ECEN4503 Random Signals Lecture #24 10 March 2014 Dr. George Scheets n Read 8.1 n Problems , 7.5 (1 st & 2 nd Edition) n Next Quiz on 28 March n Exam #1 Results Hi = 94, Low = 36, Average = 69.53, σ = A > 88, B > 78, C > 64, D > 51

ECEN4503 Random Signals Lecture #25 12 March 2014 Dr. George Scheets n Read 8.3 & 8.4 n Problems 5.5, 5.7, 5.20 (1st Edition) n Problems 5.10, 5.21, 5.49 (2nd Edition)

Random Number Generator n Uniform over [0,1] n Theoretical E[X] = 0.5 σ X = (1/12) 0.5 = n Actual E[X] = σ X =

Random Number Generator n Addition of 2 S.I. Uniform Random Numbers → Triangular PDF n Each Uniform over [0,1] n Theoretical E[X] = 1.0 σ X = (2/12) 0.5 = n Actual E[X] = σ X =

Random Number Generator n Addition of 3 S.I. Uniform Random Numbers → PDF starting to look Bell Shaped n Each Uniform over [0,1] n Theoretical E[X] = 1.5 σ X = (3/12) 0.5 = 0.5 n Actual E[X] = σ X =

R XY for age & weight X = Age Y = Weight R XY ≡ E[XY] = 4322 (85 data points)

R XY for age & middle finger length X = Age Y = Finger Length (cm) R XY ≡ E[XY] = (54 data points)

Cov(X,Y) for age & weight X = Age Y = Weight R XY ≡ E[XY] = 4322 E[Age] = E[Weight] = E[X]E[Y] = 4309 Cov(X,Y) = (85 data points)

Cov(X,Y) for age & middle finger length X = Age Y = Finger Length R XY ≡ E[XY] = E[Age] = E[Length] = E[X]E[Y] = Cov(X,Y) = (54 data points)

Age & Weight Correlation Coefficient ρ X = Age Y = Weight Cov(X,Y) = σ Age = years σ Weight = ρ = (81 data points)

Age & Weight Correlation Coefficient ρ X = Age Y = Weight Cov(X,Y) = σ Age = years σ Weight = ρ = (85 data points)

Age & middle finger length Correlation Coefficient ρ X = Age Y = Finger Length Cov(X,Y) = σ Age = years σ Length = ρ = (54 data points)

SI versus Correlation n ρ ≡ Correlation Coefficient Allows head-to-head comparisons (Values normalized) ≡ E[XY] – E[X]E[Y] σ X σ Y = 0? → We say R.V.'s are Uncorrelated = 0 < ρ < 1 → X & Y tend to behave similarly = -1 < ρ < 0 → X & Y tend to behave dissimilarly n X & Y are S.I.? → ρ = 0 → X & Y are Uncorrelated n X & Y uncorrelated? → E[XY] = E[X]E[Y] u Example Y = X 2 ; f X (x) symmetrical about 0 u Here X & Y are dependent, but uncorrelated u See Quiz 6, 2012, problem 1e

Two sample functions of bit streams.

Random Bit Stream. Each bit S.I. of others. P(+1 volt) = P(-1 volt) = 0.5 x volts f X (x) +1 1/2

Bit Stream. Average burst length of 20 bits. P(+1 volt) = P(-1 volt) = 0.5 x volts f X (x) +1 1/2 Voltage Distribution of this signal & previous are the same, but time domain behavior different.

Review of PDF's & Histograms n Probability Density Functions (PDF's), of which a Histogram is an estimate of shape, frequently (but not always!) deal with the voltage likelihoods Time Volts

Discrete Time Noise Waveform 255 point, 0 mean, 1 watt Uniformly Distributed Voltages Time Volts 0

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

Volts Bin Count Time Volts 0 The histogram is telling us which voltages were most likely in this experiment. A histogram is an estimate of the shape of the underlying PDF.

Discrete Time Noise Waveform 255 point, 0 mean, 1 watt Exponentially Distributed Voltages Time Volts 0

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

Discrete Time Noise Waveform 255 point, 0 mean, 1 watt Gaussian Distributed Voltages 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