scope

[esra]

Hello

I have a Lecroy 9304A oscilloscope with 8-bit vertical resolution.
I have collected my signal with 10 Msamples/sec, with time/division = 0.5 ms. so it collects 50000 samples. On y axis I have voltage and on x-axis I have time.

I was wondering how I can calculate the y-axis and x-axis resolution?

Like what is the minimum voltage I can measure on y axis and what is the minimum time I can measure on x-axis?

Thank you all

[Dennis_Knutson]

This is all basic arithmetic. The minimum time is 1/sample rate. The minimum voltage is range/256 (2*8). You have not specified the vertical range you are using.

[DFGray]

You may also want to look at the spec sheet of your scope.  The digital resolution (resolution you can get from your data) is often different from, and usually higher than, the analog front-end accuracy.  Your specs should give you the accuracy of the scope (vertical and horizontal) at the various settings the scope is capable of.

[esra]

Hi

Thank you for your response, it ssays vertical resolution is 8-bit. so what does this correspond in voltage??

[esra]

HI

Thank you for your response. I have attached a graph that I collected from scope. Im wondering why this graph is not fully resolved. The y axis is voltage in Volts and x axis is # of samples/sec.

Sample rate was 10 Msamples/sec and the time/div was 0.5 ms.

Thank you

[esra]

attached is th graph.

[dansch]

Hi esra,

It is hard to tell just by looking at the graph because it is not clear if the scale of the graph corresponds to the voltage range being used. According to the manual for the LeCroy 9304A (found here), the sensitivity of the scope is 2 mV/div up to 5 V/div. Assuming there are 8 vertical divisions, that corresponds to ranges of 16 mVpp up to 40 Vpp. It is not clear what the discrete range levels actually are. Depending on the range you used, as Dennis pointed out, you would divide that value by 2^8 to get what the best-case resolution is.

Since that probably won't help you based on what you are trying to figure out, perhaps you can approximate what you need looking at the graph. It won't give you an exact or accurate value, but zooming in it appears that there are approximately 6 or 7 possible levels (probably 6.5) of samples between each vertical scale (which corresponds to 0.002 of whatever units), so doing the math it appears that the "resolution" being shown is about 0.00031 of whatever units the y axis represents. Obviously, without knowing the exact levels of the samples, I cannot give an exact number.

Best of luck,

[esra]

thank you for your answer, so since y axis is in Volts, the y-axis resolution is 0.3 mV. Is this the reason why the graph is not fully resolved?? I mean the graoh looks like a line at some points is this why.

Also, what about x-axis because I can see the same lines in the horizontal direction,too. How can I calculate the resolution in x direction if the x-axis is in seconds? Say the axis is from 0 to 3 seconds.

Thank you

[dansch]

Well, even that 0.3 mV is a very approximate value. I am not really sure what you mean by it being "full resolved". There is definitely some behavior causing all of the samples to occur at the same steps in time, whether it be due to the vertical resolution limitations or something else.

As far as the horizontal resolution, it is not as easy to approximate like we did for the y-axis because some of the lines are completely solid (such as around 0.004 V), and it is hard to count how many points there are in the horizontal direction. Furthermore, the x-axis does not appear to be a unit of time anyway (and thus not a reading directly from a scope). In that graph, it is a number of samples per second. I wish I could give you even an approzimate answer, but I don't really see any way to. If you are assuming that the full rate of the scope was being used, and you absolutely had to have some kind of time resolution, the answer would be 10 ns as your period (time resolution). However, that cannot be the case for this graph because that would mean that in those 3 seconds you mentioned (x-axis being 0 to 3 seconds), there would be 300,000,000 possible points.

Edit: I remember your first post that you mentioned having a rate of 10 MS/s. Thus, if you acquire for 3 seconds, you would have 30,000,000 total samples. Clearly the graph you linked is not similar to what the actual scope reading is.

To be honest, the graph is hard for me to understand because it is implying that for a given signal intensity (y-axis), there can be multiple different values for number of samples per second (x-axis). That is like saying there are 10,000 samples per second at 0.003 V but there are also 11,000, 12,000, 15,000, etc. samples per second at 0.003 V; that cannot be, there has to be one number of samples per second and only one number. Perhaps I am slightly misunderstanding the data.

Best of luck,

[esra]

ok, so I did the graph again with the values right out of the scope

x-axis is time (in seconds)

y-axis is voltage (in Voltage)

sample rate is 10 Msamples/sec.

By saying not fully resolved , I mean the horizintal and vertical lines in both axis. I need to explain why this is happening. and the only way to explain is the resolution. so, you are saying the resolution in y axis would be 0.3 mV, so that will be the smallest I can measure. since I corrected the x-axis, what would be the x-axis resolution?

Thank you so much