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Is there a rule of thumb for a scenario for deciding minimum required scope bandwidth whew we have a known pulse train of 5MHz repetition rate and each pulse has 1ns pulse duration?

I know that these may depend on how much pulse distortion we can live with but at least how can we quantify the needed minimum scope bandwidth to observe the pulse shape in scope. Is thee a quick way to roughly estimate?

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  • \$\begingroup\$ 1ns pulse duration means it must also have some non-zero rise and fall time so it can't be a 1ns square pulse. Is the 1ns from 50% of the rising edge to 50% of the falling edge? Or how is the 1ns defined? \$\endgroup\$
    – Justme
    Mar 9 at 18:44
  • \$\begingroup\$ lets say it is gaussian fwhm \$\endgroup\$
    – pnatk
    Mar 9 at 18:48
  • \$\begingroup\$ BW = 1/(rise time). So if you can live with a 100ps rise time, 10GHz. Note that this means that if you see a 100ps rise time in the scope, all you'll know is that the pulse is much faster but absolutely positively not by how much. \$\endgroup\$
    – TimWescott
    Mar 9 at 18:50
  • \$\begingroup\$ @TimWescott So I can then take the pulse triangular which means rise time in my case is 1/2 = 0.5ns. This means min around 500MHz scope is needed? \$\endgroup\$
    – pnatk
    Mar 9 at 18:56
  • \$\begingroup\$ 1/0.5ns = 2GHz. 500MHz is definitely too little. Even with 2GHz, you'd only be able to detect the presence of pulses; you wouldn't be able to get much detail. For instance, you wouldn't be able to tell if a smaller-than-normal pulse is too short in duration but full amplitude, or full duration but weak in amplitude. If 2GHz is all you can afford, it may make sense -- but expect struggle. \$\endgroup\$
    – TimWescott
    Mar 9 at 19:01
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Square waves are composed of higher-order harmonics of the fundamental rep rate. The faster the transition time (that is, rise/fall time), the more harmonics you need to get to give a usable picture of the wave. So it’s more about that than the actual clock frequency.

More here: https://www.fluke.com/en-us/learn/blog/oscilloscopes/what-is-the-relationship-between-oscilloscope-bandwidth-and-waveform-rise-time

The article uses the following relation:

  • Bandwidth = 0.35 / risetime(s)

So for that 1ns pulse, assuming a risetime of say, 300ps, we get 1.16 GHz as the required bandwidth.

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  • \$\begingroup\$ But the original signal is not a 50% duty square wave. It is 1ns pulses repeated at rate of 5 MHz. I don't think a 50 MHz scope would see 1ns pulses that great. \$\endgroup\$
    – Justme
    Mar 9 at 18:48

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