Siglent SDS2000X Plus

[Peter_O]

so announced 'Affordable' scope HP54504A from the 90s: 50 Ohms switchable, 7pf.

[2N3055]

Thankfully not true. If the input capacitance of 17 pF would still be present even in 50 ohms mode, it could hardly be useful at frequencies above 100 MHz.

This is also the reason why an external terminator  on a high-Z scope input can never replace the genuine 50 ohms input mode.

It is still true that the 50 ohms input mode is not perfect. Yet the VSWR stays below 1:1.5 at the specified bandwidth (500 MHz in case of the SDS2000X Plus).

Hmm. Yes, that must be true, 17 pF is rather high and only about 94 Ohms at 100 MHz . I wonder what tricks they do in the front end to reduce the capacitance and improve the match in 50 Ohms mode? 

I also looked up the ~20 year old Tek TDS3054, which is specified for 13 pF, and is bit disappointed that a 15-20 year newer scope has such a high capacitance 
(Though, it is mostly an issue when not using compensated probes...)

On 1 MΩ input, Keysight 3104T (1GHz version) is 14 pF. SDS6104H12 is 15pF.  SDS2000X HD is 16pF.

There are attenuators, switches, dual path amplifiers, input protection.....

[gf]

On 1 MΩ input, Keysight 3104T (1GHz version) is 14 pF. SDS6104H12 is 15pF.  SDS2000X HD is 16pF.
There are attenuators, switches, dual path amplifiers, input protection.....

And if you measure it with a network analyzer, then likely you even see a weird curve on the smith chart (particularly at higher frequencies) which cannot be modeled with a single lumped capacitor at all, but an equivalent circuit were rather complex.

[thinkfat]

On 1 MΩ input, Keysight 3104T (1GHz version) is 14 pF. SDS6104H12 is 15pF.  SDS2000X HD is 16pF.
There are attenuators, switches, dual path amplifiers, input protection.....

And if you measure it with a network analyzer, then likely you even see a weird curve on the smith chart (particularly at higher frequencies) which cannot be modeled with a single lumped capacitor at all, but an equivalent circuit were rather complex.

I remember I did exactly that. With a NanoVNA only, but for these low frequencies that wouldn't matter much. I kind of remember the input impedance in 50 Ohm mode definitely wasn't "real 50", not even "complex 50" but at least it was close, though slightly capacitive.

[Martin72]

Hi,

Measured the bodnar pulse with internal 50 ohm termination and with external termination, a very good (and expensive) one from huber+suhner.
The overshoots when using the internal are the results of reflecting effects and doesn´t have to do with the input capacitance when using 1M, right ?
BTW, I´ve stumbled over a "paper" from siglent, determing the bandwith of an oscilloscope - Didn´t notice this so far..

https://www.siglenteu.com/operating-tip/determine-bandwidth-scope-require-application/?gclid=Cj0KCQjwyMiTBhDKARIsAAJ-9VtTWOkYkjCQzlYbWoGPkYQIbrKcR8-9IhRSgbyJIcqNY_aindOG4bkaAr0TEALw_wcB

[gf]

Measured the bodnar pulse with internal 50 ohm termination and with external termination, a very good (and expensive) one from huber+suhner.

Btw, is it actually possible to display something like FFT(dX/dt) as math trace?
If so, it were interesting how that looks, when applied to Bodnar traces, for a 0...1000MHz span, with flat top window.
(Connecting the tips of the comb's teeth should approximately show the corresponding frequency response.)

[Martin72]

  (because when we buy it, only one thing is certain)

In this case I knew it before..

Edit:

Btw, is it actually possible to display something like FFT(dX/dt) as math trace?

Tomorrow

[2N3055]

Btw, is it actually possible to display something like FFT(dX/dt) as math trace?
If so, it were interesting how that looks, when applied to Bodnar traces, for a 0...1000MHz span, with flat top window.
(Connecting the tips of the comb's teeth should approximately show the corresponding frequency response.)

Yes you have advanced math... You pretty much write just that..

[gf]

Yes you have advanced math... You pretty much write just that..

More background. The question just came into my mind when I saw yet another Bodnar trace:
Is it possible to display the frequency response directly with a math function, if the stimulus is an (almost ideal) square wave from a Bodnar pulser?

The fourier spectrum of an ideal 10MHz square wave is a comb, with teeth at 10, 30, 50,... MHz (-> classical square wave harmonics).
The magnitude of the teeth follows a 1/f response. Multiplication of each tooth by its frequency would therefore result in a "flat comb", with teeth of equal height.
It the spectrum of the stimulus is a "flat comb", then the deviation of the tips from a horizontal line (after the frontend) reflects the frequency response.
Since it is a linear system, the frequency-dependent scaling of the stimulus spectrum can also be done after the DUT.

This leads to the following 4 approaches, and I wonder which of them could be realized via supported math expressions:

FFT(X) * f       [i.e.multiplication of each frequency bin by its freqency, in linear space, not dB]
FFT(X) * Y      [where Y is a constant, pre-defined, "hand-crafted" frequency domain trace, and the multiplication happens in linear space]
FFT(X) + Z     [where Z is a constant, pre-defined frequency domain trace in dB, and the addition happens in dB space]
FFT(dX/dt)     [since multiplication by frequency in the frequency domain is equivalent to differentiation in the time domain]


Potential sources of inaccuracy/untertainty are still:
* potential aliasing (if >= 1GHz is not yet sufficiently attenuated)
* The exact implementatin of dX/dt is not known, and it is supposed to be a numeric approximation
* Bodnar pulse is fast, but still not an ideal rectangle, so the stimulus spectrum already has a small a priori roll-off from the ideal 1/f comb spectrum in the region of interest

[2N3055]

Yes you have advanced math... You pretty much write just that..

More background. The question just came into my mind when I saw yet another Bodnar trace:
Is it possible to display the frequency response directly with a math function, if the stimulus is an (almost ideal) square wave from a Bodnar pulser?

The fourier spectrum of an ideal 10MHz square wave is a comb, with teeth at 10, 30, 50,... MHz (-> classical square wave harmonics).
The magnitude of the teeth follows a 1/f response. Multiplication of each tooth by its frequency would therefore result in a "flat comb", with teeth of equal height.
It the spectrum of the stimulus is a "flat comb", then the deviation of the tips from a horizontal line (after the frontend) reflects the frequency response.
Since it is a linear system, the frequency-dependent scaling of the stimulus spectrum can also be done after the DUT.

This leads to the following 4 approaches, and I wonder which of them could be realized via supported math expressions:

FFT(X) * f       [i.e.multiplication of each frequency bin by its freqency, in linear space, not dB]
FFT(X) * Y      [where Y is a constant, pre-defined, "hand-crafted" frequency domain trace, and the multiplication happens in linear space]
FFT(X) + Z     [where Z is a constant, pre-defined frequency domain trace in dB, and the addition happens in dB space]
FFT(dX/dt)     [since multiplication by frequency in the frequency domain is equivalent to differentiation in the time domain]


Potential sources of inaccuracy/untertainty are still:
* potential aliasing (if >= 1GHz is not yet sufficiently attenuated)
* The exact implementatin of dX/dt is not known, and it is supposed to be a numeric approximation
* Bodnar pulse is fast, but still not an ideal rectangle, so the stimulus spectrum already has a small a priori roll-off from the ideal 1/f comb spectrum in the region of interest

As I said FFT(dX/dt) you can directly type in into math channel. And it will show you curve gain of frequency response.  dX/dt implementation will, of course, be numeric implementation and it works fine.  Problem with any dX/dt function is that it "amplifies" high frequency noise (because obvious reasons) so it will be "nervous". That can be "tamed" by using FFT averaging.

[Martin72]

Hi,
I type this:

FFT(dX/dt)

With X = Channel 1 into the line.
Then you will see this on the attached pic below.
Maybe it´s the wrong quotation (I´m not into FFT math) because when you switch for example the bandwith of C1 from full to 20Mhz nothing siginificant will happen.

[2N3055]

FFT(Average(d(C1)/dt))
Make note  that you want to look at spectrum of single pulse...
So you need to put only one rise edge on the screen..
Play with timebase for best results..

[Martin72]

Play with timebase for best results..

This is what I get on different timebases...
Note the information(Sa, curr. pts, delta-f) inbetween the fft area on the righter side

[gf]

Then you will see this on the attached pic below.

Is this really full bandwidth? Looks like a digital filter was applied (4-tap moving avarage?). Or is it an artifact of the dX/dt approximation if it is not applied to interpolated+upsampled data? How does does FFT(C1) look for comparison (without dX/d)t?

[mawyatt]

@2N3055,

Agree, as only one rise edge when the derivative is taken should be the Impulse Function from an ideal squarewave edge. The FFT output of an Impulse Function as the stimulus for an arbitrary linear Network revels the Transfer Function of said Network. Here the scope is the Network since it's displaying the output waveform. This is Linear Network, and the Network excited by a derivative waveform is the same as the applying the waveform, then taking the derivative of the Network output as is the case for scope, also one must also take into account the sampling function (Sinc).


Best,

[Martin72]

How does does FFT(C1) look for comparison (without dX/d)t?

Like this...

(General question: Only 10 markers avaible?)

[gf]

Like this...

This looks more or less as expected. Then it is likely the dX/dt operator which garbles the spectrum

[Martin72]

But why ?

[tautech]

(General question: Only 10 markers available?)

Yes just 10 however rather than assigning them to descending Peaks you can assign each and every one to any frequency of choice.
Check out the Marker menu for options. 

[Martin72]

Thanks, Rob.
Do you have a clue why I´m not getting the same results like 2N3055, although I mostly use the same settings?

[rf-loop]

Thanks, Rob.
Do you have a clue why I´m not getting the same results like 2N3055, although I mostly use the same settings?

You want same result... Do you also have same SDS6104 H12 Pro (the model is not even available outside of China and will never be available according to current knowledge. Its price in China is 192680 RMB) 

[gf]

Do you have a clue why I´m not getting the same results like 2N3055, although I mostly use the same settings?

A flat frequency response up to 1Ghz is indeed strange.
I'd expect at least a noticeable amount of roll-off between 500MHz and 1GHz.
Maybe you can zoom in vertically to see more details?

20ns/div seems to be the point where it starts to interpolate (at least the displayed FFT sample rate of 50Gsa/s suggests that).
25x oversampling is hopefully more than enough, in order that the observed low-pass filter effect of the dX/dt operator becomes negligible (but who knows for sure).
(Ideally, sinc interpolation should not generate any frequencies >1GHz, so the 1..25Ghz range of the oversampled spectrum should be empty, except for noise that happens to be distributed).

Frequency resolution is a bit poor. At 50GSa/s, up to 4096 FFT points could be used, leading to a window size of 4096/50e9=81.92ns then, which is still small enough to cover only a single impulse. This leads to a resolution (frequency bin spacing) of ~12MHz then.

For further debugging, you could also display the differentiated edge, Average(d(C1)/dt), as 2N3055 did.
Maybe also d(Average(C1))/dt, to see if that makes a difference?

[Martin72]

Hi,

(nearly) same settings before, plus F1= d(c1)/dt and playing around with the bandwith.
Note that there is no significant difference inbetween full, 200Mhz, 20Mhz.

[gf]

Note that the size of your FFT window is only ~10ns, since you only use 512 FFT points @ 50 GSa/s. Not much headroom for mis-placement.
Can't recognize the grid in the screenshot - is the impulse completely inside the FFT window, and approximately centered in the FFT window? Can you use more FFT points, say 4096?
[ 2N3055 had a ~80ns window (2048 points @ 25 GSa/s), therefore exact placement is not so critical. But it should not be much more than 80ns either, if the aim is to capture a single impulse only. ]
Basically The F1 trace looks resonable. To see more details you should display it with a larger amplitude.zeof ~80ns

EDIT: And for 20MHz you definitively need the max. feasible window size of 80ns. Note how long the edge of the square wave needs to settle to a stable level. The whole settling time is part of the "pulse width" of the F1 trace, which must fall into the FFT window.

[Martin72]

Hi,

Can you use more FFT points, say 4096?

Yep, when increasing the timebase. (Preadjustments : Max memory size 2M for FFT), doing this will have the known result of having a "McDonalds M" symbol on the screen.

Interesting thing(for me) : At 50ns/div FFT got 512pts, at 20ns/div rises to 1024pts, at 10ns falling back to 512pts...

Unfortunately, F-"channels" do not allow vertical fine-adjustement.

The F1 trace looks resonable. To see more details you should display it with a larger amplitude.

Here it is..