EC 2351 Digital Communication

8/21/2019 EC 2351 Digital Communication

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Digital CommunicationsChapter 2: Deterministic and Random Signal Analysis

Po-Ning Chen, Professor

Institute of Communications Engineering

National Chiao-Tung University, Taiwan

Digital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 1 / 106


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2.1 Bandpass and lowpass signal representation



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Definition (Bandpass signal)

A bandpass signal x (t ) is a real signal whose frequency content is located around central frequency f 0, i.e.X

(f

) = 0 for all

∣f

±f 0

∣ > W

X (f )

f 0f 0 −W f 0 +W −f 0 −f 0 +W −f 0 −W

f 0 may not be thecarrier frequency f c !

The spectrum of a bandpass signal is Hermitian symmetric ,

i.e., X (−f ) = X ∗(f ). (Why? Hint: Fourier transform.)Digital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 3 / 106


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Since the spectrum is Hermitian symmetric, we only need

to retain half of the spectrum X +(f ) = X (f )u −1(f )(named analytic signal or pre-envelope) in order toanalyze it,

where u −1(f ) = ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

1 f

> 0

1

2 f = 00 f


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2.1 Bandpass and lowpass signal representation -

Baseband and bandpass signals

Definition (Baseband signal)A lowpass or baseband ( equivalent ) signal x (t ) is a complex signal (because it is not necessarily Hermitian symmetric!) whose spectrum is located around zero frequency, i.e.

X (f ) = 0 for all ∣f ∣ > W It is generally written as

x (t ) = x i (t ) + ı x q (t )where

x i

(t

) is called the in-phase signal

x q (t ) is called the quadrature signal Digital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 5 / 106


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Baseband signal

Our goal is to relate x

(t

) to x

(t

) and vice versa



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From lowpass equivalent x (t ) to x (t )

Definition of bandwidth. The bandwidth of a signal is onehalf of the entire range of frequencies over which the spectrum

is (essentially) nonzero. Hence, W is the bandwidth in thelowpass signal we just defined, while 2W is the bandwidth of the bandpass signal by our definition.



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Analytic signal

Let’s start from the analytic signal x +

(t

).

x +(t ) = ∞−∞ X +(f )e ı2πft df

= ∞

−∞

X

(f

)u −1

(f

)e ı2πft df

= F −1 {X (f )u −1(f )} F −1 Inverse Fourier transform= F −1 {X (f )} ⋆ F −1 {u −1(f )}=

x

(t

) ⋆ 1

2δ

(t

) + ı

1

2πt

= 12x (t ) + ı 12 x̂ (t ),where x̂

(t

) = x

(t

) ⋆ 1πt

= ∫ ∞−∞

x (τ )π(t −τ )d τ is a real-valued signal.



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Appendix: Extended Fourier transform

F −1 {2u −1(f )} = F −1 {1 + sgn(f )}

= F −1

{1

} + F −1

{sgn

(f

)} = δ

(t

) + ı

1

πt

Since ∫ ∞−∞ ∣sgn(f )∣ = ∞, the inverse Fourier transform of sgn(f )does not exist in the standard sense! We therefore have toderive its inverse Fourier transform in the extended sense!

(∀ f )S (f ) = limn→∞S n(f ) and (∀ n) ∞−∞ ∣S n(f )∣df < ∞

⇒F−1

{S

(f

)} = limn→∞

F−1

{S n

(f

)}.



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Appendix: Extended Fourier transform

Since lima↓0 e −a

∣f

∣sgn(f ) = sgn(f ),lima↓0 ∞−∞ e −a∣f ∣sgn(f )e ı2πft df = lima↓0 −

0

−∞e f (a+ ı2πt )df +

∞

0

e f (−a+ ı2πt )df = lima↓0 − 1a + ı 2πt + 1a − ı 2πt

= lima↓0 ı 4πt

a2 + 4π2t 2 = ⎧⎪⎪⎨⎪⎪⎩0 t

= 0

ı 1πt t ≠ 0Hence, F −1 {2u −1(f )} = F −1 {1} +F −1 {sgn(f )} = δ (t ) + ı 1πt .


F ( ) ( )


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From x +(t ) to x (t )

X (f )

f 0f 0 − W f 0 + W −f 0−f 0 + W −f 0 − W ⇒

X (f )

0−W W

We then observeX (f ) = 2X +(f + f 0).

This implies

x (t ) = F −1{X (f )}= F −1{2X +(f + f 0)}

= 2x +

(t

)e − ı2πf 0 t

= (x

(t

) + ı x̂

(t

))e − ı2πf 0 t



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As a result,

x

(t

) + ı x̂

(t

) = x (

t

)e ı2πf 0t

which gives:

x

(t

) = Re

{x

(t

) + ı x̂

(t

)} = Re

x

(t

)e ı2πf 0t

By x (t ) = x i (t ) + ı x q (t ),

x (t ) = = Re (x i (t ) + ı x q (t ))e ı2πf 0t = ×i (t ) cos(2πf 0t ) − x q (t ) sin(2πf 0t )Digital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 12 / 106

F X (f ) X (f )


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From X (f ) to X (f )From x

(t

) = Re

{x

(t

)e ı2πf 0 t

}, we obtain

X (f ) = ∞−∞ x (t )e − ı2πft dt

= ∞

−∞Re

x

(t

)e ı2πf 0t

e − ı2πft dt

= ∞−∞ 12 x (t )e ı2πf 0t + x (t )e ı2πf 0t ∗ e − ı2πft dt = 12

∞

−∞x (t )e − ı2π(f −f 0)t dt

+1

2 ∞

−∞x ∗ (t )e

− ı2π(f +f 0)t dt

= 12 [X (f − f 0) +X ∗ (−f − f 0)]

X ∗ (−f ) = ∫

∞−∞ (x (t )e

− ı2π(−f )t

)∗

dt = ∫ ∞−∞

x ∗ (f )e

− ı2πft dt Digital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 13 / 106

S


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Summary

Terminologies & relations

Bandpass signal⎧⎪⎪⎨⎪⎪⎩

x

(t

) = Re

{x

(t

)e ı2πf 0t

}X

(f

) = 12

X

(f

−f 0

) +X ∗

(−f

−f 0

)Analytic signal or pre-envelope x +(t ) and X +(f )Lowpass equivalent signal or complex envelope

⎧⎪⎪⎨⎪⎪⎩x (t ) = (x (t ) + ı x̂ (t ))e − ı2πf 0 t X (f ) = 2X (f + f 0)u −1(f + f 0)Digital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 14 / 106

U f l t k


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Useful to know


From x (t ) = x i (t ) + ı x q (t ) = (x (t ) + ı x̂ (t ))e − ı2πf 0 t ,

⎧⎪⎪⎨⎪⎪⎩x i

(t

) = Re

(x

(t

) + ı x̂

(t

)) e − ı2πf 0t

x q (t ) = Im (x (t ) + ı x̂ (t )) e − ı2πf 0t Also from x (t ) = (x (t ) + ı x̂ (t ))e − ı2πf 0 t ,

⎧⎪⎪⎨⎪⎪⎩x

(t

) = Re

x (

t

) e ı2πf 0t

x̂ (t ) = Im x (t ) e ı2πf 0t Digital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 15 / 106

U f l t k


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Useful to know


From x (t ) = x i (t ) + ı x q (t ) = (x (t ) + ı x̂ (t ))e − ı2πf 0 t ,

⎧⎪⎪⎨⎪⎪⎩x i

(t

) = Re

(x

(t

) + ı x̂

(t

)) e − ı2πf 0t

x q (t ) = Im (x (t ) + ı x̂ (t )) e − ı2πf 0t Also from x (t ) = (x (t ) + ı x̂ (t ))e − ı2πf 0 t ,

⎧⎪⎪⎨⎪⎪⎩x

(t

) = Re

(x i (

t

) + ı x q (

t

)) e ı2πf 0t

x̂ (t ) = Im (x i (t ) + ı x q (t )) e ı2πf 0t Digital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 16 / 106

Us f l t k


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Useful to know


pre-envelope x +(t )complex envelope x (t )envelope

∣ ×

(t

)∣ =

x 2i

(t

) +x 2q

(t

) = r

(t

)

phase θ

(t

) = arctan

[x q

(t

)/x i

(t

)]Digital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 17 / 106 Modulaor/demodulator and Hilbert transformer


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Modulaor/demodulator and Hilbert transformer

Usually, we will modulate and demodulator with respect to

carrier frequency f c , which may not be equal to the centerfrequency f 0.

x

(t

) → x

(t

) = Re

{x

(t

)e ı2πf c t

} = modulation

x (t ) → x (t ) = (x (t ) + ı x̂ (t ))e − ı2πf c t = demodulationThe modulation requires to generate x̂ (t ), a Hilberttransform of x (t )



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Hilbert transform is basically a 90-degree phase shifter.

H (f ) = F 1πt = − ı sgn(f ) = ⎧⎪⎪⎪⎨⎪⎪⎪⎩ −ı , f > 00, f = 0

ı , f 00 f 0−X (f ) f 0−1 f


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Energy considerations

Definition (Energy of a signal)

The energy E s of a (complex) signal s (t ) is E s = ∞−∞ ∣s (t )∣2 dt Hence,

E x = ∞−∞ ∣x (t )∣2 dt

E x +

= ∞

−∞ ∣x +

(t

)∣2

dt

E x = ∞−∞ ∣x (t )∣2 dt We are interested in the connections among

Ex ,

Ex

+, and

Ex .



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From Parseval’s Theorem we see

E x = ∞−∞ ∣x (t )∣2 dt =

∞−∞ ∣X (f )∣2 df

Parsevals theorem: ∫ ∞−∞ x (t )y ∗(t )dt = ∫ ∞−∞ X (f )Y ∗(f )df Rayleigh’s theorem: ∫ ∞−∞ ∣x (t )∣2dt = ∫ ∞−∞ ∣X (f )∣2df Secondly

X (f ) = 12 X (f − f c )=X +(f )

+ 12X ∗ (−f − f c )=X ∗+ (−f )

Thirdly, f c ≫ W andX (f − f c )X ∗ (−f − f c ) = 4X +(f )X ∗+ (−f ) = 0 for all f Digital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 21 / 106


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It then shows

E x

= ∞

−∞ 1

2X

(f

−f c

) + 1

2X ∗

(−f

−f c

)2

df

= 14E x + 14E x = 12E x

and

E x = ∞−∞ ∣X +(f ) +X ∗+ (−f )∣2 df = E x + + E x + = 2E x +Theorem (Energy considerations)E x = 2E x = 4E x +


Extension of energy considerations


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Extension of energy considerations

Definition (Inner product)We define the inner product of two (complex) signals x (t ) and

y (t ) as ⟨x

(t

), y

(t

)⟩ = ∞

−∞

x

(t

) y ∗

(t

)dt .

Parseval’s relation immediately gives

⟨x

(t

), y

(t

)⟩ = ⟨X

(f

), Y

(f

)⟩.

E x = ⟨x (t ), x (t )⟩ = ⟨X (f ), X (f )⟩E x = ⟨x (t ), x (t )⟩ = ⟨X (f ), X (f )⟩Digital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 23 / 106


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We can similarly prove that

⟨x

(t

), y

(t

)⟩= ⟨X (f ), Y (f )⟩

= 1

2X

(f

−f c

) + 1

2X ∗

(−f

−f c

), 1

2Y

(f

−f c

) + 1

2Y ∗

(−f

−f c

)= 1

4 ⟨X (f − f c ), Y (f − f c )⟩ + 1

4 ⟨X (f − f c ), Y ∗ (−f − f c )⟩=0

+1

4

⟨X ∗

(−f

−f c

), Y

(f

−f c

)⟩=0 +1

4

⟨X ∗

(−f

−f c

), Y ∗

(−f

−f c

)⟩= 14⟨x (t ), y (t )⟩ + 1

4(⟨x (t ), y (t )⟩)∗ = 12 Re{⟨x (t ), y (t )⟩} .


Corss-correlation of two signals


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Corss correlation of two signals

Definition (Cross-correlation)

The cross-correlation of two signals x (t ) and y (t ) is defined as ρx ,y

= ⟨x (t ), y (t )⟩

⟨x

(t

), x

(t

)⟩ ⟨ y

(t

), y

(t

)⟩ = ⟨x (t ), y (t )⟩ E

x

E y

.

Definition (Orthogonality)

Two signals x

(t

) and y

(t

) are said to be orthogonal if ρx ,y

= 0.

The previous slide then shows ρx ,y = Re {ρx ,y }.ρx ,y 0

⇒ ρx ,y 0 but ρx ,y 0

/⇒ ρx ,y 0


2 1-4 Lowpass equivalence of a bandpass system


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2.1 4 Lowpass equivalence of a bandpass system

Definition (Bandpass system)

A bandpass system is an LTI system with real impulse response h(t ) whose transfer function is located around afrequency f c

Using a similar concept, we set the lowpass equivalentimpulse response as

h

(t

) = Re

h

(t

)e ı2πf c t

and H (f ) = 12 [H (f − f c ) +H ∗ (−f − f c )]


Baseband input-output relation


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Baseband input output relation

Let x

(t

) be a bandpass input signal and let

y (t ) = h(t )⋆x (t ) or equivalently Y (f ) = H (f )X (f )Then, we knowx

(t

) = Re

x

(t

)e ı2πf c t

h(t ) = Reh(t )e ı2πf c t

y (t ) = Re y (t )e ı2πf c t and

Theorem (Baseband input-output relation)

y (t ) = h(t ) ⋆ x (t ) ⇐⇒ y (t ) = 12

h(t ) ⋆ x (t )Digital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 27 / 106


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Proof:

For f ≠ −f c (or specifically, for u −1(f + f c ) = u 2

−1(f + f c )),Note 12 = u −1(0) ≠ u 2−1(0) = 14 .Y

(f

) = 2Y

(f

+f c

)u −1

(f

+f c

)= 2H (f + f c )X (f + f c )u −1(f + f c )= 12 [2H (f + f c )u −1(f + f c ) ] ⋅ [2X (f + f c )u −1(f + f c )]

= 1

2H

(f

) ⋅X

(f

)The case for f = −f c is valid since Y (−f c ) = X (−f c ) = 0.



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The above applies to deterministic system. How aboutstochastic system?

x (t ) y (t ) h(t) ⇓

X(t ) Y(t ) h(

t

)

The text abuses the notation by using X (f ) as the spectrumof x

(t

) but using X

(t

) as the stochastic counterpart of x

(t

).



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2.7 Random processes


Random Process


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DefinitionA random process is a set of indexed random variables

{X

(t

), t

∈ T }, where

T is often called the index set.

Classification

1 If T is a finite set ⇒ Random Vector2 If

T =Z or Z+

⇒ Discrete Random Process

3 If T = R or R+ ⇒ Continuous Random Process4 If T = R2,Z2,⋯,Rn,Zn ⇒ Random FieldDigital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 31 / 106

Examples of random process


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p p

Example

Let U be a random variable uniformly distributed over [−π, π).ThenX(t ) = cos (2πf c t +U)

is a (continuous) random process.

Example

Let B be a random variable taking values in

{−1, 1

}. Then

X(t ) = cos(2πf c t ) if B = −1sin(2πf c t ) if B = +1 = cos 2πf c t − π4 (B + 1)is a (continuous) random process.


Statistical properties of random process


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p p p

For any integer k

> 0 and any t 1, t 2,

⋯, t k

∈ T , the

finite-dimensional cumulative distribution function (cdf) for

X(t ) is given by:F X (t 1,⋯, t k ; x 1,⋯, x k ) = Pr {X(t 1) ≤ x 1,⋯, X(t k ) ≤ x k }

As event

[X

(t

) ≤ ∞] (resp.

[X

(t

) ≤ −∞]) is always regarded

as true (resp. false),

limx s →∞F X

(t 1,

⋯, t k ; x 1,

⋯, x k

)= F X (t 1,⋯, t s −1, t s +1, t k ; x 1,⋯, x s −1, x s +1,⋯, x k )andlim

x s →−∞F X

(t 1, , t k ; x 1, , x k

) 0



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Definition

Let X(t ) be a random process; then the mean function is mX(t ) = E[X(t )],

the (auto)correlation function is

R X(t 1, t 2) = E [X(t 1)X∗(t 2)] ,and the (auto)covariance function is

K X(t 1, t 2) = E (X(t 1) −mX(t 1)) (X(t 2) −mX(t 2))∗ Digital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 34 / 106

Definition


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Let X(t ) and Y(t ) be two random processes; then the cross-correlation function is

R X,Y(t 1, t 2) = E [X(t 1)Y∗(t 2)] ,and cross-covariance function is

K X,Y(t 1, t 2) = E (X(t 1) −mX(t 1)] [Y(t 2) −mY(t 2))∗ Proposition

R X,Y(t 1, t 2) = K X,Y(t 1, t 2) +mX(t 1)m∗Y(t 2)R Y,X(t 2, t 1) = R ∗X,Y(t 1, t 2) R X(t 2, t 1) = R ∗X(t 1, t 2)K Y,X

(t 2, t 1

) K ∗X,Y

(t 1, t 2

) K X

(t 2, t 1

) K ∗X

(t 1, t 2

)Digital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 35 / 106 Stationary random processes


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Definition

A random process X(t ) is said to be strictly or strict-sense stationary (SSS) if its finite-dimensional joint distributionfunction is shift-invariant, i.e. for any integer k > 0, any t 1,

⋯, t k

∈ T and any τ ,

F X (t 1 −

τ,

⋯, t k −

τ ; x 1,⋯, x k ) =

F X (t 1,⋯

, t k ; x 1,⋯, x k )Definition

A random process X

(t

) is said to be weakly or wide-sense

stationary (WSS) if its mean function and (auto)correlation

function are shift-invariant, i.e. for any t 1, t 2 ∈ T and any τ ,mX(t − τ ) = mX(t ) and R X(t 1 − τ, t 2 − τ ) = R X(t 1, t 2).The above condition is equivalent to

mX

(t

) constant and R X

(t 1, t 2

) R X

(t 1 t 2

).


Wide-sense stationary random processes


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Definition

Two random processes X(t ) and Y(t ) are said to be jointly wide-sense stationary if

Both X

(t

) and Y

(t

) are WSS;

R X,Y(t 1, t 2) = R X,Y(t 1 − t 2).Proposition

For jointly WSS X

(t

) and Y

(t

),

R Y,X(t 2, t 1) = R ∗X,Y(t 1, t 2) ⇒ R X,Y(τ ) = R ∗Y,X(−τ )K Y,X(t 2, t 1) = K ∗X,Y(t 1, t 2) ⇒ K X,Y(τ ) = K ∗Y,X(−τ )


Gaussian random process


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Definition

A random process {X(t ), t ∈ T } is said to be Gaussian if for any integer k > 0 and for any t 1,⋯, t k ∈ T , the finite-dimensional joint cdf

F X(t 1,⋯, t k ; x 1,⋯, x k ) = Pr [X(t 1) ≤ x 1,⋯, X(t k ) ≤ x k ]is Gaussian.

RemarkThe joint cdf of a Gaussian process is fully determined by itsmean function and its (auto)covariance function.




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Definition

Two real random processes {X(t ), t ∈ T X } and {Y(t ), t ∈ T Y }are said to be jointly Gaussian if for any integers j , k > 0 and for any s 1,

⋯, s j

∈ T X and t 1,

⋯, t k

∈ T Y, the finite-dimensional

joint cdf

Pr [X(s 1) ≤ x 1,⋯, X(s j ) ≤ x j , Y(t 1) ≤ y 1,⋯, Y(t k ) ≤ y k ]is Gaussian.

DefinitionA complex process is Gaussian if the real and imaginary processes are jointly Gaussian.




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Remark

For joint (in general complex) Gaussian processes,“uncorrelatedness”, defined as

R X,Y

(t 1, t 2

) = E

[X

(t 1

)Y∗

(t 2

)]= E[X(t 1)]E[Y∗(t 2)] = mX(t 1)m∗Y(t 2),implies “independence”, i.e.,

Pr [X(s 1) ≤ x 1,⋯, X(s j ) ≤ x j , Y(t 1) ≤ y 1,⋯, Y(t k ) ≤ y k ]= Pr [X(s 1) ≤ x 1,⋯, X(s k ) ≤ x k ]⋅Pr [Y(t 1) ≤ y 1,⋯, Y(t k ) ≤ y k ]Digital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 40 / 106

Theorem


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If a Gaussian random process X(t ) is WSS, then it is SSS.Proof:

For any k > 0, consider the sampled random vector

⃗Xk

=⎡⎢⎢⎢⎢⎢⎢⎢⎣

X(t 1)X

(t 2

)⋮X(t k )⎤⎥⎥⎥⎥⎥⎥⎥⎦

The mean vector and covariance matrix of

⃗Xk are respectively

m⃗Xk = E[⃗Xk ] = ⎡⎢⎢⎢⎢⎢E[X(t 1)]E[X(t 2)]⋮E

[X

(t k

)]

⎤⎥⎥⎥⎥⎥ = mX(0) ⋅ ⃗1Digital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 41 / 106


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and

K ⃗X = E[⃗Xk ⃗XH k ] = ⎡⎢⎢⎢⎢⎢⎣K X

(0

) K X

(t 1

−t 2

) ⋯K X(t 2 − t 1) K X(0) ⋯⋮ ⋮ ⋱⎤⎥⎥⎥⎥⎥⎦It can be shown that for a new sampled random vector

⎡⎢⎢⎢⎢⎢⎢⎢⎣

X(t 1 + τ )X(t 2 + τ )

⋮X

(t k

+τ

)

⎤⎥⎥⎥⎥⎥⎥⎥⎦the mean vector and covariance matrix remain the same.

Hence, X(t ) is SSS.Digital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 42 / 106

Power spectral density


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Definition

Let R X(τ ) be the correlation function of a WSS randomprocess X(t ). The power spectral density (PSD) or power spectrum of X

(t

) is defined as

S X(f ) = ∞−∞ R X(τ )e − ı2πf τ d τ

Let R X,Y(τ ) be the cross-correlation function of two jointly WSS random process X

(t

) and Y

(t

); then the cross spectral

density (CSD) is

S X,Y(f ) = ∞−∞ R X,Y(τ )e − ı2πf τ d τ Digital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 43 / 106

Properties of PSD


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PSD (in units of watts per Hz) describes thedistribution/density of power as a function of frequency.

Analogously, probability density function (pdf) describesthe distribution/density of probability as a function of

outcome.The integration of PSD gives power of the randomprocess over the considered range of frequency.Analogously, the integration of pdf gives probability over

the considered range of outcome.


Theorem


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Theorem

S X

(f

) is non-negative and real (which matches that the power

of a signal cannot be negative).

Proof: S X(f ) is real becauseS X

(f

) = ∞

−∞

R X

(τ

)e − ı2πf τ d τ

= ∞−∞ R X(−s )e ı2πfs ds (s = −τ )

= ∞

−∞R ∗X

(s

)e ı2πfs ds

= ∞−∞ R X(s )e − ı2πfs ds ∗= S ∗X(f )Digital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 45 / 106

S X(f ) is non-negative because of the following (we only provethis based on that T ⊂ R and X(t) = 0 outside [−T , T ]).


46/106

this based on that T R and X(t ) 0 outside [ T , T ]).S X

(f

) = ∞−∞

E

[X

(t

+τ

)X∗

(t

)]e − ı2πf τ d τ

= EX∗(t ) ∞−∞ X(t + τ )e − ı2πf τ d τ (s = t + τ )

= E

X∗

(t

) ∞−∞

X

(s

)e − ı2πf (s −t ) ds

= E X∗(t )X̃(f )e ı2πft In notation, X̃(f ) = F{X(t )}Since the above is a constant with respect to t (by WSS),S X

(f

) = 1

2T

T

−T E

X∗

(t

)X̃

(f

)e ı2πft

dt

= 12T E X̃(f ) T −T X∗(t )e ı2πft dt 1

2T E

X̃

(f

)X̃∗

(f

) 1

2T E

∣X̃

(f

)∣2

0.


Wiener-Khintchine theorem


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Theorem (Wiener-Khintchine)

Let {X(t ), t ∈ R} be a WSS random process. Define XT (t ) = X(t ) if t ∈ [−T , T ]0, otherwise.

and set

X̃T (f ) = ∞−∞ XT (t )e − ı2πft dt = T −T X(t )e − ı2πft dt .If S X(f ) exists (i.e., R X(τ ) has a Fourier transform), then

S X(f ) = limT →∞

1

2T EX̃T (f )2


Variations of PSD definitions


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Power density spectrum : Alternative definitionFourier transform of auto-covariance function (e.g.,Robert M. Gray and Lee D. Davisson, RandomProcesses: A Mathematical Approach for Engineers,p. 193)

I remark that from the viewpoint of digitalcommunications, the text’s definition is more appropriatesince

the auto-covariance function is independent of a

mean-shift; however, random signals with differentmeans consume different “powers.”



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What can we say about, e.g., the PSD of stochasticsystem input and output?

x (t )x

(t

) y (t )

y

(t

) h(t )

12 h

(t

)

⎧⎪⎪⎨⎪⎪⎩

◻(t ) = Re{◻(t )e ı2πf c t }

◻

(t

) = (◻(t

) + ı ˆ

◻(t

))e − ı2πf c t

where “

◻” can be x , y or h.

⇓X

(t

)X(t ) Y

(t

)Y(t ) h

(t

)1

2

h(t )

⎧⎪⎪⎨⎪⎪⎩◻(

t

) = Re

{◻

(t

)e ı2πf c t

}◻

(t

) = (◻(t

) + ı ˆ

◻(t

))e − ı2πf c t

where “◻” can be X, Y or h.



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2.9 Bandpass and lowpass random processes


Definition (Bandpass random signal)

A b d (WSS) t h ti i l X(t) i l d


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A bandpass (WSS) stochastic signal X(t ) is a real randomprocess whose PSD is located around central frequency f 0, i.e.

S X(f ) = 0 for all ∣f ± f 0∣ > W

S X

(f

)f 0f 0

−W f 0

+W

−f 0

−f 0

+W

−f 0

−W

f 0 may not be thecarrier frequency f c !

We know

⎧⎪⎪⎨

X(t ) = Re {X(t )e ı2πft }X t X t ı X̂ t e − ı2πf 0 t


Assumption


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The bandpass signal X

(t

) is WSS.

In addition, its complex lowpass equivalent process X

(t

) is

WSS. In other words,Xi (t ) and Xq (t ) are WSS.Xi

(t

) and Xq

(t

) are jointly WSS.

Under this fundamental assumption, we obtain thefollowing properties:

P1) If X

(t

) zero-mean, both Xi

(t

) and Xq

(t

) zero-mean

because mX = mXi cos(2πf c t ) −mXq sin(2πf c t ) .P2)

⎧⎪⎪⎨

R Xi (τ ) = R Xq (τ )R Xi ,Xq τ R Xq ,Xi τ


Proof of P2):

R


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R X

(τ

)= E

[X

(t

+τ

)X

(t

)]= E ReX(t + τ )e ı2πf c (t +τ )ReX(t )e ı2πf c t = E [(Xi (t + τ ) cos(2πf c (t + τ )) −Xq (t + τ ) sin(2πf c (t + τ )))(

Xi

(t

)cos

(2πf c t

) −Xq

(t

)sin

(2πf c t

))]= R Xi (

τ

) +R Xq (

τ

)2 cos(2πf c τ )+R Xi ,Xq (τ ) −R Xq ,Xi (τ )2

sin(2πf c τ )+

R Xi

(τ

) −R Xq

(τ

)2 cos(2πf c (2t + τ )) (= 0)−R Xi ,Xq (τ ) +R Xq ,Xi (τ )2

sin(2πf c (2t + τ )) (= 0)Digital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 53 / 106

P3) R X(τ ) = Re 12 R X(τ )e ı2πf c τ .


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) X( ) 2 X( ) Proof : Observe from P2),

R X(τ ) = E [X(t + τ )X∗ (t )]= E [(Xi (t + τ ) + ı Xq (t + τ ))(Xi (t ) − ı Xq (t ))]=

R Xi

(τ

) +R Xq

(τ

) − ı R Xi ,Xq

(τ

) + ı R Xq ,Xi

(τ

)= 2R Xi (τ ) + ı 2R Xq ,Xi (τ ).Hence, also from P2),R X

(τ

) = R Xi

(τ

)cos

(2πf c t

) −R Xq ,Xi

(τ

)sin

(2πf c t

)= Re12 R X(τ )e ı2πf c τ Digital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 54 / 106


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P4) S X

(f

) = 14

S X

(f

−f c

) +S ∗X

(−f

−f c

).

Proof : A direct consequence of P3).

◻Note :Amplitude X̃(f ) = 12 X̃(f − f c ) + X̃∗ (−f − f c )Amplitude sequare

∣X̃(f )∣2 = 14 X̃(f − f c ) + X̃∗ (−f − f c )2

= 14 ∣X̃(f − f c )∣2 + X̃∗ (−f − f c )

2

Wiener-Khintchine: S X(f ) ≡ ∣X̃(f )∣2.



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P5) Xi

(t

) and Xq

(t

) uncorrelated if one of them has

zero-mean.Proof : From P2),

R Xi ,Xq

(τ

) = −R Xq ,Xi

(τ

) = −R Xi ,Xq

(−τ

).

Hence, R Xi ,Xq (0) = 0 (i.e.,E[Xi (t )Xq (t )] = 0 = E[Xi (t )]E[Xq (t )]).

◻Digital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 56 / 106


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P6) If S X

(−f

) = S ∗X

(f

), Xi

(t

+τ

) and Xq

(t

) uncorrelated for

any τ if one of them has zero-mean.

Proof : From P3),

R X

(τ

) = 2R Xi

(τ

) − ı 2R Xq ,Xi

(τ

).

S X(−f ) = S ∗X(f ) implies R X(τ ) is real;hence, R Xq ,Xi (τ ) = 0 for any τ . ◻

We next discuss the PSD of a system.


X(t ) Y(t ) h(t )

Y(t ) = ∞−∞ h(τ )X(t − τ )d τ ∞

h d


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mY

= mX

−∞

h

(τ

)d τ

R X,Y(τ ) = EX(t + τ ) ∞−∞ h(u )X(t − u )du ∗

= ∞

−∞h∗

(u

)R X

(τ

+u

)du

= ∞

−∞h∗

(−v

)R X

(τ

−v

)dv

= R X(τ ) ⋆ h∗(−τ )R Y(τ ) = E

∞

−∞h(u )X(t + τ − u )du

∞

−∞h(v )X(t − v )dv ∗

= ∞

−∞ h(u ) ∞

−∞ h∗(v )R X((τ − u ) + v )dv du = ∞−∞ h(u )R X,Y(τ − u )du R X,Y τ h τ R X τ h

∗ τ h τ .



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Thus,

S X,Y(f ) = S X(f )H ∗(f ) since ∞−∞ h∗(−τ )e − ı2πf τ d τ = H ∗(f )and

S Y(f ) = S X,Y(f )H (f ) = S X(f )∣H (f )∣2.


White process


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Definition (White process)

A (WSS) process W(t ) is called a white process if its PSD is constant for all frequencies:S W

(f

) = N 0

2

This constant is usually denoted by N 02 because the PSDis two-sided. So, the power spectral density is actually N 0

per Hz (N 0/2 at f = −f 0 and N 0/2 at f = f 0).The autocorrelation function R W(τ ) = N 02 δ (⋅), where δ (⋅)is the Dirac delta function.


Why negative frequency?

A “sample answer”:


61/106

A sample answer :

It is just an imaginary convenient way created by Human

to correspond to the “imaginary” domain of a complexsignal (that is why we call it “imaginary part”).

By giving respectively the spectrum for f 0 and

−f 0 (which

may not be symmetric), we can tell the amount of real

part and imaginary part in time domain corresponding tothis frequency.

For example, if the spectrum is conjugate symmetric, weknow imaginary part (in time domain)

= 0.

Notably, in communications, imaginary part is the partthat will be modulated by (or transmitted with carrier)sin(2πf c t ); on the contrary, real part is the part that willbe modulated by (or transmitted with carrier) cos 2πf c t .


Why δ (⋅) function?D fi i i (Di d l f i )


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Definition (Dirac delta function)

Define the Dirac delta function δ

(t

) as

δ (t ) = ∞, t = 0;0, t

≠ 0

,

which satisfies the replication property , i.e., for every continuous point of g (t ),g

(t

) = ∞−∞

g

(τ

)δ

(t

−τ

)d τ.

Hence, by replication property,

∞−∞

δ u du

∞−∞

δ t τ d τ

∞−∞

1 δ t τ d τ 1.


Not that it seems δ (t ) = 2δ (t ) = ∞, t = 0;0, t ≠ 0 ; but with


63/106

g 1

(t

) = 1 and g 2

(t

) = 2 continuous at all points,

1 = ∞−∞ g 1(t )δ (t )dt ≠ ∞−∞ g 2(t )δ (t )dt = 2.So, mathematician does not “like” this function as itcontradicts intuition:

f (t ) = g (t ) for t ∈ R except for countably many points

⇒ ∞

−∞

f

(t

)dt

= ∞

−∞

g

(t

)dt

if

∞

−∞

f

(t

)dt is finite

.

Hence, δ (t ) and 2δ (t ) are two “different” Diract deltafunctions by definitions. (The multiplicative constantcannot be omitted!) Very “artificial” indeed.


Comment: x a y a x y is incorrect if a


64/106

Comment: x

+a

= y

+a

⇒ x

= y is incorrect if a

= ∞.

As a result, saying

∞ = ∞ (or δ

(t

) = 2δ

(t

) ) is not a

“rigorously defined” statement.

Summary: The Dirac delta function, like “

∞”, is simply

a concept defined only through its replication property .

Hence, a white process W(t ) that has autocorrelationfunction R W

(τ

) = N 0

2 δ

(τ

) is just a convenient and

simplified notion for theoretical research about real world

phenomenon. Usually, N 0 = KT , where T is the ambienttemperature in kelvins and k is Boltzman’s constant.Digital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 64 / 106

Discrete-time random processes

The property of a time-discrete process X n n Z+


65/106

The property of a time discrete process

{X

[n

], n

∈Z

}can be “obtained” using sampling notion via the Dirac

delta function.X[n] = X(nT ), a sample at t = nT from atime-continuous process X(t ), where we assume T = 1for convenience.The autocorrelation function of a time-discrete process isgiven by:

R X[m] = E{X[n +m]X[n]}= E

{X

(n

+m

)X

(n

)}= R X(m), a sample from R X(t ).

R X(0)R X(1)

R X(2)R X(3)R X(4)R X(5)

R X(6)R X(7)R X(8)R X(9)

R X(10)R X(11)

R X(12)R X(13)



66/106

S X

[f

] = ∞

−∞ ∞

n=−∞R X

(t

)δ

(t

−n

)e − ı2πft dt

= ∞n=−∞

∞−∞ R X(t )e − ı2πft δ (t − n)dt

= ∞

n=−∞R X

(n

)e − ı2πfn (Replication Property)

= ∞n=−∞

R X[n]e − ı2πfn (Fourier Series)Hence, by Fourier sesies,

R X[n] = 12−12 S X[f ]e ı2πfmdf = R X(n) = ∞−∞ S X(f )e ı2πfmdf .Digital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 66 / 106


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2.8 Series expansion of random processes


2.8-1 Sampling bandLimited random process


68/106

Deterministic case

A deterministic signal x (t ) is called band-limited if X (f ) = 0 for all ∣f ∣ > W Shannon-Nyquist theorem: x

(t

) can be perfectly

reconstructed if the sampling rate f s > 2W andx (t ) = ∞

n=−∞x n

2W sinc 2W t − n

2W

Note that the above is only sufficient, not necessary.


Stochastic case

A WSS h i X i id b b d li i d


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A WSS stochastic process X

(t

) is said to be band-limited

if its PSD S X(f

) = 0 for all

∣f

∣ > W

It follows that

R X

(τ

) = ∞

n=−∞R X

n

2W sinc

2W

τ

− n

2W In fact, the random process X(t ) can be reconstructed byits (random) samples in the sense of mean square.

Theorem

E X(t ) − ∞n=−∞

X n2W sinc 2W t − n

2W 2 = 0


The random samples

Problems of using these random samples


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Problems of using these random samples.

These random samples X n2W ∞n=−∞

are in general

correlated unless X(t ) is zero-mean white.EX n

2W X∗ m

2W = R X n − m

2W

≠ E

X

n

2W E

X∗

m

2W = mX

2.

If X(t ) is zero-mean white,EX n

2W X∗ m

2W = R X n − m

2W = N 0

2 δ n − m

2W

= EX n2W EX∗ m2W = mX2 = 0 except n = m.

Thus, we will introduce the uncorrelated KL expansionsin slide 87.



71/106

2.9 Bandpass and lowpass random processes (revisited)


Definition (Filtered white noise)

A N i ll d fil d hi i if i PSD l


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A process N

(t

) is called a filtered white noise if its PSD equals

S N(f ) = N 02 , ∣f ± f c ∣


73/106

P0 1) By fundamental assumption on slide 52, we obtainthat X

(t

) and X̂

(t

) are jointly WSS.

R X,X̂(τ ) and R ̂X(τ ) are only functions of τ because X̂(t ) is theHilbert transform of X(t ), i.e., R

X,X̂(τ ) = R X(τ ) ⋆ h∗(−τ ) =

−R X(τ )⋆h(τ ) (since h∗(−τ ) = −h(τ )) and R ̂X(τ ) = R X,X̂(τ )⋆h(τ ).P0-2) X

i (t ) = Re (X(t ) + ı X̂(t ))e −

ı2πf c t

is WSS byfundamental assumption.P2 ′

) ⎧⎪⎪⎨⎪⎪⎩

R X

(τ

) = R ̂X

(τ

)R X,X̂(

τ

) = −R ̂X,X(

τ

)(X(t ) + ı X̂(t ) is the “lowpassequivalent” signal of Xi (t )!)(Xi (

t

)+ı Xq (

t

) is the lowpass

equivalent signal of X(t )!)Also, R ̂X,X(τ ) = R̂ X(τ ), where R̂ X(τ ) is the Hilbert transformoutput due to input R X τ .


P3 ′) R Xi (τ ) = Re 12 R (X+ ı X̂)(τ )e − ı2πf c τ 1 2 f


74/106

R Xi

(τ

) = Re

1

2R (X+ ı X̂)

(τ

)e − ı2πf c τ

= Re(R X(τ ) + ı R ̂X,X(τ ))e − ı2πf c τ = R X(τ ) cos(2πf c τ ) + R̂ X(τ ) sin(2πf c τ )Note that Ŝ X

(f

) = S X

(f

)H Hilbert

(f

) = S X

(f

)(−ı sgn

(f

)).

P4 ′) S Xi (f ) = S X(f − f c ) + S X(f + f c ) = S Xq (f )) for ∣f ∣


75/106


● ⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

R X(τ ) = Re 12 R X(τ ) e ı2πf c τ (P 3)R ̂X,X(τ ) = R X(τ ) ⋆ hHilbert(τ )

P0-1

= Im 12 R X(τ ) e ı2πf c τ ● Then: 12 R X(τ ) = R Xi (τ ) + ı R Xq ,Xi (τ )

Proof of P 3

= (R X(τ ) + ı R ̂X,X(τ ))e − ı2πf c τ

● ⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪R Xi (τ ) = Re (R X(τ ) + ı R ̂X,X(τ )) e − ı2πf c τ (P 3′)

R Xq ,Xi (τ ) = Im (R X(τ ) + ı R ̂X,X(τ )) e − ı2πf c τ = R Xi (τ ) ⋆ hHilbert(τ )Digital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 75 / 106


76/106

Proof: Hence,

R Xq ,Xi (τ ) = Im(R X(τ ) + ı R ̂X,X(τ ))e − ı2πf c τ = −

R X

(τ

)sin

(2πf c τ

) +R ̂X,X

(τ

)cos

(2πf c τ

)= −R X(τ ) sin(2πf c τ ) + ˆR X(τ ) cos(2πf c τ ).The property can be proved similarly to P4 ′).



77/106

2.2 Signal space representation


Key idea & motivation


78/106

The low-pass equivalent representation removes thedependence of system performance analysis on carrierfrequency.

Equivalent vectorization of the (discrete or continuous)signals further removes the “waveform” redundancy inthe analysis of system performance.


Vector space concepts

I d t n ∗ H


79/106

Inner product:

⟨v1, v2

⟩ = ∑ni =1 v 1,i v ∗2,i

= vH 2 v1

(“H ” denotes Hermitian transpose)

Orthogonal if ⟨v1, v2⟩ = 0Norm:

∥v

∥ = ⟨v, v

⟩Orthonormal: ⟨v1, v2⟩ = 0 and ∥v1∥ = ∥v2∥ = 1Linearly independent:

k i =1

ai vi = 0 iff ai = 0 for all i Digital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 79 / 106

Vector space concepts

Triangle inequality


80/106

∥v1 + v2∥ ≤ ∥v1∥ + ∥v2∥Cauchy-Schwartz inequality∣⟨

v1, v2

⟩∣ ≤ ∥v1

∥ ∥v2

∥.

Equality holds iff v1 = av2 for some aNorm square of sum:

∥v1

+v2

∥2

= ∥v1

∥2

+ ∥v2

∥2

+ ⟨v1, v2

⟩ + ⟨v2, v1

⟩Pythagorean: if ⟨v1, v2⟩ = 0, thenv1 v2

2v1

2v2

2


Eigen-decomposition


81/106

1 Matrix transformation w.r.t. matrix A

v̂ = Av2 Eigenvalues of square matrix A are solutions

{λ

} of

characteristic polynomialdet(A − λI ) = 0

3 Eigenvectors for eigenvalue λ is solution v of

Av = λvDigital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 81 / 106

Signal space concept

How to extend the signal space concept to a (complex)


82/106

How to extend the signal space concept to a (complex)function/signal z

(t

) defined over

[0, T

) ?

Answer: We can start by defining the inner product forcomplex functions.

Inner product: ⟨z 1(t ), z 2(t )⟩ = ∫ T 0 z 1(t )z ∗2 (t )dt Orthogonal if ⟨z 1(t ), z 2(t )⟩ = 0.Norm:

∥z

(t

)∥ = ⟨z

(t

), z

(t

)⟩Orthonormal: ⟨z 1(t ), z 2(t )⟩ = 0 and ∥z 1(t )∥ = ∥z 2(t )∥ = 1.Linearly independent: ∑k i =1 ai z i (t ) = 0 iff ai = 0 for allai ∈ C


Triangle Inequality

z 1 t z 2 t z 1 t z 2 t


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∥ ( ) + ( )∥ ≤ ∥ ( )∥ + ∥ ( )∥Cauchy Schwartz inequality∣⟨z 1(t ), z 2(t )⟩∣ ≤ ∥z 1(t )∥ ⋅ ∥z 2(t )∥Equality holds iff z 1

(t

) = a

⋅z 2

(t

) for some a

∈C

Norm square of sum:

∥z 1(t ) + z 2(t )∥2 = ∥z 1(t )∥2 + ∥z 2(t )∥2+ ⟨

z 1

(t

), z 2

(t

) ⟩+ ⟨z 2

(t

), z 1

(t

)⟩Pythagorean property: if ⟨z 1(t ), z 2(t )⟩ = 0,z 1 t z 2 t

2z 1 t

2z 2 t

2



84/106

Transformation w.r.t. a function C

(t , s

)ẑ (t ) = T

0C (t , s )z (s )ds

This is in parallel to

v̂ v̂ t = ns =1

At ,s v s = Av.

Digital Communications: Chapter 2 Ver 2012.09.29 Po-Ning Chen 84 / 106

Eigenvalues and eigenfunctions


85/106

Given a complex continuous function C (t , s ) over [0, T )2, theeigenvalues and eigenfunctions are

{λk

} and

{ϕk

(t

)} such

that

T 0

C (t , s )ϕk (s )ds = λk ϕk (t ) (In parallel to Av = λv)


Mercer’s theorem


86/106

Theorem (Mercer’s theorem)Give a complex continuous function C (t , s ) over [0, T ]2 that is Hermitian symmetric (i.e., C

(t , s

) = C ∗

(s , t

)) and

nonnegative definite (i.e.,

∑i

∑ j ai C

(t i , t j

)a∗ j

≥ 0 for any

{ai

}and {t i }). Then the eigenvalues {λk } are reals, and C (t , s )has the following eigen-decompositionC

(t , s

) = ∞

k =1λk ϕk

(t

)ϕ∗k

(s

).


Karhunen-Loève theorem

Theorem (Karhunen-Loève theorem)


87/106

Let

{Z

(t

), t

∈ [0, T

)} be a zero-mean random process with a

continuous autocorrelation function R Z(t , s ) = E[Z(t )Z∗(s )].Then Z(t ) can be written as Z

(t

) M2

= ∑∞k =1 Zk

⋅ϕk

(t

) 0

≤ t

< T

where “ =” is in the sense of mean-square,Zk = ⟨Z(t ), ϕk (t )⟩ = ∫ T 0 Z(t )ϕ∗k (t )dt and {ϕk (t )} are orthonormal eigenfunctions of R Z(t , s ).

Merit of KL expansion: {Zk } are uncorrelated. (Butsamples {Z(k /(2W ))} are not uncorrelated even if Z(t )is bandlimited!)

Digital Communications: Chapter 2 Ver 2012 09 29 Po-Ning Chen 87 / 106

Proof.


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E[Zi Z∗ j ] = E T 0 Z(t )ϕ∗i (t )dt T 0 Z(s )ϕ∗ j (s )ds ∗

= T

0

T

0R Z

(t , s

)ϕ j

(s

)ds

ϕ∗i

(t

)dt

= T 0 λ j ϕ j (t )ϕ∗i (t )dt

= ⎧⎪⎪⎨⎪⎪⎩

λ j if i = j 0

=E

[Zi

]E

[Z∗ j

] if i

≠ j


LemmaFro a given orthonormal set {φk (t )}, how to minimize the energy of error signal e t s t ŝ t for ŝ t spanned by (i d li bi ti f) φ t ?


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( ) = ( ) − ( ) ( )(i.e., expressed as a linear combination of)

{φk

(t

)}?

Assume ŝ (t ) = ∑k ak φk (t ); then

∥e

(t

)∥2

= ∥s

(t

) −ŝ

(t

)∥2

= ∥s

(t

) −∑k ∶k ≠i ak φk

(t

) −ai φi

(t

)∥2

= ∥s (t ) −∑k ∶k ≠i ak φk (t )∥2 + ∥ai φi (t )∥2−⟨s (t ) −∑k ∶k ≠i ak φk (t ), ai φi (t )⟩− ⟨

ai φi

(t

), s

(t

) −∑k ∶k ≠i ak φk

(t

)⟩= ∥s (t ) −∑k ∶k ≠i ak φk (t )∥2

+ ∣ai ∣2

−a∗i ⟨s (t ), φi (t )⟩ − ai ⟨φi (t ), s (t )⟩By taking derivative w.r.t. Re ai and Im ai , we obtainai s t , φi t .


Definition

If every finite energy signal s t (i.e., s t 2

) satisfies


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y gy g

( )(

∥ ( )∥ < ∞)

∥e (t )∥2 = s (t ) −k

⟨s (t ), φk (t )⟩φk (t )2 = 0(equivalently,

s (t ) L2= k

⟨s (t ), φk (t )⟩φk (t ) = k

ak ⋅ φk (t )(in the sense that the norm of the difference between

left-hand-side and right-hand-side is zero), then the set of orthonormal functions {φk (t )} is said to be complete .

Digital Communications: Chapter 2 Ver 2012 09 29 Po Ning Chen 90 / 106

Example (Fourier series)⎧⎪⎪ 2T

cos2πkt

T ,

2

T sin

2πkt

T 0 k Z

⎫⎪⎪


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⎨⎪⎪⎩

∶ ≤ ∈ ⎬⎪⎪⎭is a complete orthonormal set for signals defined over [0, T )with finite number of discontinuities.

For a complete orthonormal basis, the energy of s

(t

) is

equal to ∥s (t )∥2 = j

a j φ j (t ),k

ak φk (t )= j k

a j a∗k

⟨φ j

(t

), φk

(t

)⟩= j

a j a∗

j

j

a j 2

Digital Communications: Chapter 2 Ver 2012 09 29 Po Ning Chen 91 / 106

Given a deterministic function s (t ), and a set of completeorthonormal basis φk t (possibly countably infinite),


92/106

{ ( )}s

(t

) can be written as

s (t ) L2= ∞k =0

ak φk (t ), 0 ≤ t ≤ T where

ak = ⟨s (t ), φk (t )⟩ = T 0

s (t )φ∗k (t )dt .In addition,

∥s (t )∥2 = k ∣ak ∣2.Digital Communications: Chapter 2 Ver 2012 09 29 Po Ning Chen 92 / 106

Remark


93/106

Remark

In terms of energy (and error rate):A bandpass signal s (t ) can be equivalently “analyzed”through lowpass equivalent signal s

(t

) without the

burden of carrier freq f c ;

A lowpass equivalent signal s (t ) can be equivalently“analyzed through (countably many) {ak = ⟨s (t ), φk (t )⟩}without the burden of continuous waveforms.

Digital Communications: Chapter 2 Ver 2012.09

.29 Po Ning Chen 93 / 106

Gram-Schmidt procedure

Given a set of functions v 1 t , v 2 t , , v k t


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( ) ( ) ⋯ ( )1 φ1(t ) = v 1(t )∥v 1(t )∥2 Compute for i

= 2, 3,

⋯, k (or until

∥φi

(t

)∥ = 0),

γ i (t ) = v i (t ) − i

−1

j =1 ⟨v i (t ), φ j (t )⟩φ j (t )and set φi

(t

) = γ i (t )∥γ i (t )∥ .

This gives an orthonormal basis φ1(t ), φ2(t ),⋯, φk ′(t ), wherek ′ ≤ k .



Example

Find a Gram-Schmidt orthonormal basis of the following signals


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signals.



Sol.φ1(t ) = v 1(t )∥v 1(t )∥ = v 1(t )√ 2


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γ 2(t ) = v 2(t ) − ⟨v 2(t ), φ1(t )⟩φ1(t )= v 2(t ) − 30

v 2(t )φ∗1(t )dt φ1(t ) = v 2(t )Hence φ

2(t ) = γ 2(t )∥γ 2(t )∥ =

v 2(t )√ 2

.

γ 3

(t

) = v 3

(t

) − ⟨v 3

(t

), φ1

(t

)⟩φ1

(t

) − ⟨v 3

(t

), φ2

(t

)⟩φ2

(t

)= v 3(t ) −√ 2φ1(t ) − 0 ⋅ φ2(t ) = −1, 2 ≤ t


97/106

γ 4(t ) = v 4(t ) − ⟨v 4(t ), φ1(t )⟩φ1(t ) − ⟨v 4(t ), φ2(t )⟩φ2(t )−⟨v 4(t ), φ3(t )⟩φ3(t )=

v 4

(t

)−(−√ 2

)φ1

(t

) − (0

)φ2

(t

) −φ3

(t

) = 0

Orthonormal basis=

{φ1

(t

), φ2

(t

), φ3

(t

)}, where 3

≤ 4.

Digital Co icatio s Cha te 2 Ve 2012.09

.29 Po Ni g Che 97 / 106

Example Represent the signals in slide 95 in terms of the orthonormal basis obtained in the same example.


98/106

Sol.

v 1

(t

) = √

2φ1

(t

) +0

⋅φ2

(t

) +0

⋅φ3

(t

) ⇒ √

2, 0, 0

v 2(t ) = 0 ⋅ φ1(t ) +√ 2 ⋅ φ2(t ) + 0 ⋅ φ3(t ) ⇒ 0,√ 2, 0v 3(t ) = √ 2φ1(t ) + 0 ⋅ φ2(t ) + ⋅φ3(t ) ⇒ √ 2, 0, 1v 4

(t

) = −√

2φ1

(t

) +0

⋅φ2

(t

) +1

⋅φ3

(t

) ⇒ −√

2, 0, 1

The vectors are named signal space representations orconstellations of the signals.

Di it l C i ti Ch t 2 V 2012.

09.

29 P Ni Ch 98 / 106

Remark

The orthonormal basis is not unique.For example for k 1 2 3 re define


99/106

For example, for k

= 1, 2, 3, re-define

φk (t ) = 1, k − 1 ≤ t


100/106

s 1(t ) ⇒ (a1, a2,⋯, an) for some complete basiss 2(t ) ⇒ (b 1, b 2,⋯, b n) for the same complete basisd 12 = Euclidean distance between s 1(t ) and s 2(t )= n

i =1(ai − b i )2

= ∥s 1(t ) − s 2(t )∥ = T

0 ∣s 1(t ) − s 2(t )∣2dt Di it l C i ti Ch t 2 V 2012

.

09.

29 P Ni Ch 100 / 106

Bandpass and lowpass orthonormal basis

Now let’s change our focus from 0, T to ,


101/106

[ ) (−∞ ∞)A time-limited signal cannot be bandlimited to W .A bandlimited signal cannot be time-limited to T .

Hence, in order to talk about the ideal bandlimited signal, wehave to deal with signals with unlimited time span.

Re-define the inner product as:

⟨f (t ), g (t )⟩ = ∞−∞ f (t )g ∗(t )dt Digital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 101 / 106

Let s 1,(t ) and s 2,(t ) be lowpass equivalent signals of thebandpass s 1 t and s 2 t


102/106

( ) ( )S 1,(f ) = S 2,(f ) = 0 for ∣f ∣ > f B s i (t ) = Re s i ,(t )e ı2πf c t where f c ≫ f B Then, as we have proved in slide 24,

⟨s 1(t ), s 2(t )⟩ = 12 Re {⟨s 1,(t ), s 2,(t )⟩} .Proposition

If ⟨s 1,(t ), s 2,(t )⟩ = 0, then ⟨s 1(t ), s 2(t )⟩ = 0.Digital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 102 / 106

PropositionIf {φn,(t )} is a complete basis for the set of lowpass signals,then

φn t Re 2φn, t e ı2πf c t


103/106

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩ ( ) = √ ( ) φ̃n(t ) = −Im√ 2φn,(t ) e ı2πf c t = Re ı√ 2φn,(t ) e ı2πf c t is a complete orthonormal set for the set of bandpass signals.

Proof: First, orthonormality can be proved by

⟨φn

(t

), φm

(t

)⟩ = 1

2Re

√

2φn,

(t

),√

2φm,

(t

) =

⎧⎪⎪⎨⎪⎪⎩

1 n = m0 n

≠ m

φ̃n(t ), φ̃m(t ) = 12

Re ı√ 2φn,(t ), ı√ 2φm,(t ) = ⎧⎪⎪⎨1 n = m0 n m


and φ̃n(t ), φm(t ) = 12

Re ı√ 2φn,(t ),√ 2φm,(t ) Re ı φn, t , φm, t


104/106

= { ⟨ ( ) ( )⟩}= ⎧⎪⎪⎨⎪⎪⎩Re{ ı} = 0 n = m0 n ≠ m

Now, with⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

s (t ) = Re {s (t )e ı2πf c t }ŝ (t ) = Re {ŝ (t )e ı2πf c t }ŝ

(t

) L2

= n

an,φn,

(t

) with an,

= ⟨s

(t

), φn,

(t

)⟩∥s (t ) − ŝ (t )∥2

= 0we haves t ŝ t 2

1

2s t ŝ t

2 0


and

ŝ t Re ŝ t e ı2πf c t


105/106

( ) = ( ) = Ren an,φn,(t )e ı2πf c t

= n

Re

an,

√ 2

Re

√ 2φn,

(t

)e ı2πf c t

+Iman,√ 2 Im−√ 2φn,(t ) e ı2πf c t

= n Re

an,

√ 2φn

(t

) +Im

an,

√ 2 φ̃n

(t

) ◻Digital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 105 / 106

What you learn from Chapter 2


106/106

Random processWSSautocorrelation and crosscorrelation functionsPSD and CSD

White and filtered white

Relation between (bandlimited) bandpass and lowpassequivalent deterministic signalsRelation between (bandlimited) bandpass and lowpass

equivalent random signalsProperties of autocorrelation and power spectrum density

Role of Hilbert transform

Signal space conceptDigital Communications: Chapter 2 Ver. 2012.09.29 Po-Ning Chen 106 / 106

EC 2351 Digital Communication

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