How to obtain Discrete-Time (DT) response from a switched-cap (SC) circuit (integrator)?

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 statement: “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

bernd2700 said:

Is there maybe another way what I do not think of in this moment?

There will be no problem with your approach if all input sources are in the same Nyquist zone, so you don't need one simulation per input frequency, just one per Nyquist zone. And of course this method will only work for a linear circuit.

bernd2700 said:

Or maybe this ca. 3 dB difference (from -19.82 dB to -22.98) already close to "DFT fs/2" comes from another effect?

Without analyzing this in further detail, I would guess that it is the same effect that you see at fs/2 in this plot:

Frank Wiedmann said:

Here are the simulation results:

bernd2700 said:

And then, why I see a constant line with ca. 0 dB for “PAC sampled edge ris. or fal.” (not shown in this graph)?

If you perform Dirac sampling at the output of an ideal S&H with the same sampling frequency, you will get exactly the same samples as if you perform Dirac sampling at the input of the ideal S&H.

bernd2700 said:

Or, as my question from previous, I may repeat: Is statement: “PAC sampled, time avg.” = “normal PAC”, true or false?

True, as far as I know (Andrew, please correct me if I'm wrong).

Dear Frank ( & others),

As said, thank you for pointing this out with the zero-stuffing. For all: Here is the proof, at least now for case B) : In the graphs below, the blue curve is the sampled data (with Cadence’s calculator “sample” command) with period = 100us (and thus the same values are repeated 10 times, because the period of the circuit’s switches is 1ms), and we do NOT see a DT spectrum. That is, because it was filtered away by the inherent comb filter which exists when doing repeating. If I do zero-stuffing (so as Frank proposed: set the other 9 values to 0) and thus only do “Ideal Dirac sampling”, I really get my expected DT response (red curve = symm. around the circuit’s fs/2 = 500 Hz, repeating pattern with 1 kHz)! As theory predicted, by subsequent comb-filtering the red curve, I finally get the magenta curve, which is again identical to the blue, q.e.d. .

And if we compare this to the internal Cadence PAC (green and cyan), everything now fits nicely!

It was a lot of work to come to this point, but everything definitely was worth it to do, and so: “Thanks Frank!” Without your help, I would not have been able to do all this completely alone by myself.

Side info: You have to be careful with the things: DFT, zero-stuffing, comb-filtering, shifting, to do all this very properly (otherwise you easily can get sinc. behaviour at wrong frequencies, etc.). But also here is the advantage of doing this completely by oneself: You understand what's going on behind the scenes, because you know exactly what you did. (I cannot show all the details here, since it would explode the frame of this post, but at least this hint from my side.)

I will now apply the same procedure for the other cases (for case A) and for the purely analog RC-LPF) and check, if finally everything fits to my expectation. 

Many nice greetings so far,

Bernd2700

Dear Frank ( & others),

As said, I did now the same game for test-case “A)”. And I like it! : My manually applied procedure also here is fitting pretty well to the internal Cadence PAC, as you can see below: The blue curve is the sampled data at fs_DFT = 10kHz with repeated 10x values, because the circuit’s “fs_S&H = 1kHz. The red curve is the zero-stuffed out of the blue, so “Ideal Dirac sampling” done and thus fits to the internal “Cadence PAC sampled: ris. or fal. edge”. The cyan is the comb-filtered red and thus perfectly fits to the blue again (as previously explained). And I think, the red line can be called a DT response (everywhere 0dB is thus also symm. around fs/2!)

What fits perfectly at lower frequencies and deviates more and more apart as frequency is increasing, is the (blue or cyan) compared to the green PAC sampled, time averaged. Around 4.5 kHz, I get a difference of 3 dB (ca. -20 dB for the blue and cyan, and -23 dB for the green). But as you said, Frank, it might be the same what you were  showing, when going close to DFT fs/2 = 5 kHz, and things would maybe resolve, if I do the game with fs_DFT sampling up to 100 kHz and thus getting the spectrum up to 50 kHz, instead of only up to 5 kHz. (Then, the transient sim. would be probably very long, so I did not do it. However what I quickly did, is to do a PAC only until 10 kHz (thus showing a spectrum until 5 kHz), but it is exactly the green curve, just plotted until 5 kHz.)

                                                      (click to enlarge)

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For my curiosity, I now input instead of a MT, only exactly 1 tone, namely with ca. 4559 Hz, and the result is:

Here, everything sounds logical: I get alias components at fs – fin = 5 kHz – ca. 4559 Hz = ca. 441 Hz, and all its replica. As it should be now, exactly as the Cadence PAC does, only 1 input tone, I was expecting to see at 4559 Hz also this -23 dB. But instead, I see -20.07 dB.

I also now tried to look what the PAC does. As you proposed, I take the output from the transient analysis with only that input tone of ca. 4559 Hz (in steady state condition, so close at the end of the sim. at 104.8 s) and again do “(max-min)/2” from the values. As we also see at one glance, I roughly get this 0 dB (exactly being: 20*log10(1.9556/2) = -0.19 dB), as the PAC sampled: ris. / fal. edge delivers (the straight line at 0 dB). But how does the PAC sampled, option “time averaged” work here? What to measure to get the -23 dB instead? For sure, there is also a way, but how?

Thanks,

bernd2700

bernd2700 said:

But as you said, Frank, it might be the same what you were  showing, when going close to DFT fs/2 = 5 kHz, and things would maybe resolve, if I do the game with fs_DFT sampling up to 100 kHz and thus getting the spectrum up to 50 kHz, instead of only up to 5 kHz. (Then, the transient sim. would be probably very long, so I did not do it.

I don't think that you actually need to run the simulation. Just modify your sampled signal so that you now have 100 identical samples instead of 10 indentical samples in a row.

bernd2700 said:

But how does the PAC sampled, option “time averaged” work here?

This is just the nomal PAC analysis without any sampling, so you simply need to run an FFT on the output signal in order to get the result in the frequency domain.

Dear Frank,

In principle, a good idea to do every 100th sample, but I think it will not work, as I really do not produce input tones above 5 kHz, and so my AC analysis will divide then just mathematical “noise”. Moreover, I think if I set just the parameter “fs = 100 kHz”, all the frequency vectors with the real injected sinusoids will get misaligned then (or if I increase all by factor 10, everything is just shifted by factor 10, or… ).

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What is much more important, is:

The following procedure EVERYTIME perfectly (almost) fits to the Cadence PAC sampled, time averaged analysis: Generate multi-sinusoids, then run a tran. analysis of the DUT, then sample the transient data (calc. command “sample”), then doing the DFTs and divide output/input. So this is my blue curve. And, I think I can summarize and state with confidence now:

     [ My manual MT AC analysis over the 2 DFTs ( = blue curve) ] matches [ PAC sampled, time averaged ]

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However, what did only work out for test-case “A)” and B), is: By further manually doing zero-stuffing out of the blue curve, and so doing “Ideal Dirac sampling”, this (my red curves) then fits to the “Cadence PAC sampled, e.g. rising edge”, because:

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I now tried to apply the same procedure - as completely successfully done with test-case “A)” and B) - also now for the purely analog RC-LPF circuit, and it does not fit to the PAC sampled, e.g. rising edge. The results can be seen in below graph:

My blue curve again fits perfectly to the Cadence PAC sampled (green), e.g. time-averaged option. But this time, all options of PAC sampled produce the very same result: “rising edge”, “falling edge”, “time averaged”.

And so, by blindly applying the same procedure again as done with the previous test cases A and B, (Sampling, DFTs, then zero-stuffing, then comb filtering), this time I get garbage (red and cyan), because the circuit this time does NOT have an (inherent) ZOH at its output where the data is repeated 10 times the same value.

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My problem now is: I of course cannot include “switch - case” statements in my analysis, depending on a circuit! A la: If there is a ZOH-like-thing in the circuit, then the red line will fit to PAC sampled rising edge; if not, the blue curve already will fit to PAC sampled rising edge. The analysis itself is stupid and cannot know the circuit (and the circuit’s fs), and thus, the procedure has to work for EVERY circuit analyzed with it, as the Cadence PAC does!

=> So what can I do about that I not only get a more or less perfect matching to the “PAC sampled, time avg. option” with my own analysis, but also for the “PAC sampled, ris. or fal. edge option”, regardless of the DUT I have?

Pleeeeeeaaase, give me an advice.

Thanks a lot,

bernd2700

Sorry, guys, this phenomenon, I just saw, has nothing to do with the PAC sampled, it also occurs when I stay purely in Matlab.

I did the following test: I got the idea and simulated the same pure analog RC-LPF, but now completely inside Matlab (over Matlab's "lsim"), and subsequently applied the very same procedure (just with other fs and number of samples) as always done now, and I get below result:

I expect that the cyan curve will be the same as the blue, but it isn’t!

So please, for the moment being, forget my question of the previous post (that applying the same procedure does not fit for Cadence PAC sampled, ris. or fal. edge).

I quickly wanted to update you now with this info.

So I just have to debug it what’s going on here ...

Sorry for having asked too early, before I did this test case over pure Matlab, but my idea came just now.

bernd2700

Case “Purely analog RC-LPF”: Matlab internal comparison

First of all, case “Pure analog RC-LPF” from just above post: I did not find any errors in my scripts, so I think when doing zero stuffing, I just lose information (because concerned samples then replaced with the value “0”), and it can be proven, that the less I do zero-stuffing, the more the cyan or red line resembles the orig. blue line (graphs not shown to not overload this post).

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Case A) = “Ideal S&H”: Comparison with extended frequency range

The sim. results are here from my MT DFT analysis, now in extended frequency range. Originally, I sampled up to 5kHz, and now until up to 50 kHz. As expected, the results do fit perfectly also until approx. a factor 10 more in frequency, as we can see from below graphs: On the left, there is a repeated 1:1 copy from the old post just for remembering, on the right, also including the new yellow extended freq. analysis result. We see, the blue orig. 3 dB deviation at 4.5 kHz to the green Cadence PAC, time avg. is completely gone now. Now, the yellow curve fits perfectly to the green until approx. 10 kHz.

    Case A) “Ideal S&H”:

    Left figure: Repeated 1:1 copy from prev. post   ;         Right fig.: Now + yellow curve with extended MT freq. range.

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Case “Purely analog RC-LPF”: Comparison Cadence PAC <=> My MT AC analysis

I extended the frequency range for my MT analysis also for this case now up to 50 kHz.

As we see, the results from my MT AC analysis below (blue), and from the Cadence PAC sampled, time avg. (also below, also blue line) PERFECTLY match!

With my analysis’ frequency range increase from 5kHz to 50 kHz, you can see from the result below, the cyan line perfectly fits the blue line now also up to some kHz, as expected.

Although I now additionally also used a 2nd period only out of the base frequency of some mHz and thus have steady-state condition for my DFT, we still must clearly say from approx. 5 kHz up to DFT fs/2 = 50 kHz, the cyan line does not fit the blue line. And my red line does not fit the PAC sampled, rising edge (red and orange).

For the Cadence PAC sampled, rising edge, this performs somehow better when looking to the higher frequencies, say from 10 kHz onwards up to its fs/2: The internal deviation from the PAC sampled, ris. edge (red and orange) is not so big to its time averaged option (blue and black), as my internal deviation over my MT AC analysis between red and blue - this is bigger.

We can compare the both red curves, one from my above MT AC analysis and one from the Cadence PAC sampled rising edge, and they should approx. yield the same result. But they do not.

Maybe the Cadence PAC does some internal additional calibration before and / or does higher oversampling even though not requested (I have reduced, as you see, also the maxacfreq. for this purpose. But also for case A) the 2 time avg. results are really look-alike up to even fs/2 ; graph not shown here).

=> I don’t know, but maybe you, Frank, why the internal Cadence PAC sampled, ris. edge (red) is closer to its time-avg. option (black), than my MT AC analysis “Ideal Dirac sampled” (red) to my “normal DFT” (blue)? So, in other words: Why Cadence PAC sampled, rising edge seems to do not lose so much information in doing also “Ideal Dirac” sampling as mine? Which factor Cadence PAC sampled does? I here used a factor of 10 (so 1x the value, 9x zero).

Thanks a lot,

bernd2700

The sampled analyses of Spectre always sample exactly once per period (as given by the PSS fundamental frequency). You only needed to do the "zero-stuffing" because in your transient analysis, you were sampling the output with a higher frequency than the frequency of the sample-and-hold (which is not possible in sampled PAC). When you do the "zero-stuffing" on the time-continuous circuit, you effectively (down-)sample the output at a lower frequency, resulting in aliasing effects in your red curve.

Dear Frank,

Many thanks for your answer! Unhappily, I did not understand all parts of it. Starting with the unclear parts. I always like examples.

So, watching my simulated circuit below again: For a test I have changed now the sampling frequency of the “Ideal S&H” from orig. 1 kHz to now 10 kHz (period=100u) to not be the same as the clock source “V0” (per=1m). This still defines my fundamental beat frequency unchanged with 1 kHz in the PSS.

You wrote: “you were sampling the output with a higher frequency than the frequency of the sample-and-hold (which is not possible in sampled PAC).”

Yes, with my MT AC analysis, I sampled the output with a higher freq., namely with 100 kHz. But also, the Cadence PAC sampled, I tell: “Please plot me the freq. response until 100 kHz”. And so, as far as I understood correctly, when reaching this frequency, 1 sinusoid only with 100 kHz is injected into the DUT (with 1 V, I have specified). As a consequence, the PAC must sample the output with at least 200 kHz to properly fulfill Shannon (Nyquist). And so: [Output sample freq. with 200 kHz with sampled PAC] >> [Freq. of “Ideal S&H” with 10 kHz]. This is possible. But possibly, you were meaning something else, but what??

You write: “The sampled analyses of Spectre always sample exactly once per period (as given by the PSS fundamental frequency).”

So again my example: If I want to see the freq. response at, say, 100 kHz, the “PAC sampled” must sample with >= 200 kHz. If this is, as you write “once per period”, then this period is 5us. But the PSS fundamental freq. is given with period 1ms. So it is 200 times per period. So for me, it is either not once per period or not given by the PSS fundamental period. So also here, sorry, I have no clue, what you mean. For me, the PSS is there to find a large-signal valid operating point of the circuit.

You write: “When you do the "zero-stuffing" on the time-continuous circuit, you effectively (down-)sample the output at a lower frequency, resulting in aliasing effects in your red curve.”

Ok, yes, this part is clear, because I did not allow even a still higher sampling frequency for my RC-LPF DFT result. I somehow wanted to compare my MT AC analysis to PAC sampled: And if I tell the PAC (or PSS in that case) “maxacfreq. = 100 kHz”, then I expected also comparable results. Indeed, we see, in repeated below graphs, that also the red PAC sampled curve with ”PSSmaxacfreq=100k” is indeed “noisy”, especially above 20 kHz, and the spikes of the red curve deviate to the blue line by also really a lot of dB, but in-between, the red curve reaches the blue again, and so I feel, the PAC sampled performs still “better” than my MT AC analysis, where between the peaks, my red curve does not reach the blue curve again (see yellow circles).

       Left: Cadence PAC sampled                          Right:      My MT AC analysis

       Case “Purely analog RC-LPF”, output node "Vo":

Explanation for all the other readers (which seem to be quite a lot for this special topic, as I saw now with 2330 views, wow!):

I have chosen “fs_DFT = 100kHz” (as said to be comparable to Cadence PAC sampled with maxacfreq.=100k), so my samples are e.g. [y1 y2 y3 ...] for the CT “RC-LPF” circuit. As you have proposed, Frank (but for case B) ), I do zero-stuffing also here with the same procedure [ y1 zeros y11 zeros y21 zeros ... ], so I throw away the samples y2 until y10 and so on. This means from signal-processing point of view, I did a “brute-force” decimation (by factor 10), I mean, without an additional anti-aliasing LPF, therefore Frank has written “... resulting in aliasing effects in your red curve”.

So I will change now my procedure for the zero-stuffing method and will try in the following: [ y1 zeros y2 zeros y3 zeros ... ]. But this means, I have to effectively increase the sample rate now to fs_DFT = 1MHz (if I keep factor 10). And I will check, if I will obtain a DT for cases A) and B) and a CT response for the case “purely analog RC-LPF”.

Short Summary about CT and DT response of a circuit:

Because for cases A) and B), it’s the circuit which delivers more times the same value and in the case “RC-LPF” not. That’s why you, Frank, have written “If the circuit isn’t DT, you won’t get a DT response”. Now you made it clear to me, Frank (1000 Thanks!!!), but originally, it completely confused me from Ken Kundert’s statement: “You must decide upfront if you are interested in the CT or DT behavior of a circuit”. And so, I thought then, I can freely chose, which is not true. Ken Kundert means: If you have any purely CT circuit (e.g. this RC-LPF) and want to see its DT response (because e.g. it gets sampled later by a customer, so by a subsequent, additional CIRCUIT), you have to do steps if you also want to see what will this DT behavior be in the end (so you have to modify your CIRCUIT, e.g. install an additional “Ideal S&H” block). But in any case, you have to change your circuit, to see a DT for a purely CT DUT, like the RC-LPF. If you only change your analysis type (e.g. from “PAC sampled, option time averaged” to “PAC sampled, option rising edge”) for a purely CT DUT, you should see every time the same Bode plot, because a CT circuit does not have a DT answer. Correct, Frank?

And this I will check if it is true not only for the Cadence PAC for all types of CT and DT DUTs, but also over my MT AC analysis procedure, now with “real” zero-stuffing (up-sampling) (but it will take some time)...

Many thanks and nice greetings,

bernd2700