EEE306 Digital Communication 2013 Spring Slides 07

8/16/2019 EEE306 Digital Communication 2013 Spring Slides 07

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DEPARTMENT OF ELECTRICAL &ELECTRONICS ENGINEERING

DIGITAL COMMUNICATION

Spring 2010

Yrd. Doç. Dr. Burak Kelleci

OUTLINE

Probability of Error due to Noise


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PROBABILITY OF ERROR DUE TO NOISE

Since the matched filter is the optimum detector

,

probability of error rate due to the noise can be

calculated.

Let’s consider polar NRZ signaling, where

symbols 1 and 0 are represented by positive and

negative pulses by equal amplitude.

( ) ( )( )⎩

⎨⎧+−++=

sentwas0symbolsentwas1symbol

t w A

t w At x


Let’s assume that the receiver knows the starting

.

The receiver samples the output of the matched

filter at t=Tb and decides whether the

transmitted symbol is 0 or 1 from the received

noisy signal x(t)


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In decision device, the sampled value is compared

.

If the sampled signal is above λ the receiver assumesthat the transmitted symbol is 1.

If the sampled signal is below λ the receiver assumesthat the transmitted symbol is 0.

If the sampled signal is equal λ the receiver makes aguess so that the outcome does not alter the average

pro a i ity o error.

There are two possible kinds of errors

Symbol 1 is chosen when actually 0 was transmitted.

Symbol 0 is chosen when actually 1 was transmitted.

To determine the probability of error, these two

situations should be analyzed separately.


Suppose that symbol 0 was sent, then the

The signal at the matched filter output sampled

at t=Tb becomes

( ) ( ) bT t t w At x ≤≤+−= 0

=bT

dt t x

which represents the random variable Y

( )∫+−=bT

b

dt t wT

A0

0

1


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Since the additive noise is Gaussian distributed,

( )

( ) ( )∫ ∫

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡=

+=

b b

b b

T T

T T

b

Y

dtduuwt w E T

AY E

0 0

22

1

1

σ

RW(t,u) is the autocorrelation function of the

white noise w(t)

( )∫ ∫=b bT T

W

b

b

dtduut RT

T

0 0

0 0

,1


Since the power spectral density of the white

, ,

The variance of Y becomes

( ) ( )ut N

ut RW −= δ 2

, 0

T T N b b1

b

b

Y

T

N

dtduut T

2

2

0

0 0

=

−= δ σ


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The probability density function of the random

Therefore the probability density function of the

random variable Y. iven that s mbol 0 has sent

( )( )

2

2

2

2

1 X

X x

X

X e x f σ

μ

σ π

−−

=

( )( )

bT N

A y

b

Y eT N

y f /

0

0

2

/

10|

+−

=π


The probability density function is plotted below/

e ,

given that symbol 0 has sent.

The shaded area from λ to infinity corresponds tothe range of values that are assumed as symbol 1.

In the absence of noise the decision is always

symbol 0 for the sampled value of –A.


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The probability of error on the condition that

( )

( )∫+

∞

=

>=

λ

λ

dy y f

yPP

A

Y

e0

2

0|

sentwas0symbol|

∫ −

=

λ π

dyeT N

bT N

b

/

0

0

/

1


The value of he threshold λ can be assigned with

and p1.

It is clear that the sum of p0 and p1 must be 1.

If we assume that symbols 0 and 1 are occurred

with equal probabilities (p0=p1=0.5), it reasonable

to assume the threshold halfway between +A

– .

0=λ


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The conditional probability of error when symbol

Let’s define a new variable z

( )

∫∞ +−

=0

/

0

00

2

/

1dye

T N P b

T N

A y

b

eπ

A y +

bT N 2/0


Let’s reformulate Pe0

where Eb is the transmitted signal energy per bit

∫∞

−

=0

2

/2

20

2

1

N E

z

e

b

dzePπ

T A E 2=


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THE Q-FUNCTION

The Q-function is used by communication

the Gaussian distribution.

Error function is the twice of the area under a

normalized Gaussian from 0 to u. The

complementary error function is defined as one

minus the error function ∞−= z d eu

22erfc

The Q-function and complementary errorfunction is related to each other

uπ

( ) ⎟ ⎠

⎞⎜⎝

⎛ =

2erfc

2

1 uuQ


The conditional probability of error Pe0 can be

- .

For the conditional probability of error Pe1, the

mean is +A

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ =

0

0

2

N

E QP be

( )( )

bT N

A y

b

Y eT N

y f /

0

0

2

/

11|

−−

=π


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The probability of error Pe1 is area from -∞ to λ.

( )

( )

∫

∫

∞−

−−

∞−

=

=

λ

λ

π dye

T N

dy y f

bT N

A y

b

Y

e

/

0

1

0

2

/

1

1|

Setting the threshold to λ=0 and putting

results in Pe1=Pe0

bT N A y z2/0

−−=


The average probability of symbol error Pe is

Since Pe0=Pe1 and p0=p1=0.5

1100 eee P pP pP +=

⎟ ⎞

⎜⎛

=

== 10

2 E QP

PPP

b

e

eee

The average probability of symbol error with

binary signaling only depends on Eb/N0, the ratio

of the transmitted signal energy per bit to the

noise spectral density.

0


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10-1

Matlab Code:

10-4

10-3

10-2

P r o b a b i l i t y

o f E r r o r P

e

clear all;

close all;

EbN0_dB=0:1e-3:10;

EbN0=10.^(EbN0_dB/10);

Pe=0.5*erfc(sqrt(2*EbN0)/sqrt(2));

figure(1);

' '

0 2 4 6 8 1010

-6

10-5

Eb/N

0 (dB)

_ , , ,

xlabel('E_b/N_0 (dB)');

ylabel('Probability of Error P_e');grid;

This waterfall curve makes the digital systems superior compared to analog

systems if Eb/N0 is above few dBs.

INTERSYMBOL INTERFERENCE

Intersymbol Interference (ISI) happens when the

spectrum.

The simplest case is band-limited channel.

For example, a brick-wall band-limited channel

passes all frequencies |f|W.

Consider a olar si nalin where

⎩⎨⎧

−

+=

sentwas0symbol1

sentwas1symbol1k a


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Once the impulse modulator output is passed

response of the pulse g(t), the transmitted signal

becomes

s t asses throu h the channel with im ulse

( ) ( )∑ −=k

bk kT t gat s

response of h(t) added white Gaussian noise and

received by the receiver. The receiver passes the received signal through

the receive filter, which is the matched filter for

the optimum noise performance.


The output of the receive filter is

sampled at t=Tb and the decision device decides

the bits.

The receive filter output is

t nkT t pat yk

bk +−= ∑

( ) ( )[ ] ( )

( )[ ] ( )iik

k

bk i

i

k

bk i

t nT k i paa

t nT k i pat y

+−+=

+−=

∑

∑∞

≠−∞=

∞

−∞=

μ

μ


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The first term is the contribution of the ith

.

The second term represents the residual effect of

all other transmitted bits on the decoding of ith

bit.

This residual effect due to the occurrence of

pulses before and after the sampling instant ti is

.

The last term n(ti) represents the noise sample at

the time ti

When the signal-to-noise ratio is high, the system

is limited by ISI rather than noise.

THE TELEPHONE CHANNEL

The frequency response of a telephone channel is

.


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The properties of this channel

-

3.5KHz. Therefore, we should use a line code with a

narrow spectrum, so we can maximize the data rate.

The channel does not pass DC. It is preferable to use

a line code that has no DC component such as bipolar

signaling or Manchester code.

These two requirements are contradictory.

The polar NRZ has narrow spectrum but has DC

The Manchester has no DC, but requires higher

bandwidth.


For the data rate 1600bps, the signals are shown below.

a is NRZ line code and b is Manchester line code.

Since NRZ requires DC, the signal drifts to zero volt

especially when there is long strings of the symbols of

the same polarity.

Manchester code has no DC drift but has distortion.


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For the data rate 3200bps, the DC drift in NRZ is

distortion.

The Manchester code has significant spectral

components outside the band limits of this

channel.

Documents

EEE306 Digital Communication 2013 Spring Slides 07