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A ccuracy of
1
Standard
Deviation
Measurements
Example 1
Example 2
An interval that does not vary still can be shown to have a large
standard deviation due to the sampling process. The error in the
displayed standard deviation depends on the size of two elements:
the portion of the interval that exceeds the multiple of the
sample periods; and
the portion of the interval that includes complete sample
periods.
Assume the timing analyzer is measuring a time interval of exactly
25.0 ns. It makes 10 executions using a 10 ns sample period. Five
of the executions show the interval to last 30 ns (three sample
clocks), and five of the executions show the interval to last 20 ns
(two sample clocks). Even though the input signal has a true
standard deviation of 0.0 ns, the timing analyzer will calculate the
standard deviation of this signal to be 5.28 ns and display this
standard deviation on-screen. When the sampled standard
deviation is less than one sample period, its value is determined
mainly by the sampling process.
Assume the timing analyzer is measuring a time interval that varies
from 5 to 8 us. The timing analyzer is operating with a 10 ns
sample period. After a series of repetitive measurements, the
timing analyzer shows a standard deviation of 1.5 us for the interval
being measured. This dispersion is determined by variations in the
time interval itself, and not by the sampling process. In this case,
the standard deviation is much larger than one sample period.
Accurate Timing Measurements D-5
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