ECEN4523 Commo Theory Lecture #38 16 November 2015 Dr. George Scheets n Read 8.6 & 8.7 n Problems: 8.6-1, 3, & 4 n Quiz #9, 20 November (Live) [Chapter 8] u Remote DL students < 27 November n Design Problem u Reworks due 2 weeks after return

ECEN4523 Commo Theory Lecture #39 18 November 2015 Dr. George Scheets n Section 9.3; Read 523 – 528, Scan Remainder. Read 9.5 n Problems: n Quiz #9, 20 November (Live) [Chapter 8] u Remote DL students < 27 November n Design Problem u Reworks due 2 weeks after return u +10 & +5 Extra Credit still available n Quiz 8 Results u Hi = 10, Low = 4.3, Ave = 7.81, σ = 2.47

Lizzy Borden n Alleged 1892 Ax Murderer Lizzie Borden took an axe And gave her mother forty whacks. And gave her mother forty whacks. When she saw what she had done She gave her father forty-one. I prefer Statistically Independent Random Variables.

Random Variable n Assigning numbers to experimental outcomes u X ≡ Set of all possible assigned numbers u x ≡ A specific number from the set X n RV could just as easily be "Random Voltage"… u … in many cases

24 Hour Rainfall Amounts | It Rained Stillwater, OK ≈ 6,200 days

Call Length on Cellular Telephones Source: "Primary User Behavior in Cellular Networks..." IEEE COMMUNICATIONS MAGAZINE, March 2009

Term Paper Scores

Histograms n Number of bins → ∞ n Bin size → 0 n Total Number of counts → ∞ n Bin Value = Bin Count / Total Count n Result = PDF

Noise Waveforms 255 point Zero Mean Uniform Noise Time Volts 0

15 Bin Histogram (255 points of Uniform Noise) Volts Bin Count 0

Volts Bin Count Time Volts

15 Bin Histogram (255 points of Uniform Noise) Volts Bin Count 0

15 Bin Histogram (2500 points of Uniform Noise) Volts Bin Count When bin count range is from zero to max value, a histogram of a uniform PDF source will tend to look flatter as the number of sample points increases.

15 Bin Histogram (2500 points of Uniform Noise) Volts Bin Count But there will still be variation if you zoom in.

15 Bin Histogram (25,000 points of Uniform Noise) Volts Bin Count 0 0 2,000

Noise Waveforms 255 point Zero Mean Gaussian Noise Most Real World Noise is Gaussian Time Volts 0

15 bin Histogram (255 points of Gaussian Noise) Volts Bin Count 0

15 bin Histogram (2500 points of Gaussian Noise) Volts Bin Count 0 400

Noise Waveforms 255 point Zero Mean Exponential Noise Time Volts 0

15 bin Histogram (255 points of Exponential Noise) Volts Bin Count 0

Test: What is the Voltage Histogram of this random waveform? time x(t) (volts) 10% of time voltage = 5 volts. 90% of time voltage = 0 volts. 5 0

Bin Count Voltage Histogram 05 90% 10% voltage

f X (x) Voltage PDF voltage

Bin Count You randomly sample the time waveform 100 times over a 60 minute period. Would you expect exactly 90 counts of 0 volts? voltage You should be able to calculate P(exactly ninety 0v counts | 100 random samples)? (Binomial PDF)

Thermal Noise n Weak Wideband RF Noise n Power Spectrum flat out to ≈ 1,000 GHz n Two sided spectral density N o /2 = kT o /2 watts/Hz u k ≡ Boltzmann's Constant = 1.381* w/(HzK) u T o ≡ Temperature in degrees Kelvin F ºF = (K )* degrees F = degrees Kelvin F Thermal Noise Power in 1,000 GHz = (N o /2)(2*1,000*10 9 ) = nanowatts n Gaussian distributed voltages n Can ignore in most situations n Cannot ignore on RF receivers!

Discrete Time Noise Waveforms 255 point Zero Mean Gaussian Noise Thermal Noise is Gaussian Time Volts 0

Venn Diagram for Back-to-Back BSC 1st Hop OK 1st Hop BE 2nd Hop OK2nd Hop BE 1st OK ∩ 2nd OK 1st BE ∩ 2nd OK 1st OK ∩ 2nd BE 1st BE ∩ 2nd BE

Venn Diagram for Back-to-Back BSC 1st Hop OK 1st Hop BE 2nd Hop OK2nd Hop BE 1st OK ∩ 2nd OK 1st BE ∩ 2nd OK 1st OK ∩ 2nd BE 1st BE ∩ 2nd BE P(End-to-End BE) = P(1OK ∩ 2BE) + P(1BE ∩ 2OK)

Venn Diagram for Back-to-Back BSC 1st Hop OK 1st Hop BE 2nd Hop OK2nd Hop BE 1st OK ∩ 2nd OK 1st BE ∩ 2nd OK 1st OK ∩ 2nd BE 1st BE ∩ 2nd BE P(End-to-End BE) = P(1OK ∩ 2BE) + P(1BE ∩ 2OK)

Venn Diagram for Back-to-Back BSC 1st Hop OK 1st Hop BE 2nd Hop OK2nd Hop BE 1st OK ∩ 2nd OK 1st BE ∩ 2nd OK 1st OK ∩ 2nd BE 1st BE ∩ 2nd BE P(End-to-End BE) = P(1OK ∩ 2BE) + P(1BE ∩ 2OK) - P([1OK ∩ 2BE] ∩ [1BE ∩ 2OK])

16 QAM Signal Constellations n Transmitted Constellation has dots n Received Constellation samples are random u Divide into bins u Get a histogram, in the limit get PDF source: waltertech426.blogspot.com/2013/08/matlab-m-ary-quadrature-signal.html & Wikipedia

2nd order Gaussian PDF source: