative to the fundamental, or 6.1MHz - 16MHz in absolute terms. Is this correct?

Then you ask about gains in the 6-10MHz range. So in particular, let's consider 8MHz. Your question is with a relative output frequency of 8MHz, or an absolute output frequency of 14MHz, why is it that the 0 and +1 sidebands differ? Is that correct?

In this situation the 0 sideband gives the transfer function from inputs at 14MHz to outputs at 14MHz. The +1 sideband gives the transfer function from inputs at 14+6=20MHz to outputs at 14MHz. There is no reason why these two transfer functions should be the same. Indeed, for a linear time-invariant circuit, which are incapable of converting signals from one frequency to another in steady state, the +1 sideband will always be 0 whereas the 0 sideband will usually be nonzero.

Does this answer your question? ???

Title: Re: PXF Gain
Post by emad on Nov 2^nd, 2002, 5:26am

Let me clarify my question with a numerical example.
For a fundamental tone at 6 MHz and output range of 1 kHz-10 MHz: I expect the gain from around the -1 sideband (around zero frequency) @ an 8 MHz offset to be equal to the gain from around the "0 sideband" around the fundamental from an offset of 2 MHz (6+2=8 MHz). In fact, if you choose to plot the sideband as 'in' or 'absin' isetead of the default 'out' you can see the overlape zones much easier.

Am I wrong in my expectation?

Title: Re: PXF Gain
Post by Ken Kundert on Nov 2^nd, 2002, 10:57am

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
f_out = relharmnum*f_fund + 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 f_out ranges from 6.1MHz to 16MHz. The input frequency for a particular transfer function is
f_in = sideband * f_fund + f_out

In your question, you ask about two particular transfer functions. The first is where f=8MHz and sideband=-1. In this case f_out = 1*6MHz + 8MHz = 14MHz and f_in = -1*6MHz + 14MHz = 8MHz. In the second, f=2MHz and sideband=0. In this case f_out = 1*6MHz + 2MHz = 8MHz and f_in = 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.