ecen4503 random signals lecture #24 10 march 2014 dr. george scheets n read 8.1 n problems 7.1 -...
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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
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ECEN4503 Random SignalsECEN4503 Random SignalsLecture #25 12 March 2014Lecture #25 12 March 2014Dr. George ScheetsDr. George Scheets
ECEN4503 Random SignalsECEN4503 Random SignalsLecture #25 12 March 2014Lecture #25 12 March 2014Dr. George ScheetsDr. George Scheets
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)
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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
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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
ActualActualE[X] = 1.008E[X] = 1.008σσXX = 0.4114 = 0.4114
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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
ActualActualE[X] = 1.500E[X] = 1.500σσXX = 0.5015 = 0.5015
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RXY for age & weightRXY for age & weight
X = AgeY = Weight
RXY ≡ E[XY] = 4322
(85 data points)
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RXY for age & middle finger lengthRXY for age & middle finger length
X = AgeY = Finger Length (cm)
RXY ≡ E[XY] = 194.2
(54 data points)
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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)
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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)
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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)
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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)
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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)
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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
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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
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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
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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.
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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
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Discrete Time Noise Waveform255 point, 0 mean, 1 watt
Uniformly Distributed Voltages
Discrete Time Noise Waveform255 point, 0 mean, 1 watt
Uniformly Distributed Voltages
Time
Volts
0
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15 Bin Histogram(255 points of Uniform Noise)15 Bin Histogram(255 points of Uniform Noise)
Volts
BinCount
0
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15 Bin Histogram(2500 points of Uniform Noise)15 Bin Histogram(2500 points of Uniform Noise)
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.
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15 Bin Histogram(2500 points of Uniform Noise)15 Bin Histogram(2500 points of Uniform Noise)
Volts
BinCount
0140
200
But there will still be variation if you zoom in.
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15 Bin Histogram(25,000 points of Uniform Noise)15 Bin Histogram(25,000 points of Uniform Noise)
Volts
BinCount
00
2,000
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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.
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Discrete Time Noise Waveform255 point, 0 mean, 1 watt
Exponentially Distributed Voltages
Discrete Time Noise Waveform255 point, 0 mean, 1 watt
Exponentially Distributed Voltages
Time
Volts
0
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15 bin Histogram(255 points of Exponential Noise)15 bin Histogram(255 points of Exponential Noise)
Volts
BinCount
0
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Discrete Time Noise Waveform255 point, 0 mean, 1 watt
Gaussian Distributed Voltages
Discrete Time Noise Waveform255 point, 0 mean, 1 watt
Gaussian Distributed Voltages
Time
Volts
0
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15 bin Histogram(255 points of Gaussian Noise)15 bin Histogram(255 points of Gaussian Noise)
Volts
BinCount
0
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15 bin Histogram(2500 points of Gaussian Noise)15 bin Histogram(2500 points of Gaussian Noise)
Volts
BinCount
0
400
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