Could you post a bit of the stored data and an image of the real data? This would help folks to answer the question. There is always a bit of discrepancy between real signals and digitized data. The quality of the analog front end and the ADC parameters need to be optimized to the signal content in order to get an accurate representation of the signal in digital form. If the sample rate is too slow, for instance, you will get aliasing  that makes the signal look like it has frequency content that doesn't actually exist in the real signal and cause some frequency content to be lost in the digitized form. Since the scope does not have any means to store the original analog signal, you will not be able to recover the real data.

Dear Ron

I think this wouldn't be a technical problem and the key is in excel format that I have saved. but It's the first time that I'm using saved data from the scope and the time is limited to discover. I have attached an image and an excel file from a sample waveform of function generator having Vp-p=5 V & f=1 Hz.

Dear Farzane,

After inspecting your waveform and you excel sheet i find that you had correctly acquired your signal. It is a square wave with a periodic time of one hertz and half cycle time of 500 mseconds and Vpp =+- 2.5 V. You have used an A/D converter with a resolution of 7 bits,one sign bit and 6 magnitude bits for + - 2.5 V. Therefore you get dialectal number

of 2^6= 64. You have also noise superimposed on you waveform in the order of +- 2 steps. Your sampling time is 500/25=20 mseconds, giving sampling frequency of 50Hz which is much greater than the Nyquist limit of 2Hz. 

Dear Farzane,

After inspecting your waveform and you excel sheet i find that you had correctly acquired your signal. It is a square wave with a periodic time of one hertz and half cycle time of 500 mseconds and Vpp =+- 2.5 V. You have used an A/D converter with a resolution of 7 bits,one sign bit and 6 magnitude bits for + - 2.5 V. Therefore you get dialectal number

of 2^6= 64. You have also noise superimposed on you waveform in the order of +- 2 steps. Your sampling time is 500/25=20 mseconds, giving sampling frequency of 50Hz which is much greater than the Nyquist limit of 2Hz. 

Thank you Abdelhalim.

I think that in this situation, there would be a kind of function that converts raw data of excel into real data, and in all of cases we can make this function.If this is true, How can I make this function?

Yea Farzane. 

It is the A/D relation.  The analog value V= D times Vstep, where D is the digital value and V step is the analog value of the the digit one. Vstep= VFS/ 2^n with VFS is the ful scale value and n the number of bots in digital value. In your case n is =6 and VFs=2.5 V. Then Vstep= 2.5/ 64. So, since your peak number in the excel sheet= 64 , the the analog value will be 64x 2.5/64=2.5 volt.Then the scaling factor is Vstep.

wish you success.

Hi

In previous post i had attached just one of my data from scope. But my real data were in different ranges. today the company finally answered me and I found that the key was number of points in the screen. I have attached this file for those who want to know.

thanks a lot for your attention and taking time.

Dear Farzave,

This agrees with what i said no more no less. In the third column in the spread sheet you have the the digital values D and your resolution of the A/D CONVERTER IS 2 mV per step.This means you have an A/D converter with Vstep2 mV.

So the first analog value in the table will be D Vstep  2mVx 51 102 mV and the second value will be -21x 2mV40 Mv and so on....

If you change your resolution the The Dvalue will change accordingly. E.G if the resolution is made 4mV , THE D VALUE FOR THE FIRST POINT IN THE sHEET WILL BE 25.

WISH YOU SUCCESS.