Poc Complete Notes

8/13/2019 Poc Complete Notes

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Principles of Communication The communication process:

Sources of information, communication channels, modulationprocess, and communication networks

Representation of signals and systems:

Signals, Continuous Fourier transform, Sampling theorem,

sequences, z-transform, convolution and correlation.

Stochastic processes:

Probability theory, random processes, power spectral density,

Gaussian process.

Modulation and encoding:

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Basic modulation techniques and binary data transmission:AM,

FM, Pulse Modulation, PCM, DPCM, Delta Modulation

Information theory:

Information, entropy, source coding theorem, mutual

information, channel coding theorem, channel capacity,

rate-distortion theory.

Error control coding:

linear bloc codes, cyclic codes, convolution codes

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Course Material

1. Text: Simon Haykin, Communication systems, 4th edition,

John Wiley & Sons, Inc (2001)2. References

(a) B.P. Lathi, Modern Digital and Analog Communcations

Systems, Oxford University Press (1998)(b) Alan V. Oppenheim and Ronald W. Schafer, Discrete-Time

signal processing, Prentice-Hall of India (1989)

(c) Andrew Tanenbaum, Computer Networks, 3rd edition,

Prentice Hall(1998).

(d) Simon Haykin, Digital Communication Systems, John

Wiley & Sons, Inc.

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Course Schedule*Duration:* 14 Weeks

Week 1:* Source of information; communication channels,

modulation process and Communication Networks

Week 2-3:* Signals, Continuous Fourier transform, Sampling

theorem

Week 4-5:* sequences, z-transform, convolution, correlation

Week 6:* Probability theory - basics of probability theory,

random processes

Week 7:* Power spectral density, Gaussian process

Week 8:* Modulation: amplitude, phase and frequency

Week 9:* Encoding of binary data, NRZ, NRZI, Manchester,

4B/5B

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Week 10:* Characteristics of a link, half-duplex, full-duplex,

Time division multiplexing, frequency division multiplexing

Week 11:* Information, entropy, source coding theorem, mutual

information

Week 12:* channel coding theorem, channel capacity,rate-distortion theory

Week 13:* Coding: linear block codes, cyclic codes, convolution

codes Week 14:* Revision

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Overview of the Course

Target Audience: Computer Science Undergraduates who have not

taken any course on Communication

Communication between asourceand adestinationrequires achannel.

A signal (voice/video/facsimile) is transmitted on a channel:

Basics of Signals and Systems This requires a basic understanding of signals

Representation of signals

Each signal transmitted is characterised by power.

The power required by a signal is best understood by

frequency characteristics or bandwidth of the signal:

Representation of the signal in the frequency domain -

Continuous Fourier transform

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A signal trasmitted can be either analog or digital A signal is converted to a digital signal by first

discretising the signal - Sampling theorem - Discrete-time

Fourier transform

Frequency domain interpretation of the signal is easier interms of the Z-transform

Signals are modified by Communication media, the

communication media are characterised as Systems

The output to input relationship is characterised by a

Transfer Function

Signal in communcation are characterised by Random variables

Basics of Probability

Random Variables and Random Processes

Expectation, Autocorrelation, Autocovariance, Power

Spectral Density

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Analog Modulation Schemes AM, DSB-SC, SSB-SC, VSB-SC, SSB+C, VSB+C

Frequency Division Muliplexing

Power required in each of the above

Digital Modulation Schemes

PAM, PPM, PDM (just mention last two)

Quantisation

PCM, DPCM, DM

Encoding of bits: NRZ, NRZI, Manchester

Power required for each of the encoding schemes

Information Theory

Uncertainty, Entropy, Information

Mutual information, Differential entropy

Shannons source and channel coding theorems

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Analogy between Signal Spaces and Vector Spaces

Consider two vectors V1 and V2 as shown in Fig. 1. IfV1 is to be

represented in terms ofV2

V1 =C12V2+Ve (1)

where Ve is the error.

V

VC

2

1

V12

2

Figure 1: Representation in vector space

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The error is minimum when V1 is projected perpendicularly ontoV2. In this case, C12 is computed using dot product between V1and V2.

Component ofV1 along V2 is

=

V1.V2

V2 (2)

Similarly, component ofV2 along V1 is

=V1.V2

V1 (3)

Using the above discussion, analogy can be drawn to signal spaces

also.

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Let f1(t) and f2(t) be two real signals. Approximation off1(t) by

f2(t) over a time interval t1 < t < t2 can be given by

fe(t) =f1(t) C12f2(t) (4)

where fe(t) is the error function.

The goal is to find C12 such that fe(t) is minimum over the interval

considered. The energy of the error signal given by

= 1

t2 t1

t2

t1

[f1(t) C12f2(t)]2 dt (5)

To find C12,

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C12= 0 (6)

Solving the above equation we get

C12=

1

t2t1

t2

t1

f1(t).f2(t) dt

1

t2t1

t2

t1

f22

(t) dt(7)

The denominator is the energy of the signal f2

(t).When f1(t) and f2(t) are orthogonal to each other C12= 0.

Example: sin n0tand sin m0t be two signals where m and n are

integers. When m=n

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0

0

sin n0t. sin m0t dt= 0 (8)

Clearly sin n0t and sin m0t are orthogonal to each other.

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f(t) =n

r=1

Crgr(t) (4)

fe(t) =f(t) n

r=1

Crgr(t) (5)

= 1t2 t1

t2

t1

[f(t) n

r=1

Crgr(t)]2 dt (6)

To find Cr,

C1=

C2=...=

Cr= 0 (7)

When is expanded we have

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= 1

t2 t1 t2

t1

f(t)

2f(t)n

i=1

Crgr(t) +n

r=1

Crgr(t)n

k=1

Ckgk(t) dt

(8)

Now all cross terms disappear

1

t2 t1

t2t1

Cigi(t)Cjgj(t)dt= 0, i=j (9)

since gi(t) and gj(t) are orthogonal to each other.

Solving the above equation we get

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Cj =

1t2t1

t2t1

f(t).gj(t) dt

1t2t1

t2t1

g2j (t) dt

(10)

Analogy to Vector Spaces: Projection off(t) along the signal

gj(t) =Cj

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Representation of Signals by a set of Mutually

Orthogonal Complex Functions

When the basis functions are complex. a

Ex = t2t1

|x(t)|2dt (11)

represents the energy of a signal.

Suppose g(t) is represented by the complex signal x(t)a|u+ v|2 = (u+ v)(u + v) =|u|2 +|v|2 + uv+ uv

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Ee =

t2t1

|g(t) cx(t)|2dt (12)

= t2t1

|g(t)|2

dt 1

Ex t2t1

g(t)x

(t)dt2

+ (13)c

Ex 1Ex

t2t1

g(t)x(t)dt

2

(14)

Minimising the second term yields

c= 1

Ex

t2t1

g(t)x

(t)dt (15)

Thus the coefficients can be determined by projection g(t) along

x(t).

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Fourier Representation of continuous time signals

Any periodic signal f(t) can be represented with a set of complex

exponentials as shown below.

f(t) = F0+ F1ej0t + F2e

j20t + + Fnejnnt + (1)

+ F1ej0t + F2e

j20t + Fnejnnt + (2)

The exponential terms are orthogonal to each other because

+

(ejnt)(ejmt) dt= 0, m =n

The energy of these signals is unity since

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+

(ejnt)(ejmt) dt= 1, m=n

Representing a signal in terms of its exponential Fourier seriescomponents is called Fourier Analysis.

The weights of the exponentials are calculated as

Fn =

t0+Tt0

f(t).(ejn0t) dt

t0+Tt0

(ejn0t).(ejn0t)

dt

=1

T

t0+Tt0

f(t).(ejn0t) dt

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Extending this representation to aperiodic signals:When T and 0 0, the sum becomes an integraland 0becomes continuous.

The resulting represention is termed as the Fourier Transform

(F()) and is given by

F() = +

f(t)ejt dta

The signal f(t) can recovered from F() as

f(t) = +

F()ejt db

aAnalysis equationb

Synthesis equation

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Some Important Functions

Delta function is a very important signal in signal analysis. It is

defined as

+

(t) dt= 1

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0 t

( t )

1/2

Area under the curve is always 1

Figure 1: The Dirac delta function

The Dirac delta function is also called the Impulse function.

This function can be represented as the limiting function of anumber of sampling functions:

1. Gaussian Pulse

(t) = limT0

1

Te

t

2

T2

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2. Triangular Pulse

(t) = limT0

1

T

1

|t|

T

, |t| T (3)

= 0, |t| > T (4)

3. Exponential Pulse

(t) = limT0

1

2T

e|t|T

4. Sampling Function

k

Sa(kt)dt= 1

(t) = limk

k

Sa(kt)

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5. Sampling Square function

(t) = limk

k

Sa2(kt)

The unit step function is another important function signal

processing. It is defined by

u(t) = 1, t >0

= 1

2, t= 0

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Fourier Representation of continuous time signals

Properties of Fourier Transforma

TranslationShifting a signal in time domain introduces linear

phase in the frequency domain.

f(t) F()

f(t t0) e

jt0

F()

Proof:aF and F1 correspond to the F orward and I nverse F ourier transf orms

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F() =

+

f(t t0)ejt dt

Put =t t0

F() =

+

f()ej(+t0 dt

= ejt0 +

f()ej d (1)

= F()ejt0 (2)

ModulationA linear phase shift introduced in time domainsignals results in a frequency domain.

f(t) F()

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ej0tf(t) F( 0)

Proof:

F() = +

f(t)ej0tejt dt

=

+

f(t)ej(0)t dt (3)

= F( 0) (4)

ScalingCompression of a signal in the time domain results in

an expansion in frequency domain and vice-versa.

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f(t) F()

f(at) 1

|a|F(

a)

Proof:

F() = +

f(at)ejt dt

Put =at

Ifa >0

F(f(at)) =

+

f()ej

a d

= 1

a

F(

a

)

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Ifa 0 (see Figure 1)

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f(t) = eat, t >0

F() = =

0

e(a+j)tdt

= 1a+j

t

1

1/a

+/2

/2

Magnitude

Phase

0 0

eat

Figure 1: The exponential function and its Fourier transform

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e|a|t

2/a

Magnitude

t

1

F( )

Figure 2: e|a|t and its Fourier transform

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f(t) = eat, t >0

f(t) = 1, t= 0

= eat

, t


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F() = +T

2

T

2

Aejt dt

= AejT/2 e+jT/2

j

= ATsin T

2T2

= sinc(T

2 )

The rectangular functionrect(t) and its Fourier transformF() are shown in Figure 3

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T

2

T T T

T

2 2

A

T

AT

24 2

t

F( )

Figure 3: rect(t) and its Fourier transform

Fourier transform of the sincfunction Using the duality property, the Fourier transform of the sinc

function can be determined (see Figure 4).

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42

6......... .........

2 2

f(t)

6 2

4

0 0 0 0 0 0

t

0 0

A 0 A2

Figure 4: sinc(t) and its Fourier transform

An important point is that a signal that is bandlimited is

not time-limited while a signal that is time-limited is

not bandlimited

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Continuous Fourier transforms of Periodic

Functions

Fourier transform ofejn0t Using the frequency shifting

property of the Fourier transform

ejn0t = 1.ejn0t

F(ejn0t) = F(1) shifted by 0

= 2( n0)

Fourier transform of cos 0t

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cos 0t = ej0t +ej0t

2

F(ej0t) = F(1) shifted by 0

= 2( 0)

F(cos 0t) = ( 0) +(+0)

Fourier transform of a periodic function f(t)

The periodic function is not absolutely summable.

The Fourier transform can be represented by a Fourier

series. The Fourier transform of the Fourier series representation of

the periodic function (period T) can be computed

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f(t) =

n=

Fnejn0t, 0 =

2

T

F(f(t)) = F(

n=

Fnejn0t

)

=

i=

FnF(ejn0t)

= 2

n=

Fn( n0)

Note: The Fourier transform is made up of components at

discrete frequencies.

Fourier transform of a periodic function

f(t) =

n= (t nT) (a periodic train of impulses)

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f(t) =

n=

Fnejn0t, 0 =

2T

Fn = 1

T

F(f(t)) = 1

TF(

n=

ejn0t)

=

i=

FnF(ejn0t)

= 21

T

n=

( n0)

= 0

n=

( n0)

Note: A periodic train of impulses results in a Fourier

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transform which is also a periodic train of impulses (see Figure

1).

(tnT) 0( n )

f(t) F( )

Figure 1: The periodic pulse train and its Fourier transform

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Sampling Theorem and its Importance

Sampling Theorem:

A bandlimited signal can be reconstructed exactly if it is

sampled at a rate atleast twice the maximum frequency

component in it.

Figure 1 shows a signal g(t) that is bandlimited.

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0

G( )

m m

Figure 1: Spectrum of bandlimited signal g(t)

The maximum frequency component ofg(t) is fm. To recoverthe signal g(t) exactly from its samples it has to be sampled at

a rate fs 2fm.

The minimum required sampling rate fs = 2fm is called

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Nyquist rate.

Proof: Let g(t) be a bandlimited signal whose bandwidth is fm

(m = 2fm).

g(t) G( )

0

(a) (b)

mm

Figure 2: (a) Original signal g(t) (b) SpectrumG()

T(t) is the sampling signal with fs = 1/T >2fm.

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t( )

T

(a)

s

()

(b)

Figure 3: (a) sampling signal T(t) (b) Spectrum T()

Let gs(t) be the sampled signal. Its Fourier Transform Gs() isgiven by

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F(gs(t)) = F[g(t)T(t)]

= Fg(t)+

n=

(t nT)=

1

2

G() 0

+n=

( n0)

Gs() = 1

T

+n=

G() ( n0)

Gs() = F[g(t) + 2g(t)cos(0t) + 2g(t) cos(20t) + ]

Gs() = 1

T

+n=

G( n0)

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g (t)G ( )

s

s

0 s sm m

Figure 4: (a) sampled signal gs(t) (b) Spectrum Gs()

Ifs = 2m, i.e., T = 1/2fm. Therefore, Gs() is given by

Gs() = 1

T

+n=

G( nm)

To recover the original signal G():

1. Filter with a Gate function, H2m() of width 2m.

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2. Scale it by T.

G() =T Gs()H2m().

0 m m

2m

H ( )

Figure 5: Recovery of signal by filtering with a filter of width 2m

Aliasing

Aliasing is a phenomenon where the high frequency

components of the sampled signal interfere with each other

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because of inadequate sampling s


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Oversampling

In practice signal are oversampled, where fs is significantly

higher than Nyquist rate to avoid aliasing.

0 mmss

Figure 7: Oversampled signal-avoids aliasing

Problem: Define the frequency domain equivalent of the Sampling

Theoremand prove it.

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Discrete-Time Signals and their Fourier

Transforms

Generation of Discrete-time signals

Discrete time signals are obtained by sampling a continuous

time signal.

The continuous time signal is sampled with an impulse train

with sampling period T

which is usually taken greaterthan or equal to Nyquist Rate toavoid Aliasing.

The Discrete-Time Fourier Transform (DTFT) of a discrete time

signal g(nT)is represented by

a

G(ej) =+

n=

g(nT).ejnT

aG(ej ) represents the DTFT. It signifies the periodicity of the DTFT

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In practice, it is assumed that signals are adequately sampled and

hence T is dropped to yield:

G(ej) =

+n=

g(n).ejn

The inverse DTFT is given by:

g(n) = 1

2 +

G().ejnd

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Some important Discrete-Time Signals

Discrete time impulse or unit sample functionUnit sample

function is similar to impulse function in continuous time (see

Figure 1. It is defined as follows

(n) =

1, for n= 0

0, elsewhere

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12345 0 1 2 3 4 5

1

n

( )n

Figure 1: The unit sample function

Unit step functionThis is similar to unit step function in

continuous time domain (see Figure 2) and is defined as follows

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u(n) =

1, for n 0

0, elsewhere

n

u(n)

0 1 2 3 4 5 6 7123

1

Figure 2: The unit step function

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Properties of the Discrete-time Fourier transform

Time shift property a

F(f(n n0)) F(ej)ejn0

Modulation property

F(f(n)ej0n) F(ej(0))

Differentiation in the frequency domain

F(nf(n)) dF(ej)

d

Convolution in the time domain

F(f(n) g(n)) 12F(e

j

)G(ej

)

Prove that the forward and inverse DTFTs form a pair

aA tutorial on this would be appropriate

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F(ej) =+

n=

f(n).ejn

f(n) = 1

2 +

F().ejnd

f(n) = 1

2

+

+l=

f(l).ejlejnd

=+

l=

f(l). 1

2

+

)ej(ln)d

f(n) =

+l=

f(l)(l n)

= f(n)

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Z transforms

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Z-transforms

Computation of the Z-transform for discrete-time signals:

Enables analysis of the signal in the frequency domain.

Z- Transform takes the form of a polynomial.

Enables interpretation of the signal in terms of the roots of the

polynomial.

z1 corresponds to a delay of one unit in the signal.

The Z - Transform of a discrete time signal x[n] is defined as

X(z) =

+

n=

x[n].z

n (1)

where z=r.ej

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The discrete-time Fourier Transform (DTFT) is obtained by

evaluating Z-Transform at z=ej.

or

The DTFT is obtained by evaluating the Z-transform on the unitcircle in the z-plane.

The Z-transform converges if the sum in equation 1 converges

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Region of Convergence(RoC)

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Region of Convergence(RoC)

Region of Convergence for a discrete time signal x[n] is defined as a

continuous region in z plane where the Z-Transform converges.

In order to determine RoC, it is convenient to represent the

Z-Transform as:a

X(z) = P(z)

Q(

z)

The roots of the equation P(z) = 0 correspond to the zeros of

X(z)

The roots of the equation Q(z) = 0 correspond to the poles ofX(z)

The RoC of the Z-transform depends on the convergence of theaHere we assume that the Z-transform is rational

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polynomials P(z) and Q(z),

Right-handed Z-Transform

Let x[n] be causal signal given by

x[n] =anu[n]

The Z- Transform ofx[n] is given by

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X(z) =+

n=

x[n]zn

=+

n=

anu[n]zn

=+

n=0

anzn

=+

n=0

(az1)n

=1

1 az1

=z

z a

The ROC is defined by |az

1

| |a|.

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The RoC for x[n] is the entire region outside the circle

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The RoC for x[n] is the entire region outside the circle

z=aej as shown in Figure 1.

RoC |z| > |a|

a

zplane

Figure 1: RoC(green region) for a causal signal

Left-handed Z-Transform

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Let x[n] be an anti-causal signal given by

y[n] = bn

u[n 1]

The Z- Transform ofy[n] is given by

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Y(z) =+

n=

y[n]zn

=+

n=

bnu[n 1]zn

=1

n=

bnzn

=+

n=0

(b1z)n + 1

=

1

1 zb + 1

=z

z b

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Y(z) converges when|b1z


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( ) g | | | | |

The RoC for y[n] is the entire region inside the circle

z=bej as shown in Figure 2

a

RoC |z| < |a|

zplane

Figure 2: RoC(green region) for an anti-causal signal

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Two-sided Z-Transform

Let y[n] be a two sided signal given by

y[n] =anu[n] bnu[n 1]

where, b > a

The Z- Transform ofy[n] is given by

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Y(z) =+

n=

y[n]zn

=+

n=

(anu[n] bnu[n 1])zn

=+

n=0

anzn 1

n=

bnzn

=+

n=0

(az1)n +

n=1

(b1z)n

=

1

1 az1.

1

1 zb + 1

=z

z a.

z

z b

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Y(z) converges for |b1z|


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|z| > |a| . Hence, for the signal

The ROC for y[n] is the intersection of the circle z=bej

and the circle z=aej as shown in Figure 3

RoC |a| < |z| < |b|

zplane

a b

Figure 3: RoC(pink region) for a two sided Z Transform

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Transfer function H(z)

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Consider the system shown in Figure 4.

x[n]

h[n]

X(z)

y[n] = x[n]*y[n]

H(z)

Y(z) = X(z)H(z)

Figure 4: signal - system representation

x[n] is the input and y[n] is the output

h[n] is the impulse response of the system. Mathematically,

this signal-system interaction can be represented as follows

y[n] =x[n] h[n]

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In frequency domain this relation can be written as

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Y(z) =X(z).H(z)

or

H(z) = Y(z)X(z)

H(z) is called Transfer function of the given system.

In the time domain ifx[n] =[n] then y[n] =h[n],h[n] is called the impulse response of the system.

Hence, we can say that

h[n] H(z)

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Some Examples: Z-transforms

Delta function

Z([n]) = 1Z([n n0]) = z

n0

Unit Step function

x[n] = 1, n 0

x[n] = 0, otherwise

X(z) =

1

1 z1 , |z| >1

The Z-transform has a real pole at the z = 1.

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Finite length sequence

x[n] = 1, 0 n N

x[n] = 0, otherwise

X(z) = 1 zN

1 z1

= zN1zN 1

z 1 , |z| >1

The roots of the numerator polynomial are given by:

z= 0, N zeros at the origin

and the nth roots of unity:

z=ej2kN , k= 0, 1, 2, , N 1 (2)

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Causal sequences

x[n] = (1

3)nu[n] (

1

2)nu[n 1]

X(z) = 1

1 13

z1 z1

1

1 12

z1, |z| >

1

3

The Discrete time Fourier transform can be obtained by setting

z=ej Figure 5 shows the Discrete Fourier transform for the

rectangular function.

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1

N +N

2

22 4

24

+1 +1 +1 +1

1

Figure 5: Discrete Fourier transform for the rectangular function

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Some Problems

Find the Z-transform (assume causal sequences):

1. 1, a1!

, a2

2!, a

3

3!,

2. 0, a, 0,a3

3!

, 0, a5

5!

, 0,a7

7!

,

3. 0, a, 0,a2

2!, 0, a

4

4!, 0,a

6

6!,

Hint: Observe that the series is similar to that of the exponential

series.

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Properties of the Z-transform

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1. RoC is generally a disk on the z-plane.

0 rR |z| rL

2. Fourier Transform ofx[n] converges when RoC includes the

unit circle.

3. RoC does not contain any poles.4. Ifx[n] is finite duration, RoC contains entire z - plane except

for z= 0 and z= .

5. For a left handed sequence, RoC is bounded by|z| < min(|a|, |b|).

6. For a right handed sequence, RoC is bounded by

|z| > max(|a|; |b|).

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Inverse Z-transform

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To determine the inverse Z-transform, it is necessary to know the

RoC.

RoC decides whether a given signal is causal (exists for positive

time), anticausal (exists for negative time) or both causal andanticausal (exists forboth positive and negative time)

Different approaches to compute the inverse Z-transform

Long division methodWhen Z-Transform is rational, i.e. it canbe expressed as the ratio of two polynomials P(z) and Q(z)

X(z) = P(z)

Q(z)

Then, inverse Z-transform can be obtained using long division:

Divide P(z) byQ(z). Let this be:

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X( )

i (1)

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X(z) =

i=

aizi (1)

The coefficients of the RHS of equation (1) correspond to

the time sequencei.e. the coefficients of the quotient of thelong division gives the sequence.

Partial Fraction method the Z-Transform is decomposed into partial fractions

the inverse Z-transform of each fraction is obtained

independently the inverse sequences are then added

The method of adding inverse Z-transform is illustrated below.

Let,

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X(z) =

M

k=0

bk.zk

N

k=0

ak

.zk

, M < N

=

M

k=1

(1 ck.z1)

N

k=1

(1 dk.z1)

=

N

k=1

Ak

(1 dk.z1)

where,

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Ak = (1 dk.z1)X(z)|z=dk

For s multiple poles at z=di

X(z) =MN

k=0

Br.z1 +

N

k=1,k=i

Ak

(1 dk.z1)+

s

m=1

Cm

(1 di.z1)m

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Properties of the Z-Transform

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Linearity:

a1x1[n] +a2x2[n] a1X1(z) +a2X2(z),RoC=Rx1 Rx2

Time Shifting Property:

x[n n0] zn0X(z),

RoC=Rx (except possible addition/deletion

of z= 0 or z= )

Exponential Weighting:

zn

0 x[n] X(z1

0 z),RoC= |z0|Rx

The poles of the Z-transform are scaled by |z0|

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Linear Weighting

dX(z)

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nx(n) zdX(z)

dz ,

RoC=Rx (except possible addition/deletion

of z= 0 or z= )

Time Reversal

x[n] X(z1),RoC= 1

Rx

Convolution

x[n] y[n] X(z)Y(z),RoC=Rx Ry

Multiplication

x[n]w[n] 1

2j

X(v)w(

z

v)v1dv

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Inverse Z-Transform Examples

U i l di i i C l

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Using long division: Causal sequence

1

1 az1,RoC= |z| > |a| = 1 +az1 +az2 +az3 +

IZT(1 +az1 +a2z2 +a3z3 + ) =anu[n]

Using long division: Noncausal sequence

1

1 az1 ,RoC= |z| < |a|

Here the IZT is computed as follows:

IZT( 1

1 az1

) =IZT( z

a+z

)

This results in:

IZT(a1z+a2z2 +a3z3 + ) = anu[n 1]

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Inverse Z-transform - using Power series expansion

X(z) =log(1 +az1), |z| > |a|

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Inverse Z-transform - Inspection method

anu[n] 1

1 az1, |z| > |a|

GivenX(z) = 1

1 12

z1, |z| > |

1

2|

= x[n] = (1

2

)nu[n]

Inverse Z-transform - Partial fraction method

Example 1: All-Pole system

X(z) = 1(1 1

3z1)(1 1

6z1)

, |z| > 13

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Using partial fraction method we have:

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Using partial fraction method, we have:

X(z) = A1

1 13

z1+

A2

1 16

z1,

|z| > 13

A1 = (1 1

3z1)X(z)|

z= 13

A2 = (1 16

z1)X(z)|z= 1

6

A1 = 2

A2 = 1

x(n) = 2( 13

)nu[n] 1( 16

)nu[n]

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Example 2: Pole-Zero system

X(z) = 1 + 2z1 +z2

1 32

z1 + 12

z2, |z| >1

(1 +z1)2

(1 12

z1)(1 z1)

= 2 +

1 + 5z1

(1 12z1)(1 z1)

= 2 9

1 12

z1+

8

1 z1

x[n] = 2[n] 9(1

2 )nu[n] + 8u[n]

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Example 3: Finite length sequences

X(z) = z2(1 1

2z1)(1 +z1)(1 z1)

= z

2

frac12z 1 +

1

2 z

1

= [n+ 2] 1

2[n+ 1] [n] +

1

2[n 1]

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Inverse Z-Transform Problem

1. GivenX(z) = zz1

zz2

+ zz3

, determine all the possible

sequences for x[n].Hint: Remember that the RoC must be a continuous region

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Basics of Probability Theory and Random

Processes

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Basics of probability theory a

Probability of an event Erepresented by P(E) and is given by

P(E) = NE

NS(1)

where, NS

is the number of times the experiment is performed

and NEis number of times the event Eoccured.

Equation 1 is only an approximation. For this to represent the

exact probability NS .

The above estimate is therefore referred to as RelativeProbability

Clearly, 0P(E)1.arequired for understanding communication systems

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Mutually Exclusive Events

L t S b th l h i N t E E E E

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LetSbe the sample space having N events E1, E2, E3, , EN.

Two events are said to be mutually exclusive or statistically

independent ifAi Aj = and

N

i=1 A

i =S for all i and j.

Joint Probability

Joint probability of two events A and B represented by

P(A B) and is defined as the probability of the occurence ofboth the events A and B is given by

P(A B) =

NAB

NS

Conditional Probability

Conditional probability of two events A and B represented as

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P(A|B) and defined as the probability of the occurence of

event A after the occurence ofB.

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P(A|B) =

NAB

NB

Similarly,

P(B|A) = NAB

NB

This implies,

P(B|A)P(A) =P(A|B)P(B) =P(A B)

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Chain Rule

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Let us consider a chain of events A1, A2, A3, , ANwhich are

dependent on each other. Then the probability of occurence of

the sequence

P(AN, AN1, AN2, , A2, A1)

= P(AN|AN1, AN2, , A1).

P(AN1|AN2, AN3, , A1). .P(A2|A1).P(A1)

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Bayes Rule

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AA

A

A

A

B

2

1

3

4

5

Figure 1: The partition space

In the above figure, ifA1, A2, A3, A4, A5 partition the sample space

S, then (A1 B), (A2 B), (A3 B), (A4 B), and (A5 B)

partition B.

Therefore,

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P(B) =

ni 1

P(Ai B)

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i=1

=ni=1

P(B|Ai).P(Ai)

In the example figure here, n= 5.

P(Ai|B) = P(Ai B)

P(B)

=P(B|Ai).P(Ai)

ni=1

P(B|Ai).P(Ai)

In the above equation, P(Ai|B) is called posterior probability,

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P(B|Ai) is called likelihood, P(Ai) is called prior probability andn

i=1

P(B|Ai).P(Ai) is called evidence.

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Random Variables

Random variable is a function whose domain is the sample space

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Random variable is a function whose domain is the sample space

and whose range is the set of real numbersProbabilistic description

of a random variable

Cummulative Probability Distribution:

It is represented as FX(x) and defined as

FX(x) =P(Xx)

Ifx1 < x2, then FX(x1)< FX(x2) and 0FX(x)1.

Probability Density Function:

It is represented as fX(x) and defined as

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fX(x) = dFX(x)

dx

This implies,

P(x1Xx2) =

x2x1

fX(x) dx

fX(x) = dFX(x)

dx

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Random Process

A random process is defined as the ensemble(collection) of time

f i h i h b bili l ( Fi )

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functions together with a probability rule (see Figure 2)

S

S

S

1

2

n

x (t)

x (t)

x (t)n

2

1

Sample Space

T +T

Figure 1: Random Processes and Random Variables

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x1(t) is an outcome of experiment 1

x2(t) is the outcome of experiment 2..

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..

xn(t) is the outcome of experiment n

Each sample point in S is associated with a sample function

x(t)

X(t, s) is a random process

is an ensemble of all time functions together with a

probability rule

X(t, sj) is a realisation or sample function of the random

process

Probability rules assign probability to any meaningful eventassociated with an observation An observation is a sample

function of the random process

A random variable:

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{x1(tk), x2(tk),...,xn(tk)} = {X(tk, s1), X(tk, s2),...,X(tk, sn)}

X(tk, sj) constitutes a random variable.

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Outcome of an experiment mapped to a real number

An oscillator with a frequency 0 with a tolerance of 1%

The oscillator can take values between 0(1 0.01)

Each realisation of the oscillator can take any value between

(0)(0.99) to (0)(1.01)

The frequency of the oscillator can thus be characterised by

a random variable

Stationary random processA random process is said to be

stationary if its statistical characterization is independent ofthe observation interval over which the process was initiated.

Mathematically,

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FX(t1+T)X(tk+T)=FX(t1)X(tk)

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Mean, Correlation and CovarianceMean of a stationary

random process is independent of the time of observation.

X(t) =E[X(t)] =x

Autocorrelation of a random process is given by:

RX(t1, t2) =E[X(t1)X(t2)]

= +

+

x1.x2fX(t1)X(t2)(x1, x2) dx1 dx2

For a stationary process the autocorrelation is dependent only

on the time shiftand not on the time of observation.

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Autocovariance of a stationary process is given by

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CX

(t1, t

2) =E[(X(t

1

x)(X(t

2

x)]

Properties of Autocorrelation

1. RX() =E[X(t + )X(t)]

2. RX(0) =E[X2(t)]

3. The autocorrelation function is an even function i.e,

RX() =RX().

4. The autocorrelation value is maximum for zero shift i.e,RX() RX(0).

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Proof:

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E[(X(t + ) X(t))2] 0

= E[X2(t + )] + E[X2(t)] 2E[X(t + )X(t)] 0

= RX(0) + RX(0) + 2RX() 0

= RX(0) RX() RX(0)

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A slowly varying random process

A rapidly varying random process

Figure 2: Autocorrelation function of a random process

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Random Process: Some Examples

A sinusoid with random phase

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A sinusoid with random phase

Consider a sinusoidal signal with random phase, defined by

X(t) =a sin(0t+ )

where 0 and a are constants, and is a random variable that

is uniformly distributed over a range of 0 to 2 (see Figure 1)

f() =

1

2 , 0 2= 0, elsewhere

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0sin( t+)

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Figure 1: A sinusoid with random phase

This means that the random variable is equally likely to

have any value in the range 0 to 2. The autocorrelation

function ofX(t) is

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RX(t) =E[X(t+)X(t)]

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=E[sin(0t+0) + ) sin(0t+ )]

= 12 E[sin(20t+0+ 2)] +12 E[sin(

0)]

= 1

2

2

0

1

2cos(4fct+0+ 2)] cos(0) d

The first term intergrates to zero, and so we get

RX() = 1

2cos(0)

The autocorrelation function is plotted in Figure 2.

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X(t) consisting of a random sequence of binary symbols 1 and

0.+1

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1

t delay

Figure 3: A random binary wave

1. The symbols 1 and 0 are represented by pulses of amplitude+1 and 1 volts, respectively and duration T seconds.

2. The starting time of the first pulse,tdelay, is equally likely

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to lie anywhere between zero and T seconds

3. tdelay is the sample value of a uniformly distributed randomvariableTdelay with a probability density function

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fTdelay (tdelay) = 1

T, 0 tdelay T

= 0, elsewhere

4. In any time interval (n 1)T < t tdelay < nT, where n isan interger, a 1 or a 0 is determined randomly (for example

by tossing a coin: heads = 1, tails = 0

E[X(t)] = 0, for all t since 1 and 0 are equally likely.

Autocorrelation function RX(tk, tl) is given by

E[X(tk)X(tl)], where X(tk) and X(tl) are random variables

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Case 1: when |tk tl| > T. X(tk) and X(tl) occur in differentpulse intervals and are therefore independent:

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0

= T|tktl|

0

1

T

dtdelay

= (1 |tk tl|

T ), |tk tl| < T

The autocorrelation function is given by

RX() = (1 ||

T ), || < T

= 0, || > T

This result is shown in Figure 4

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1

TT

Figure 4: Autocorrelation of a random binary wave

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Random Process: Some Examples

Quadrature Modulation ProcessGiven two random variables

X1(t) and X2(t)

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X1(t) = X(t) cos(20t+ )

X2(t) = X(t) sin(20t+ )

where 0 is a constant, and is a random variable that is

uniformly distributed over a range of 0 to 2, that is,

f() = 1

2, 0 2

= 0, elsewhere

The correlation function ofX1(t) and X2(t) is

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R12() =E[X1(t)X2(t+)]=E[X(t)cos(0t+ )X(t ) sin(20(t ) + )]

= 2

0

1

2

X(t)X(t )cos(0t+ ) sin(0(t ) + ) d

=1

2RX()sin(0)

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Random Process: Time vs. Ensemble Averages

Ensemble averages

Difficult to generate a number of realisations of a random

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process

= use time averages

Mean

x(T) = 12T

+T

T

x(t) dt

Autocorrelation

Rx(, T) = 1

2T

+TT

x(t)x(t+) dt

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ErgodicityA random process is called ergodic if

1. it is ergodic in mean:

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limT+

x(T) = X

limT+

var[x(T)] = 0

2. it is ergodic in autocorrelation:

limT+

Rx(, T) = RX()

limT+

var[Rx(, T)] = 0

where X and RX() are the ensemble averages of the same

random process.

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In Figure 1 , h[n] is an LSI system if it satisfies the following

properties LinearityThe system is called linear, if the following

equation holds for all signals x1[n] and x2[n] and any a and

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equation holds for all signals x1[n] and x2[n] and any a and

b:

x1[n] y1[n]

x2[n] y2[n]

= a.x1[n] +b.x2[n] a.y1[n] +b.y2[n]

Shift InvarianceThe system is called Shift Invariant, if the

following equation holds for any signal x[n]

x[n] y[n]

= x[n n0] y[n n0]

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The assumption is that the output of the system is linear,

in that if the input scaled, the output is scaled by thesame factor.

The system supports superposition

When two signals are added in the time domain, the

output is equal to the sum of the individual responses If the input to the system is delayed by n0, the output is

also delayed by n0.

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Random Process through a linear filter

A random process X(t) is applied as input to a lineartime-invariant filter of impulse response h(t),

( )

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It produces a random process Y(t) at the filter output as

shown in Figure 1

X(t) Y(t)

h(t)

Figure 1: Transmission of a random process through a linear filter

Difficult to describe the probability distribution of the outputrandom process Y(t), even when the probability distribution of

the input random process X(t) is completely specified for

t +.

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Estimate characteristics like mean and autocorrelation of the

output and try to analyse its behaviour. MeanThe input to the above system X(t) is assumed

stationary. The mean of the output random process Y(t) can

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( )

be calculated

mY(t) = E[Y(t)] =E

Z +

h()X(t ) d

=Z +

h(

)E

[X

(t

)]d

=

Z +

h()mX(t ) d

= mXZ +

h() d

= mXH(0)

where H(0) is the zero frequency response of the system.

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AutocorrelationThe autocorrelation function of the output

random processY

(t). By definition, we have

R (t u) = E[Y (t)Y (u)]

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RY(t, u) =E[Y(t)Y(u)]

where tand u denote the time instants at which the process isobserved. We may therefore use the convolution integral to

write

RY(t, u) = EZ

+

h(1)X(t 1) d1

Z +

h(2)X(t 2) d2

=

Z +

h(1) d1

Z +

h(2)E[X(t 1)X(t 2)] d2

When the input X(t) is a wide-stationary random process,

The autocorrelation function ofX(t) is only a function of

the difference between the observation times t 1 and

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u 2.

Putting =t u, we get

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RY() = Z +

Z +

h(1)h(2)RX( 1+ 2) d1 d2

RY(0) =E[Y2(t)]

The mean square value of the output random process Y(t)is obtained by putting = 0 in the above equation.

E[Y2

(t)] =

Z +

Z +

h(1)h(2)RX(2 1) d1 d2

=1

2

Z +

Z +

"Z +

H() exp(j1) d

#h(2)RX(2 1) d1 d2

= 12

Z +

H() dZ

+

h(2) d2Z

+

RX(2 1) exp(j21) d1

Putting =2 1

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E[Y

2(t)] =

1

2 Z +

H() d

Z +

h(2) exp(j2) d2 Z +

RX() exp(j2) d

=1

2

Z +

H() d

Z +

H() d

Z +

RX () exp(j) d

This is simply the Fourier Transform of the autocorrelation

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This is simply the Fourier Transform of the autocorrelation

function RX(t) of the input random process X(t). Let this

transform be denoted by SX(f).

SX() = +

RX() exp(j) d

SX() is called the power spectral density or power spectrum

of the wide-sense stationary random process X(t).

E[Y2(t)] = 1

2

+

|H()|2SX() df

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Themean square valueof the output of a stable lineartime-invariant filter in response to awide-sense stationary

random processis equal to the integral over all frequencies

of thepower spectral densityof the input random process

multiplied by thesquared magnitude of the transfer function

of the filter.

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Definition of Bandwidth

Bandwidth is defined as a band containing all frequenciesbetween upper cut-off and lower cut-off frequencies. (see

Figure 1)

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g )

f ful

Bandwidth

u b

3 dB

BW f = f fl

Figure 1: Bandwidth of a signal

Upper and lower cut-off (or 3dB) frequencies corresponds to the

frequencies where the magnitude of signals Fourier Transform

is reduced to half (3dB less than) its maximum value.

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Importance of Bandwidth

Bandwidth enables computation of the power required to

transmit a signal.

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Signals that are band-limited are not time-limited

Energy of a signal is defined as:

E=

+

|x(t)|2dt

Energy of a signal that is not time-limited can be

computed using Parsevals Theorema:

+

|x(t)|2dt=

+

|X()|2d

aThe power of a signal is the energy dissipated in a one ohm resistor

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Nout=

1

2

N02

+

|H()|2

d

=N02

+|H()|

2d

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2

0

H(0)

0 2 B2 B

Ideal LPF

Practical LPF

Figure 2: Ideal and Practical LPFs

For an ideal LPF,

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Nout=N0BH2(0)

=B =

1

2

+0

|H()|2d

H2(0)

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Modulation

M d l ti i th t hift i th f

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Modulation is a process that causes a shift in the range of

frequencies in a signal.

Signals that occupy the same range of frequencies can be

separated

Modulation helps in noise immunity, attentuation - depends onthe physical medium

Figure 1 shows the different kinds of analog modulation schemes

that are available

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Baseband

Modulation

Carrier

Modulation

Communication SystemCarrier Modulation

Amplitude Angle

(AM)

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( )

Frequency Phase

(FM) (PM)

Figure 1: A broad view of communication system

Amplitude ModulationIt is the process where, the amplitude of

the carrier is varied proportional to that of the message signal.

Amplitude Modulation with carrier

Let m(t) be the base-band signal, m(t) M() and c(t)

be the carrier, c(t) =Accos(ct). fc is chosen such that

fc >> W, where W is the maximum frequency component

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ofm(t).

The amplitude modulated signal is given by

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s(t) =Ac[1 + kam(t)] cos(ct)

S() = Ac

2 (( c) + (+ c)) +

kaAc

2 (M( c) + M(+ c))

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m(t)

M( )

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t

t

S( )

s(t)

f f

f f

mm

c c

2fm

A /2c

1/2 A/2 k M(0)a

Figure 2: Amplitude modulation

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Figure 2 shows the spectrum of theAmplitude Modulated

signal.

ka is a constant called amplitude sensitivity. kam(t)


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Product

Modulatorm(t)

s(t)

Figure 3: Modulation using product modulator

Demodulation in AM:An envelope detector is used to get

the demodulated signal (see Figure 4).

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+

r

R mv (t)C

Figure 4: Demodulation using Envelope detector

The voltage vm(t) across the resistor R gives the message

signal m(t)

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Double Side Band - Suppressed Carrier(DSB-SC) Modulation

In AM modulation, transmission of carrier consumes lot of

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power. Since, only the side bands contain the information

about the message, carrier is suppressed. This results in a

DSB-SC wave.

A DSB-SC wave s(t) is given by

s(t) = m(t)Accos(ct)

S() =

Ac

2 (M(

c) + M(+ c))

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s(t) S(f)

1/2 A M(0)c

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f f c c

2fm

t

Figure 5: DSB-SC modulation

Modulation in DSB-SC:Here also product modulator is used as

shown in Figure 3, but the carrier is not added. Figure 6 showsthe spectrum of the DSB-SC signal.

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c 1/2 A M(0) cos( )

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2f 2f0c c

Figure 6: Spectrum of Demodulated DSB-SC signal

Demodulation in DSB-SC:A coherent demodulator is used.

The local oscillator present in the demodulator generates a

carrier which has samefrequency and phase(i.e. = 0 in

Figure 7) a

as that of the carrier in the modulated signal (seeFigure 7)

aClearly the design of the demodulator for DSB-SC is more complex than

that vanilla AM

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s(t)

v(t) v (t)Product

ModulatorLPF

o

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( )

LocalOscillator

c cos(2 f t + )

Figure 7: Coherent detector

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v(t) = s(t) cos( t + )

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v(t) =s(t). cos(ct + )

=m(t)Accos(ct)cos(ct + )

= m(t)

2 Ac[cos(2ct + ) + cos()]

If, the demodulator (Figure 7) has constant phase, the original

signal is reconstructed by passing v(t) through an LPF.

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Single Side Band (SSB) Modulation

In DSB-SC it is observed that there is symmetry in the

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y y

bandstructure. So, even if one half is transmitted, the otherhalf can be recovered at the received. By doing so, the

bandwidth and power of transmission is reduced by half.

Depending on which half of DSB-SC signal is transmitted,there are two types of SSB modulation

1. Lower Side Band (LSB) Modulation

2. Upper Side Band (USB) Modulation

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M( )

Baseband signal

DSBSC

2 B2 B

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USB

LSB

c c

cc

c c

Figure 1: SSB signals from orignal signal

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Mathematical Analysis of SSB modulation

02 B 2 B

M( )

M ( ) M ( )

+

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+

+

00

cc c

cc

c c

ccM ( ) M ( )

M ( )M ( )

+

+

Figure 2: Frequency analysis of SSB signals

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From Figure 2 and the concept of the Hilbert Transform,

USB() = M+( c) + M(+ c)

USB(t) = m+(t)ejct + m(t)e

jct

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But, from complex representation of signals,

m+(t) = m(t) + jm(t)m(t) = m(t)jm(t)

So,

USB(t) =m(t) cos(ct) m(t)sin(ct)

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Similarly,

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LSB(t) =m(t)cos(ct) + m(t) sin(ct)

Generation of SSB signalsA SSB signal is represented by:

SSB(t) =m(t) cos(ct) m(t)sin(ct)

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m(t)

DSBSC

sin( t)

/2

+

+SSB signal

cos( t)

c

c

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DSBSC/2

Figure 3: Generation of SSB signals

As shown in Figure 3, a DSB-SC modulator is used for SSB

signal generation.

Coherent Demodulation of SSB signalsSSB(t) is multipliedwith cos(ct) and passed through low pass filter to get back the

orignal signal.

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SSB(t)cos(ct) =1

2m(t) [1 + cos(2ct)] 1

2m(t) sin(2ct)

=1

2m(t) +

1

2cos(2ct)

1

2m(t) sin(2ct)

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M( ) c cM ( + ) M ( )+

2 2 c c0

Figure 4: Demodulated SSB signal

The demodulated signal is passed through an LPF to remove

unwanted SSB terms.

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Vestigial Side Band (VSB) Modulation

The following are the drawbacks of SSB signal generation:

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1. Generation of an SSB signal is difficult.

2. Selective filtering is to be done to get the original signal

back.

3. Phase shifter should be exactly tuned to 900

.

To overcome these drawbacks, VSB modulation is used. It can

viewed as a compromise between SSB and DSB-SC. Figure 5

shows all the three modulation schemes.

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c

cB

B

0

()VSB

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Figure 5: VSB Modulation

In VSB

1. One sideband is not rejected fully.

2. One sideband is transmitted fully and a small part (vestige)

of the other sideband is transmitted.

The transmission BW is BWv =B+ v. where, v is the

vestigial frequency band. The generation of VSB signal is

shown in Figure 6

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m(t)

H ( )

( ) ( )t

cos( t)

c

iVSB

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Figure 6: Block Diagram - Generation of VSB signal

Here, Hi() is a filter which shapes the other sideband.

V SB() = [M( c) + M(+ c)] .Hi()

To recover the original signal from the VSB signal, the VSBsignal is multiplied with cos(ct) and passed through an LPF

such that original signal is recovered.

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cos( t)

LPF

H ( )

m(t) ( )t

0

c

VSB

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Figure 7: Block Diagram - Demodulation of VSB signal

From Figure 6 and Figure 7, the criterion to choose LPF is:

M() = [V SB(+ c) + V SB( c)] .H0()

= [Hi(+ c) + Hi( c)] .M().H0()

= H0() =1

Hi(+ c) + Hi( c)

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Appendix: The Hilbert Transform

The Hilbert Transform on a signal changes its phase by 900. TheHilbert transform of a signal g(t) is represented as g(t).

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g(t) = 1

+

g()

t d

g(t) = 1

+

g()

t

d

We, say g(t) and g(t) constitute a Hilbert Transform pair. If we

observe the above equations, it is evident that Hilbert transform is

nothing but the convolution ofg(t) with 1

t .The Fourier Transformof g(t) is computed from signum

function sgn(t).

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sgn(t)

2

j

=1

t jsgn()

Where,

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sgn() =

1, >0

0,

= 01,


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3. g(t) and g(t) are orthogonal over the entire interval to+.

+

g(t)g(t) dt= 0

Complex representation of signals

Ifg(t) is a real valued signal, then its complex representation g+(t)is given by

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g+(t) = g(t) + jg(t)

G+() = G() + sgn()G()

Therefore,

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G+() =

2G(), >0

G(0), = 0

0,


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G() =

2G(), 0

Essentially the pre-envelopeof a signal enables the suppression

of one of the sidebands in signal transmission.

The pre-envelopeis used in the generation of the SSB-signal.

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Angle Modulation

In this type of modulation, the frequency or phase of carrier is

varied in proportion to the amplitude of the modulating signal.

c(t)

Ac m

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t

c

Figure 1: An angle modulated signal

Ifs(t) =Accos(i(t)) is an angle modulated signal, then

1. Phase modulation:

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i(t) =ct+kpm(t)

where c = 2fc.

2. Frequency Modulation:m

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i(t) = c+kfm(t)

i(t) = t0

i(t) dt

= 2

t0

fi(t) dt+

t0

kfm(t) dt

Phase ModulationIfm(t) =Amcos(2fmt) is the message

signal, then the phase modulated signal is given by

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s(t) =Accos(ct+kpm(t))

Here,kp is phase sensitivity or phase modulation index.

Frequency ModulationIfm(t) =Amcos(2fmt) is the message

signal, then the Frequency modulated signal is given bym

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2fi(t) =c+kfAmcos(2fmt)

i(t) =ct+kfAm2fm

sin(2fmt)

here,kfAm

2 is called frequency deviation (f) andf

fmis called

modulation index (). The Frequency modulated signal is

given by

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s(t) =Accos(2fct+sin(2fmt))

Depending on how small is FM is eitherNarrowband

FM(


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oscillator

m(t)

A cos( t)

Asin( t)

NBFM signal

/2 Phase shifter

c

c

Figure 2: Generation of NBFM signal

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Wide-Band FM (WBFM)

A WBFM signal has theoritically infinite bandwidth.

Spectrum calculation of WBFM signal is a tedious process om

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Spectrum calculation of WBFM signal is a tedious process.For, practical applications however the Bandwidth of a

WBFM signal is calculated as follows:

Let m(t) be bandlimited to BHz and sampled adequately at

2BHz. If time period T = 1/2B is too small, the signal can

be approximated by sequence of pulses as shown in Figure

??

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t

mp

T com

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T

Figure 3: Approximation of message signal

If tone modulation is considered, and the peak amplitude of

the sinusoid is mp, the minimum and maximum frequencydeviations will be c kfmp and c+kfmp respectively.

The spread of pulses in frequency domain will be 2T

= 4B

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as shown in Figure ??

com

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k m k m +4 B c

pfc f p

Figure 4: Bandwidth calculation of WBFM signal

Therefore, total BW is 2kfmp+ 8B and if frequency

deviation is considered

BWfm = 1

2(2kfmp+ 8B)

BWfm = 2(f+ 2B)

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The bandwidth obtained is higher than the actual value.

This is due to the staircase approximation ofm(t).

The bandwidth needs to be readjusted. For NBFM, kf isvery small an d hence fis very small compared to B.

This implies

com

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Bfm 4B

But the bandwidth for NBFM is the same as that of AM

which is 2B

A better bandwidth estimate is therefore:

BWfm = 2(f+B)

BWfm = 2(kfmp

2 +B)

This is also calledCarsons Rule

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Demodulation of FM signals

Let fm(t) be an FM signal.

fm(t) =

A cos(ct+kf

t0

m() d)

This signal is passed through a differentiator to get com

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This signal is passed through a differentiator to get

fm(t) =A (c+kfm(t))sin

ct+kf

t

0 m() d

If we observe the above equation carefully, it is both

amplitude and frequency modulated.

Hence, to recover the original signal back an envelopedetector can be used. The envelope takes the form (see

Figure ??):

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Envelope=A (c+kfm(t))

FM signal

Envelope of FM signal

com

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Figure 5: FM signal - both Amplitude and Frequency Modulation

The block diagram of the demodulator is shown in Figure ??

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t( ) ( t)fm fm

.

d/dt

Detector

EnvelopeA( + k m(t)) c f

d.c

om

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Figure 6: Demodulation of an FM signal

The analysis for Phase Modulationis identical.

Analysis of bandwidth in PM

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i = c+kpm

(t)

mp = [m(t)]max

= kpmp

BWpm = 2(f+B)d.com

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( f )

BWpm = 2(kpm

p

2 +B)

The difference between FM and PM is that the bandwidth

is independent of signal bandwidth in FM while it is

strongly dependent on signal bandwidth in PM. a

aowing to the bandwidth being dependent on the peak of the derivative of

m(t) rather than m(t) itself

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Angle Modulation: An Example

An angle-modulated signal with carrier frequencyc = 2 10

6 is described by the equation:

EM(t) = 12 cos(ct+ 5 sin 1500t+ 10 sin 2000t)d.com

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( ) ( )

1. Determine the power of the modulating signal.

2. What is f?

3. What is ?

4. Determine , the phase deviation.

5. Estimate the bandwidth ofEM(t)?

1. P = 122

/2 = 72 units2. Frequency deviation f, we need to estimate the

instantaneous frequency:

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Noise Analysis - AM, FM

The following assumptions are made:

Channel modelld.com

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distortionless

Additive White Gaussian Noise (AWGN)

Receiver Model (see Figure 1)

ideal bandpass filter

ideal demodulator

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Modulated signal

s(t)

w(t)

x(t)

DemodulatorBPF

Figure 1: The Receiver Model ld.com

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g R v

BPF (Bandpass filter) - bandwidth is equal to the message

bandwidth B midband frequency is c.

Power Spectral Density of Noise

N0

2 , and is defined for both positive and negative frequency (seeFigure 2).

N0 is the average power/(unit BW) at the front-end of the

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receiver in AM and DSB-SC.

2

N0

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cc

B B

4 4

Figure 2: Bandlimited noise spectrum

The filtered signal available for demodulation is given by:

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x(t) = s(t) +n(t)

n(t) = nI(t)cos ct

nQ(t)sin ct

nI(t)cos ct is the in-phase component and

nQ(t)sin t is the quadrature component rld.com

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nQ(t)sin ct is the quadrature component.

n(t) is the representation for narrowband noise.

There are different measures that are used to define the Figure ofMerit of different modulators:

Input SNR:

(SN R)I=Average power of modulated signal s(t)

Average power of noise

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Output SNR:

(SNR)O =Average power of demodulated signal s(t)

Average power of noise

The Output SNR is measured at the receiver.

Channel SNR:rl

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(SN R)C=

Average power of modulated signal s(t)

Average power of noise in message bandwidth

Figure of Merit (FoM) of Receiver:

F oM= (SN R)O(SN R)C

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To compare across different modulators, we assume that (see

Figure 3):

The modulated signal s(t) of each system has the same average

power

Channel noise w(t) has the same average power in the message

bandwidthB.rl

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m(t)

message with same

power as modulated wave

n(t)

Low Pass Filter(B)

Output

Figure 3: Basic Channel Model

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Figure of Merit (FoM) Analysis

DSB-SC (see Figure 4)

s(t) = CAccos(ct)m(t)A2C2P or

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(SN R)C = A2cC

2P

2BN0

P = +2B2B S

M()d

x(t) = s(t) +n(t)

CAccos(ct)m(t)

+nI(t)cos ct+nQ(t)sin ct

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m(t)

message with same

power as modulated wave(B)

Band Pass Filter

Modulator

Product(B)

y(t)v(t)Low Pass Filter

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n(t) LocalOscillator

Figure 4: Analysis of DSB-SC System in Noise

The output of the product modulator is

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v(t) = x(t)cos(ct)

= 12

Acm(t) +12

nI(t)

+1

2[CAcm(t) +nI(t)]cos2ct

12 nQ(t)sin2ct

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The Low pass filter output is:

=12

Acm(t) +12

nI(t)

= ONLY inphase component of noise nI(t) at the output

= Quadrature component of noise nQ(t) is filtered at theoutput

Band pass filter width = 2B

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Receiver output is nI(t)2Average power ofnI(t) same as that n(t)

Average noise power = (1

2)22BN0

= 12 BN0

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(SN R)O,DSBSC = C2A2cP/4

BN0/2

= C2

A2cP

2BN0

F oMDSBSC =

(SNR)O(SNR)C

|DSBSC= 1

Amplitude Modulation

The receiver model is as shown in Figure 5

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m(t)

message with same

power as modulated wave

(B)

Band Pass Filter

Modulator

Envelopev(t)x(t)

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n(t)

Figure 5: Analysis of AM System in Noise

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s(t) = Ac[1 +kam(t)]cos ct

(SN R)C,AM = A2c(1 +k2aP)

2BN0x(t) = s(t) +n(t)

= [Ac+Ackam(t) +nI(t)]cos ct

nQ(t)sin ctw

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y(t) = envelope of x(t)

= [Ac+Ackam(t) +nI(t)]2 +n2Q(t)1

2

Ac+Ackam(t) +nI(t)

(SN R)O,AM A2ck

2aP

2BN0

F oMAM =

(SNR)O(SNR)C

|AM = k

2aP1 +k2aP

Thus the F oMAM is always inferior to F oMDSBSC

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Frequency Modulation

The analysis for FM is rather complex

The receiver model is as shown in Figure 6

m(t)

message with same

(B)

Band Pass Filterx(t)

Limiter Discriminatorw

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g

n(t)

power as modulated wave( )

Bandpasslow pass filter

y(t)

Figure 6: Analysis of FM System in Noise

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(SN R)O,FM =3A2ck

2fP

2N0B3

(SNR)C,FM = A2c

2BN0 world.com

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F oMFM =

(SN R)O(SN R)C

|FM =

3k2fP

B2

The significance of this is that when the carrier SNR is

high, an increase in transmission bandwidthBT provides a

corresponding quadratic increase in output SNR orF oMFM

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Digital Modulation

Continuous-wave(CW) modulation (recap):

A parameter of a sinusoidal carrier wave is varied

continuously in accordance with the message signal.

Amplitude

Frequency Phase

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Digital Modulation:

Pulse Modulation: Analog pulse modulation: A periodicpulse train isused as a carrier. The following parameters of

the pulse are modified in accordance with the message

signal. Signal is transmitted at discrete intervals of time.

Pulse amplitude Pulse width

Pulse duration

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Pulse Modulation: Digital pulse modulation: Message signalrepresented in a form that is discrete in both amplitude and

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time.

The signal is transmitted as a sequence of coded pulses

No continuous wave in this form of transmission

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Analog Pulse Modulation

Pulse Amplitude Modulation(PAM)

Amplitudes of regularly spaced pulses varied in proportion

to the corresponding sampled values of a continuous

message signal. Pulses can be of a rectangular form or some other

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appropriate shape.

Pulse-amplitude modulation is similar to natural sampling,

where the message signal is multiplied by a periodic train of

rectangular pulses.

In natural sampling the top of each modulated rectangular

pulse varies with the message signal, whereas in PAM it ismaintained

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