Elec3100 2014 Ch7.1 BasebandCommunicationsNoise

8/20/2019 Elec3100 2014 Ch7.1 BasebandCommunicationsNoise

1/42

Source Coding Channel CodingInformation source

and input transducerModulator

Channel

Source Decoding Channel DecodingInformation sink

and output transducer

Demodulator

(Matched Filter)

Ch7.1: Binary

Communications

Ch7.1: Baseband Communications & Noise

• Questions to be answered:

System Model: AWGN Channel

White Gaussian Noise: A Random Process

Suboptimal Receiver: Integrate-and-Dump

Performance Evaluation: How Good is the

Receiver?

Elec3100 Chapter 7.1 1


2/42

System Model

White Gaussian Noise

Suboptimal Receiver

Performance Evaluation

BER

A General Case




3/42


4/42

Baseband Digital Data Transmission

• The system model for baseband digital data

transmission is

(+ A,- A) Receiver Transmitter

n t : PSD 12 N 0

s t

y t

AWGN

Binary signal



5/42

• Example of a digital signal and transmitted waveform is

1 0 0 0 1

• Example of a noise-corrupted received signal is

s t

Baseband Digital Data Transmission



6/42

System Model


Suboptimal Receiver


BER

A General Case




7/42

• In this course, we assume internal noise sources are

much more significant than external ones and deal with

these only.

• Noise internal to a communications system arises as aresult of random motion of charge carriers within the

devices (transistors, resistors, diodes, etc)

composing the system.

• Internal noise is classified as thermal , shot,

flicker or others.

Types of Noise



8/42

• Thermal noise arises from the random motion of chargesin conducting medium (such as resistors) and is the mostsignificant noise source we need to consider in

ELEC3100.

• Essentially, by connecting a very sensitive oscilloscope

across a resistor we can observe thermal noise.

• A typical noise output would look like:

Thermal Noise



9/42

Nyquist Theorem

• The signal is completely random and cannot be

predicted. It has an average value of zero but hasassociated with it a certain mean square voltage.

• The mean square voltage associated with the thermalnoise can be found from the Nyquist Theorem:

where

T = Temperature of the resistor in Kelvink = Boltzmann’s constant 1.38 x 10-23 joule/K

B = Bandwidth of interest

R = Resistance

kTBRvn 42

[K] = [°

C] + 273.15



10/42

Noise can be modeled as a Random Process

• Definition: A random process maps a probability space S to a set of functions, (, ).

• It assigns to every outcome a time function (, ) for ∈ where is a discrete or continuous index set.

• If is discrete (e.g. integer valued), (, ) is a discrete-timerandom process.

• If is continuous, (, ) is a continuous-time randomprocess.

• For a fixed t , (, ) is a random variable.

• Basically, we can understand a random process as a

sequence of random variables.

S



11/42

Random Process: Example• Suppose that is selected at random from S = [0,1] and

consider

time

0

t t t X for)cos(),(

r a n d o m n e s s

31

3

2

1



12/42

Characterization of A RP:

Mean and Variance Functions

Mean

Variance

Note that the mean and variance may be functions of

time. However, since the randomness comes from “”, wetreat “t” as a constant in the above calculations.

( )( ) [ ( )] ( ) X X t m t E X t xf x d x



13/42

Characterization of A RP:

Autocorrelation and Autocovariance

o Autocorrelation

o Autocovariance

o Correlation coefficient

1 21 2 1 2 ( ), ( )( , ) [ ( ) ( )] ( , ) X X t X t R t t E X t X t xyf x y dxdy

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

2121

221121

t mt mt t Rt mt X t mt X E t t C

X X X

X X X

1 21 2

1 1 2 2

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

X X

X X

C t t t t C t t C t t



14/42

Wide Sense Stationary

• Definition: A process () is wide sense stationary

(WSS) if and only if its mean is constant and itsautocorrelation function (1, 2) (or autocovariancefunction) depends only upon the time difference 1 2.

• Suppose () is WSS with autocorrelation () wheret = 1 2.– (0) is the average power of the process, E[ X (t )2].

– () is an even function of t .

– |R X (t )| ≤ R X (0)(the autocorrelation function is maximum at the origin)

1 2 1 2 1 2( ) for all ( , ) ( ) for all , X X X m t m t C t t C t t t t



15/42


16/42

Gaussian Process

The additive noise in a communications system can be

modeled as a Gaussian process (by the central limit

theorem)

- at a particular time t , the noise signal amplitude will be

Gaussian distributed.

- we further assume that the Gaussian process is stationary andhas zero mean:

and its autocorrelation is

ni ,,2,1

)(2

)()()( t t t oii X N

t X t X E R ni ,,2,1

(t i) E X (t i) 0



17/42


18/42

Why White noise?



19/42


20/42

• One of the significant differences between analog anddigital communications systems is that for digital systems,

the probability of error is used as a measure ofperformance where as in analog systems SNR is used.

• We have modeled the AWGN. The next question is how to

build a receiver to obtain a good performance.

• A possible receiver structure (integrate-and-dump) fordetecting the digital transmitted signals is shown below

>0 choose +A


21/42

Integrate-and-Dump

• Not necessarily optimum in all situations.• The integrator averages out the noise received so that the

output waveform will look like



22/42

• We know that the noise at the input to the receiver is AWGN.• We can expect the output from the integrator to have a

Gaussian noise distribution.

• Putting these ideas into a mathematical framework we get

1 Decision V

AWGN

0

dt 0

T

n t ~ S N f N 0

2

s t Gaussian distribution

Linear System

t y

Integrate-and-Dump



23/42

with

s t A A

0 t T if "1" transmitted0 t T if "0" transmitted

V s t n t 0

T

dt

AT N if "1" is sent

AT N if " 0" is sent

N n t dt 0

T

N is Gaussian distributed

Linear combination of Gaussian

Integrate-and-Dump



24/42

System Model


Suboptimal Receiver


BER

A General Case




25/42

Key Figure of Merit

• The probability of receiving a bit in error

for digital systems is an important measure

of performance.

• Digital communications relies heavily on

these error calculations - VIP



26/42

Conditional Error Probability

Prior Probability

P e P (V 01) P 1 P V 00 P 0

Problem

P e P 0 received 1 sent P 1 sent P 1 received 0 P 0 sent

Assume the decision threshold is set at 0

P e P ( E 1) P 1 P E 0 P 0 Total probability theorem



27/42


28/42

• The key to estimating the error probability is to find

out the distribution of the received signal.

• This in turn relies on the distribution of the noise.

• The noise mean can be calculated as

E N E n t dt 0T

E n t 0

T

dt 0

0

Conditional Distribution



29/42

Var N E N 2 E n t dt 0

T

2

E n t n v 0

T

0

T

dtdv

R

n t

v

0

T

0

T

dtdv

N

0

2 t v

0

T

0

T

dtdv

N 02

dv0

T

N 0

T

2 2

AWGN

f

S n(f)

t - n

R(t - n )

N o /2

N o /2

Noise Variance



30/42

AT

f V v1

n

E V 1 AT Var V 1 2

f V v1 ~ N AT , 2

0

-AT

f V v 0

n

E V 0 AT Var V 0

2

f

V

v 0

~ N AT , 2

0

Shifted positive due to

The positive pulse +A

Conditional Probabilities



32/42

• Thus, we know that the output noise from theintegrator will have the following Gaussiandistribution

• Now,

N ~ N 0, 2

f n n 12 2exp n

2

2 2

P E 0 P V 00 sent f V v 0 dv0

P E 1 P V 01 sent f V v1 dv

0

Conditional Error Probability



33/42

• Assumptions:1. Symmetry,

2. “0” and “1” are equally likely, then

P E 0 P E 1

P e P E 0 P E 1 P 0 P 1 1

2

P e P E 0 f V v 0 dv0

12

e v AT

2

2 2 dv0

• We therefore have

P e P ( E 1) P 1 P E 0 P 0 12

P E 0 P E 0 P E 0

Noise distribution

is the same at “1”

and “0”

Total Error Probability

33


34/42

• Let

• where

x v AT

dx dv

P e

1

2 e

x2

2

dx AT

Q AT

Q( x) 1

2 e t

2

2 dt x

Q(.) function

VIP Transformation

Evaluation Relies on Q-Function



35/42

System Model


Suboptimal Receiver


BER

A General Case




36/42

• Next, note that one can represent as (“0”

sent) or (“1” sent).

s t

s0 t

A

t

T-A

tT

s1 t

s0 t

E 1 s12 t dt

0

T

A2T E 0 s02 t dt 0T

A2T

Energy

bit “1” bit “0”

s1 t

A General Case



37/42

37

• We can use these energy calculations to find a more

general result for the error probabilities. That is,

where

10

10

2

1

sent"1"sent"0"Energy/bit

E E

P E P E E b

But E 1 E 0 A2T .

P e Q AT

Q A2T 2

2

2

N 0T

2

A General Result

Elec3100 Chapter 7.1


38/42

• Therefore,

where

with

uu erf 1erfc

ut dt eu

0

22erf

P e Q A2

T

2

2

Q A

2

T 2

T

Q 2 E b

N 0

1

2erfc

E b N

0

Complementary

Error Function

Error Function

Signal Energy to

Noise Power Spectral

Density Ratio

Relation with SNR?

SNR vs. Eb/No

38


39/42

• A graph of P e for baseband signaling is

where

10-2

1.0

10-1

10-3

10-4

Actual

P e

10log10Z

-10 5 0 5 10

P e Q 2 Z 12

erfc Z Z E b N

0

Bit Error Rate (BER)



40/42

Example 7.1 A baseband digital Tx system sends A valued

rectangular pulses through a channel at a rate of1Mbps with amplitude 1V when the noise PSD is 10-7

W/Hz.

Answer :Q 2 E b / N 0

Q 2 A2T / N 0

T 1/1000000 106 Q 2 106 /(2 107)

Q( 10) Q(3.16)

Q(u) eu

2 / 2

u 2

Q(3.16) 0.00085

Probability of Error, Pe

40


41/42

Example 7.2

Digital data is to be transmitted through a baseband system

with N 0=10

-7

W/Hz and the received signal amplitude A=20mV.

(a) If 1000 bits per second (bps) are transmitted what is the error

probability? Ans. P e = 2.58 . 10-3.

(b) If 10000 bps are transmitted, to what value must A be adjusted

in order to attain the same error probability as in part a)?

Ans. A = 63.2 . 10-3 V = 63.2mV.



42/42

P e Q A2

T 2

2

Q A2

T

2

T

Q 2 E b N o

• In the last example, for a fixed amplitude, error probability

increases as bit rate increases.• This can be understood by looking at the error probability

expression

• Increase in bit rate smaller bit period T

higher noise power lower signal energy

N

AQ P

oe

22

Are they happening at the same time?

BER vs. Data Rate

Elec3100 2014 Ch7.1 BasebandCommunicationsNoise

Documents

Transcript of Elec3100 2014 Ch7.1 BasebandCommunicationsNoise