Communication Theory II

Lecture 10: Stochastic Process (cont’d)

Ahmed Elnakib, PhD

Assistant Professor, Mansoura University, Egypt

1March 11th, 2015

Midterm Degrees

2

Exam Type of Assignment Max. Final

Oral* Oral Presentation 10 10

Sections/Lectures/Online Quiz Quiz 15

40Matlab© Homework Code 10

Mid term 4 Questions, e ach 6 marks 20

Extra Credit Code/Quiz/Participation 10 (5 each)

Total Maximum -- -- 50**

*Note that the oral and final term degrees are added to determine the success degree, you should get at least 30% to be eligible for success (i.e., 33/110) otherwise you will be “راسب نظرى”** Note that you should get at least 60/150 to be eligible for “رفع”, otherwise you will get a score of “ضعيف”

Oral: Open Access

o EURASIP Journal on Wireless Communications and Networking http://www.springer.com/engineering/signals/journal/13638 Impact factor=0.8 http://jwcn.eurasipjournals.com/content

o Example: In the content slides:

• In 2013, an estimated number of mobile subscribers in rural areas is about 1 billion out of a total of 6.8billion[1]

In the reference silde:[1] M. Alsharif et al., “Energy optimisation of hybrid off-grid system for remote telecommunication base stationdeployment in Malaysia,” vol. 64, 2015 (impact factor=0.8, number of citations=xx)

oAny references is OK, new or old, journals, books, conferences,…,etc Put the number of citations: as the reference is older, the number of citations should be larger If you know the impact factor (only journals have impact factor), put it, otherwise, no problem

o You can select any topic in communications provided that you understand it well (anyproblem solved or unsolved). Just add your comments (may be in the aid of aconclusion/future-work section of a given paper) 3

Number of Citations

4

Lecture Outlines

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o Stochastic Process:Power Spectral Density

Transmission of A Process in LTI system

is the 1st-order probability density function of the process X(t), observed at time t

Statistical Parameters of a Process

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is the 2nd-order probability density function of the process X(t), observed at t1and t2

1. Mean

2. Autocorrelation

3. Autocovariance

For a weakly process

For a weakly process

What About the Frequency Content of a Process?

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o We need to define a quantity that defines the frequency content of the process

o Since the process theoretically exists over the entire time interval -∞ to ∞: we can call it apower signal

o For a wide sense stationary process:

A power spectral density (PSD) is defined as a weighted mean of the PSD of all sample functionof the process (ensemble average of all the PSDs of all sample functions):

𝑆𝑋 𝑤 = lim𝑇→∞

|𝑋𝑇 𝑤 |2

𝑇,

𝑋𝑇 𝑤 is the Fourier transform of a truncated random process 𝑋 𝑡 rect(𝑡

𝑇)

𝑆𝑋 𝑤 is always nonnegative

o For a nonstationary process:

PSD may differ and/or may not be exist from one sample function to anotherThe PSD does not exist

o Wiener-Khinchine theorem:

The PSD of a wide-sense stationary process is the Fourier transform of its autocorrelation function

𝑆𝑋 𝑤 𝑅𝑋𝑋(𝜏)

𝑆𝑋 𝑤 = −∞∞

𝑅𝑋𝑋(𝜏) 𝑒−𝑖𝜔𝜏𝑑𝜏 𝑅𝑋𝑋 𝜏 =

1

2𝜋 −∞∞

𝑆𝑋𝑋(𝑤) 𝑒𝑖𝜔𝜏𝑑𝑤

o The power of a random process:

𝑃𝑋= 𝐸 𝑋2 𝑡 = 𝑅𝑋𝑋 (0)=

1

2𝜋 −∞∞

𝑆𝑋(𝑤) 𝑑𝑤

Is PSD a new quantity?

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The ensemble mean of the square amplitudes of the sample functions at any instants

The area under the power spectral density (PSD)

Since 𝑑𝑤 = 2𝜋 𝑑𝑓 and PSD is an even function

𝑃𝑋= 2 0

∞

𝑆𝑋(𝑤) 𝑑𝑓, 𝑓 =𝑤

2𝜋`

o Mean Square value 𝑃𝑋= 2 0∞𝑆𝑋(𝑤) 𝑑𝑓=

o Symmetry (even functions) 𝑆𝑋 𝑤 = 𝑆𝑋 −𝑤 ,

o Maximum magnitude at zero shift:

o Normalized autocorrelation/PSD: 0≤ρXX≤1

𝑝𝑋 𝑤 is a pdf

𝐹𝑋 𝑓 is a spectral distribution function 𝐹𝑋 𝑓 = −∞𝑥

𝑆𝑋(𝑤) 𝑑𝑓

Autocorrelation /PSD Properties for a Weakly Process

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𝑆𝑋 𝑤 = −∞∞

𝑅𝑋𝑋(𝜏) 𝑒−𝑖𝜔𝜏𝑑𝜏

0 ≤ 𝑃𝑋 𝑤 ≤ 1 𝑝𝑋 𝑤 =𝑆𝑋 𝑤

−∞∞

𝑆𝑋(𝑤) 𝑑𝑓

o Consider a sinusoidal signal with random phase, defined by

Example 1: Sinusoidal Wave with Random Phase

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o The random variable is equally likely to have any value in the interval [– ]

o Each value of corresponds to a point in the sample space S of the stochastic process X(t)

o The process X(t) represents a locally generated carrier in the receiver of a communication

system, which is used in the demodulation of a received signal

o Determine the PSD, and the power of the process

o 𝑅𝑋𝑋(𝜏) =𝐴2

2cos(2𝜋𝑓𝑐 𝜏)

o 𝑆𝑋 𝑤 =F(𝑅𝑋𝑋(𝜏))=𝐴2

2𝜋[𝛿 𝑤 + 𝑤𝑐 + 𝛿 𝑤 − 𝑤𝑐 ]

o 𝑃𝑋= 𝑅𝑋𝑋(0) =𝐴2

2or 𝑃𝑋=2 0

∞𝑆𝑋(𝑤) 𝑑𝑓=2 𝜋

𝐴2

2 0∞𝛿 2𝜋𝑓 + 2𝜋𝑓𝑐 𝑑𝑓=

𝐴2

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Solution

o White noise is a stationary process whose power spectral density has a constant value

across the entire frequency interval

o 𝑆𝑋 𝑤 =𝑁𝑜

2 𝑅𝑋𝑋(𝜏)=

𝑁𝑜

2𝛿 (𝜏)

Example 2: White Noise

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𝑆𝑋𝑋 𝑤𝑅𝑋𝑋(𝜏)

w

Transmission of A weakly Stationary Process in a LTI System

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o 𝜇𝑌 = 𝐻 0 𝜇𝑋, where 𝐻 0 =

o 𝑅𝑌𝑌 𝜏 = ℎ 𝜏 ∗ ℎ −𝜏 ∗ 𝑅𝑋𝑋 𝜏 𝑆𝑌 𝑤 = 𝐻 𝑤 2 𝑆𝑋 𝑤

Since the mean of a weakly stationary

process is constant

Proof 1

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as follows:

Proof 2

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𝑅𝑌𝑌(𝜏)

𝑅𝑋𝑋

𝑅𝑌𝑌 𝜏 = ℎ 𝜏 ∗ ℎ −𝜏 ∗ 𝑅𝑋𝑋 𝜏

Sum of Two Weakly Stationary Random Processes

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𝑅𝑋𝑋(𝜏) 𝑅𝑌𝑌(𝜏)

𝑅𝑍𝑍(𝜏)

𝑅𝑍𝑍(𝜏)

(orthogonal or coherent)

𝑅𝑍𝑍(𝜏) 𝑅𝑋𝑋(𝜏) 𝑅𝑌𝑌(𝜏)

𝑅𝑋𝑋(𝜏) 𝑅𝑌𝑌(𝜏)

Example

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K2K1

Solution

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K2

K1

Solution

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K2K1

+

𝑅𝑌𝑌 𝜏

𝑅𝑌1𝑌1𝜏 𝑅𝑌

2𝑌2𝜏

K1 K2

12

K1 K2

12𝑅𝑌𝑌 0

Hints on Using Fourier Transform

oWhich to use? frequency (f) or angular frequency (w) equations? Both are OK

Both must give the same answer

You can easily switch your solution at the end

Use the proper matched tables (either frequency (f) tables or angular frequency (w) tables)

oWhat is forbidden: Improper use of unmatched tables (e.g., a frequency argument with a

property of angular frequency)

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‘w’ Table

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‘f ’ Tables

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𝑑𝑤 = 2𝜋𝑑𝑓

o We can use an angular frequency argument (w) and matched tables

Example: Characteristic function

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∅𝑋 𝑣 =E(𝑒−𝑖𝑣𝑥)= −∞∞𝑓𝑋 𝑥 𝑒

𝑖𝑣𝑥d𝑥 𝑓𝑋 𝑥 =1

2π −∞∞∅𝑋 𝑣 𝑒

−𝑖𝑣𝑥d𝑣

Calculate the characteristic function φY(v) of the standard Cauchy distribution 𝑓𝑋 𝑥 =1

𝜋

1

1+𝑥2, −∞ < 𝑥 < ∞

Example 2: lecture 7, page 6

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Solution

From the table

g(t) G(w)

G(t) 2𝜋 g(-w)Using duality

w2

2𝜋 𝑒−𝑎|𝑤|2𝑎

𝑎2+ 𝑡2

Put a=1 2𝜋 𝑒−|𝑤|2

1 + 𝑡2

Divide by 2π 1

𝜋

1

1 + 𝑡2𝑒−|𝑤|

φX(v)=F{𝑓𝑋 𝑥 }|v=-w =𝑒−|𝑣|

Where 𝑣 is an angular frequency argument

‘w’ Table

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Calculate the characteristic function φY(v) of the standard Cauchy distribution 𝑓𝑋 𝑥 =1

𝜋

1

1+𝑥2, −∞ < 𝑥 < ∞

Example 2: lecture 7, page 6, another solution

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Other way:

φX(u)=F{ 𝑓𝑋 𝑥 }|u=-f =1

2𝜋𝑒−|𝑢| , Where 𝑢 is a

frequency argument

using scaling property; scaling factor=1/2𝜋

φX(2𝜋𝑢)=φX(v)=2𝜋1

2𝜋𝑒−|2𝜋𝑢|=𝑒−|𝑣|𝑓𝑋

𝑥

2𝜋

Solution

From the table

g(t) G(f)

G(t) g(-f)Using duality

w2

𝑒−𝑎|𝑓|2𝑎

𝑎2+ 𝑡2

Put a=1 𝑒−|𝑓|2

1 + 𝑡2

Divide by 2π 1

𝜋

1

1 + 𝑡21

2𝜋𝑒−|𝑓|

φX(v)=F{𝑓𝑋 𝑥 }|v=-w =𝑒−|𝑣|

Where 𝑣 is an angular frequency argument

∅𝑋 𝑢 = −∞∞𝑓𝑋 𝑥 𝑒

𝑖𝑢𝑥d𝑥 𝑓𝑋 𝑥 = −∞∞∅𝑋 𝑢 𝑒

−𝑖𝑢𝑥d𝑢 , 𝑢 =𝑣

2𝜋

‘f ’ Tables

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Questions

28