Dear Frank, Aaahhhhhhhh, Yeeeesssssssss, clear!! How easy! As you proposed, I will check with 1 freq. at a time only. ( I would like to also do this freq. sweep automatically, but for this, I would also need then hundreds of transient sims in Cadence (1 takes currently ca. 1 hour) + DFTs + manually export the data hundreds of times + combine all this into 1 plot. Is there maybe another way what I do not think of in this moment? ) Also one other thing is now clear (For the other readers, here clearly stated): By sampling with 10 kHz an already sampled signal with 1 kHz, I get, as Frank correctly said, 10 times the same value, before it changes to the next one. That can be called a “Repeater”. And a “repeater” is mathematically identical, as far as I know, with zero stuffing ( = as you propose, set the other 9 identical values to 0 ) PLUS a comb filter of first order, so “(1+z^-1+z^-2 ... z^-9)^1”, which causes this sinc. behaviour. I will do this ( need a bit of time to do properly ). So I obviously INADVERTEDLY created an additional, 2nd sinc. filter by doing the repeater (the first one is already inside the “DUT”), killing my expected repeating spectrum at 1 kHz at the output. Things are getting much clearer now! However, what I also need to think about in more detail, is, when now looking quickly at once glance to the following: With the 2 times in series sinc. filter for e.g. (also?) circuit A), I would expect either: *) Because of my 2nd inadvertedly indroduced sinc filter by doing repeating upsampling, shouldn’t for case A) the blue curve be attenuated more than the green one from the PAC (see zoomed fig. below), or: *) The result from the Cadence “PAC sampled time averaged” (green) should exactly match the blue, if e.g. the “PAC sampled time average” does the same effect, so to also introduce a sinc (I don’t know what does “time average” exactly) Or maybe this ca. 3 dB difference (from -19.82 dB to -22.98) already close to "DFT fs/2" comes from another effect? (Case A, zoomed) And then, why I see a constant line with ca. 0 dB for “PAC sampled edge ris. or fal.” ( not shown in this graph )? Because 1x sinc. filter should still be present from the DUT, only the 2nd one from upsampling is gone when performing the internal Cadence “PAC sampled ris./fal. edge”, because this is doing, I learned from you “ideal Dirac sampling” which doesn't show sinc behaviour. You wrote “ This is also the reason why you get a straight line at 0 dB if you do a sampled PAC analysis at the output of an ideal S&H (and the result of the normal PAC analysis is simply the sinc function in this case) ”. But I solely did a “PAC sampled”, not a “normal PAC”, and I still can chose to either get a “sinc.” shape (with option “time averaged” or approx. a straight line (with option rising or falling edge). But BOTH are “PAC sampled ” analysis (only sub-options to select). Or, as my question from previous, I may repeat: Is “PAC sampled, time avg.” = “normal PAC”, true or false? So ??? But one step after the other ... Anyway, 1000 thanks - Very nice, Frank, that you bring it to the point!! bernd2700
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