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.....
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.
Measured the bodnar pulse with internal 50 ohm termination and with external termination, a very good (and expensive) one from huber+suhner.
(because when we buy it, only one thing is certain)
Btw, is it actually possible to display something like FFT(dX/dt) as math trace?
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..
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
Then you will see this on the attached pic below.
How does does FFT(C1) look for comparison (without dX/d)t?
Like this...
(General question: Only 10 markers available?)
Thanks, Rob.
Do you have a clue why I´m not getting the same results like 2N3055, although I mostly use the same settings?
Do you have a clue why I´m not getting the same results like 2N3055, although I mostly use the same settings?
Can you use more FFT points, say 4096?
The F1 trace looks resonable. To see more details you should display it with a larger amplitude.