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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
= 0, t
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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|
Using the Power Series expansion for log(1 +x), |x|
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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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E[X(tk)X(tl)] =E[X(tk)]E[X(tl)] = 0,f o r|tk tl| > T
Case 2: when |tk tl| < T, with tk = 0 and tl < tk. X(tk) and
X(tl) occur in the same pulse interval provided tdelay satisfies
the condition tdelay < T |tk tl|.
E[X(tk)X(tl)|tdelay] = 1, tdelay < T |tk tl|
= 0, elsewhere
Averaging this result over all possible values oftdelay, we get
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E[X(tk)X(tl)] =
T|tktl|0
fTdelay (tdelay) dtdelay
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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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In NBFM
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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 ??
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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
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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
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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