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Edited by Oleg10011001 at 2016-7-22 00:23
I'm sorry, I do not quite understand how it is possible that in this case it is not clear what the dots (mean of 800 and 4000) ... As you can see, in both cases prescribed scan time - 2ns. At this sampling frequency of the signal - 1Gs/sec. Thus, it appears that we must have only two dots on the cell. The number of cells we have, on the width of the screen, not much different. In one mode, the window 19 cells, and in the lower window when two windows 16 cells. Thus, one can easily calculate that the width of the screen we can have only 38 dots and 32 dots, respectively; their density does not differ significantly. More dots can only be used in an equivalent sample 25Gs/sec, but it is not used here. What is it I do not mean right?
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