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I have the oscilloscope R&S® RTC1000 at 300MHz / 2GS/s / 500Mpts (Mega Points) with 12 horizontal divisions on the screen.

1st Example:
If I have a signal 1Hz the total samples that the oscilloscope will take according to the length of entire displayed signal will be:

f = 1Hz
T = 1sec
T/div = 1sec/12divisions≈0,833sec/div = 83,3ms/div
Total Recorder Samples = Samples Rate * Time/div * Screen Horizontal Divisions
= 2GS/s * 83,3ms/div * 12div ≈ 2GS = 2.000.000.000 samples,

but oscilloscopes memory can maximum store 1.000.000.000 samples per channel, so if I want to measure two signals of 1Hz at each channel the oscilloscope will need sto store 2.000.000.000 samples per channel, so the oscilloscope will decrease the sampling rate at 1.000.000.000 in order the total number of samples can be stored in his memory where is 1.000.000.000 samples (1Mpts) per channel?

2nd Example:
f = 300MHz
T = 3,33ns
T/div = 3,33ns/12divisions ≈ 0,278ns/div
Total Recorder Samples = 2GS/s * 0,278ns/div * 12div ≈ 5.56 samples,

So (if I am correct to my calculations) the signal on the oscilloscopes screen will be distorted and not accurate because of the two little samples(*1)? And then an oscilloscope with 300MHz bandwidth (or whatever bandwidth value) what signals can measure with a decent accuracy?

*1: According to Nyquist's theorem the minimum sampling frequency is fsmin ≥ 2fsignal and for this particular oscilloscope the minimum sampling frequency is 2GS/300MHz = 6,66 times over the oscilloscopes bandwidth value, but in practice I don’t believe that this is enough for a decent accurate recreation of a signal.

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  • \$\begingroup\$ Where was oscilloscopes memory can maximum store 500.000.000 samples defined? The R&S®RTC1000 OSCILLOSCOPE Specifications contains Memory depth per channel : 2 × 1 Msample or 1 × 2 Msample for the Base unit Horizontal system. I.e. I read that as 1.000.000 samples per channel with using two channels, or 2.000.000 samples when using a single channel. \$\endgroup\$ Commented Jun 2, 2024 at 15:16
  • \$\begingroup\$ I apologise. My fault is 2 Mpts, so 2000000 samples. \$\endgroup\$ Commented Jun 2, 2024 at 15:26

1 Answer 1

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There are really two things going on. Sampling into the memory buffer for each channel, and rendering for display.

Scope Paramters

The parameters you quote seem to be confusing Mega and Giga when it comes to memory depth, and also partially differ from the datasheet, which shows 2GSps (1GSps) with 2Mpt (1Mpt) memory depth for 1ch (2ch) mode. I base this answer on the datasheet presented values.

Acquisition capabilities

Acquisition Parameters

You are correct at 2GSps, you will only capture ~6.7 samples per cycle of your 300MHz waveform.

You are also correct that to capture a full cycle of a 1Hz waveform, your scope will have to drop the sample rate to compensate for memory limits, giving a max sample rate of 2MSps to get a full 1 second capture.

If you were to download the raw data from the oscilloscope and look at it, you will find 2Mpts for the 1Hz capture. You will find probably much more than 7 samples for the 300MHz capture, because although only the one cycle is displayed on-screen, the scope probably recorded the full 2Mpts buffer length so will have many cycles worth.

You are also correct that the oversampling factor is plenty to capture the full information from a 300MHz bandwidth-limited signal as indicated by the Nyquist theorem.

Display

However this is where things get interesting. How does the oscilloscope display the waveform on-screen? The scope display resolution is after-all fixed at a few hundred to a couple thousand pixels wide. From the datasheet again, your scope has a 640x480 pixel display, so it can show at most 640 samples on-screen.

  • For your 1Hz waveform with 2Mpts of data, there is no way that every single sample can be rendered on-screen.

    Instead the scope will decimate the data (either intelligently with a multi-rate filter, or by simply keeping every Nth sample).

  • For your 300MHz cycle, there are much fewer samples than can be displayed on-screen. So somehow the scope needs to convert from samples to pixels with interpolation.

    A low-end scope will do this with simple linear interpolation. Basically space the few samples it has equally across the screen, then draw a straight line between them. This will as you predict look very low resolution and nothing at all like your real waveform. Welcome to the world of pushing cheap test equipment to the limit.

    A higher-end scope is much more intelligent. Remember Nyquist? Assuming a bandwidth limited signal, we have all the information we need to recovery the real signal exactly (not quite, that statement neglects quantisation errors and noise). From the few samples the scope has, it can use more advanced interpolation methods to bump up the sample rate and recover the time-continuous signal. A common method is using sinc(x) interpolation, a.k.a. Whittaker-Shannon interpolation. By increasing the sample rate high enough to get one sample per display pixel, you get a very nice smooth, albeit frequency limited, view of the original signal.

    If there is any frequency content beyond the bandwidth limit of the scope and interpolation, this information is lost (either through aliasing or low-pass filtering depending on the scope and signal amplitudes), so you will not see any of this high frequency content - e.g. a 300MHz square wave will look very sinusoidal.

Looking at your scope datasheet again, it does support what it calls sin(x)/x interpolation for display. This is the same thing as sinc interpolation. It also supports the lower end linear interpolation presumably through some sort of display setting.

Display modes for R&S RTC1000 Scope

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  • \$\begingroup\$ Hi Tom and I have no words to thank you for your time and for your detailed answer. If you please, I would like to ask you some more things. 1st: What is the “sample & hold” interpolation technic (I didn’t find something on the internet – maybe has a different official term?)? 2nd: Why to have a 2.000.000.000samples/sec (with the using of one channel only) if we can only store 2.000.000samples? Many Thanks again. \$\endgroup\$ Commented Jun 3, 2024 at 21:03
  • \$\begingroup\$ Also, I tried to measure a sinus signal at 100uHz. I had set the T/div=100s and the oscilloscope at the upper side of the screen show that his sample rate was 5samples/sec. So, the total samples number that he must took will be 100s*12divisions*5samples/sec=6000samples. Can you imagine why he drop the samples so down? His memory has capacity of 2Mpts (2.000.000samples), so in theory he couldn’t take more samples? The interpolation mode that the oscilloscopes use was the “sin(x)/x” and for this I believe I saw a smooth (just atiny pixeling) sinus signal. i.postimg.cc/XqtXXYX0/IMG323.png \$\endgroup\$ Commented Jun 3, 2024 at 21:04
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    \$\begingroup\$ @DarkKnight the sample and hold mode is probably straircase style view. So hold the current value (horizontal line) until the next sample (draw vertical line to new value) forming a step shape waveform. \$\endgroup\$ Commented Jun 3, 2024 at 21:19
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    \$\begingroup\$ Sample rate and memory are two only weakly linked specifications of the system. Having a high sample rate ensures you can see high frequency signals and get a much more accurate view of any waveform. Having a large sample memory allows storing long duration waveforms so you can see how things change over time (e.g. a modulated data signal). There's inevitably a trade off - it's hard to make very fast very deep memory. \$\endgroup\$ Commented Jun 3, 2024 at 21:28
  • \$\begingroup\$ Any idea about why in sinus signal at 100uHz that I tried to measure the samples where to few? \$\endgroup\$ Commented Jun 4, 2024 at 16:31

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