DC Digital Communication MODULE I

8/7/2019 DC Digital Communication MODULE I

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DIGITAL COMMUNICATION

Module I

variables and random processes Detection and -

interpretation of signals Response of a bank ofcorrelators to noisy input Detection of known

signals in noise probability of error correlationand matched filter receiver detection of signals.

Estimation concepts criteria: MLE estimator

filter for wave form estimation Linear

rediction.

Compiled by MKP for CEC S6-EC, DC, Dec 2008


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Ortho onal functions Consider a set of functions defined

over the interval1 2 3( ), ( ), ( ),........, ( ),........nt t t t

Let these functions satisfy the condition

2t

A set of functions which has this property is said to be orthogonal

1

i jt

1 2 .

Suppose we have an arbitrary function s(t) and we are interested in s(t)

only in the interval t1 tot2 .where the set of functions (t) are

or ogona .

Now we can express s(t) as a linear sum of the functions

n(t).

where s1,s2,s3 ,,sn are constants.

( ) ( ) ( )1 1 2 2( ) ...... ...... (1)n n s t s t s t s t = + + + +



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Ortho onal functions If such an expansion is possible, the orthogonality of the s makes it

eas to com ute the coefficients s .

To evaluate sn we multiply both sides of equation (1) by n(t) and

integrate over the interval of orthogonality.

( ) ( )2 2 2 2

1 1 1 1

2

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

t t t t

n n n n nt t t t

s t t s t t dt s t t dt s t dt = + + + +

equation become zero except a single term.

2 2 2t t

1 1n n n

t t

2 ( ) ( )t s t t dt 1

2

1

2,

( )

t

n t

nt

st dt

=

Then



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Ortho onal functions If we select the functions n(t) such that

2

1

2( ) 1t

nt

t dt =

When orthogonal functions are selected with the condition,

1

, ( ) ( )n nt

s s t t dt =Then2

2( ) 1t

n t dt =they are said to be normalized to have unit energy.

The use of normalized functions has the advantage that sns can be

1t

.

A set of functions which are both orthogonal and normalized is

.

We can represent any real valued signals as linear combinations of

orthonormal basis functions.



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Gram-Schmidt Orthogonalization Procedure

We may represent a given set of real-valued energy signals

s t s t s t .. s t each of duration T seconds usin

orthonormal basis functions as given below.

0N t T 1

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

i jj ijs

i M s t t

=

==

ij

1,2,.....,T i M==

The functions are orthonormal which

0 1,2,....,j N=1 2 3( ), ( ), ( ),........, ( )Nt t t t

,

0

0

1( ) ( )

T

i j

for i j

ort t d

it

==



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The first condition states that the basis functionsare orthogonal with respect to each other over the interval 0 to T.

1 2( ), ( ),....., ( )Nt t t

The second condition states that the basis functions are normalized tohave unit energy.

Given the set of coefficients {sij}, j=1,2,,N .we can generate thesignal si(t), i=1,2,.M using a scheme as shown in figure (1).

,its own basis function, followed by a summer.

Conversely, given the set of signals si(t), i=1,2,. , operating asinput, we can generate the coefficients {sij}, j=1,2,,N using the

scheme given in figure (2)

It consists of a set of N product-integrators or correlators with acommon input, and with each supplied with its own basis function.



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sT

( )1 t

0

2is0

T

d t2 tis t

iNs

t

0d t

(2)Figure0

1, 2, ,( ) ( ) ,

T

ij i j

i M s s t t dt

==

=


, , ,


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If the given set of signals s1(t), s2(t), s3(t),.. sM(t) are linearly

dependent, then there exists a set of coefficients a1,a2,..,a not all

zero, such that we may write

1 1 2 2

( ) ( ) ( ) 0M M

a s t a s t a s t + + + = aM sM

1 2 11 2 1( ) ( ) ( ) ( )

MM M

a a a s t s t s t s t

= + + +

It implies that sM(t) may be expressed in terms of the remaining (M-1)signals.

M M M

Next consider the set of signals s1(t), s2(t), s3(t),.. sM-1(t) . If this set

is linearly dependent there exists a set of numbers b1,b2,..,bM-1 not alle ual to zero such that

1 1 2 2 1 1( ) ( ) ( ) 0M Mb s t b s t b s t + + + =



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Suppose that bM-1 0 . Then we may express sM-1(t) as a linear combination of the remaining M-1 signals.

1 2 21 1 2 2

1 1 1

( ) ( ) ( ) ( )MM MM M M

b b b s t s t s t s t

b b b

= + + +

es ng e se o s gna s s1 t , s2 t , s3 t ,.. sM-2 t or near

dependencies, and continuing in this fashion, we will eventually end

up with a linearly independent subset s1(t), s2(t), s3(t),.. sN(t) , NM

of the original signal. It is important to note that each member of the original set of signals

s t s t s t .. s t ma be ex ressed as a linear combination of

this subset of N signals s1(t), s2(t), s3(t),.. sN(t).



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Now we may express the linearly independent functionss1(t), s2(t), s3(t),.. sN(t) in terms of orthonormal functions

using1 2 3( ), ( ), ( ), ........, ( )Nt t t t

0N t T 1 1,2,. . .,. .

i jj ij i N= =

1 11 1 12 2 1( ) (t)+ (t)+ + (t) (1 .)N N s t s s s a =

2 21 1 22 2 2.N N

1 1 2 2( ) (t)+ (t)+ + (t) (1 .) N N N NN N s t s s s n =



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In equation (1 a.), set to zero all coefficients except s11. We then have

Since is to be a normalized function,1(t)

1 11 1

2

11 10

( )T

s s t dt = 1E=

11

11

( )(t) s ts

=1

1

( )s tE

=

In the next step we set to zero all the coefficients except the first two

s21

and s22

in equation (1b)

2 21 1 22 2( ) (t)+ (t) (3) s t s s =



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Multiply both sides of equation (3) by and integrating over theinterval 0-T

1(t)

2

2 1 21 1 22 1 20 0 0

( ) (t) (t)+ (t) (t)T T T

s t s s =

2 1 21

0( ) (t)

T

s t s = 21 2 10 ( ) (t)T

s s t =

To evaluate s22, we rewrite equation (3) as

2 21 1 22 2( ) (t) (t) (4) s t s s =

[ ]2 2 2 2

2 21 1 22 2 22( ) (t) (t)dt = (5)T T

s t s dt s s = Compiled by MKP for CEC S6-EC, DC, Dec 2008


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2( ) (t)

T

s s t s dt =

Again re-writing equation (3)

0

[ ]2 2 21 122

(t) = ( ) (t) s t ss

21 12

22 11

1 (t)= ( ) s ss ts s

21 12 2

22 11(t)= ( )

s ss ts s

Continuing in this manner we re-write equation(1c) as

3 31 1 32 2 32 2( ) (t)+ (t)+ (t) s t s s s =



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Manipulating the equation as above we getT

31 3 10

=

T

32 3 20

t s s t =

[ ]33 3 31 1 32 20 ( ) (t) (t) s s t s s dt = 3 31 1 32 2

3

( ) (t) (t)

( )

s t s s

t

=



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We continue in this manner until we have used all the N equationsand obtained all the orthonormal functions 1 2 3( ), ( ), ( ), ........, ( )Nt t t t

an a e coe c en s sij nee e o express e unc ons

s1(t), s2(t), s3(t),.. sN(t) in terms of 1 2 3( ), ( ), ( ), ........, ( )Nt t t t

Since all of the derived subset of linearly independent signals

s1(t), s2(t), s3(t),.. sN(t) may be expressed as a linear combination of

,

that each one of the original set of signals s1(t), s2(t), s3(t),.. sM(t)may be expressed as a linear combination of the same set of basis

1 2 3, , ,........, N

.



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1

11

)()(

E

tst =

)()()( 111111 tstEts ==

T

)()()(

)()(

12122

01221

tststg

tttss

=

=

= T

dttg

tgt

0

2

2

22

)(

)()(

T

)()()()( 12120

2

22 tstdttgtsT

+=

=

1

0)()(

i

jiij dtttss )()( 121222 tsts +=

=T

ii

dtt

tgt

2

)()(

=

1j

jijii


0


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Geometric inter retation of si nals

0N t T

1

1,2,.....,i jj ij i M= =

=

Each si nal in the set S t is com letel determined b the vector of

0

, ,....( )

.,

1,2,....,( )ij i j s s t t dt

j N

==

its coefficients as given by

1iS

2

1, 2, 3, ,

i

i

S

s i M

= = i

S

i



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The vectorSi is called signal vector. We ma now ex and the conventional idea of two and

three dimensional Euclidean spaces to an N-dimensional

Euclidean space. We can visualize the set of signal vectors { Si },

i=1,2,3,,M as defining a corresponding set of M points

n an - mens ona uc an space.

The N mutually perpendicular axes are labeled 1, 2,

3,.. N

This N-dimensional Euclidian space is called signal space.

-

three signals, that is, N=2 and M=3.



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2 32, 3N M= =

2

s

1

1s1

2

0

3



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Geometric inter retation of si nals The length of a vectorsi is defined by the symbol The dot product of any vector with itself gives the squared length.

iS

( )

2

i i i s s s=

2 (4)N

ijs=

The cosine of the angle between the vectors Si and Sj is given by

1j=

( )cos i js s =

The two vectors are orthogonal if their dot product is zero.

i js s



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Geometric inter retation of si nals The energy of a signalsi (t) of duration T is defined as

2

0( )i i E s t dt =

1( ) ( )

N

i jj ijt tss

==

2

0( )i s t dt =

T

0 i is t s t t =

NT N

Interchanging the order of summation and integration,

110ik kkjj i j ==

1 1 0( ) ( )

N N

ij ik j

T

j ki kts s t dtE

= ==



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=T

0 j k

N N2

N

s= 1 1i ij ik j k= = 1 j=

i

is equal to the squared length of the signal vectorsi representing it.



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DC Digital Communication MODULE I