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  3. PAC and PXF inconsistency?

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PAC and PXF inconsistency?

Yevgeny
Yevgeny over 10 years ago

Good day.


I am trying to simulate power supply noise propagation through an amplifier stage. The amplifier is driven by a large signal, which is the PSS beat. The circuit is slightly non-linear (THD around 30dB). I use harmonic balance for PSS engine. Now I add power supply interference (small signal); I try it in two different ways:

1) Run PAC analysis, with the only PAC source being power supply voltage source, look at the output spectrum (differential voltage).

2) Run PXF (differential voltage output) and look at the transfer function from the said power supply voltage source.

I deem the two analyses should have given same spectra, but they don't. In fact, the base harmonic (i.e. number "0") is exactly identical, all the others as not.

Can anyone please explain that to me?

UPD: I have tried to verify this with transient, i.e. ran a long TRAN simulation with both the input (LO) and the power supply (SUP) active; checked the output spectrum. Thus I validate a single frequency point of the SUP. It came out very close to PAC (and very far from PXF). So I better rephrase my question: where is my understanding of PXF went wrong? The image below shows these results. Peaks are come from the transient spectrum (1st harmonic). The brown and green curves are PAC results. The two remaining curves (flat ones) are PXF.

Many thanks,

Yevgeny.

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  • Andrew Beckett
    Andrew Beckett over 10 years ago

    Yevgeny,

    Your understanding of pxf is incorrect. Shawn's description is right, but maybe I can clarify by saying it again (maybe slightly differently).

    When you do a pxf sweep, you are sweeping the output frequency, not the input frequency (so the opposite of pxf). The simulator then gives you the transfer function from an input (default, from any source) at each sideband to the output. So for example, if you have an output at Fout, you will get transfer functions from input frequencies of k*Flo+Fout where k is the sideband number (so -5, -4, -3 ... 0 ..., 3, 4, 5 etc). Note that these are transfer functions, not signal levels caused by a particular size input signal, and also they are the transfer function caused by mixing with that multiple of the PSS fundamental.

    For pac, you are specifying an input frequency, and then getting the output signal level (not transfer function) at each output sideband. So that's an input at Fin, and then a set of output signals at k*Flo+Fin. For a given frequency sweep, the only place the results would match is the 0th sideband.

    Now, in your case, you have an amplifier, so it's presumably not a LO frequency but some large signal interferer that you wish to look at intermodulation effects from a small signal on the supply. Say that large signal interferer was at 100MHz (just to keep it simple), and you were injecting a noise frequency of 10MHz. With 3 sidebands in the PAC analysis, you would see the amount of this signal at 10MHz, 90MHz,110MHz,190MHz,210MHz,290MHz, 310MHz.

    If you did a pxf analysis, with the output frequency at 10MHz, it would tell you the transfer function from a noisy input on the supply (or anywhere else) at 10MHz, 90MHz,110MHz,190MHz,210MHz,290MHz, 310MHz. So in other words, you're finding out what the various means of getting a 10MHz output would be, as opposed to applying a 10MHz input.

    Assuming that you are OK with converting the signal levels to transfer functions (commonly done by setting the PAC input signal to 1), if you were to run PAC with an input at 90MHz, and then look at the output at 10MHz (the -1 sideband), it should give the same as PXF would with a 10MHz output (the frequency on the form) and the transfer function probed at 90MHz (the -1 sideband). (the shape would be flipped though).

    Hope that helps!

    Regards,

    Andrew

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  • Andrew Beckett
    Andrew Beckett over 10 years ago

    Yevgeny,

    Your understanding of pxf is incorrect. Shawn's description is right, but maybe I can clarify by saying it again (maybe slightly differently).

    When you do a pxf sweep, you are sweeping the output frequency, not the input frequency (so the opposite of pxf). The simulator then gives you the transfer function from an input (default, from any source) at each sideband to the output. So for example, if you have an output at Fout, you will get transfer functions from input frequencies of k*Flo+Fout where k is the sideband number (so -5, -4, -3 ... 0 ..., 3, 4, 5 etc). Note that these are transfer functions, not signal levels caused by a particular size input signal, and also they are the transfer function caused by mixing with that multiple of the PSS fundamental.

    For pac, you are specifying an input frequency, and then getting the output signal level (not transfer function) at each output sideband. So that's an input at Fin, and then a set of output signals at k*Flo+Fin. For a given frequency sweep, the only place the results would match is the 0th sideband.

    Now, in your case, you have an amplifier, so it's presumably not a LO frequency but some large signal interferer that you wish to look at intermodulation effects from a small signal on the supply. Say that large signal interferer was at 100MHz (just to keep it simple), and you were injecting a noise frequency of 10MHz. With 3 sidebands in the PAC analysis, you would see the amount of this signal at 10MHz, 90MHz,110MHz,190MHz,210MHz,290MHz, 310MHz.

    If you did a pxf analysis, with the output frequency at 10MHz, it would tell you the transfer function from a noisy input on the supply (or anywhere else) at 10MHz, 90MHz,110MHz,190MHz,210MHz,290MHz, 310MHz. So in other words, you're finding out what the various means of getting a 10MHz output would be, as opposed to applying a 10MHz input.

    Assuming that you are OK with converting the signal levels to transfer functions (commonly done by setting the PAC input signal to 1), if you were to run PAC with an input at 90MHz, and then look at the output at 10MHz (the -1 sideband), it should give the same as PXF would with a 10MHz output (the frequency on the form) and the transfer function probed at 90MHz (the -1 sideband). (the shape would be flipped though).

    Hope that helps!

    Regards,

    Andrew

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