random process in analog communication systems

7/28/2019 Random Process in Analog Communication Systems

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88

Introduction to Random Processes

Definition: Let (Ω,ℑ ,P) be a probability space. Let X be the mapping from the

sample space Ω to a space of functions called sample functions. Then X is called a

random process (r.p.) if at each time t i the mapping X is a random variable (r.v.),

i.e., X (t i, ζ ) belongs to ℑ for each fixed t i, t i ∈ R (the real line).

Example: A continuous-time r.p. with a finite sample space Ω = ζ 1, ζ 2 … ζ n is

illustrated on the next slide.

Observations:

1. For each fixed ζ i, X (t ,ζ i) is an ordinary time function

2. For each fixed t i, X (t i,ζ i) is an ordinary random variable.

As a consequence of these observations, at each t i the random process can be

described by either the probability density function (pdf), if exists, or by the

cumulative distribution function (cdf).



89



90

Let X (t ,ζ) be a random process (r.p.), then its mean, mean-square, and variance at

time t i are defined by

For the sake of conciseness, let X (t i) ≡ X (t i, ζ).

A measure of the statistical likeness (dependence) of the process from one instant

to another is the autocorrelation function. Mathematically, for arbitrary X (t )

or

),()( ζ µ ii X t X E t =

22 ),()( ζ ii t X E t X =

[ ][ ]22

2

)()(

),(),(),(),()(

i X i

i X ii X ii X

t t X

t t X t t X E t

µ

ζ µ ζ ζ µ ζ σ

−=

−−= ∗

,)()(),( 2*121 t X t X E t t R X ≡

( )1

2

1 2 1 1 2 2 1 2

1 2

1 2 1 1 2 2

( , ; , ) , ( ) is a continuous-valued r. p.

( , )( ) , ( ) , ( ) is a discrete-valued r. p.

x

R

X

R

x x f x t x t dx dx X t

R t t x x P X t x X t x X t

∗

∗

= = =

∫∫

∑∑



91

Let the ac component of X (t ) be defined by

then, the covariance of X (t ) is defined by

Definition: A random process X (t ) is stationary if it has the same nth order

distribution function as X (t + T ), ∀ T ∈ R, that is, for

The first-order probability density function of X (t ), if it exists, is defined by

( ) ( ) ( ),c i i X i X t X t t µ ≡ −

1 2 1 2

1 2 1 2 1 2 1 2

1 2 1 2

1 2 1 2

( , ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( , ) ( ) ( )

X c c

X X X X

X X

X X X

C t t E X t X t

E X t X t X t t t X t t t

E X t X t t t

R t t t t

µ µ µ µ

µ µ

µ µ

∗

∗ ∗ ∗ ∗

∗ ∗

∗

≡

= − − +

= −

= −

[ ]1( ) ( ) ,

T

n X t X t = X

1 1 1 1( , ; ; , ) ( , ; ; , ), .

n n n nF x t x t F x t T x t T t R= + + ∀ ∈

X X

1 1 1 1 1 1 1( , ) ( , ) ( ,0) ( ) , for . X X X X

f x t f x t T f x f x t T = + = = = −



92

Clearly, the first-order pdf does not depend on time. Hence,

which are constants.

Example: The first-order probability density function of a wide-sense stationary

Gaussian r.p. is given by

The second-order density function of a stationary r.p. X (t) is given by

where

Hence,

1 1( ) ( ) , and X X t E X t µ µ = =

( ),

2exp

2

1),(

2

2

111

−−=

X

X

X

X

xt x f

σ

µ

σ π

22

11

2 ))(()( X X X t X E t σ µ σ =−=

1 1 2 2 1 1 2 2

1 2 2 1 1

1 2 2 1 1

( , ; , ) ( , ; , )

( ,0; , ),

( , ; ), ,

f x t x t f x t T x t T

f x x t t t T

f x x t t t T

= + +

= − = −

= − = −

X X

X

X

[ ]1 2( ) ( ) .T

X t X t = X

)()()(),(

122121t t Rt X t X E t t R

X X −== ∗



93

A random process (r.p.) X (t ) is wide sense stationary (WSS) if

1. Its mean is constant (does not depend on time)

2. Its autocorrelation depends on the time difference only.

Mathematically,

and for all t 1 and t 2,

Remark Strict-sense stationary (SSS) implies WSS. However, the converse is not

always true.

Let t 2 = t 1 + τ , then if X (t ) is WSS, we get

X t X E µ =)(

),()()( 1221 t t Rt X t X E X −=∗

1 2 1 1

1 1

( ) ( ) ( ) ( )

( )

( ).

X

X

E X t X t E X t X t

R t t

R

τ

τ

τ

∗ ∗= +

= + −

=



94

Observation: If t 1 = t 2, then

is equal to the mean square value (total average power) of X (t ).Furthermore, assuming X (t ) is a real-valued random process, we get

This implies that

and These two conditions imply

Equivalently

2

1( ) (0) 0. X E X t R= ≥

[ ]

( ) 0)()0(2

)0()(2)0(

)()()(2)()()( 222

≥±=

+±=

++±+=±+

τ

τ

τ τ τ

X X

X X X

R R

R R R

t X t X t X t X E t X t X E

)0()( X X R R −≥τ

).0()( X X

R R ≤τ

),0()()0( X X X R R R ≤≤− τ

).0()( X X R R ≤τ



95

Let t 1 = t 2 + τ , then

Consequently,

Observation: The autocorrelation function is an even function of its argument τ .

Example: Consider the random binary signal shown below

where T b is the bit interval and t d is the value that the random delay T d takes on.

).()(),( 1221 τ −=−= X X X Rt t Rt t R

( ) ( ), an even function of its argument. X X

R Rτ τ = −



96

Assumptions:

1. During any bit interval where n is an arbitrary integer,

2. The time delay is uniformly distributed, namely,

Let us first compute the mean µ x(t).

To compute the autocorrelation function we must consider several situations. Let

t 1 > t 2, and t 1 and t 2 lie in two different bit intervals, then X (t 1) is statistically

independent of X (t 2). Mathematically,

,b

nT

bit bit 1/ 2P A P A= = = − =

≤≤

=otherwise,0

0,/1)(

bd b

d T

T t T t f

d

.0

2

1

2

1

bit)( bit)()(

=

⋅−⋅=

−=⋅−+=⋅==

A A

AP A AP At X E t X µ

000)()()()(),(212121

=⋅=⋅== t X E t X E t X t X E t t R X



97

Likewise, if t 2 > t 1 and t 1 and t 2 lie in two different bit intervals, the autocorrelation

of X (t) is equal to zero, i.e.,

In conclusion, if t 1 and t 2 lie in two different bit intervals,

On the other hand, if t 1 and t 2 occur in the same bit interval and then

This condition will hold when

1. For arbitrary n,

.0),( 21 =t t R X

,0),( 21 =t t R X

,12 bT t t <−

( )

21 2 1 2

2

1 2

2

1 2

2

1 2

2

1 2

( , ) and lie in thesame bit interval ( )

and liein thesame bit interval ( )

1and liein thesame bit interval

2

1 ( ) and liein thesame bit interval2

and liein thesamebit i

X R t t A P t t X t A

A P t t X t A

A P t t

A P t t

A P t t

= ⋅ ∩ =

+ − ⋅ ∩ = −

= ⋅

+ − ⋅

= ⋅ nterval

d bd b t nT t t t T n +<≤≤+− 21)1(



98

Since n is arbitrary, let n = 1, then

or

Thus,

2. t 1 and t 2 will also lie in the same interval if

or

d bd t T t t t +<≤≤ 21

.12 t t T t d b ≤<−

[ ]

1

2

1 2 2 1

1 2

2 12 1.

and liein thesame bit interval

1

1

1 ,

b

b d

t

d

bt T

b

b

b

P t t P t T T t

dt T

t T t T

t t t t

T

−

= − < ≤

=

= + −

−= − ≥

∫

d bd t T t t t +<≤≤ 12

.21 t t T t d b ≤<−



99

In this case,

Putting these two results together, we get

Example: Let a random process X (t ) be described by

where ωc is a known constant and

Ais a zero-mean random variable.

[ ]

2

1

1 2 1 2

2 1

2 1

1 2.

and liein thesame bit interval

1

1

1 ,

b

b d

t

d

bt T

b

b

b

P t t P t T T t

dt T

t t T T

t t t t

T

−

= − < ≤

=

= − +

−= + ≥

∫

<−

−−

=

otherwise,0

interval bitsametheinlieand ,,1),( 2112

122

21

t t T t t T

t t A

t t R b

b X

t At X cω cos)( =



100

Then,

and

Define the mean square value of A by

then the autocorrelation of X

(t ) is found to be

Clearly, this r.p. is neither SSS nor WSS.

0

cos

cos

)()(

0

=

=

=

=

t A E

t A E

t X E t

c

c

X

ω

ω

µ

1 2 1 2

1 2

21 2

( , ) ( ) ( )

cos cos

cos cos

X

c c

c c

R t t E X t X t

E A t A t

E A t t

ω ω

ω ω

=

= ⋅

=

2 2 A E A≡

[ ] .,,)(cos)(cos2

coscos),( 211221

2

21

2

21 t t t t t t A

t t At t R cccc X ∀−++== ω ω ω ω



101

Example: Let X (t ) be a r.p. described by

with A a known constant amplitude,ω

c a known constant frequency and θ be a

uniformly distributed r.v., i.e.,

Let A = 1, then

But,

[ ]π π ,~ −Θ U

( ) ( )cos( )

cos cos sin sin

cos cos sin sin .

X

c

c c

c c

t E X t E t

E t t

t E t E

µ ω

ω ω

ω ω

== + Θ

= Θ − Θ

= ⋅ Θ − ⋅ Θ

.0sin2

1

2

1coscos ==⋅=Θ

−−∫

π

π

π

π

θ π

θ π

θ d E

.0cos2

1

2

1sinsin =−=⋅=Θ

−−

∫π

π

π

π

θ π

θ π

θ d E

),cos()( Θ+= t At X cω



102

Hence,

Let us now compute its autocorrelation.

Again,

and

.,0)( t t X X ∀== µ µ

( ) ( )

( ) ( ) ( )

( ) ( ) ( )

1 2 1 2

1 2

1 2 1 2

1 2 1 2 1 2

1 2 1 2 1 2

( , ) ( ) ( )

cos( ) cos( )

1cos ( ) 2 cos ( )

2

1cos ( ) cos 2 sin ( ) sin 2 cos ( )

2

1cos ( ) cos 2 sin ( ) sin 2 cos ( )

2

X

c c

c c

c c c

c c c

R t t E X t X t

E t t

E t t t t

E t t t t t t

t t E t t E t t

ω ω

ω ω

ω ω ω

ω ω ω

=

= + Θ ⋅ + Θ

= + + Θ + −

= + Θ − + Θ + −

= + Θ − + Θ + −

1 1

cos 2 cos 2 sin 2 0.2 4

E d

π π

π π

θ θ θ π π −−

Θ = ⋅ = =∫

1 1

sin 2 sin 2 cos 2 0.

2 4

E d

π π

π π

θ θ θ

π π −−

Θ = ⋅ = − =∫



103

Hence, the autocorrelation of X (t ) is

Consequently, X (t) is WSS because the autocorrelation is a function of the time

difference.

Cross-Correlation Function

The cross-correlation function of two random processes X (t ) and Y (t ) is defined

by

If X (t ) and Y (t ) are jointly WSS, then

and

( ) Rt t t t t t R c X ∈∀−= 211221 ,,)(cos2

1),( ω

1 2 1 2

1 2 1 2

( , ) ( ) ( )

( , ) ( ) ( )

XY

YX

R t t E X t Y t

R t t E Y t X t

∗

∗

=

=

)(),( 1221 t t Rt t R XY XY −=

).()( 1221 t t Rt t R YX XY −=− ∗



104

Definition: A WSS r.p. X (t ) is ergodic in the mean if the time average of its sample

functions converges to the ensemble (statistical) average E X (t ) = µ X in the

mean-square sense, i.e.,

in the mean square sense, as T → ∞.

Observation: Note that M X (T ) is a random variable, thus we can compute its

statistical moments, i.e., its mean, variance, etc..Let us begin with the calculation of its mean.

This means that M X (T ) is an unbiased estimate of µX.

Suppose now that X (t ) is a real-valued random process, then

( ) X

T

T X dt t X T T M µ →≡ ∫− )(2

1

( ) 1 1 1 1

( ) ( ) ( )2 2 2 2

T T T T

X X X

T T T T

E M T E X t dt E X t dt E X t dt dt T T T T

µ µ − − − −

= = = = =

∫ ∫ ∫ ∫



105

where C X (t 2 – t 1) is the auto covariance (ac component) of the process X (t ).

If Var ( M X (T )) is small, then the time average M X (T ) is a good estimate of the

mean µ X . Furthermore, if this variance converges to zero as T → ∞, i.e.,

then M X (T ) is a consistent estimate of µ X .

( ) ( )( ) ( )

[ ] ,)(4

1)(

4

1

)(4

1

)(2

1

)(2

1

2112221

2

122

2

21122

2

2211

222

dt dt t t C T

dt dt t t RT

dt dt t t RT dt t X T dt t X T E

T M E T M E T M Var

T

T

T

T

X

T

T

T

T

X X

T

T

X

T

T

X X

T

T

T

T

X X X X X

∫ ∫∫ ∫

∫ ∫∫∫

− −− −

− −−−

−=−−=

−−=−

⋅=

−=−=

µ

µ µ

µ µ

( ) lim 0, X T Var M T →∞ =



106

Definition

A WSS r.p. X (t ) is ergodic in correlation at the time shift τ if and only if

in the mean-square sense.

If the condition is true for all τ then X (t) is ergodic in correlation.

Remark: Although ergodicity holds in most practical cases, care must be

exercised to make sure the required conditions for ergodicity are met before

proceeding to use time averages instead of ensemble averages.

Let us consider what happens to the output of a continuous-time bounded-input

bounded-output (BIBO) stable linear and time invariant (LTI) system described by

the impulse response h(t ), when its input is a WSS r.p. X (t ). We know that

)()()(21lim τ τ X

T

T T

Rdt t X t X T

=+∫−

∗∞→

.)()()()()( ∫∫∞

∞−

∞

∞−

−=−= λ λ λ λ λ λ d ht X d X t ht Y



107

Thus, the mean value of the output is given by

where H (ω) = Fh(t ).

Observation: H (0) exists if and only if the system is BIBO stable.

Now, the autocorrelation of the output can be computed as follows:

),0()()(

)()()()()()(

H d hd h

d ht X E d ht X E t Y E t

X X X

Y

µ λ λ µ λ λ µ

λ λ λ λ λ λ µ

===

−=

−==

∫∫

∫∫∞

∞−

∞

∞−

∞

∞−

∞

∞−

1 2 1 2 1 2

2 1 2 1

( , ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ).

Y

X Y

R t t E Y t Y t h h E X t X t d d

h h R t t d d R t t

α β α β α β

α β α β α β

∞ ∞∗ ∗ ∗

−∞ −∞

∞ ∞∗

−∞ −∞

= = − −

= − + − = −

∫ ∫∫ ∫



108

Likewise, the cross-correlation between X (t ) and Y (t ) is given by

Example: Consider the change detector T [·], i.e, Y (t) = T [ X (t )], such that

Y (t ) = X (t ) – X (t -1), where X (t ) is a real-valued WSS r.p. with zero mean.

Change detector.

).()()()()(

)()()()()()(

)()()()()(),(

12121212

2121

212121

t t Rt t ht t Rd t t Rh

d t X t X E hd t X t X h E

d t X ht X E t Y t X E t t R

X XY X

XY

−∗−=−=−−=

−=

−=

−==

∗∞

∞−

∗

∞

∞−

∗∗∞

∞−

∗∗

∞

∞−

∗∗∗

∫

∫∫

∫

α α α

α α α α α α

α α α



109

Let us first compute the mean value of the output of the detector,

To detect a change, let us compute the autocorrelation of the output, i.e.,

For the detector to be effective, the variance of Y (t) should be large (the power of the difference X (t ) – X (t -1) should be large).

Let the autocorrelation of the input be described by

( ) ( ) ( 1)( )

( ) ( 1) 0.

Y

X X

t E E X t X t Y t

E X t E X t

µ

µ µ

= = − −

= − − = − =

[ ][ ]

).()1()1()(2

)()1()1()(

)1()1()()1()1()()()(

)1()1()()1()1()()()(

)1()()1()()()(),(

21121212

12121212

21212121

21212121

22112121

t t Rt t Rt t Rt t R

t t Rt t Rt t Rt t R

t X t X E t X t X E t X t X E t X t X E

t X t X t X t X t X t X t X t X E

t X t X t X t X E t Y t Y E t t R

Y X X X

X X X X

Y

−=+−−−−−−=

−++−−−−−−=−−+−−−−=

−−+−−−−=

−−−−==

,0,)(

122

21 >=−

−−

α σ

α t t

X X et t R



110

Then, the autocorrelation of the output is

Since the mean of the output is zero, the autocovariance (ac power) of the output is

the same as its autocorrelation (total power), i.e.,

Power spectral density (PSD)

Definition: Let X (t ) be a WSS r.p. with autocorrelation function R X (τ ). Then the

power spectral density S X (ω) is the Fourier transform of R X (τ ), i.e.,

Moreover, under certain general conditions,

.2),(12122

21121212 +−−−−−−− −−= t t

X

t t

X

t t

X Y eeet t Rα α α

σ σ σ

( ).122)0( 22222 α α α σ σ σ σ −−− −=−−== eee RY X X X X Y

( ) ( ) ( ) . j

X X X S R R e d

ωτ ω τ τ τ ∞

−

−∞

= ℑ = ∫

1 1( ) ( ) ( ) .

2

j

X X X R S S e d ωτ τ ω ω ω

π

∞−

−∞

= ℑ = ∫



111

Properties of the PSD:

1. S X (ω) is real-valued since R X (τ ) is such that R X (τ ) = R X * (-τ ) .

2. If X (t) is real-valued, then S X (ω) is an even function of ω since R X (τ ) is real and

even. Otherwise, S X (ω) is not an even function of ω.

3. S X (ω) ≥ 0, ∀ ω.

4.

Example: Let X (t ) have autocorrelation function

What is the power spectral density of X (t )?

).()()1( 22

2

ω ω τ

τ X n

F

n X

n

n S d Rd ⇔−

.0,)( 2 >= −α σ τ

τ α e R X X

2 2( ) ( )j j

X X X X S R e e d e d

α τ α τ ωτ ωτ ω τ σ τ σ τ ∞ ∞

− − −−

−∞ −∞

= ℑ = ⋅ =∫ ∫



112

or

Practical approach to calculate S X (ω)

Define the truncated (windowed) version of the WSS r.p. X (t ) by

Let

22

2

22

2

2

0

0

2

0

0

22

2

)10(1

)01(1

11

)(

α ω ασ

α ω ω α ω α σ

ω α ω α σ

ω α ω α σ

τ τ σ τ σ ω

ωτ ατ ωτ ατ

ωτ ατ ωτ ατ ωτ τ α

+=

+−++=

−

+−−

−=

−−+−=

+==

∞

−−

∞−

−

∞−−

∞−

−∞

∞−

−−

∫∫∫

X X

X

j j X

j j

X

j

X X

j j

j j

e j

e j

d ed ed eS

[ ]

−∈

≡elsewhere,0

,),()(

T T t t X t X T

( ) ( ) ( ) .T

j t

T T

T

X t X t e dt ω χ ω −

−

≡ ℑ = ∫



113

Then,

Since is a random variable, we can compute its ensemble average, i.e,

Let τ = t – s and s = s, then

where

.)()()( )(2

∫ ∫− −

−−∗=T

T

T

T

st j

T dtdses X t X ω ω χ

2)(ω χ T

.)(

)()()(

)(

)(2

∫ ∫

∫ ∫

− −

−−

− −

−−∗

−=

=

T

T

T

T

st j

X

T

T

T

T

st j

T

dtdsest R

dtdses X t X E E

ω

ω ω χ

∫ ∫∫ ∫ℜ

−

− −

−− =−

),(

)( )(),()(

s

j

X

T

T

T

T

st j

X dsd e Rs J dtdsest R

τ

ωτ ω τ τ τ

110

11=

−=

∂

∂

∂

∂∂∂

∂∂

=

s

s

t

sst J

τ τ



114

Graphically, the map is described by

Space transformation



115

Equivalently,

Therefore,

and

.)(

2

12

)()2()()2(

)()()(),(

2

2

0

2

2

0

0

2

2

0),(

∫

∫∫

∫ ∫∫ ∫∫ ∫

−

−

−

−−

− −−

−−

−

−

ℜ

−

−=

++−=

+=

T

T

j

X

T

j

X

T

j

X

T

T

T

j

X

T T

T

j

X

s

j

X

d e R

T

T

d e RT d e RT

dsd e Rdsd e Rdsd e Rs J

τ τ τ

τ τ τ τ τ τ

τ τ τ τ τ τ τ

ωτ

ωτ ωτ

τ

ωτ

τ

ωτ

τ

ωτ

∫−

−

−=

T

T

j

X T d e R

T

E

T

2

2

2)(

2

1)(

2

1τ τ

τ ω χ ωτ

21lim ( ) ( ) ( )

2

j

T X X T

E R e d S T

ωτ χ ω τ τ ω ∞

−

→∞−∞

= =∫



116

where the Fourier transform of the truncated signal is

Therefore,

Hence S X (ω) is real, nonnegative and is related to the average power at frequency

ω.

Example: Let X (t ) have the autocorrelation function R X (τ ) = σ 2δ (t ). Then

Consider again the output of a BIBO stable LTI system excited by a WSS r.p. X (t )Then

∫−

−=T

T

t j

T dt et X .)()( ω ω χ

2)(

2

1lim)( ω χ ω T T

X E T

S ∞→

=

.,)()()( 22 ω σ τ τ δ σ τ τ ω ωτ ωτ ∀=== ∫∫∞

∞−

−∞

∞−

−d ed e RS

j j

X X

∫ ∫ ∫

∫ ∫ ∫∫∞

∞−

∞

∞−

∞

∞−

−

∞

∞−

∞

∞−

∞

∞−

−∞

∞−

−

−+=

−+==

.)()()(

)()()()()(

β α τ β α τ β α

τ β α β α τ β α τ τ ω

ωτ

ωτ ωτ

d d d e Rhh

d ed d Rhhd e RS

j

X

j

X

j

Y Y



117

Let λ = τ + α - β . With respect to the innermost integral, d λ = d τ and τ = λ + β - α .

Thus,

or

since

It is clear from the above result that

Average total power of Y (t )

=

=

∫∫∫

∫ ∫ ∫∞

∞−

−∞

∞−

−∞

∞−

∞

∞−

∞

∞−

∞

∞−

−+−

λ λ β β α α

β α λ λ β α ω

ωλ ωβ ωα

α β λ ω

d e Rd ehd eh

d d d e RhhS

j

X

j j

j

X Y

)()()(

)()()()( )(

),()(

)()()()(

2ω ω

ω ω ω ω

X

X Y

S H

S H H S

=

= ∗

*( ) ( ) ( ) ( ), for real ( ). j j jh e d h e d h e d H h

ωα ωα ωα α α α α α α ω α

∗ ∗∞ ∞ ∞∗ − −

−∞ −∞ −∞

= = =

∫ ∫ ∫

.,0)( ω ω ∀≥Y S

).0()(

2

Y Rt Y E =



118

Now,

This implies that the total average power of Y (t ) (mean square value of Y (t )) is

Example: Consider an LTI system with impulse response h(t ) = 2e-5t u(t ), excited by

a r.p. X (t ) with mean µ X and autocorrelation R X (τ ) = 10δ (τ ). What are the analyticalexpressions of RY (τ ) and S Y (ω)?

We know that

Hence,

and

1 1( ) ( ) ( )

2

j

Y Y Y R S S e d

ωτ τ ω ω ω π

∞−

−∞

= ℑ = ∫

2 2 21 1( ) (0) ( ) ( ) ( ) ( ) ( ) .

2 2Y Y X X

E Y t R S d H S d H f S f df ω ω ω ω ω π π

∞ ∞ ∞

−∞ −∞ −∞

= = = =∫ ∫ ∫

2 5 2

( ) ( ) ( ) with ( ) 2 ( )5

t

Y X S H S H e u t

jω ω ω ω

ω

−= = ℑ =+

ω ω ω ω ω

ω ∀+=−+=⋅+= ,25

40)5)(5(

40105

2)( 2

2

j j jS Y

1 1

2 2 2

1 1 40 40 2 5( ) ( ) 4

2 2 25 25 25

j j

Y Y R S e d e d ωτ ωτ τ ω ω ω

π π ω ω ω

∞ ∞− −

−∞ −∞

• = = = ℑ = ℑ + + + ∫ ∫



119

or

Gaussian processes

Let Z be a Gaussian random variable. Then its pdf has the form

Consider now the situation where1) X 1, X 2, ..., X N is a sequence of statistically independent r.v.’s.

2) The X i’s have the same pdf with mean µ X and variance

If the X i’s meet these two criteria, then they are independent and identically

distributed (iid).

Furthermore, if

τ τ τ ∀= −

,4)(5

e RY

.,2

)(exp

2

1)(

2

2

z z

z f Z

Z

Z

Z ∀

−−=

σ

µ

σ π

.2

X σ

∑=

−=

N

i X

X i

N

X

N

Y

1

,1

σ

µ



120

Then, as N → ∞, Y N becomes a zero-mean Gaussian r.v. with unit variance, i.e.

This is the so-called central limit theorem.

It is important to note that if the X i’s are iid, then the pdf of Y N is such that if

An n-fold convolution, where f Z ( z) is the common pdf of the Z i’s.

Properties of a Gaussian Processes

1) Let X (t ) be a real Gaussian r.p. that is input to a BIBO stable LTI system withimpulse response h(t ). Then Y (t ) = h(t ) ∗ X (t ) is also a Gaussian r.p.

2) Let X (t ) be a Gaussian r.p. and t 1, t 2 be two known sampling instants of the

process X (t ). Then the random variables X (t 1) and X (t 2) are jointly Gaussian

ye y f y

Y N ∀→ − ,

2

1)( 2/2

π

( )1

1, / , 1, ,

N

N i i i X X i

Y Z Z X i N N

µ σ =

= = − =∑

1 2

1times

( ) ( ) ( ) ( ) ( ) ( ) ( ) N N Y Z Z Z Z Z Z

N

f z f z f z f z f z f z f z−

= ∗ ∗ ∗ = ∗ ∗ ∗



121

with pdf

where

3) Let X (t) be a Gaussian r.p. and t 1, t 2 and be two known sampling instants. If the

r.v.’s X (t 1) and X (t 2) are uncorrelated, i.e,

then the r.v.’s X (t 1) and X (t 2) are independent.

( )1 1

1

( ), ( ) 1/2

1 1exp ( ) ( ) ,

2 2

T

X t X t f

π

− = − − − ∆ x x μ Σ x μ

=

=

2

1

2

1

)(

)(

X

X

t X

t X X

( )

( )

11 1

2 2 2

( )

( )

E X t X t E E

X t E X t

µ

µ

= = = =

μ X

[ ]

−−

−

−= 2211

22

11)()(

)(

)(µ µ

µ

µ t X t X

t X

t X E Σ

Σdet=∆

( )( ) ,0)()( 2211 =−− µ µ t X t X E



122

Uncorrelatedness means that

and

Hence,

where

Thermal noise

Let V TN (t ) be the thermal noise voltage that appears across the terminals of a

resistor due to the random motion of electrons. Then the mean square voltage

observed over a bandwidth of Δ f can be approximated by

,det0

0 2

2

2

12

2

2

1 σ σ σ

σ ⋅=⇒

= ΣΣ

−−

−−=

2

2

2

22

2

1

2

11

21 2

)(

2

)(exp

2

1)(

σ

µ

σ

µ

σ πσ

x x f xX

),()(2

)(exp2

12

)(exp21)( 2)(1)(2

2

2

22

2

2

1

2

11

1

21 x f x f x x f t X t X ⋅=

−−⋅

−−=

σ µ

σ π σ µ

σ π X X

.)(

)(

2

1

=

t X

t X X



123

where k is Boltzmann’s constant (1.38x10-23 Joules/Kelvin), T is the absolute

temperature in degrees Kelvin, and R is the resistance in Ohms.

It should be clear that

We can think of V TN (t ) as a r.p. that has a constant psd over the range of

frequencies

or

2 2( ) 4 Joules / sec 4 volt ,TN

E V t kTR f kTR f ≅ ∆ ⋅Ω = ∆

Wattsor Joules/sec4)(1 2 f kT t V

R E TN ∆≅

,2

,2

∆∆−

f f

∆∆−∈

≅elsewhere,0

2,

2,/Hzvolts4

)(2 f f

f kTR f S

TN V



124

Graphically,

PSD of thermal noise.

This is true because

White Noise

It is an extension of the concept of thermal noise, that is, if W (t ) is a white noise

process, then

2

2

2

( ) (0) 4 4TN

f

TN V

f

E V t R kTRdf kTR f

∆

∆−

= = = ⋅ ∆∫

f kT

f S e

W ∀= ,2

)(



125

where k is Boltzmann’s constant and T e is the equivalent noise temperature of the

receiver.

Let N 0 = kT e, then and

White noise autocorrelation and its psd.

Observations

1. , are uncorrelated

2. If W (t ) is Gaussian, then W (t 1) and W (t 2), t 1 ≠ t 2, are also statistically independent.

3. The white noise model is good approximation whenever the bandwidth of the

noise is much larger that the bandwidth with an LTI system excited by such

a noise.

f

N

f S W ∀= ,2)(0

).(2)(0

τ δ τ

N

RW =

21210 ),(and )()(

2)( t t t W t W N RW ≠⇒= τ δ τ



126

Example: Let an LTI system be described by

What are the autocorrelation and the psd of the output Y (t) when the input to the

system X (t) is white noise?

We already know that

Hence,

Now,

≤

=elsewhere,0

,1)(

W H

ω ω

).()()(2

ω ω ω x y S H S =

02 0

,( ) ( ) 2

20, elsewhere

Y

N W N S H

ω ω ω

≤= =

[ ] ( )( )

( )τ π τ

τ

π

τ πτ πτ πτ ω π τ

τ τ

ω

ω

ωτ ωτ

W W N

W

W W N

W

N

ee j

N

e j

N

d e

N

R

jW jW

W

W

j

W

W

j

Y

sinc2

sin

2

sin24422

1

)(

00

0000

=

=

=−===

−

=

−=−∫



127

Narrowband noise

In order to optimize communication system performance in the presence of noise,

filtering operations around the carrier frequency are performed at the front end of

the receiver. The result of these operations is that the receiver input wideband

noise is transformed into narrowband noise centered around the carrier frequency.

An example of a narrow-band noise signal is shown below

PSD of narrowband noise.

In-phase and quadrature representation: Let N I (t ) and N Q(t ) be the in-phase and

quadrature components, respectively, of the narrowband noise N (t ).



128

Then the bandpass narrowband noise can be written as

Properties of N I

(t) and N Q

(t):

1)

2) If N (t) is Gaussian, then N I (t) and N Q(t) are jointly Gaussian.

3) If N (t) is stationary, then N I (t) and N Q(t) are jointly stationary.

4)

5)

6)

Example: Consider a receiver whose prefilter is described by

].,[~),2sin()()2cos()()( π π φ φ π φ π −+−+= U t f t N t f t N t N cQc I

0)()( == t N E t N E Q I

≤++−==

elsewhere,0

),()()()(

B f f f S f f S f S f S c N c N

N N Q I

)()()( t N Var t N Var t N Var Q I ==

[ ] ≤−−+=−= elsewhere,0

,)()()()( B f f f S f f S j f S f S c N c N N N N N I QQ I

+≤≤−=

elsewhere,0

,1)(

B f f B f f H cc



129

Let the received signal be described by

where sT (t ) is the transmitted signal and w(t ) is white Gaussian noise.

Let us consider the contribution of the white Gaussian noise to the output of the

filter only. In this case, if N (t ) is the output due to the noise component, then

Moreover,

)()()( t wt st s T R +=

+≤≤−=

elsewhere,0

,2)(

0

cc N

f B f B f N

f S

1 2 2 20 0

0 c

( ) ( ) ( )2 2

2 sinc(2B )cos(2 f ).

c c

c c

f B f B

j f j f j f

N N N

f B f B

N N R S f S f e df e df e df

N B

π τ π τ π τ τ

τ π τ

− + +∞−

−∞ − − −

= ℑ = = +

=

∫ ∫ ∫



130

Finally,

PSD of in-phase and quadrature noise components.

<<−

==elsewhere,0

,)()(

0 B f B N f S f S

Q I N N



131

Noise performance of CW modulation

Let us now investigate the effects that channel imperfections have on the

performance of a CW modulated communication system.We will first focus on the effect of noise. Consider the following receiver model

Generic receiver block diagram

where sT (t ) is the transmitted signal

w(t ) is white noise with psd N 0/2, ∀ f

s R(t ) = sT (t ) + w(t ) is the received signal

x(t ) is the filtered signal available for demodulation

xdem(t ) is the estimate of m(t )



132

At the output of the bandpass filter, we get

where and

If we assume that the bandpass filter is ideal with center frequency f c then the

noise is bandpass, i.e.,

Noise component psd at output of BPF

where the bandpass noise can be mathematically described by

n I (t ) and nQ(t ) are the in-phase and quadrature noise components.

)()()( t nt st x +=

)()()( t st ht s T BPF ∗= )()()( t wt ht n BPF ∗=

( ) ( ) cos(2 ) ( ) sin(2 ) I c Q c

n t n t f t n t f t π φ π φ = + − +



133

Definition: The input signal-to-noise ratio (SNR)i is the ratio of the average power

of the filtered modulated signal s(t ) to the average power of the filtered noise n(t ).

Let Ps ≡ Average power of s(t )

Pn ≡ Average power of n(t )

Then,

At the modulator output, assuming the noise component is additive,

where sdem(t ) contains information about the message signal m(t ) and ndem(t ) is the

noise contribution at the output of the demodulator.

Definition: The output signal-to-noise ratio (SNR)o is the ratio of the average power

of sdem(t ) to the average power of the noise component ndem(t ).

Remark: Both sdem(t ) and ndem(t ) depend on the type of modulation scheme used.

.)(0 T

s

n

s

i B N

P

P

PSNR ==

),()()( t nt st x demdemdem +=



134

To establish a fair performance comparison between different modulation schemes

we shall assume the following:

1. sT

(t ) has the same average power, regardless of the modulation scheme

2. w(t ) has the same power in the bandwidth W m

Definition: The channel signal-to-noise ratio (SNR)c is the ratio of the average

power of the modulated signal to the average power of channel noise in the

message bandwidth W m

Let the communication system performance figure of merit be

Consider first the performance of DSB using a coherent detector at the receiver , i.e,

.)/()( co SNRSNR



135

From the block diagram

where and w(t ) is AWGN with psd Let m(t ) have zero mean and Then

),()()( t wt st s DSB R +=

)()cos()( t mt At s cc DSB ⋅+= φ ω ,2/)( 0 N f S w = . f ∀[ ].2,0~ π φ U

( )[ ] ( )[ ]

( ) ( ) ( )

( )

( ) ( )

( )

( ) )()(cos)()(

)()(cos)(

cos)(cos)()(

)()(cos)()(

)()(cos)(cos)(cos)()(

)(cos)()()(cos)()()(

00

00

2

2

t wt w E t E t m E t w E A

t w E t E t m E A

t t E t mt m E A

t wt wt t mt w A

t wt t m A

t t t mt m A E

t wt t m At wt t m A E t st s E

cc

cc

ccc

cc

cc

ccc

cccc R R

τ φ ω τ

φ τ ω τ

φ ω φ τ ω τ

τ φ ω τ

φ τ ω τ φ ω φ τ ω τ

φ ω τ φ τ ω τ τ

+++++

++++

++++=

+++++

++++ ++++=

+++++++=+



136

If m(t ) is WSS, we get

Let the ideal bandpass filter have gain c in the frequency range

).()(2

)(cos)(2

1

)()(cos)(2

1

)()(cos2

1)22(cos

2

1)(

),()(cos2

1)22(cos

2

1)()()(

02

2

0

2

2

τ τ δ τ ω τ

τ τ ω τ

τ τ ω φ τ ω ω τ

τ τ ω φ τ ω ω τ τ

Rscmc

wcmc

wcccmc

wcccmc R R

R N

R A

R R A

Rt E R A

Rt E R At st s E

=+=

+=

+

+++=

+

+++=+

mcmc

W f f W f +≤≤−



137

Then

Now,

Therefore,

+≤≤−

==elsewhere,0

),()()()(

22 mcmcs

s BPF X

W f f W f f S c f S f H f S R

R

[ ]

2 0

2 0

2 2 0

1( ) ( ) ( ) cos(2 ) ( )

2 2

1 1( ) ( ) ( )

2 2 2

1 1( ) ( ) .

4 4 2

R Rs s c m c

c m c c

c m c c m c

N S f R A R f

N A S f f f f f

N A S f f A S f f

τ τ π τ δ τ

δ δ

= ℑ = ℑ + ℑ

= ∗ + + − +

= + + − +

+≤≤−+−++=elsewhere,0

,2

)(4

)(4)(

0

22222

mcmccmc

cmc

X W f f W f

N c f f S

Ac f f S

Ac f S



138

So, the psd of the message signal component s(t ) is

,

whereas the psd of the noise component n(t ) is

.

Let Pm be the average power of m(t ), i.e,

then the message signal component s(t ) has average power

[ ]

+≤≤−−++=elsewhere,0

,)()(4)(

22

mcmccmcm

c

s

W f f W f f f S f f S Ac

f S

+≤≤−=elsewhere,0

,2)(

0

2

mcmcn

W f f W f N c

f S

,)( df f S Pm

m

W

W

mm ∫−

=

−++= ∫∫

+

−

+−

−−

mc

mc

mc

mc

W f

W f

cm

W f

W f

cmc

s df f f S df f f S Ac

P )()(4

22



139

or

The average power of n(t ) is

or

At the output of the mixer,

But, and

Hence,

.2

24

)()(4

222222mc

mc

W

W

m

W

W

mc

s

P AcP

Acdf f S df f S

AcP

m

m

m

m

=⋅=

+= ∫∫

−−

.22

)( 0

20

2

m

W f

W f

W f

W f

nn W N cdf df N c

df f S Pmc

mc

mc

mc

=

+== ∫∫∫

+

−

+−

−−

∞

∞−

m

mc

m

mc

n

s DSBiW N

P A

W N c

P Ac

P

PSNR

0

2

0

2

22

,42

2/)( ===

[ ] ).cos()()()cos()()( φ ω φ ω ++=+= t t nt st t xt v cc

)cos()()( φ ω += t t mcAt s cc .)sin()()cos()()( φ ω φ ω +−+= t t nt t nct n cQc I

)cos()sin()()(cos)()(cos)()( 22 φ ω φ ω φ ω φ ω ++−+++= t t t cnt t cnt t mcAt v ccQc I cc



140

or

At the output of the lowpass filter, assuming an ideal response,

Since m(t ) is uncorrelated (and therefore orthogonal) to n I (t ) then

and

So, the average power of the component that contains the signal is

( )[ ] ( )[ ] ( ).)(2sin2

)()(2cos1

2

)()(2cos1)(

2)( φ ω φ ω φ ω +−+++++= t

t cnt

t cnt t m

cAt v c

Q

c

I

c

c

2

)()(

2)(

t cnt m

cAt x I c

dem +=

)(4

)(4

)(222

τ τ τ I dem nm

c

x Rc

R Ac

R +=

)(4

)(4

)(222

τ I dem nm

c

x S c

f S Ac

f S +=

mc

W

W

m

c

m

c

os

P Ac

df f S Ac

df f S Ac

Pm

m

4

)(4

)(4

22

2222

=

== ∫∫−

∞

∞−



141

and the average power of the component that contains the noise is

Therefore, the output SNR is

Consequently the figure of merit (using the SNR at the output of the BPF) is

A ratio of 2 would mean that the demodulator gain is equal to 2, however, the pre-

detection bandpass filter has a bandwidth which is twice that of the message

bandwidth. Therefore, the noise bandwidth is doubled which implies that the noise

power is doubled at the demodulator input. The demodulator gain is necessary

to overcome this effect.

24)(

4

0

2

0

22 N W c

df N c

df f S c

P m

W

W

non

m

m

I === ∫∫

−

∞

∞−

0

2

0

2

22

,22/

4/)(

N W

P A

N W c

P AcSNR

m

mc

m

mc

DSBo ==

24/

2/

)(

)(

0

2

0

2

,

, == N W P A

N W P A

SNR

SNR

mmc

mmc

DSBi

DSBo

Let us now compute the figure of merit in terms of the channel SNR.



142

Let us now compute the figure of merit in terms of the channel SNR.

The received signal power spectral density is still given by

2 2 01 1( ) ( ) ( ) .

4 4 2

Rs c m c c m c

N S f A S f f A S f f = + + − +

Thus, the signal component average power is

2 2

2 2

( ) ( ) ( ) ( )4 4

24 2

c m c m m m

c m c m m m

f W f W W W

c cs m c m c m m

f W f W W W

c c mm

A AP S f f df S f f df S f df S f df

A A PP

− + +

− − − − −

= + + − = +

= ⋅ =

∫ ∫ ∫ ∫

and the noise average power in the message signal bandwidth is ( )002

2m m

N W N W = .

Hence, the channel SNR is given by ( )

2

2

,0 0

2

2

c m

c m

c DSB

m m

A P A PSNR

N W N W = = . The figure of merit is now

2, 0

2

, 0

( ) / 21

( ) / 2

o DSB c m m

c DSB c m m

SNR A P W N

SNR A P W N = = .



143

Noise in AM receivers using Envelope detection

Recall that

where m(t ) has average power Pm and . Now

Therefore, the autocorrelation of the transmitted waveform is

Thus, the received signal power is

If the message bandwidth is W m then the noise power in the message bandwidth is

assuming AGWN with psd

( ) [1 ( )]cos( ) AM c a c

s t A k m t t ω φ = + +

~ [0 2 ]U φ π ,

( ) cos( ) ( ) cos( ) AM c c c a c

s t A t A k m t t ω φ ω φ = + + +

2 2 2

( ) cos( ) ( )cos( )2 2 AM

c c as c m c

A A k R Rτ ω τ τ ω τ ⇒ = +

2 2 2 22

(0) 1 , (0).

2 2 2 AM

c c a cs m a m m m

A A k AP R k P P R = + = + =

002 ,

2n m m

N P W W N

= =

( )0

2

N

W S f =



144

The channel signal-to-noise ratio is

or

Consider the following receiver model

AM envelope detector receiver

2 21

2

0( )

c a m A k P

c AM mSNR W N

+

, =2 2

0

1( )

2

c a m

c AM

m

A k PSNR

W N ,

+ =



145

Let the bandpass filter be ideal, with unity gain and bandwidth 2W m. Then, at its

output,

or

where

and

Therefore, at the output of the envelope detector, we have

( ) ( ) ( ) AM

x t s t n t = +

[ ]

[ ]

( )

( ) ( )

( ) 1 ( ) cos( ) ( ) cos( ) ( )sin( )

( ) ( ) cos( ) ( )sin( )

( ) , ( ) ( )

( )cos ( )

c

c a c I c Q c

c a c I c Q c

j t j t

c

x t A k m t t n t t n t t

A k A m t n t t n t t

Re x t e x t a t e

a t t t

ω φ ψ

ω φ ω φ ω φ

ω φ ω φ

ω φ ψ

+

= + + + + − +

= + + + − +

= =

= + +

[ ]122 2( ) ( ) ( ) ( )

c c a I Qa t A A k m t n t n t = + + +

1( )

( ) tan

( ) ( )

Q

c c a I

n t t

A A k m t n t

ψ − =

+ +

122

( )( ) ( ) [ ( ) ( )] 1

( ) ( )

Q

c a c I

c a c I

n t y t a t A k A m t n t

A k A m t n t

= = + + +

+ +



146

Remark: Phase information is irrelevant at the output of an envelope detector !

For an envelope detector to work properly, the carrier power must be much larger

than the average noise power Pn. If this were the case, then

Note: The DC term does not contribute anything to the demodulation of m(t ) and

can be removed by a DC blocking capacitor.

Let the output of the DC blocking capacitor be

Thus, the output signal-to-noise ratio is given by

, since the in-phase noise has its energy in the frequency

interval [-W m, W m].

Finally, assuming less than 100% of modulation, we get the following figure of

merit:

( ) ( ) ( ).c a c I y t A k A m t n t ≅ + +

noisesignal +=+≅ )()()( t nt m Ak t y I ca

2 2

0

( ) 2a c mo AM

m

k A PSNR W N

, =



147

Figure of merit

It is clear that the performance of AM with envelope detection is inferior to that of

DSB with coherent detection. This is due to explicit transmission of the carrier in

AM.

Example: Consider an AM system in AWGN with psd

Let m(t ) have a bandwidth of 4 kHz and pdf

2 2

0

2 2

0

22

212

( )1

( ) 1

a c m

m

c a m

m

k A P

W N o AM a m

A k Pc AM a m

W N

SNR k P

SNR k P

,

+ ,

= = = <+

( ) triangle2

m

x f x

=

.,102)(

120 f Hz

Watts N f S

W ∀==

−



148

If the receiver signal is demodulated by an envelope detector and appropriate

postdetection filtering, assuming k a = 0.9 find the minimum value of the carrier

amplitude Ac that will yield (SNR)0,AM ≥ 40 dB.

Solution: We know that

Now,

thus,

[ ]( ) 1 0 9 ( ) cos( ), AM c c

s t A m t t ω φ = + . + ~ [0 2 ]U φ π ,

1 0 1

1 1 0

1 1

0 0

( ) tr ( 1) ( 1)2

( 1) ( 1)

0, as expected.

x E m t x dx x x dx x x dx

x x dx x x dx

− − = = + + − +

= − − + + − +

=

∫ ∫ ∫

∫ ∫

1 1

2 2 2

1 0

4 3

( ) 2 ( 1)2

1 1 1 12 2 2

4 3 4 3 12 6

m

xP E m t x tr dx x x dx

x X

−

= = = − +

− = + = − + = =

∫ ∫



149

So,

Now,

Consider now the case when the channel SNR is less than 1. At the input of the

envelope detector, we have (assuming φ = 0 since envelope detector ignores phase)

where r (t ) is the envelope of the noise.

( )

( )

2 22 2 16

120

2

(0 9)( )

2 2(4000) 2 10

8437500

ca c mo AM

m

c

Ak A PSNR

W N x

A

, −

.= =

=

10

4

10

10log ( ) 40dB

log ( ) 4 ( ) 10

o AM

o AM o AM

SNR

SNR SNR

,

, ,

≥

≥ ⇒ ≥

4

2 2

10volts8437500

0 034 volts

c

c

A

A

⇒ ≥

⇒ ≥ .

[ ]

[ ] ( )

( ) 1 ( ) cos ( )

1 ( ) cos ( )cos ( ) ,

c a c

c a c c

x t A k m t t n t

A k m t t r t t t

ω

ω ω ψ

= + +

= + + +



150

Now, means that almost all the time since r (t )

is a r.p.. Let us illustrate this using a time-varying phasor diagram.

( ) 1c AM SNR , < [ ]( ) 1 ( )

c ar t A k m t > +

[ ]( ) 1 ( ) 0 ( ) ( )oc a x t A k m t r t t ψ = + ∠ + ∠

1

( )( ) tan ( )

Q

I

n t t

n t ψ −

=

It is clear from the picture that (using the direction of r(t) as the reference) ( )x t



151

It is clear from the picture that (using the direction of r (t ) as the reference)

is equal to

and that

However,

Since

This means that the information has, on average, been lost and it makes no sense to

talk about any further or the possibility to demodulate m(t )!

( ) x t

[ ] [ ]( ) ( ) 1 ( ) cos ( ) 1 ( ) sin ( )c a c a

x t r t A k m t t jA k m t t ψ ψ = + + − +

[ ]( ) ( ) 1 ( ) cos ( ).c a

y t r t A k m t t ψ ≅ + +

( ) ~ [0 2 ] ( ) ( )t U E y t r t ψ π , ⇒ ≅

cos ( ) 0 E t ψ =

( )o AM

SNR ,

N i i FM i



152

Noise in FM receivers

Consider the following FM receiver model

FM frequency discriminator receiver.

where the transmitted signal is

and w(t ) is WGN with zero mean and psd N 0/2, ∀ f

At the output of the bandpass filter,

0

( ) cos ( )

t

FM c cs t A t k m d ω ω τ τ

= +

∫

( ) ( ) ( )FM

x t s t n t = +



153

where the noise component is described by

Moreover, where

Thus,

Graphically, assuming (SNR)c,FM

>> 1

( )( ) ( ) cos( ) ( )sin( ) ( ) cos ( ) I c Q c c

n t n t t n t t r t t t ω ω ω ψ = − = +

( )( ) cos ( ) ,FM c cs t A t t ω φ = +

0

( ) ( )t

t k m d ω φ τ τ = ∫( ) ( )( ) cos ( ) ( ) cos ( )

c c c x t A t t r t t t ω φ ω ψ = + + +

( ) ( ) ( ) ( ).c

x t A t r t t φ ψ ⇒ = ∠ + ∠

From the abo e phasor diagram sing the signal component as the reference



154

From the above phasor diagram, using the signal component as the reference,

where

Now, a high (SNR)c,FM means that Ac >> r (t ) for almost all t . Therefore,

The job of the limiter is to remove the amplitude variations. Hence,

and

[ ] [ ]( )

( ) ( )cos ( ) ( ) ( )sin ( ) ( )

( )

c

j t

x t A r t t t jr t t t

x t eθ

ψ φ ψ φ = + − + −

=

[ ][ ]

1( )sin ( ) ( )

( ) ( ) tan( ) cos ( ) ( )c

r t t t t t

A r t t t

ψ φ θ φ

ψ φ

− −

= + + −

( ) ( )( ) cos ( ) ( ) and ( ) ( ) sin ( ) ( )c c c A r t t t A x t A j r t t t ψ φ ψ φ + − ≅ ≅ + −

[ ]1 ( )( ) ( ) tan sin ( ) ( )

c

r t t t t t

Aθ φ ψ φ −

≅ + −

( )c

l t A≅ [ ]( )

( ) ( ) sin ( ) ( )c

r t t t t t

Aθ φ ψ φ ≅ + −

From our previous FM derivations the output of an ideal discriminator is



155

From our previous FM derivations, the output of an ideal discriminator is

where

is the noise contribution to the output of the discriminator.

At any given time t = t 0,

since , the statistics of are determined by ψ(t).

Hence,

The Fourier Transform of the last equation yields

[ ]( )1 12 2 2

( )( ) ( ) sin ( ) ( ) ( ) ( ),

k d t

d dt

c

d r t v t t t t m t n t

dt A

ω θ

π π π φ ψ φ

= ≅ + − = +

( )12

( ) ( )sin ( ) ( )c

d d A dt

n t r t t t π

ψ φ = −

( ) 0 0 0sin ( ) ( ) ( ) 0 E t t t ψ φ φ − | =

0( ) ~ [0 2 ]t U ψ π , [ ]sin ( ) ( )t t ψ φ −

( )1 1( ) [ ( )sin ( )] .

2 2

Q

d

c c

dn t d n t r t t

A dt A dt

ψ

π π

≅ =

)()()()(22

1)(

2

1)( f N f H f N

A

f j f fN j

Adt

t dnF

A f N QeqQ

c

Q

c

Q

c

d ===

= π π π

This is equivalent to passing the noise component through the filter



156

This is equivalent to passing the noise component through the filter

So, at the output of the discriminator, we have

Let Pm be the average message power. Then the power of the message component

at the output of the discriminator is k f 2 Pm . The psd of the noise component, on the

other hand, is

This means that the power of the noise component is

( )c

f

eq A H f j=

( ) ( ) ( ) ( ) ( ) ( ) ( )2

eq Q f eq Q

k v t m t h t n t k m t h t n t ω

π ≅ + ∗ = + ∗

( ) ( )

2

2 02,

2

0, elsewhereQ

T

eq N c

f B N f

H f S f A

| |≤

=

2

2

20

2.

BT

BT

noise

c

N P f df

A−

= ∫



157

However, in general, and the average noise power can be decreased by

passing v(t ) through an ideal lowpass filter with bandwidth W m , i.e.

At the output of the lowpass filter,

The average power of the signal component is still k f 2Pm .The psd of the noise

component is, however,

2T B

mW >

( ) 1,0, elsewhere

mlpf f W H f | |≤=

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

f lpf lpf eq Q

f lpf eq Q

y t k m t h t h t h t n t

k m t h t h t n t

≅ ∗ + ∗ ∗

= + ∗ ∗

( ) ( ) ( )2 2

Qlpf eq N H f H f S f =

20

2,

0, elsewhere

m

c

N f f W A

| |≤

and the average power of the noise component at the output of the lowpass filter is



158

and the average power of the noise component at the output of the lowpass filter is

so,

At the input of the bandpass filter, and the channel SNR in the

message bandwidth W m is given by

So, the figure of merit for FM is

2 30 0

2 2

2

3

m

m

W

m

c cW

N N f df W

A A−

=∫

30

2

2 2 2

320

3

3( )

2m

c

f m c f m

o FM N W m

A

k P A k PSNR

N W , = =

( ) ( ) ( )FM s t s t w t = +

22

2

0 0

( )2

c A

cc FM

m m

ASNR

W N W N , = =

2 2

30

2

0

3 22

2 2

2

2

3( )

( )

c f m

m

c

m

A k P

W N f mo FM

Ac FM m mW N

k PSNR f D

SNR W W

,

,

∆= = ∝ =

Si i d B lt i i d fi f it ! Thi hf k∆



159

Since increased BT results in an improved figure of merit ! This happens

with FM only!

Now, recall that

This means that the high frequency components of the noise contribution get

emphasized. This effect can be reduced by first emphasizing the high frequency

components of the message signal before modulation at the transmitter and then de-

emphasizing both message and noise at the output of the receiver, i.e.,

, f

f k ∆ ∝

20

2,

( ) 2

0, elsewhered

T

c N

N B f f AS f

| |≤=

where the de-emphasis filter is described by



160

where the de-emphasis filter is described by

In principle, the average power of the message at the output of the receiver is not

affected by the pre-emphasis-de-emphasis operation. The noise component, on the

other hand, will have its spectral characteristics modified by the de-emphasis filter,

i.e.,

The average power of the noise component at the output of the post-detection

lowpass filter is given by

( )( )

1de

pe

H f H f

= m f W ≤

( ) ( )( ) 220

2 2,

0, elsewhered

de m

cde N

N f H f f W A H f S f

| |≤=

( )220

,2.

m

m

W

de n de

cW

N f H f df P

A−

=∫

Clearly H (f) must have highpass characteristics which implies that H (f) has



161

Clearly, H pe( f ) must have highpass characteristics which implies that H de( f ) has

lowpass chartacteristics.

Example: Let the pre-emphasis filter be described by

and

( ) ( )1

11

pe de H f jaf H f jaf

= + ⇒ =+

( )

( )

2

0

, 2 2 2

0

2 2 2 2 2

102 2

10

2 2

1

1 1 1

1

1 tan

22 tan

m

m

m

m

m

m

W

n de

c W

W

c W

W

W c

m m

c

N f

P df A a f

N df

A a a a f

N f af A a a

N W aW

a A a

−

−

−

−

−

= +

= − +

= −

= −

∫

∫

So2

( ) f m

k PSNR =



162

So,

or

The improvement with de-emphasis is therefore

0

2 2

2 11( )

( )c

o FM de N

m maa A

SNRW tan aW

, , −=

−

( )

( )

3 2

1

2

2

3 3tan

13 3 tan

m f m

m m

f m

m

a W k P

aW aW de m

k Pm m

W

F M a W

F M aW aW

−−

−

. .= =

. . −

0

2

1

0

223

2

2for meritof Figure

N W

A

)aW (tanaW N

Pk Aa

FM

m

c

mm

m f c

de

][

−−=⇒

)aW (tanaW Pk W aFM

mm

m f mde 1

23

for meritof Figure −−=



163

Channel Propagation Losses

Simple Channel Model

where ( )T s t is the transmitted signal, α is the attenuation coefficient, ( )0 1, w t α < < isWGN and ( )r t is the received signal.

Free space line of sight propagation:



164

Free space line-of-sight propagation:

Let d be the separation distance between transmitter and receiver. Then the received

signal power ( )r P d is described by the Friis free space equation

( )2

,4

T r r T

G GP d P

d L

λ

π

=

where T P is the transmitted power,T

G and r G are the transmitter and receiver antenna

gains, respectively, L is the system loss factor (not related to propagation, 1 L ≥ ), and λ is the wavelength of the carrier in meters.

Now,2

4,e

e

AG A

π

λ = is the effective aperture of the antenna, namely, the ratio of the

available power at the terminals of the receiving antenna to the power flux density of a

plane wave incident on the antenna from that direction, and 0/ ,c f λ = is the speed of

light in meter/s and 0 f is the carrier frequency in Hz.

The loss factor L is due to transmission line attenuation, filter losses, and antenna

losses. Clearly, 1 L = ⇒ a lossless system.

Friis propagation model is valid for d in the far-field (Fraunhofer region) of the



165

transmitting antenna.

The Fraunhofer distance is given by22 , f

Dd λ

= where D is the largest physical

linear dimension of the antenna. Furthermore, to be in the far-filed region, f d D and

. f d λ

Since does not hold for 0,d = large-scale propagation models use a close-in

distance0 f d d ≥ as a known received power reference point.

Hence, ( ) ( )2

00 0, .r r f

d P d P d d d d

d

= ≥ ≥

The dynamic range of r

P can change by many orders of magnitude over a typical

coverage area of several square kilometers. As a result, dBm or dBW units are used to

express the received power levels, e.g.,

( )( )0 0

10 10in 10 log 20 log ,

1

r

r

P d d P d dBm

mW d

= +

( )0r P d is in mW units.



166

For practical systems that use low-gain antennas in the 1-2 GHz region, 0 1md = for

indoor environments and 0 100md = or

0 1kmd = for outdoor environments.

Example: Find f d for an antenna with maximum dimension of 1 m and operating

frequency 900 MHz.

Solution: D = 1 m, f = 900 MHz8

6

3x10 m/s 10.333 m

900x10 Hz 3

c

f λ ⇒ = = = =

( ) ( )2 2

2 2x 16 m

0.333 f

Dd

λ ⇒ = = =

Example: If the transmitted signal power is 5 Watts, express this power in units of dBm

and dBW. If power is applied to a unity-gain antenna with a 900 MHz carrier frequency,

find the received power in dBm at a free space distance of 100 m from the antenna.

What’s ( )10 kmr P ? Assume unity gain for the receiver antenna.

Solution:



167

3

10 10

10 10

in mW 5x10 mWin dBm 10 log 10 log 37 dBm

1 mW 1 mW

in W 5 Win dBW 10 log 10 log 7 dBW1 W 1 W

t t

t t

PP

PP

= = =

= = =

Assuming a lossless system and unity-gain antennas,

( )( )

( )( ) ( )

( ) ( )

224

2 2

5 1 1 1/ 3100 3.5x10

4 400 1

t T r r

PG GP mW

d L

λ

π π

−= = =

( )( )

4

10

10 10 4

3.5x10 mW

in dBm 10 log 34.5 dBm1 mW

100 m in mW 10010 km in dBm 10 log 20 log

1 mW 10

34.5 40 74.5 dBm

r

r

r

P

PP

−

= = −

= +

= − − = −

Eff i N i T d N i Fi



168

Effective Noise Temperature and Noise Figure

A thermal noise source has power spectral density ( ) [ ]S Watts Hz , .2

n

kT f f = ∀

When constrained to a frequency bandwidth of B Hz, the noise power is

x2 Watts.2

n

kT P B kTB= =

Consider a thermal noise source connected to an amplifier and a load, i.e.,

Thermal noise

source

Amplifier

H(f)

Load

The power spectral density at the output of the amplifier is ( ) ( ) ( )2

. x nS f H f S f =

The noise power at the output of the amplifier is



169

p p p

( ) ( ) ( )2 2

2 x n

kT P H f S f df H f df

∞ ∞

−∞ −∞

= =∫ ∫

Define the noise equivalent bandwidth by

( )21

,2

neq B H f df G

∞

−∞

≡ ∫

where ( ) 2

max , f

G H f ≡ then

x neqP GkTB=

Now, a practical amplifier generates additional noise internally, i.e.,

0

.

i i

i

x x x neq x

x

neq

neq

P P P GkTB P

PGkB T

GkB

= + = +

= +

Define the noise equivalent temperature of the amplifier by ,i x

e

PT

GkB≡ then



170

neqGkB

( )0

1 e x neq e neq

T P GkB T T GkTB

T

= + = +

Define the noise figure as 1 eT F

T ≡ + , then

0. x neqP GkTB F =

But, an input signal ( )is t with power isP to an amplifier with power gain G produces an

output signal ( )0s t with power

0.

is sP GP=

Hence, the output power SNR is given by

0

00

1i is s s

i x neq n

P GP PS S

N P GkTB F P F F N

= = = =

and ( )10 10 10

0

10 log 10 log 10 log ,i

S S F

N N

= −

where 10 log ( )F is the SNR loss due to the additional noise introduced by the



171

where1010 log ( )F is the SNR loss due to the additional noise introduced by the

amplifier.

Example: Obtain the noise figure of 2 amplifiers in cascade with gains G1 and G2,

respectively.

Solution: The noise output powers of the two amplifiers are described by

( )

( )

01

2

02 01 2 2

1 1

2 1 2 1 1 2 1

1 2

i

i i

x neq e

x x x x neq e x neq e

neq

P G kB T T

PP G P P G G kB T T P G G kB T T GG kB

= +

= + = + + = + +

Let 2

2

2

,i x

e

neq

PT

G kB= then

02

2 1 21 2 1 1 2

1 1

1 2

11

,

e e e x neq e neq

neq Total

T T T P G G kB T T G G kTB

G T G T

G G kTB F

= + + = + +

=



172

where

1 2

1

1

1e e

Total

T T

F T G T

= + + = ( )1 2

1 21 1

1 1

1 1 1 1 .e e

T T

F F T G T G

+ + + − = + −

and

2, 1

1

ee Total e

T T T

G

= +

In general, for M amplifiers in cascade,

321

1 1 2 1 2 1

2, 1

1 1 2 1

11 1 M

Total

M

e eM e Total e

M

F F F F F

G G G G G G

T T T T

G G G G

−

−

−− −= + + + +

= + + +



173

Propagation Losses:

The signal undergoes attenuation because of the physical medium, operating

frequency and the separation between transmitter and receiver.

Define the dimensionless propagation loss as T

r

P

P≡L or in dB as

( ) ( )10 1010 log 10 log .dB T r

P P= −L

For wireline transmission, the propagation loss is specified in dB per unit length, e.g.,

dB/km. For line-of-sight radio communications,

24 d π

λ

≡

L



174

Example: A signal is transmitted through a 100 km coaxial cable with 1 dB/km loss. T P

in dBW is -20 dBW (equivalent to 10 mW). What isr P and the output power of a LNA

with 25 dB?dBG =

Solution: dB =L 100 km x 1 dB/km = 100 dBW

, ,r dB T dB dBP P= − =L -20-100 = -120 dBW

0 0, ,r dB dB r dBP GP P G P= ⇒ = + = 25-120 = -95 dBW

Analog Repeaters:

Consider the following communication system with a repeater:

Transmitter



175

At the output of the repeater,0 .T P

P G=L

If 0, .T G P P= =L

Leti

S N

be the input SNR of the amplifier, then

0

1 1 1 1.r T T

ia a neq a neq a neq

P P PS S

N F N F kTB F kTB F kTB

= = = = L L

This system can be viewed as the cascade of networks with noise figures L and .aF

Hence, the overall noise figure is

1.a

Total

F F

G

−= +L

Now, if 1

G =L

, then1

.1a

Total a

F F F

−= + =L L

L

Link Budget:



176

Link Budget:

We already know that in free-space propagation

2

0,4

T T R R f

P G GP d d d

d L

λ

π

= ≥ ≥

. The

productT T

P G is also referred to as the equivalent isotropic radiated power (EIRP) and the

term

2

10 1010log 20log4 4d d

λ λ

π π

=

is the free-space loss in decibels.

In dBs,

, 10 ,20log4

R dBW dBW R dB dBP EIRP G Ld

λ

π

= + + −

.

Example: Consider a communication system that uses a transmitter with directional

antenna with gain 30T G dB= and transmits at the power level of 100 Watts , and a

receiver whose antenna gain is 3 R

G dB= . If the transmit frequency is 100 MHz, the

antennas are separated by 100 km and the system loss is 5 dB , calculate the receive

power in dBW.

83 10c ×



177

Solution: The wavelength λ is8

3 103

10

c

f

×= = meters

10 5

320log 112.4

4 10 free space L dB

π −

= = − ×

,

1030 10log (100) 30 20 50 EIRP dBW = + = + =

, 112.4 50 3 5 64.4 R dbW P dBW = − + + − = − .

The noise power seen by the receiver antenna is described by ( )n ant e

P k T T kTB= + = ,

where 231.38 10 /k Joules Kelvin−= × ,

ant eT T T = + , and B is the bandwidth of interest.

The antenna temperatureant

T is not a physical temperature, but depends on where the

antenna is pointed and the frequency band of the received signal. For example, if the

antenna is mounted on a satellite and pointed towards the earth, a typical value is

300ant T K = . Furthermore, 0 ( 1)eT T F = − , where 0 290T K = . Hence, the received signal-to-noise (SNR) power ratio in the frequency band B is given by

2

0,4

R T T R f

n

P P G Gd d d

P d LkTB

λ

π

= ≥ ≥

.



Example: Consider again the communication system of the previous example. Suppose

now that the receiver noise figure is5 3.162

F dB

= =, the antenna temperature is

50ant

T K = and the receiver bandwidth 30 kHz. What is the receiver SNR?

Solution: 290(3.162 1) 627.1eT K = − = , 50 627.1 677.1ant eT T T K = + = + = , and

23 4 161.38 10 (677.1)(3 10 ) 2.8 10nP Watts− −= × × = × .

Now,16

, 10

2.8 1010log 155.5

1n dBW

WattsP dBW

Watts

− ×= = −

and the SNR in dBW is

, ,64.4 ( 155.5) 91.1 R

R dBW n dBW

n dBW

PP P dBW

P

= − = − − − =

.

random process in analog communication systems

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