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[OPTIMUM RECEIVER DESIGN AND EVALUATION FOR AWGN] Dr. Muqaibel

1 Dr. Ali Muqaibel Digital Communications I

Optimum Receivers for the AWGN

channel

ContentsOptimum Receiver ........................................................................................................................................ 2

Optimum Demodulator ................................................................................................................................ 2

Correlation Demodulator .......................................................................................................................... 2

Matched filter Demodulator ..................................................................................................................... 5

Frequency domain interpretation of matched filter ............................................................................ 8

The Optimum Detector ............................................................................................................................... 11

Minimum Distance Detection ................................................................................................................. 12

The Maximum Likelihood Sequence Detector ........................................................................................ 15

A symbol by symbol MAP detector for signals with memory. ................................................................ 17

Performance of the Optimum Receiver for Memory less Modulation ....................................................... 17

Probability of Error for Antipodal Binary Modulation ............................................................................ 17

For Binary orthogonal signals ................................................................................................................. 18

Comparison of Digital Modulation Methods .............................................................................................. 20

Performance analysis for wire-line and radio communication Systems .................................................... 21Regenerative repeaters .......................................................................................................................... 21

Link Budget analysis in radio comm. Systems ........................................................................................ 21

AWGN: Additive White Gaussian Noise

Objective :

Receiver design Performance evaluation (memory, No memory)

Note: These notes are preliminary and are posted by the request of the students. Please,report to me all the mistakes that you find in the document. ([email protected]).This material is for the sole purpose of in class usage. Please observe the copyright of theoriginal authors for any content in these notes.
mailto:[email protected]:[email protected]:[email protected]:[email protected]


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Optimum Receiver

+

r(t)

n(t)

( )ns t

( ) ( ) ( ) 0

{ ( ), 1,2,..., }

m

m

r t s t n t t T

s t m M

= +

=

0

1( ) /

2nm

S f N W Hz=

Objective:Upon observation of r(t), what is the optimum receiver design in terms ofprobability of making error.

demodulator detector

receiver

Demodulator :converts the received waveform r(t) into an N-dimensional vector

1 2[ ]Nr r r r = whereNis the dimension of the transmitted signal.

Detector:Is to decide which of the M possible signals waveforms was transmitted based on r

.Types of detectors

1. The optimum detector2. Maximum likelihood sequence detector3. Symbol by symbol MAP

Optimum Demodulator

1. Correlators2. Matched filters

Correlation Demodulator

Decomposes the received signal-to-noise ratio into N-dimensional vectors. Linearly weighted orthogonal basis functions {fn(t)}, where {fn(t)} spans the

signal space but not the noise space.

We can show that the noise terms that falls outside the signal space is irrelevant to the detection.


3/23



0

( )T

dt

1( )f t 1r

0

( )T

dt

2 ( )f t 2r

0

( )T

dt

( )Nf t Nr

( )r tto the detector

0 0

0

0

' '

1 1 1

'

1

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

1,2,

( ) ( ) , 1, 2,

( ) ( ) , 1, 2,

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

( ) ( ) ( )

T T

k m k

k mk k

T

mk m k

T

k k

N N N

mk k k k k k

k k k

N

k k

k

r t f t dt s t n t f t dt

r s n k N

where

s s t f t dt k N

n n t f t dt k N

r t s f t n f t n t r f t n t

n t n t n f t

= = =

=

= +

= + =

= =

= =

= + + = +

=

' ( )n t is zero mean Gaussian noise process. It is the unrepresented part of noise inherent.' ( )n t is Gaussian because it is the sampled output of a linear filter excited by a Gaussian input.

0

0 0

0

0 0

0 0

0

( ) [ ( )] ( ) 0

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

1 ( ) ( ) ( )2

1 1( ) ( )

2 2

1

T

n k

T T

k k k m

T T

k m

T

k m mk

mk

E n E n t f t dt for all k

E n m E n n f t f dtd

N t f t f dtd

N f t f dt N

= =

=

=

= =

=

1mk = , when m=k and zero otherwise.


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{ }k

n are zero mean uncorrelated random variable with common covariance

20

1

2n

N =

2 20

[ ] [ ]

1

2

k mk k nk

r n

E r E s n s

N

= + =

= =

Gaussian uncorrelated implies statistically independent.

1 2[ ]Nr r r r

=

1

2

00

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

( )1

( / ) exp[ ] , 1,2,...

N

m mk

k

k mk

mk

p r s p r s m M

r s

when p r s k N NN

=

= =

= =

The joint conditional pdf

2

/ 210 0

( )1( / ) exp[ ], 1,2,...,

( )

Nk mk

m Nk

r sp r s m M

N N

=

= =

As a final step we can show that 1 2( , ,..., )Nr r r are sufficient statistics.

No additional relevant information can be extracted from ' ( )n t .

'[ ( ) ] 0kE n t r = uncorrelated proof p 235

Gaussian and uncorrelated implies statistically independent which implies ignore ' ( )n t

Example

M-ary PAM

a

0 T

g(t)

t

2 2 2

0 0

( )T T

gE g t dt a dt a T= = =

PAM one basis function


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0 T

f(t)

t

1

T

2

1 1/ 0( ) ( )0

T t Tf t g totherwisea T

= =

The output of the correlator

0 0

1( ) ( ) ( )

T T

r r t f t dt r t dt T

= =

The correlator becomes a simple integrator when f(t) is rectangular .

0 0 0

2

0 0 0 0

00

0 0

1 1{ [ ( ) ( )} [ ( ) ( ) ]

[ ] 0

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

1( )

2 2

T T T

m m

m

T T T T

n

T T

r s t n t dt s t dt n t dt

T Tr s n

E n

E n t n dtd E n t n dtdT T

Nt dtd N

T

= + = +

= +

=

= =

= =

pdf of the sampled output

2

00

( )1

( / ) exp[ ]m

m

r s

p r s NN

=

Matched filter Demodulator

We use N filters

( ) 0( )

0

k

k

f T t t Th t

otherwise

=

0

0

( ) ( ) ( )

( ) ( ) 1,2,...

t

k k

t

k

y t r h t d

r f T t d k N

=

= + =

If we sample at the end the period t=T


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0

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

k k ky T r f d r k N = = =

Matched filter:A filter whose impulse response h(t)=s(T-t) or s(t) where s(t) is assumedto be confined to the time interval 0tT.

y(T)

T 2T

T

Tt

s(t)

A

( )r t

1( )f T t

2 ( )f T t

( )Nf T t

Property of Matched filter: If a signal s(t) is corrupted by AWGN, the filter with an impulseresponse matched to s(t) maximizes the output signal to noise ratio (SNR).Proof :


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0 0 0

0 0

2

0 2

2

2

0 0

0

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

( ) ( ) ( ) ( ) ( )

( ) ( )

( )

[ ( )]

[ ( )] var

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

1

2

t t t

T T

s n

s

n

n

T T

n

y t r h t d s h t d n h t d at t T

y T s h T d n h T d

y T y T

y TSNR

E y T

where E y T noise iance

E y T E n n t h T h T t dtd

N

= = + =

= +

= +

=

=

=

0 0

20

0

2 2

0 00

2 20 0

0 0

( ) ( ) ( )

1( )

2

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

1 1( ) ( )

2 2

T T

T

T T

T T

t h T h T t dtd

N h T t dt

s h T d h s T d

SNR

N h T t dt N h T t dt

=

= =

Can we maximize the numerator while the denominator is held constant.

Cauchy-Schwarz inequality:

2 2 21 2 1 2[ ( ) ( ) ] ( ) ( )g t g t dt g t dt g t dt

With equality when 1( )g t =c 2 ( )g t

1( )g t = h(t) , 2 ( )g t =s(T-t) more when h(t) = c s(T-t), c2drop from the numerator and

denominator

20

0 00

2 2( )

TE

SNR s t dt N N

= =

Property: The output SNR from the matched filter depends on the energy of the waveform s(t)but not on the details (shape) of s(t).


8/23



Frequency domain interpretation of matched filter

2 2

22 2 2

2 2

( ) ( )

( ) ( ) ( )

( ) ( ) ( ) '

j ft

j ft j ft j ft

s

s

Y f S f e

Y t Y f e df S f e e df

at t T

Y T S f df s t dt E Pasaval s relation

=

= =

=

= = =

Noise at the output of the matched filter

2

0

2 2

0 0 0

22

0

00

1( ) ( )

2

1 1 1( ) ( ) ( )

2 2 2

2( )

1

2

n

ss s

n

f H f N

P f df N H f df N S f df EN

P E EP Y T SNR

P NEN

=

= = = =

= => = = =

Matched filter and correlator at equivalent at t=T but matched filter is immune to time jitters.

Example 5.1.2M=4 bi-orthogonal signals

2

T

2

T T

1( )f t

t

2

T

2

TT

2( )f t

t

2

T

2

T T

1 1( ) ( )h t f T t =

t

2

T

2

T T

t

2 2( ) ( )h t f T t =

T

t

21

2A T

1 ( )sy t

T

t

21

2A T

2 ( )sy t

2

T

Note the response to s1(t) is evaluated at T


9/23



1 2 1 2

2

00

[ ] [ ]

( ) 2

1

2

r r r E n n

E ESNR

NN

= = +

= =

Additional Example (Matched Filters)Consider the signal s(t)

2

A

2

T T

t

2

A

a) Determine the impulse response of a filtered matched to this signal and sketch it.

h(t)=s(T-t)

2

A

2

T T

t

2

A

b) Determine the matched filter output as function of time.c) This is done by convolution, we may ask about the values at the peak (figure)

A2T/4 = E2

T

t

21

4A T

2

A

21

8A T

3

4

T2

T

A pair of pulses that are orthogonal to each other over the interval [0,T], are used for twodimensional matched filter

Filtered match to s1(t)

Filtered match to s2(t)

input

Output 1

Output 2

2

A

2

T T

t

2

A

2

A

4

T Tt

2

T

2

A

What is the output of the matched filter at T?


10/23



a) When s1(t) is applied to the two branches.b) When s2(t) is applied to the two branches.

First branch is shown in problem 4.1Second branch is zero [ generalize to all orthogonal signals].

Additional Problems on Matched filters.

Another method for approximating realization of matched filter is the (RC) low pass filter[integrator].

R

input outputC

0

0

1 1( ) ,21

H f ff RCj

f

= =+

The input signal is rectangular of pulse amplitude A and duration T.

Objective:Optimize the selection of 3-dB cutoff frequency f0of the filter. So that the peakoutput SNR is maximized.

Show that f0=0.2/T is the optimum. Compared to matched filter 1dB loss.

The peak value of the output power is

s(t)

T

0[1 exp( 2 )]A f T

2 20[1 exp( 2 )]outP A f T =

f0is the 3-dB cutoff frequency of the RC filter.

0 0 0

2

0

2 21 ( )out

N N fdfN

ff

= =+

The corresponding value for the SNR2

0

0 0

2( ) [1 exp( 2 )]out

ASNR f T

N f

=

Differentiating with respect to (f0T) and setting the result equal to zero.The maximum value of (SNR)outis at f0=0.2/T.


11/23



The Optimum Detector

Detectorr

assume no memory

Objective:Maximize the probability of correct decisions.Posterior probabilities.P(signal ms was transmitted| r ) = ( | )mP s r where m =1,2,..M

Hence the name maximum a posteriori probability (MAP)

Using Bayes rule:( | ) ( )

( | )( )m m

m

P r s P sP s r

P r=

( )m

P s a priori probability of the mthsignal.

1( ) ( | ) ( )

M

m mmP r P r s P s=

=

When the M-signals are equally probable ( )mP s

=1/M for all m.

The same rule that maximizes ( | )mP s r

is equivalent to maximizing ( | )mP r s

Likelihood function: is the conditional PDF ( | )mP r s

or any monotonic function of it.

Maximum likelihood (ML) criterion.MAP = ML if {

ms

} is equi-probable.

For AWGN , the likelihood function is given by 5.1.122

/ 210 0

( )1( | ) exp

( )

Nk mk

m Nk

r sP r s

N N

=

=

or

20

10

1 1ln ( | ) ln( ) ( )

2

N

m k mk

k

P r s N N r sN

=

=

Maximizing ln ( | )mP r s

is equivalent to minimizing

2

1

( | ) ( )N

m k mk

k

D r s r s

=

=

where

( | )mD r s

is Euclidean distance and m= 1,2,M


12/23



Minimum Distance Detection

Another interpretation of ML criterion.

2 2

1 1 1

D( r, ) 2N N N

m n n mn mn

n n n

s r r s s

= = =

= + 2 2

2 . m mr r s s

= + where m = 1,2,3M

2' ( , ) 2 .m mD r s r s s

= +

Let '( , ) ( , )m m

C r s D r s

=

To get rid of the minus. Now, we maximize C rather than minimize D.2

( , ) 2. .m m mC r s r s s

=

2

ms

can be eliminated if energy is fixed for all m =1,2,M, but cannot if signals have uequal

energy (PAM).

0

( , ) 2 ( ) ( )T

m m mC r s r t s t dt E

= m= 0,1,2,,M

An alternative realization of the optimum AWGN receiver.

0

( )T

dt

1( )s t

0

( )T

dt2 ( )s t

0

( )T

dt

( )Ns t

( )r t Selectthe

Largest

1

1

2E

2

1

2E

1

2 NE

Summary:Optimum ML

1) compute ( , )mD r s or '( , )mD r s distance metrics and chooses the smallest or


13/23



2) compute ( , )m

C r s

correlator metrics and choose the largest

3) ML= MAP if eqiprobable {m

s

}, otherwise

( , ) ( | ) ( )m m mPM r s P r s P s

=

Example:Let us assume that we are sending one of two levels either 2 or 6 as shown in the figure. Toillustrate the importance of subtracting the energy of the symbol. We will consider two cases.The first one will assume that the received amplitude is 4 and then we will consider theamplitude to be 3.

Tt

2

2

T

s1(t)

Tt

6

2

T

s2(t)

2

( , ) 2. .m m mC r s r s s

=

For the middle case with amplitude 4 we get a fair comparison afterremoving the bias

2

( , ) 2. .m m mC r s r s s

=

1( , ) 2.(4).(2) 4 16 4 12C r s = = =

2( , ) 2.(4).(6) 36 48 36 12C r s

= = = For the case that the received signal with amplitude of 3 we get the

wrong decision unless we remove the bias

1( , ) 2.(3).(2) 4 12 4 8C r s

= = =

2( , ) 2.(3).(6) 36 36 36 0C r s

= = =

Example:

In binary PAM s1= -s2= bE

Prior probabilities P(s1) = , P(s2) = 1-

Determine the metrics of the optimum MAP for AWGN.

Tt

3

2

T

r(t)

Tt

4

2

T

r(t)


14/23



( )b nr E y T = + zero mean and2

0

1

2n

N =

Note: the variance of the sampled noise isN0/2. In general the noise power, P=N0B, According toNyquistB=1/(2T), When looking at the energy weE=PT

2

1 2

( )1( | ) exp 22

b

nn

r E

P r s

=

2

2 2

( )1( | ) exp

22

b

nn

r EP r s

+=

( ) 21 1 2

( , ) ( | ) exp22

b

nn

r EPM r s p r s

= =

( ) 22 2 2

1( , ) (1 ) ( | ) exp

22

b

nn

r EPM r s p r s

+ = =

If 1 2( , ) ( , )PM r s PM r s> then choose s11

2

1

2

( , ) 1

( , )

s

s

P M r s

P M r s

>

=

=

successive symbols are interdependent.

The maximum likelihood (ML) sequence detector: searches for the minimum Euclidean

distance path through the trellis that characterizes the memory in the transmitted sequence.

Example NRZI

(PAM) (zero : like before, one : flip)

s1= -s2= bE = +

2

1 2

( )1( | ) exp

22

k b

k

nn

r EP r s

=

2

2 2( )1( | ) exp 22k b

k

nn

r EP r s

+=

( ) 2

( ) ( )1 2 2

11

( )1( , ,... | ) ( | ) exp

22

kmK K

m m k kk k k

kk nn

r sP r r r s P r s

==

= =

Maximize the above probability.By taking the logarithm and consider only those relevant term.

( ) ( ) 2

1

( , ) ( )K

m m

k k

k

D r s r s=

=

For binary we need to search 2ksequences, where kis the sequence length.

Viterbi algorithm: is a sequential trellis search algorithm for performing ML sequence detection Used for decoding convolutional codes Assume initial state s0The example of Binary NRZI


16/23



0/bE 0 / bE0 /

bE

0 / bE

1/b

E

1/ bE1/

bE

t 2TT 3T

1/bE

At 2Tand so on there are two arrows entering the state, we choose the minimum distancesurvivor.

For state 1 we do similar

2 21

2 21 1

1 2

2

(1,0) ( ) ( )

(0,1) ( ) ( )

b

b

b

b

D

D r E r E

r E r E = + = + +

At t=3T Suppose the survivors are (0,0) and (0,1)2

0 0 3

20 1 3

(0,0,0) (0,0) ( )

(0,1,1) (0,1) ( )

b

b

D D r E

D D r E

= + +

= + +

and2

1 0 3

21 1 3

(0,0,1) (0,0) ( )

(0,1,0) (0,1) ( )

b

b

D D r E

D D r E

= +

= +

Using Viterbi in this example the number of path searched is reduced by a factor of two ateach stage.The memory length isL(L=1 in the previous example)You make a decision when the survivor path agree. Variable delay negative ??Practically at 5Land then make decision at keven then before k-5Lis almost identical. Make a decision.

2 20

2 20 1

1 2

2

(1,1) ( ) ( )

(0,0) ( ) ( )

b

b

b

b

D

D r E r E

r E r E = + +

= + + +


17/23



Example (in the previous NRZI)

LetEb=1 and the demodulated sequence is (0.9,-0.8,0.3,-1.1,1.2,1.5,-0.7)

2 2 2 20

2 2 2 20

(1,1) (0.

(0,0) (0.9 1) ( 0.8 1) 1.9 0.

9

2 3.61

1) ( 0.8 1) ( 0.1) 0.2 0.01 0.04 0.0

0.04

5

3.65D

D = +

= + +

+ =

+

+ = + =

= + = + =

D1(0,1)=(0.9+1)2+(-0.8-1)2=1.92+1.82=3.61+3.24=6.85

D1(1,0)=(0.9-1)2+(-0.8-1)2=0.12+1.82=0.01+3.24=3.25

20

20 0

0(1,1,0) (1,1) (0.3 1) 0.05 1.

(1,0,1) (1,0) (0.3 1) 3.2

69

5 0.49 3.74

1.74

D

D

D

D= + + = + =

= + = + =

1

21

2

(1,1,1) (1,1) (0.3 1) 0.05 0.

(1,0,0) (1,0) (0.3 1) 3.25 1.69 4.9

9

4

4 0.54

D

D

D

D= + = + =

= + + = + =

A symbol by symbol MAP detector for signals with memory.

Optimum (minimize symbol error) If M is large => large computational complexity. Used for convolution codes and turbo coding. Beyond the scope.

Performance of the Optimum Receiver for Memory less Modulation

Probability of Error for Antipodal Binary Modulation

2s 1s

bE

bE

h 1R2R

PAM antipodal s1(t)= -s2(t)

1 br s n E n= + = + AWGN


18/23



20

20

20

( ) /

1

0

( ) /

2

0

0 0 ( ) /

1 1

0

1( | )

1( | )

1( | ) ( | )

b

b

b

r E N

r E N

r E N

P r s eN

P r s eN

P e s P r s dr e dr N

+

+

=

=

= =

By replacing variables0/ 2

br E

xN

= , 0

0

2

2

Ndx dr dr dx

N= => =

0

,

20 b

x r

Er x

N

= => =

= => =

02

2

0

2

2 /

/ 2

/ 2

02 /

/ 2

1

2

21

2

1( )

2

b

b

E N

x

x b

E N

x

x

e dx

Ee dx Q

N

Q x e dx

=

= =

=

Assuming s1and s2are equiprobable

1 2

0

21 1( / ) ( / )

2 2b

b

EP P e s P e s Q

N

= + =

Note Pedepends on the ratio0

bE

N

212 12

212

0

12

4

2

b b

b

d E E d

dP Q

N

= => =

=

For Binary orthogonal signals

0

1

s

2s

bE

bE

2 bE


19/23



1

2

[ 0]

[0 ]

b

b

s E

s E

=

= 12 2 bd E=

If s1 is transmitted1 2

2

2 1 2

1 1 2

[ ]

( , ) 2 .

( , ) 2.[ ].[0 ] (1)

( , ) 2.[ ].[ 0] (2)

b

m m m

b b b

b b b

r E n n

C r s r s s

C r s E n n E E

C r s E n n E E

= +

=

= +

= +

(1)Can be further simplified to 12 b bn E E and(2)Can be further simplified to 1 12 2 2b b b b bE n E E E n E+ = + Probability of error

2 1

1 2 1

1 2

1 2

2 1

1 2 1

( , ) ( , )

[ | ] [ ( , ) ( , )]

2 2 2

( ) 0

( | ) [ ]

b b b b b

b b

b

b

C r s C r s

P e s P C r s C r s

E n E E n E E

E n n E

n n E

P e s P n n E

>

= >

+ <

+

= >

n1 andn2are zero mean Gaussian random variables

x= n2-n1is zero mean Gaussian random variable with variance =2

0N =

2 20

0

/ 2 / 22 1

00 /

1 1( )

2 2b b

x N x bb

E E N

EP n n E e dx e dx Q

NN

> = = =

Because of the symmetry s2is the same.

b = SNR/bit

Compare orthogonal with antipodal (factor of 2 increase in energy)2

10 12

212

10log 2 3 2

4

b

b

dB d E for orthogonal

d E for antipodal

= =

=


20/23



Probabilityoferro

rPb

( )b bP Q =

( )2b bP Q =

SNR per bit, ( )b dB

Figure 5.2.4110

0r

=

1r =

Explain the concept of Union bound

In addition to the above antipodal and orthogonal examples, we can extend the analsys to othermodulation techniques. For example, Many orthogonal signals Bi-orthogonal Simplex M-ary PAM

2

0

220

2( 1)

6(log )2( 1)

( 1)

g

M

bav

d EMP Q

M N

M EMQ

M M N

=

=

Because 22

6

1g avd E P T

M=

M-ary PSK QAM

Comparison of Digital Modulation Methods

P 226-229 . Power spectral Efficiency!

Reading Material Quiz


21/23



Performance analysis for wire-line and radio communication Systems

Regenerative repeaters

n(t)

s(t)r(t)

Mathematical model for channel with attenuation and additive noise is( ) ( ) ( )r t s t n t = +

0

2bb

EP Q

N

=

PAM Binary.

kRepeaters assuming single error at a time.

0

2 bEP kQN

at repeater.. Why not exact equal sign? (error cancellation)

Analog: RequiredEb/N0reduced by k

0

2b

EP Q

kN

At decision receivernote receiver is connected k= 1+ repeater.

Example

1000 km 10 km repeater k=100 10-5

10-5= 1000

2b

EQ

N

=> 10-7=0

2b

EQ

N

=> SNR =11.3dB

10-5=0

2b

EQ

N

=> SNR =29.6 dB 29.6-11.3 = 18.3 dB

Link Budget analysis in radio comm. Systems

Microwave line-of-sight Transmission Link budget analysis

2 (1)

4T T R

R

P G AP

d=

The design should specify PT, size of antenna (transmit and receiver)


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SNR required to achieve a given performance and data rate.

GT: antenna transmit gain 24T T

P G

d, GT =1 for isotropic antenna.

T TP G : Effective radiated power.

ERPorEIRPcompared with isotropic antenna

AR: effective area of the antenna,2

(2)4R

R

GA

=

c f= , c= 3X108m/s

Substitute (2) in (1)

2(4 / )T T R

R

P G GP

d =

Free space path loss factor2

4

sL

d

=

Additional losses atmospherica

L .

R T T R s aP P G G L L=

Calculation of the antenna gain is antenna specific and depends on (dimensions) diameterD,Illumination efficiency factorEffective areaARAreaABeam width B

Dish, horn etc

(PR)dB= (PT)dB+ (GT)dB+ (GR)dB+ (Ls)dB+ (Lo)dB

Example

A geosynchronous satellite orbit (36,000 km)Power radiated is 100 W 20 dB above 1W -- 20dBWTransmit antenna gain 17dB ERP = 37 dBW3-m parabolic antenna at 4 GHz downlink = 0.5 efficiency factor

GR=2

39D dB

=

2

195.64

sL dB

d

= =

(PR)dB= 20 + 17 + 39 195.6 = -119.6 dBWPR= 1.1 X 10

-12Wis this low or high ?


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What matter is the SNR.

Noise is flat for up to 10-12HzN0=kBT0 W/Hz

KB: is Boltzmanns constant 1.38X10

-23

Total noise NW

Performance is dependent on0 0 0

1b b R RE T P P

N N R N= =

0 0 Re

bR

q

EPR

N N

=

Example for the same previous example.

RP =1.1 X 10-12W (-119.6 dBW)

N0= 4.1 X 10-21W/Hz, Pr=N0W = KBT0W

= -203.9 dBW/Hz

0

RP

N= -119.6 + 203.9 = 84.3 dB Hz

0

bE

NSNR is 10dB

RdB = 84.3-10 = 74.3 dB with respect to 1bit/sec= 26.9 Mbps

420 PCM (64000 bps)

The introduced safety margin

RdB = 0 0

= () + () + () + () + ()0

Download - DC1_Receiver Muqaibel.pdf

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