How does averaging impact oscilloscope's bandwidth?

[Muxr]

So on my HP 54610B there is a large disparity between the analog bandwidth and single shot "digital" bandwidth. It's two orders of magnitude different. Something like 500Mhz vs 5Mhz (Scope's sample rate is 20Ms/s).

For repetitive signals I assume I am getting 500Mhz. But what happens when I enable averaging? Again for a repetitive stable signal. The analog signal has to be sampled, does the bandwidth drop?

Observing a signal without averaging:

Signal with averaging x256 after about 30 seconds:

[tggzzz]

So on my HP 54610B there is a large disparity between the analog bandwidth and single shot "digital" bandwidth. It's two orders of magnitude different. Something like 500Mhz vs 5Mhz (Scope's sample rate is 20Ms/s).
For repetitive signals I assume I am getting 500Mhz. But what happens when I enable averaging? Again for a repetitive stable signal. The analog signal has to be sampled, does the bandwidth drop?

No, it doesn't drop. The waveform is sampled at full bandwidth, and then the Nth sample in one sweep is averaged with the Nth samples in other sweep. If the Nth sample in one sweep was averaged with the (N+1)th sample in the same sweep[1] then the bandwidth would drop.

Try looking at a digital PRBS. The first bit after the trigger will look like a "normal" bit, but subsequent bits will be 50% high and 50% low, so they will be averaged towards the mean value.

[1] if the trigger is not stable w.r.t. the repetitive signal, then the Nth signal in one sweep may be averaged with the (N+1)th sample in the next sweep, which also implies a reduced bandwidth.

[Muxr]

So on my HP 54610B there is a large disparity between the analog bandwidth and single shot "digital" bandwidth. It's two orders of magnitude different. Something like 500Mhz vs 5Mhz (Scope's sample rate is 20Ms/s).
For repetitive signals I assume I am getting 500Mhz. But what happens when I enable averaging? Again for a repetitive stable signal. The analog signal has to be sampled, does the bandwidth drop?

No, it doesn't drop. The waveform is sampled at full bandwidth, and then the Nth sample in one sweep is averaged with the Nth samples in other sweep. If the Nth sample in one sweep was averaged with the (N+1)th sample in the same sweep[1] then the bandwidth would drop.

Try looking at a digital PRBS. The first bit after the trigger will look like a "normal" bit, but subsequent bits will be 50% high and 50% low, so they will be averaged towards the mean value.

[1] if the trigger is not stable w.r.t. the repetitive signal, then the Nth signal in one sweep may be averaged with the (N+1)th sample in the next sweep, which also implies a reduced bandwidth.

Ahh that makes sense. Thanks for the explanation!

[Kleinstein]

If the Signal is perfectly stable, more averaging only improves S/N ratio. So no change in bandwidth or similar change in the displayed waveform.

If the Signal is not perfectly stable, one can get a distorted signal: e.g. if the resonance drifts significantly the apparent damping of the ringing may increase with averaging. Also trigger errors tend to smooth out the signal, reducing the amplitude at higher frequencies. Usually the effect is small. Larger effects can appear if the is some kind of interference of line frequency and a signal that is nearly (but no perfectly) in sync with the line signal.

[T3sl4co1l]

Averaging narrows the bandwidth to spikes centered around the trigger rate and its harmonics, ideally with equal weighting, up to the analog bandwidth of the channel (i.e., including aliasing at and above the Nyquist frequency).  The spikes have a sinc(f) form, because the average is uniformly weighted (i.e., a "sliding average" filter, ideally -- though a first order IIR is probably most common).  The (-3dB) width of each spike is something like Fs/(2*N) for N averages, and the maximum attenuation (valley between spikes) goes as 1/N.  (Incoherent random noise adds as RMS, so that the expected SNR gain goes as sqrt(N).)

So, it filters noise at frequencies aperiodic/anharmonic to the trigger rate.

Note that the trigger itself must be very repeatable to get a stable averaged waveform.  If the trigger itself is influenced by noise (a trigger comparator is also a mixer, so jitter is introduced by triggering off a noisy waveform), then the resulting sample bandwidth will be smeared by the same amount -- this is especially noticeable at the high harmonics, where the jitter in sample rate is multiplied by N.

An illustration of that might be, a switching waveform with a very sharp rise time and ringing; if averaging attenuates the leading edge and ringing, then the frequency components in that edge are being smeared out by the jitter.

Tim

[Muxr]

Great answers guys, thank you!