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ECEN4503 Random Signals
Week 9
Dr. George Scheets
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Read 8.1, 8.3, & 8.4
Problems 7.1 - 7.3, 7.5; 5.10, 5.21, & 5.49
Next Quiz covers Chapter 5
Cov(X,Y) for age & weight
X = Age
Y = Weight
RXY ≡ E[XY] = 4322
E[Age] = 23.12
E[Weight] = 186.4
E[X]E[Y] = 4309
Cov(X,Y) = 13.07
(85 data points)
Cov(X,Y) for age & middle finger length
X = Age
Y = Finger Length
RXY ≡ 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)
Age & Weight Correlation Coefficient ρ
X = Age
Y = Weight
Cov(X,Y) = 13.07
σAge = 1.971 years
σWeight = 39.42
ρ = 0.1682
(85 data points)
Age & middle finger length Correlation Coefficient ρ
X = Age
Y = Finger Length
Cov(X,Y) = 0.05822
σAge = 2.218 years
σLength = 0.5746
ρ = 0.04568
(54 data points)
SI versus Correlation
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ρ ≡ Correlation Coefficient
Allows head-to-head comparisons (Values normalized)
≡ E[XY] – E[X]E[Y]
σX σY
= 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
X & Y are S.I.? → ρ = 0 → X & Y are Uncorrelated
X & Y uncorrelated? → E[XY] = E[X]E[Y]
 Example Y = X2; fX(x) symmetrical about 0
 Here X & Y are dependent, but uncorrelated
 See Quiz 6, 2012, problem 1e
Two sample functions of bit
streams.
1.25
x
1
i
0
1
1
0
20
40
60
0
1.25
x
i
80
100
i
100
1
0
1
1
0
0
50
100
150
200
i
250
300
350
400
400
Random Bit Stream. Each bit S.I. of others.
P(+1 volt) = P(-1 volt) = 0.5
1.25
x
i
1
0
1
1
0
20
40
0
60
80
i
100
fX(x)
1/2
-1
100
+1
x volts
Bit Stream. Average burst length of 20 bits.
P(+1 volt) = P(-1 volt) = 0.5
1.25
x
i
1
0
1
1
0
50
100
0
150
200
250
300
350
i
Voltage Distribution of
this signal & previous
are the same, but time
domain behavior different.
400
fX(x)
1/2
-1
400
+1
x volts
Review of PDF's & Histograms
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Probability Density Functions (PDF's), of
which a Histogram is an estimate of shape,
frequently (but not always!) deal with the
voltage likelihoods
Volts
Time
Discrete Time Noise Waveform
255 point, 0 mean, 1 watt
Uniformly Distributed Voltages
Volts
0
Time
15 Bin Histogram
(255 points of Uniform Noise)
Bin
Count
0
Volts
15 Bin Histogram
(2500 points of Uniform Noise)
Bin
Count
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.
0
0
Volts
15 Bin Histogram
(2500 points of Uniform Noise)
Bin
Count
200
But there will still be variation if you zoom in.
140
0
Volts
15 Bin Histogram
(25,000 points of Uniform Noise)
2,000
Bin
Count
0
0
Volts
Volts
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.
Volts
Bin
Count
0
Time
Discrete Time Noise Waveform
255 point, 0 mean, 1 watt
Exponentially Distributed Voltages
Volts
0
Time
15 bin Histogram
(255 points of Exponential Noise)
Bin
Count
0
Volts
Discrete Time Noise Waveform
255 point, 0 mean, 1 watt
Gaussian Distributed Voltages
Volts
0
Time
15 bin Histogram
(255 points of Gaussian Noise)
Bin
Count
0
Volts
15 bin Histogram
(2500 points of Gaussian Noise)
400
Bin
Count
0
Volts
Random Processes
X(t) = set of possible experimental waveforms
 x(t) = specific waveform from this set

 a.k.a.
Sample Function
X = set of possible experimental numbers
 x = specific number from this set
 Theoretically
X = X(t1); Get X by sampling X(t) at time t1
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 Can
generate voltage PDF estimate from these
Can generate E[X], E[X2], etc.
 Statistical Expected Values from X(t)
Random Processes
X(t) = set of possible experimental waveforms
 x(t) = specific waveform from this set

 a.k.a.
Sample Function
X = set of possible experimental numbers
 x = specific number from this set
 Practically
Get X by sampling x(t) at different times
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 Can
generate voltage PDF estimate from these
Can generate A[x(t)], A[x(t)2], etc.
 Time Averages from x(t)
Ergodic Process X(t) volts
Ergodic → A[x(t)] = E[X(t1)]
 E[X] = A[x(t)] volts
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 Mean
 Average
 Average
Value
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Vdc on multi-meter
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E[X]2 = A[x(t)]2 volts2
 (Normalized)
D.C. power watts
Ergodic Process
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E[X2] = A[x(t)2] volts2
 2nd
Moment
 (Normalized) Average Power watts
 (Normalized) Total Power watts
 (Normalized) Average Total Power watts
 (Normalized) Total Average Power watts
Ergodic Process
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E[X2] - E[X]2 volts2
= A[x(t)2] - A[x(t)]2
σ2X
 (Normalized) AC Power watts
 Variance

E[(X -E[X])2] = A[(x(t) -A[x(t)])2]
Deviation σX
AC Vrms on multi-meter volts
 Standard
Some Waveforms Aren't Ergodic
Random DC voltage
 x(t) might be, say, +2.3556 volts DC
 X(t) might be the set of all DC waveforms
from, say, -5 vdc to +5 vdc
 E[X(t)] would = 0 vdc
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 If
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waveforms equally likely
A[x(t)] would = 2.3556 vdc
Given some waveform x(t)...
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To find voltage PDF
 Randomly
sample waveform
Visualize resulting Histogram
 Mapping
from 1 R.V. to Another
Map time (t) → voltage (x)
Treat time as Uniformly Distributed R.V.
Treat waveform x(t) as mapping g(t)
x (voltage) = g(t)