ECEN4503 Random Signals Lecture #24 10 March 2014 Dr. George Scheets n Read 8.1 n Problems 7.1 -...

ECEN4503 Random SignalsECEN4503 Random SignalsLecture #24 10 March 2014Lecture #24 10 March 2014Dr. George ScheetsDr. George Scheets

ECEN4503 Random SignalsECEN4503 Random SignalsLecture #24 10 March 2014Lecture #24 10 March 2014Dr. George ScheetsDr. George Scheets Read 8.1Read 8.1 Problems 7.1 - 7.3, 7.5 (1Problems 7.1 - 7.3, 7.5 (1stst & 2 & 2ndnd Edition) Edition) Next Quiz on 28 MarchNext Quiz on 28 March

Exam #1 ResultsExam #1 ResultsHi = 94, Low = 36, Average = 69.53, Hi = 94, Low = 36, Average = 69.53, σσ = 13.41 = 13.41A A >> 88, B 88, B >> 78, C 78, C >> 64, D 64, D >> 51 51



Read 8.3 & 8.4Read 8.3 & 8.4 Problems 5.5, 5.7, 5.20 (1st Edition)Problems 5.5, 5.7, 5.20 (1st Edition) Problems 5.10, 5.21, 5.49 (2nd Edition)Problems 5.10, 5.21, 5.49 (2nd Edition)

Random Number GeneratorRandom Number Generator Uniform over [0,1]Uniform over [0,1]

TheoreticalTheoreticalE[X] = 0.5E[X] = 0.5σσXX = (1/12) = (1/12)0.50.5 = 0.2887 = 0.2887

ActualActualE[X] = 0.5003E[X] = 0.5003σσXX = 0.2898 = 0.2898

Random Number GeneratorRandom Number Generator Addition of 2 S.I. Uniform Random Numbers → Triangular PDFAddition of 2 S.I. Uniform Random Numbers → Triangular PDF Each Uniform over [0,1]Each Uniform over [0,1] TheoreticalTheoretical

E[X] = 1.0E[X] = 1.0σσXX = (2/12) = (2/12)0.50.5 = 0.4082 = 0.4082


Random Number GeneratorRandom Number Generator Addition of 3 S.I. Uniform Random Numbers → PDF starting to look Bell ShapedAddition of 3 S.I. Uniform Random Numbers → PDF starting to look Bell Shaped Each Uniform over [0,1]Each Uniform over [0,1] TheoreticalTheoretical

E[X] = 1.5E[X] = 1.5σσXX = (3/12) = (3/12)0.50.5 = 0.5 = 0.5


RXY for age & weightRXY for age & weight

X = AgeY = Weight

RXY ≡ E[XY] = 4322

(85 data points)

RXY for age & middle finger lengthRXY for age & middle finger length

X = AgeY = Finger Length (cm)

RXY ≡ E[XY] = 194.2

(54 data points)

Cov(X,Y) for age & weightCov(X,Y) for age & weightX = AgeY = Weight

RXY ≡ E[XY] = 4322E[Age] = 23.12E[Weight] = 186.4E[X]E[Y] = 4309Cov(X,Y) = 13.07

(85 data points)

Cov(X,Y) for age & middle finger lengthCov(X,Y) for age & middle finger length

X = AgeY = Finger Length

RXY ≡ E[XY] = 194.2E[Age] = 22.87E[Length] = 8.487E[X]E[Y] = 194.1Cov(X,Y) = 0.05822

(54 data points)

Age & Weight Correlation Coefficient ρAge & Weight Correlation Coefficient ρ

X = AgeY = Weight

Cov(X,Y) = 13.07σAge = 1.971 yearsσWeight = 39.42ρ = 0.1682

(81 data points)

Age & Weight Correlation Coefficient ρAge & Weight Correlation Coefficient ρ

X = AgeY = Weight

Cov(X,Y) = 13.07σAge = 1.971 yearsσWeight = 39.42ρ = 0.1682

(85 data points)

Age & middle finger length Correlation Coefficient ρAge & middle finger length Correlation Coefficient ρ

X = AgeY = Finger Length

Cov(X,Y) = 0.05822σAge = 2.218 yearsσLength = 0.5746ρ = 0.04568

(54 data points)

SI versus CorrelationSI versus Correlation ρρ ≡ Correlation Coefficient ≡ Correlation Coefficient

Allows head-to-head comparisons (Values normalized)Allows head-to-head comparisons (Values normalized) ≡ ≡ E[XY] – E[X]E[Y]E[XY] – E[X]E[Y] σσXXσσY Y

= 0? → We say R.V.'s are Uncorrelated = 0? → We say R.V.'s are Uncorrelated = 0 < = 0 < ρρ < 1 → X & Y tend to behave similarly < 1 → X & Y tend to behave similarly = -1 < = -1 < ρρ < 0 → X & Y tend to behave dissimilarly < 0 → X & Y tend to behave dissimilarly

X & Y are S.I.? → X & Y are S.I.? → ρρ = 0 → X & Y are Uncorrelated = 0 → X & Y are Uncorrelated X & Y uncorrelated? → E[XY] = E[X]E[Y]X & Y uncorrelated? → E[XY] = E[X]E[Y]

Example Y = XExample Y = X22; f; fXX(x) symmetrical about 0(x) symmetrical about 0

Here X & Y are dependent, but uncorrelatedHere X & Y are dependent, but uncorrelated See Quiz 6, 2012, problem 1eSee Quiz 6, 2012, problem 1e

Two sample functions of bit streams.Two sample functions of bit streams.

0 20 40 60 80 1001

0

11.25

1

xi

1000 i

0 50 100 150 200 250 300 350 4001

0

11.25

1

xi

4000 i

Random Bit Stream. Each bit S.I. of others.P(+1 volt) = P(-1 volt) = 0.5Random Bit Stream. Each bit S.I. of others.P(+1 volt) = P(-1 volt) = 0.5

0 20 40 60 80 1001

0

11.25

1

xi

1000 i

x volts

fX(x)

+1-1

1/2

Bit Stream. Average burst length of 20 bits.P(+1 volt) = P(-1 volt) = 0.5Bit Stream. Average burst length of 20 bits.P(+1 volt) = P(-1 volt) = 0.5

x volts

fX(x)

+1-1

1/2

0 50 100 150 200 250 300 350 4001

0

11.25

1

xi

4000 i

Voltage Distribution ofthis signal & previousare the same, but time

domain behavior different.

Review of PDF's & HistogramsReview of PDF's & Histograms Probability Density Functions (PDF's), of which a Probability Density Functions (PDF's), of which a

Histogram is an estimate of shape, frequently (but not Histogram is an estimate of shape, frequently (but not always!) deal with the voltage likelihoods always!) deal with the voltage likelihoods

Time

Volts

Discrete Time Noise Waveform255 point, 0 mean, 1 watt

Uniformly Distributed Voltages


Uniformly Distributed Voltages

Time

Volts

0

15 Bin Histogram(255 points of Uniform Noise)15 Bin Histogram(255 points of Uniform Noise)

Volts

BinCount

0


Volts

BinCount

00

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.


Volts

BinCount

0140

200

But there will still be variation if you zoom in.

15 Bin Histogram(25,000 points of Uniform Noise)15 Bin Histogram(25,000 points of Uniform Noise)

Volts

BinCount

00

2,000

Vol

ts

Bin

Cou

nt

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.


Exponentially Distributed Voltages


Exponentially Distributed Voltages

Time

Volts

0

15 bin Histogram(255 points of Exponential Noise)15 bin Histogram(255 points of Exponential Noise)

Volts

BinCount

0


Gaussian Distributed Voltages


Gaussian Distributed Voltages

Time

Volts

0

15 bin Histogram(255 points of Gaussian Noise)15 bin Histogram(255 points of Gaussian Noise)

Volts

BinCount

0

15 bin Histogram(2500 points of Gaussian Noise)15 bin Histogram(2500 points of Gaussian Noise)

Volts

BinCount

0

400

ECEN4503 Random Signals Lecture #24 10 March 2014 Dr. George Scheets n Read 8.1 n Problems 7.1 -...

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