2020 Solution 06

Total jitter (Tj) is calculated by the following formula.
​
Tj = Dj + QBER Rj
​
Where,
Tj is the total jitter
Dj is the deterministic jitter component
Rj is the random jitter component
QBER is a multiplier.
Choose the best response to the following statement.​
​"Peak to peak random jitter will vary with the number of samples."
(a) True. Random jitter is "unbounded", with tails in the distribution. Therefore measured pk-to-pk values will change with the number of samples.​
​(b) False​
​(c) Indeterminate​
​(d) Unlikely​
​(e) Likely​
​(f) .......
Your answer

Verify

The following is a trivial verification. First generate a list of random numbers.
In[]:=
myRandomNumbers=RandomVariate[NormalDistribution[0,0.3],10^3];
It is essential to to verify that the dataset meets the expectations about boundedness. Plot the dataset to see if it looks like a Gaussian distribution with tails.
In[]:=
Histogram[myRandomNumbers,AxesOrigin{0,0},PlotRangeAll]
Out[]=
Pick up random samples of varying lengths from the list of random numbers. You will see three sublists with 20, 50 and 80 elements.
In[]:=
randomSamples=RandomSample[myRandomNumbers,#]&/@{20,50,80}
Out[]=
{{0.143469,0.33215,-0.0933257,-0.00717707,-0.44774,0.477221,0.141959,-0.120462,0.0682826,0.1866,0.394421,-0.0262532,0.14604,-0.140918,0.646917,0.0272508,0.674578,-0.468153,0.165312,-0.118361},{0.611546,-0.127389,-0.167028,-0.0819431,-0.113884,0.141233,-0.358543,0.3892,0.374876,0.205589,-0.547886,0.296499,0.159787,-0.116407,0.0471571,0.232443,-0.298479,-0.142546,-0.0392723,0.232682,-0.0273524,0.00667336,0.137234,-0.238624,-0.325941,0.111439,-0.586394,0.362974,-0.29943,0.101678,0.411517,0.18324,-0.032499,-0.521616,-0.307438,-0.0100416,-0.0467681,0.180776,-0.26121,0.146324,0.173702,0.100197,-0.157089,0.506017,-0.245983,0.0484572,-0.426327,-0.00965429,0.159299,0.0375794},{0.0490753,-0.157289,0.139494,0.319027,-0.215311,-0.0569092,-0.0529944,0.395318,-0.063641,-0.440002,-0.384882,0.0385878,-0.130095,0.437084,0.0933442,0.236701,0.0756806,-0.538021,-0.307438,0.118944,0.21126,0.326494,0.153812,0.0883184,-0.418382,-0.150329,0.571329,0.0371754,-0.0167733,-0.553166,-0.342943,-0.377584,-0.0341578,0.296499,0.0652008,0.353415,0.313709,-0.133556,0.50361,0.341605,0.306505,0.44596,0.0913112,-0.20209,-0.013103,-0.127389,0.248897,0.119589,-0.327647,0.217366,-0.134043,-0.0103086,0.027102,-0.348384,-0.315566,0.37618,-0.216593,-0.608052,-0.251285,0.13965,0.0375794,-0.0433797,-0.310569,0.18324,-0.670764,-0.554966,-0.0707858,-0.345559,0.18966,-0.294505,0.031725,-0.132071,0.13022,0.224113,-0.237211,-0.242634,0.6817,0.224452,0.255429,0.257515}}
Test each sublist to find the peak-to-peak value within that sublist. Does it vary with the number of elements sampled? The result is of the form {no.of samples, pk-pk}.
In[]:=
samplesPkpk={Length@#,Max[#]-Min[#]}&/@randomSamples
Out[]=
{{20,1.14273},{50,1.19794},{80,1.35246}}
Its always a good idea to plot.
In[]:=
ListPlot[samplesPkpk,FrameTrue,FrameStyle16,FrameLabel{"Samples, n","pk-pk, Unitless"},GridLinesAutomatic]
Out[]=
The peak-to-peak value varies with the number of samples.