To begin with,I am not designig an oscilloscope but am curious about how certain parameters are determined. My questions might sound very trivial but I would be grateful if someone could answer them convincingly.

  1. I have seen oscilloscopes with say a 50 MHz bandwidth but 1 GS/s sampling. My question is, why do I need such a high sampling rate when the signal is restricted to 50 MHz? I read on an oscilloscope manufacturers website that a good rule of thumb is that to visualize a signal of say X MHz, you should have an oscilloscope between 4X to 6X MHz bandwidth. What is the justification for such high sampling rates then (When compared to the bandwidth)?

  2. I have also seen probes with a bandwidth limitation. Is it a must to use these if the scope has a filter as well? On the other hand, can there be an oscilloscope which leaves the filtering entirely to the probe and has no filtering of its own?

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    \$\begingroup\$ 1GS/s is only 20 samples for a 50MHz signal. An analog scope gives you much better fidelity. \$\endgroup\$
    – Kartman
    Mar 3, 2021 at 11:39
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    \$\begingroup\$ @Kartman nonsense; Shannon-Nyquist literally says you can reproduce the exact 50 MHz bandwidth with only >100 MHz sampling rate. What is much better might be the fidelity of operation if bandlimiting to 50 MHz isn't exactly achieved. Also, some people design their oscilloscopes (or configure them) to connect sample values with straight lines, which is mathematically incorrect; but that's nothing to do with analog vs digital, but with "appropriate" and "incorrectly designed". \$\endgroup\$ Mar 3, 2021 at 11:58
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    \$\begingroup\$ @MarcusMüller "We mustn't give in to the naysayers who claim analog scopes are any better." - Oh no, I agree. And in my experience 10 samples/cycle is plenty to match or beat analog scopes. My point was more that "you only need 2 samples/cycle because Shannon-Nyquist" is not true in practice. \$\endgroup\$
    – marcelm
    Mar 3, 2021 at 13:18
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    \$\begingroup\$ @JohnB But what if the signal isn't a perfect sine wave? Or you don't even know the shape of the signal? What if your sine wave has noise? Or amplitude or frequency modulation? Yes, if the signal is a perfect sine wave and you know this, 2 samples/cycle is all you need. But in that case, why use an oscilloscope? A frequency counter would suffice. \$\endgroup\$
    – marcelm
    Mar 3, 2021 at 13:21
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    \$\begingroup\$ @marcelm if the signal isn't a perfect sine wave, then it's not the highest frequency component in the signal, and therefore your sampling rate doesn't meet the Nyquist criteria for that signal. \$\endgroup\$
    – John B
    Mar 3, 2021 at 13:23

1 Answer 1


When an oscilloscope says it has a 50 MHz bandwidth, that usually means the frequency response is 3dB down (half the power or 71% of the voltage) at 50 MHz.

Although Nyquist says you only need to sample at >100 MHz to reproduce a signal with a 50 MHz bandwidth, that means a signal with no energy above 50 MHz. The usual definition of 3dB down leaves plenty of energy above that.

Sampling at 1 GS/s, we would like no energy above 500 MHz. I'm sure our 50 MHz 'scope could achieve that filtering comfortably, if it wanted to. It probably achieves it accidentally, at least in cheaper 'scopes, by only designing the amplifiers for 50 MHz and letting them roll off at higher frequencies.

By default, most oscilloscopes connect the sampled dots together with straight lines. A 20:1 ratio between the frequency of the displayed waveform and the sample rate is just about enough to get a good-looking sinewave.

The bandwidth limitation on a 'scope probe is a similar beast to that of a scope, it should work up to that frequency, and you don't know what's going to happen above that. Don't assume either a scope or a probe is going to implement a proper filter above its bandwidth specification, don't assume it's going to work to specification above it.


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