1. ECEN4503 Random Signals Lecture #24 10 March 2014 Dr. George Scheets n Read 8.1 n Problems 7.1 - 7.3, 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 = 13.41 A > 88, B > 78, C > 64, D > 51

2. 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)

3. Random Number Generator n Uniform over [0,1] n Theoretical E[X] = 0.5 σ X = (1/12) 0.5 = 0.2887 n Actual E[X] = 0.5003 σ X = 0.2898

4. 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 = 0.4082 n Actual E[X] = 1.008 σ X = 0.4114

5. 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] = 1.500 σ X = 0.5015

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

7. R XY for age & middle finger length X = Age Y = Finger Length (cm) R XY ≡ E[XY] = 194.2 (54 data points)

8. Cov(X,Y) for age & weight X = Age Y = Weight R XY ≡ E[XY] = 4322 E[Age] = 23.12 E[Weight] = 186.4 E[X]E[Y] = 4309 Cov(X,Y) = 13.07 (85 data points)

9. Cov(X,Y) for age & middle finger length X = Age Y = Finger Length R XY ≡ E[XY] = 194.2 E[Age] = 22.87 E[Length] = 8.487 E[X]E[Y] = 194.1 Cov(X,Y) = 0.05822 (54 data points)

10. Age & Weight Correlation Coefficient ρ X = Age Y = Weight Cov(X,Y) = 13.07 σ Age = 1.971 years σ Weight = 39.42 ρ = 0.1682 (81 data points)

11. Age & Weight Correlation Coefficient ρ X = Age Y = Weight Cov(X,Y) = 13.07 σ Age = 1.971 years σ Weight = 39.42 ρ = 0.1682 (85 data points)

12. Age & middle finger length Correlation Coefficient ρ X = Age Y = Finger Length Cov(X,Y) = 0.05822 σ Age = 2.218 years σ Length = 0.5746 ρ = 0.04568 (54 data points)

13. 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

14. Two sample functions of bit streams.

15. 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

16. 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.

17. 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

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

19. 15 Bin Histogram (255 points of Uniform Noise) Volts Bin Count 0

20. 15 Bin Histogram (2500 points of Uniform Noise) Volts Bin Count 0 0 200 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.

21. 15 Bin Histogram (2500 points of Uniform Noise) Volts Bin Count 0 140 200 But there will still be variation if you zoom in.

22. 15 Bin Histogram (25,000 points of Uniform Noise) Volts Bin Count 0 0 2,000

23. 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.

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

25. 15 bin Histogram (255 points of Exponential Noise) Volts Bin Count 0

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

27. 15 bin Histogram (255 points of Gaussian Noise) Volts Bin Count 0

28. 15 bin Histogram (2500 points of Gaussian Noise) Volts Bin Count 0 400