Chapter 3: Data Transmission

COE 341: Data & Computer Communications (T071)Dr. Radwan E. Abdel-Aal

2

Remaining Six Chapters:

Physical Layer

Transmission Medium

Data Link

Chapter 4: Transmission Media

Chapter 3: Signals, their representations, their

transmission over media, Resulting impairments

Chapter 5: Encoding: From data to signals

Chapter 7: Data Link: Flow and Error control,

Link management

Chapter 6: Data Communication: Synchronization,

Error detection and correction

Chapter 8: Improved utilization: Multiplexing

3

Agenda Concepts & Terminology Signal representation:

Time and Frequency domains Bandwidth and data rate Decibels and Signal Strength (Appendix 3A ) Fourier Analysis (Appendix B ) Analog & Digital Data Transmission Transmission Impairments Channel Capacity

4

Terminology (1)

Transmission system: Components Transmitter Receiver Medium

Guided media e.g. twisted pair, coaxial cable, optical fiber

Unguided media e.g. air, water, vacuum

5

Terminology (2)

Link Configurations: Direct link

No intermediate ‘communication’ devices (these exclude repeaters/amplifiers)

Two types: Point-to-point

Only 2 devices share link Multi-point

More than two devices share the same link, e.g. Ethernet bus segment

Amplifier

A

B

C

6

Terminology (3)Transmission Types (ANSI Definitions) Simplex

Information flows in one direction only all the timee.g. Television, Radio broadcasting

Duplex Information flows in both directions Two types:

Half duplex Only one direction at a time e.g. Walki-Talki

Full duplex In both directions at the same time e.g. telephone

7

Frequency, Spectrum and Bandwidth Time domain concepts Analog signal

Varies in a smooth, continuous way in both time and amplitude

Digital signal Maintains a constant level for sometime and then changes to

another constant level (i.e. amplitude takes only a finite number of discrete levels)

Periodic signal Same pattern repeated over time

Aperiodic signal Pattern not repeated over time

8

Analogue & Digital Signals

Only a few amplitude levels allowed

- Binary signal: 2 levels

All values on the time and amplitude axes are allowed

9

PeriodicSignals

S (t+nT) = S (t); 0 t TWhere:t is time over first periodT is the waveform periodn is an integer

T

Temporal Period

t t+2Tt+1T

Signal behavior over one perioddescribes behavior at all times

…

10

Aperiodic (non periodic) Signals in time

s(t)

0

1

+ X/2- X/2 t

11

Continuous and Discrete RepresentationsAvailability of the signal over the horizontal axis

Continuous:

Signal is defined at all points on the horizontal axis

Discrete:

Signal is defined Only at certain points on the horizontal axis

Sampling with a train of delta function

(Time or Frequency)

12

Sine Wave s(t) = A sin(2f t +)

Peak Amplitude (A) Peak strength of signal, volts

Repetition Frequency (f) Measures how fast the signal varies with time Number of waveform cycles per second (Hz) f = 1/ T(xx sec/cycle) = yy cycles/sec = yy Hz

Angular Frequency () = radians per second = 2 f = 2 /T

Temporal (time) Period, T = 1/f Phase ()

Determines relative position in time, radians (how to calculate?)

T (Period)

A (Amplitude)= A sin ()

13

Varying one of the three parameters of a sine wave carriers(t) = A sin(2ft +) = A sin(t+)

Varying A

Varying

Varying f

Can be used to convey information…!

M o d u l a t I o n

FMPM

AM

14

Sine Wave Traveling in the +ive x directions(t) = A sin (k x - t]

Direction of wave travel, at velocity v

k = Wave Number= 2 /

= Angular Frequency= 2 f = 2 / T

x

Distance, x

t = 0

t = t

Spatial Period = Wavelength

xFor point p on the wave:

Total phase at t = 0: kx - (0) = kx

Total phase at t = t: k(x+ x) - (t)

Same total phase, kx = k(x+ x) - (t)k x = t

Wave propagation velocity v = x / t v = /k = /T = f

v = f

p

Show that the wave s(t) = A sin (k x + t]travels in negative x direction

V is constant for a given wave type and medium

+ ive x direction

15

Wave Propagation Velocity, v m/s Constant for:

A given wave type (e.g. electromagnetic, seismic, ultrasound, ..) and a given propagation medium (air, water, optical fiber)

For all types of waves: v = f

For a given wave type and medium (given v): higher frequencies correspond to shorter wavelengths and vise versa:Electromagnetic waves:

long wave radio (km), short wave radio (m), microwave (cm)… light (nm) For electromagnetic waves:

In free space, v speed of light in vacuum v = c = 3x108 m/sec Over other guided media (coaxial cable, optical fiber, twisted

pairs): v is lower than c

Shorter wavelength .. Higher frequency

16

Wavelength, meters) Is the Spatial period of the wave:

i.e. distance between two points in space on the wave propagation path where the wave has the same total phase

Also: Distance traveled by the wave during one temporal (time) cycle:

dT = v T = (f) T =

17

Frequency Domain Concepts Response of systems to a sine waves is easy to analyze But signals we deal with in practice are not all sine

waves, e.g. Square waves Can we relate waves we deal with in practice to sine

waves? YES! Fourier analysis shows that any signal can be treated as

the sum of many sine wave components having different frequencies, amplitudes, and phases (Fourier Analysis: Appendix B)

This forms the basis for frequency domain analysis For a linear system, its response to a complex signal will

be the sum of its response to the individual sine wave components of the signal.

Dealing with functions in the frequency domain is simpler than in the time domain

18

Addition of Twofrequency Components

A = 1*(4/)frequency = f

A = (1/3)*(4/)frequency = 3f

+

=

Fundamental

3rd harmonic

Approaching a square wave

Frequency Domain: S(f) vs f Time Domain: s(t) vs t

Discrete Function in f

t3

Periodic function in t

Fourier Series

f

Frequency Spectrum

1/3 rd the Amplitude3 times the frequency

Fourier Series

19

Asymptotically approaching a square wave by combining the fundamental + an infinite number of odd harmonics at prescribed amplitudes

-1.5

-1

-0.5

0

0.5

1

1.5

0 2 4 6 8 10t

sq

ua

re s

ign

al,

sw

(t)

What is the highestHarmonic added?

Topic for a programming assignmentAdding more

higherharmonics

20

More Frequency Domain Representations: A single square pulse (Aperiodic signal)

Time Domain: s(t) vs tFrequency Domain: S(f) vs f

• What happens to the spectrum as the pulse gets broader … DC ?

• What happens to the spectrum as the pulse gets narrower … spike ?

Continuous Function in f Aperiodic function in t

Fourier Transform

0

1

+ X/2- X/2t

s(t)

timefrequency

1/X

Sinc(f) = sin(f)/f

To

To To

Fourier Transform

21

Spectrum & Bandwidth of a signal Spectrum of a signal

Range of frequencies contained in a signal Absolute (theoretical) Bandwidth (BW)

Is the width of spectrum = fmax- fmin But in many situations, fmax = !

(e.g. a square wave), so: Effective Bandwidth

Often called bandwidth Narrow band of frequencies containing most of

the signal energy Somewhat arbitrary: what is “most”?

f 3f 5f 7f f

S(f)

….

22

Signals with a DC Component

t

1V DC Level

1V DC Component

t

+

+

+

_

NO DC Component, Signal average over a period = 0

DC Component Component at zero frequency

Determines if fmin = 0 or not

23

Bandwidth for these signals:fmin fmax Absolute

BWEffective

BW

1f 3f 2f 2f

0 3f 3f 3f

0 1/X ?

= (fmax- fmin)

24

Bandwidth of a transmission system

Is the Range of signal frequencies that are adequately passed by the system

Effectively, the transmission system (TX, medium, RX) acts as a filter Poor transmission media, e.g. twisted pairs, have

a narrow filter bandwidth This cuts off higher frequency signal components

poor signal quality at receiver And limits the signal frequencies (Hz) that can be

used for transmission

limits the data rates used (bps)

f 3f 5f 7f f

S(f)

….

25

Limiting Effect of System Bandwidth

)2sin(k

14)(

1 k odd,

kfttsk

f 3f

f 3f 5f

f 3f 5f 7f

f 3f 5f 7f ……

BW = 2f

BW = 4f

BW = 6f

BW =

…

Better reception requires larger B

W

Mo

re d

iffic

ult

rece

ptio

n w

ith s

ma

ller

BW

1,3

1,3,5

1,3,5,7

1,3,5,7 ,9,…

1

2

3

4Fourier Series for a Square Wave

Var

ying

Sys

tem

BW

Received Waveform

26

System Bandwidth and Achievable Data Rates Any transmission system supports only a limited range of frequencies (bandwidth) for satisfactory transmission

For example, this bandwidth is largest for expensive optical fibers and smallest for cheap twisted pair wires

So, bandwidth is money Economize in its use Limited system bandwidth degrades higher frequency

components of the signal transmitted poorer received waveforms more difficult to interpret the signal at the receiver (especially with noise) Data Errors

More degradation occurs when higher data rates are used (signal will have more components at higher frequency )

This puts a limit on the data rate that can be used with a given signal to noise requirement, receiver type, and a specified error performance Channel capacity issues

27

Bandwidth and Data Rates

2B = 4f’

B = 4f

f’ 3f’

f 3f 5f

Period T = 1/f

1 0 1 0

Data Element =Signal Element

T/2Data rate = 1/(T/2) = (2/T) bits per sec = 2f bps

Given a bandwidth B,Data rate = 2f = B/2

To double the data rate you need to double f: Two ways to do this…

1. Double the bandwidth with same received waveform (same RX conditions & error rate)

f’ 3f’ 5f’

New bandwidth: 2B,Data rate = 2f’ = 2(2f)= 4f = B

2. Same bandwidth, B, but tolerate poorer received waveform (needs better receiver, higher S/N ratio, or tolerating more errors in data)

B = 2f’

Bandwidth: B,Data rate = 2f’ = 2(2f) = 4f = B

1 1 1 10 0 0 0

1 1 1 10 0 0 0

B

2B

B

Data

5f’

1 1 1 10 0 0 0X 2

X 2

28

Bandwidth & Data Rates: Tradeoffs… Compromises Increasing the data rate (bps) while keeping BW the same (to economize) means working with inferior (poorer) waveforms at the receiver, which may require: Ensuring higher signal to noise ratio at RX (larger signal

relative to noise): Shorter link distances Use of more en-route repeaters/amplifiers Better shielding of cables to reduce noise, etc.

More sensitive (& costly!) receiver Suffering from higher bit error rates

Tolerate them? Add more efficient means for error detection and correction- this

also increases overhead!.

29

Appendix 3A: Decibels and Signal Strength The decibel notation (dB) is a logarithmic measure of the ratio

between two signal power levels NdB = number of decibels P1 = input power level (Watts) P2 = output power level (Watts)

e.g. Amplifier gain Signal loss over a link

Example: A signal with power level of 10mW is inserted into a transmission line Measured power some distance away is 5mW Power loss in dBs is expressed as

NdB =10 log (5/10)=10(-0.3)= -3 dB

- ive dBs: P2 < P1 (Loss), +ive dBs: P2 > P1 (Gain)

1

210log10

P

PNdB

Amplifier

P1 P2 P3

Lossy Link

30

Relationship Between dB Values and Power ratio (P2/P1)Power RatioPower Ratio dBdB Power RatioPower Ratio dBdB

1 0

101 10 10-1 -10

102 20 10-2 -20

103 30 10-3 -30

104 40 10-4 -40

105 50 10-5 -50

106 60 10-6 -60

2 3 1/2 -3

31

Decibels and Signal Strength

Decibel notation is a relative, not absolute, measure: A loss of 3 dB halves the power (could be 100 to 50, 16 to 8, …) A gain of 3 dB doubles the power (could be 5 to 10, 7.5 to 15, …)

Will see shortly how we can handle absolute levels Advantage:

The “log” allows replacing: Multiplication with Addition

C = A * B

Log C = Log A + Log B and Division with Subtraction

A = C / B

Log A = Log C - Log B

32

Decibels and Signal Strength

]4

PowerSignal Received[log1013 10 mW

Gain: 35 dB4 mW

Loss: 12 dB Loss: 10 dBTransmitted Signal

Amplifier

ReceivedSignal

?

Net power gain over transmission path:+ 35 – 12 – 10 = + 13 dB (+ ive means there is net gain)

Received signal power = (4 mW) log10-1(13/10) = 4 x 101.3

= 4 x 101.3 mW = 79.8 mW

Example: Transmission line with an intermediate amplifier

]10

13[log]

4

PowerSignal Received[ 10

1mW

Still we use some multiplication!

33

How to represent absolute power levels?Decibel-Watt (dBW) and Decibel-mW (dBm)

As a ratio relative to a fixed reference power level With 1 W used as a reference dBW

With 1 mW used as a reference dBm

Examples: Power of 1000 W is 30 dBW, 1 W = ? dBW –10 dBm represents a power of 0.1 mW, 1 mW = ? dBm X dBW = (X + ?) dBm

W

PowerLevelPower W

dBW 1log10_ 10

mW

PowerLevelPower mW

dBm 1log10_ 10

Caution!: Must be same units attop and bottom

Caution!: Must be same units attop and bottom

WK 4

34

dBs & dBms are added algebraically

mW

mW

P

P

1

2

GP1 P2

G = power ratio =

G dBs = 10 log10 G =

dBmdBm

mWmW

mWmW

mW

mW

mW

mW

PP

mWPmWP

mWPmWP

mWP

mWP

P

P

12

)]1/1([log10)]1/2(log[ 10

)]1/1(log)1/2(log[ 10

1/1

1/2log10

1

2log10

1010

1010

1010

Similarly for dBs & dBWs

G is Positive for gain Negative for attenuation

35

Decibels and Signal Strength Example: Transmission line with an intermediate amplifier

Gain: 35 dB4 mW

Loss: 12 dB Loss: 10 dBTransmitted Signal

Amplifier

ReceivedSignal

?

Net power gain over transmission path:+ 35 – 12 – 10 = + 13 dB (+ ive means actual net gain)

RX signal power (dBm) = 6.02 + 13 =19.02 dBm Check: 19.02 dBm = 10 log (RX signal in mW/1 mW)

RX signal = log-1 (19.02/10) = 79.8 mW

TX Signal Power in dBm = 4 mW = 10 log (4/1) = 6.02 dBm

• If all ratios are in dBs and all levels are in dBm solve by algebraic addition Same for {dBs and dBWs} (No need for any multiplication/division)

As inSlide 31

36

Decibels & Voltage ratios Power decibels can also be expressed in terms

of voltage ratios Power P = V2/R, assuming same R

Relative:

Absolute: dBV and dBmV Decibel-millivolt (dBmV) is an absolute unit,

with 0 dBmV being equivalent to 1mV. Also dBV

1

22

1

22

1

2 log20/

/log10log10

V

V

RV

RV

P

PNdB Note that this is still a power ratio…

But expressed in terms of voltages

mV

VoltageN mV

dBmV 1log20 Caution!: Must be same units at

top and bottom

37

Appendix B: Fourier AnalysisSignals in Time

Periodic Aperiodic

Discrete Continuous Discrete Continuous

DFS FS FTDFT

FS : Fourier SeriesDFS : Discrete Fourier SeriesFT : Fourier TransformDFT : Discrete Fourier Transform

Use Fourier TransformUse Fourier Series

…

38

Fourier Series for periodic continuous signals Any periodic signal x(t) of period T and repetition

frequency f0 (f0 = 1/T) can be represented as an infinite sum of sinusoids of different frequencies and amplitudes – its Fourier Series. Expressed in Two forms:

1. The sine/cosine form:

1

000 )2sin()2cos(

2)(

nnn tnfBtnfA

Atx

T

dttxT

A0

0 )(2

T

n dttnftxT

A0

0 )2cos()(2

T

n dttnftxT

B0

0 )2sin()(2

If A0 is not 0,x(t) has a DC component

DC Component

f0 = fundamental frequency = 1/T Where:

Two components at each frequency

Frequencies are multiples

of the fundamentalfrequency f0

= f(n)

= f’(n)

All integrals overone period only

39

Fourier Series: 2. The Amplitude-Phase form: Previous form had two components at each frequency (sine, cosine i.e. in quadrature) : An, Bn coefficients

The equivalent Amplitude-Phase representation has only one component at each frequency: Cn, n

Derived from the previous form using trigonometry:

cos (a) cos (b) - sin (a) sin (b) = cos [a +b]

1

00 )2cos(

2)(

nnn tnfC

Ctx

00 AC 22nnn BAC

n

nn A

B1tan

Now components have different amplitudes, frequencies, and phases

Now we have Only one component at each frequency nf0

The C’s and ’s are obtained from the previous A’s and B’s using the equations:

40

Fourier Series: General Observations

FunctionFunction SeriesSeries

No DC A0 = 0

Even Function x(t) = x(-t)

Symmetric about Y axis

Bn = 0;

for all n

Odd Function x(t) = - x(-t)

Symmetric about the origin

An = 0;

for all n

1

000 )2sin()2cos(

2)(

nnn tnfBtnfA

Atx

0)(1

0

T

dttxT

DC Even Function Odd FunctionFunction

Fourier Series Expansion

41

Correction

42

Fourier Series Example

1

-1

1/2-1/2 1 3/2-3/2 -1 2

T

0}][]{[2

1212)(2)(2

2)(

2

12/1

2/10

1

2/1

2/1

0

1

0

2

00

0

tt

dtdtdttxdttxdttxT

AT

x(t)

Note: (1) x(– t)=x(t) x(t) is an even function(2) f0 = 1 / T = ½ Hz

Note: A0 by definition is 2 x the DC content

43

1

0

0

2/

0

0

0

0 )2cos()(2)2cos()(4

)2cos()(2

dttnftxdttnftxT

dttnftxT

ATT

n

2/

2/

0

0

0 )2sin()(2

)2sin()(2 T

T

T

n dttnftxT

dttnftxT

B

2/

0

0

0

2/

0 )2sin()(2

)2sin()(2 T

T

dttnftxT

dttnftxT

0)2sin()(2

)()2sin()(2 2/

0

0

2/

0

0 TT

dttnftxT

tdtnftxT

Replace t by –t Swap limitsin the first integral

Contd…

x(t), since x(t) is an even function

= 0 for n even

= (4/n) sin (n/2) for n odd

1

-1

1/2-1/2 1 3/2-3/2 -1 2

T

a function of n only

f0 =1/2

- sin(2nf t) dt Then Bn = 0 for all n

44

1

000 )2sin()2cos(

2)(

nnn tnfBtnfA

Atx

tnn

ntx

oddn

cos2

sin4

)(,1

4 4 4 4( ) cos cos3 cos5 cos 7 ...

3 5 7x t t t t t

4 1 1 1( ) cos cos3 cos5 cos 7 ...

3 5 7x t t t t t

Contd…

A0 = 0, Bn = 0 for all n,

An = 0 for n even: 2, 4, … = (4/n) sin (n/2) for n odd: 1, 3, …

Cosine is an even function

Amplitudes, n odd

Original x(t) is an even function!

f0 = ½, so 2 f0 =

2 3 (1/2) t

3rd Harmonic

2 (1/2) t

Fundamentalf0 = ½

45

Another Example

1

-1

1-1 2

T

-2

x1(t)

Note that x1(-t)= -x1(t) so, x(t) is an odd function

Also, x1(t)=x(t-1/2)

2

17 cos

7

1

2

15 cos

5

1

2

13 cos

3

1

2

1 cos

4)(1 tttttx

This waveform is the previous waveform shifted right by 1/2

Previous Example

46

Another Example, Contd…

2

77 cos

7

1

2

55 cos

5

1

2

33 cos

3

1

2 cos

4)(1

tttttx

...7in

7

1 5sin

5

1 3in

3

1 in

4)(1 tsttststx

tt sin2

cos

tt 3sin

2

33 cos

tt 5sin2

55 cos

tt 7sin

2

77 cos

Because:

Sine is an odd function

)2sin(k

14)(1 0

1 k odd,

tkftxk

As given before for the square wave on slide 25.

47

Fourier Transform For aperiodic (non-periodic) signals in time, the spectrum

consists of a continuum of frequencies (not discrete components) This spectrum is defined by the Fourier Transform For a signal x(t) and a corresponding spectrum X(f), the

following relations hold

sincos je j

dfefXtx ftj 2 )()(

dtetxfX ftj 2 )()(

Forward FT (from time to frequency) Inverse FT (from frequency to time )

X(f) is always complex (Has both real & Imaginary parts), even for x(t) real.

sincos je j

je

Real

Imaginary

nf0 f 1T/2

Express sin and cos je

48

Sinc function

Sinc2 function

(non-periodic in time)

(Continuous in Frequency)

49

Fourier Transform Example

x(t)A

22

dtetxfX ftj 2 )()(

2/

2/

22/

2/

2

2 )(

ftjftj efj

AdteAfX

f

fA

f

ff

f

A

j

ee

f

A fjfj )sin()sin(

12

2/22/2

j

ee jj

2sin

2cos

jj ee

Sin (x) / xi.e. “sinc” function

Area of pulseIn time domain

50

Fourier Transform Example, contd.

f

fAfX

)sin()(

A

2/

= A

f

Study the effect of the pulse width

Sin (x) / x“sinc” functionLim x0 (sin x)/x = (cos x)x=0/1 =1

First zero in the Frequency spectrum: sin f = 0ff =

51

The narrower a function is in one domain, the wider its transform is in the other domain

The Extreme Cases

0

52

Power Spectral Density (PSD) & Bandwidth

Absolute bandwidth of any time-limited signal is infinite But luckily, most of the signal power will be

concentrated in a finite band of lower frequencies Power spectral density (PSD) describes the

distribution of the power content of a signal as a function of frequency

Effective bandwidth is the width of the spectrum portion containing most of the total signal power

We estimate the total signal power in the time domain

53

Signal Power in the time domain Signal is specified as a function s(t) representing signal voltage or

current Assuming resistance R = 1

Instantaneous signal power (t) = v(t)2/1= i(t)2*1 = |s(t)|2

Signal power can be obtained as the average of the instantaneous signal power over a given interval of time = constant

For periodic signals, this averaging is taken over one period, i.e.

This measure in the time domain gives the total signal power Effective BW is then determined such that it contains a specified

portion (percentage) of this total signal power

22

1

1( )

2 1

t

t

x t dtt t

T

Total dttsT

P0

2)(

1

TotalP (1)s

54

Signal Power in the Frequency Domain: Periodic signals For periodic signal we have a discrete spectrum (the F Series):

For a DC component, Power = Vdc2

For AC components Power = Vrms2= Vpeak

2 (use eqn. 1 on prev. slide)

Power spectral density (PSD) is a discrete function of frequency:

Where (f) is the Dirac delta function: Total signal power (watts) up to the j th harmonic is:

2

1

j

nnjUpto CCP

1

220Component th the 2

1

4

1

1

00 )2cos(

2)(

nnn tnfC

Ctx

220

1

1( ) ( )

4 2 n on

CPSD f C f nf

1 =00 0( ) f

ff

1

0 f0 2f0 3f0

…

(f-nf0)

f

(A function of frequency)

(A quantity, summation of PSD components- not a function of a frequency)

55

Example

Consider the following signal

The PSD is: (A function of Frequency)

The signal power is: (A quantity)

7in

7

1 5sin

5

1 3in

3

1 in1)( tsttststx

watt586.0 49

1

25

1

9

11

2

1] 7,5,3,1

harmonicsthPower

)]5.3(]7

1[)5.2(]

5

1[)5.1(]

3

1[)5.0(1[

2

1)( 2222 fffffPSD

(No DC)

56

Continuous (not discrete) frequency spectrum PSD (Power spectrum density) function, in Watts/Hz,

is a continuous function of frequency: S(f), Total signal power contained in the frequency band

f1< f < f2 (in Watts) is given by:

(Integration, instead of summation, over frequency)

2

1

)(2f

f

dffSP

Signal Power in the Frequency Domain: Aperiodic signals

Components exist in both negative and positive frequencies

Watts/Hz

57

Complete Fourier Analysis Example Consider the half-wave rectified cosine signal, Figure B.1 on page 793:

1. Write a mathematical expression for s(t) over its period T

2. Compute the Fourier series for s(t) (Amplitude & Phase form)

3. Get an expression for the power spectral density function for s(t)

4. Find the total power of s(t) from the time domain

5. Find the order of the highest harmonic n such that the Fourier series for s(t) contains at least 95% of the total signal power

6. Determine the corresponding effective bandwidth for the signal

58

Example (Cont.)

1. Mathematical expression for s(t):

cos(2 ) , -T/4 T/40 , T/4 3T/4( ) oA f t t

ts t

-T/4-3T/4 +3T/4+T/4

T/2

Where f0 is the fundamental frequency, f0 = (1/T)

59

Example (Cont.)2. Fourier series

Before we start… what to expect?

-T/4-3T/4 +3T/4+T/4

• DC Component?• Even or odd function?• A0 ?• An ?• Bn ?

1

000 )2sin()2cos(

2)(

nnn tnfBtnfA

Ats

Sine/cosine form of the Fourier Series

To get to the amplitude-phase form of the Fourier series, we must first obtain the sine-cosine form

60

Example (Cont.)Fourier Analysis:

1 )2/sin( as , 2

)2/sin(2)2/sin()2/sin(

)2/sin()2/sin(/2

)/2sin(2

)2cos(2

)(2

4/

4/

4/

4/

4/

4/

0

A

AA

A

T

Tt

T

A

dttfT

Adtts

TA

Tt

Tt

T

T

o

T

T

T

dttxT

A0

0 )(2

-T/4-3T/4 +3T/4+T/4

f0 = (1/T)

DC = ?

61

Example (Cont.)

2. Fourier Analysis (cont.):

/ 4 / 4

/ 4 / 4

/ 4

/ 4

2 2( )cos(2 ) cos(2 )cos(2 )

sin(2 ( 1) ) sin(2 ( 1) )2 , for 1

4 ( 1) 4 ( 1)

cos( / 2) cos( / 2) , for

( 1) ( 1)

T T

n o o o

T T

T

o o

o o T

AA s t nf t dt f t nf t dt

T T

n f t n f tAn

T n f n f

A n nn

n n

1

2

sin( ) sin( ) cos( )cos( ) , and

2( ) 2( )

sin( ) cos(

Note:

)

ax bx ax bxax bx dx

a b a b

x x

T

n dttnftxT

A0

0 )2cos()(2

f0 = (1/T)

n = 1 will be treatedSeparately later

From integral tables

62

Example (Cont.)Fourier Analysis (cont.):

n 1

2 2

2 2

2

( ) ( )

( ) ( )

( )

2

0 , for and 1

( 1) ( 1)

( 1) ( 1)

( 1) ( 1) ( 1)( 1) ( 1)

( 1)( 1)

( 1) ( 1) ( 1)( 1)

( 1)

oddn n

n n

n

n

n

A n n

AA

n n

A n n

n n

An n

n

2(1 )

2

2 ( 1) , for

( 1)even

n

An

n

, for n even

63

Example (Cont.)Fourier Analysis (cont.):For n = 1, A1 is obtained separately

Note: cos2 = ½(1 + cos 2)

64

Example (Cont.)

Fourier Analysis (cont.):

/ 4 / 4

/ 4 / 4

/ 4

/ 4

2 2( )sin(2 ) cos(2 )sin(2 )

cos(2 ( 1) ) cos(2 ( 1) )2 , for 1

4 ( 1) 4 ( 1)

0

T T

n o o o

T T

T

o o

o o T

AB s t nf t dt f t nf t dt

T T

n f t n f tAn

T n f n f

, for 1n

cos( ) cos( ) sin( )cos(Note

): )

2( 2( )

ax bx ax bxax bx dx

a b a b

T

n dttnftxT

B0

0 )2sin()(2

-

65

Example (Cont.)Fourier Analysis (cont.):For n = 1, B1 is obtained separately

/ 4 / 4

1

/ 4 / 4

/ 4

/ 4

/ 4

/ 4

2 2( )sin(2 1 ) cos(2 )sin(2 )

sin(4 )

cos(4 ) cos( ) cos( )4 4

0

T T

n o o o

T T

T

o

T

T

o T

AB s t f t dt f t f t dt

T T

Af t dt

T

A Af t

i.e. Bn = 0 for all n (our function is even!)

66

Example (Cont.)Fourier Analysis (cont.):

00 AC n 0 ince ,22 allforBsABAC nnnnn

Note: n are not required for PSD and power calculations

67

Example (Cont.)

3. Power Spectral Density function (PSD):

220

1

2 2 2

2 2 2 22,4,6,...

1( ) ( )

4 2

( )2 ( ) ( )

8 ( 1)

n on

oo

n

CPSD f C f nf

f nfA A Af f f

n

n = 1n = 0 (DC) n = Even

1

00 )2cos(

2)(

nnn tnfC

Ctx

For large n, power decays (1/n4)… Good or bad?

68

Example (Cont.)4. Total Power:

(From the time domain)3 / 4 / 42

2 2

/ 4 / 4

/ 42

/ 4

2

1( ) cos (2 )

sin(4 )

2 8

4

T T

s o

T T

T

o

o T

AP s t dt f t dt

T T

f tA t

T f t

A

Note: cos2 = ½(1 + cos 2)

= Half the power of a full

sine wave

= 0.25 A2

-T/4-3T/4 +3T/4+T/4

Zero

69

Example (Cont.)5. Finding n such that we get at least 95% of the

total power:

2 2 220

0 2 2

2

2

For

40.1014

4 4

0.1014% 40.5%

0.25

0

n

C A APSD A

APower

A

n

(Only the DC component)

Power

% of total power in this component

70

Example (Cont.)Finding n such that we get at least 95% of the

total power, contd.:

2 2 2 220 1

1 2

2

2

For

0.2264 2 8

0.226% 90.5%

0.

1

25

n

C C A APSD A

APower

n

A

(DC + first harmonic)

Power

% of total power in these two components

71

Example (Cont.)Finding n such that we get at least 95% of the

total power, Contd.:

2 2 2 2 2 220 1 2

2 2 2

2

2

For

20.2485

4 2 2 8 9

0.2485% 99.41

2

2

0. 5%

n

C C C A A APSD A

AP wer

A

n

o

(DC + first harmonic + second harmonic)

n = 2, and

6. the effective bandwidth is: Beff = fmax – fmin

Beff = 2f0 – 0 = 2f00 f0 2f0 3f0

…

f

Beff

DC

Power

OK! 95%

72

Bandwidth about a Center Frequency So far we have considered signals in their base band form (without modulation)

Data is often sent as variations in a high frequency carrier signal having a frequency fc (modulation)

So, bandwidth (BW) of this signal occupies a range of frequencies centered about fc

The larger fc, the larger the BW obtainable

Largest BW obtainable for a given center frequency fc is 2

fc

With Amplitude Modulation, For each component of the modulating signal:

0 fc f

BW

])(2cos[])(2cos[)2cos()2cos(2 tfftfftftf mcmcmc

CarrierModulating

Signal

Carrier

73

Analog and Digital Data Transmission Data

Entities that convey meaning Signals

Electric or electromagnetic representations of data

Data Transmission Communication of data

through propagation and processing of signals that represent them

WK5

74

Data types in nature: Analog and Digital Data Analog Data Continuous values within some interval Examples: audio, video Typical bandwidths:

Speech: 100Hz to 7kHz Voice over telephone: 300Hz to 3400Hz Video: 4MHz

Digital Data Discrete values (not necessarily binary) Examples: integers, text characters, mixture:

2347, “text”, SDR054

75

Analog and Digital Signals

Means by which data get transmitted over various media, e.g. wire, fiber optic, space

Analog signal: Continuously variable in time and amplitude

Digital signal: Uses a few (two or more) DC levels

76

Analog Signal Example 1: Speech Data

Frequency range for human hearing: 20Hz-20kHz Almost fully utilized by music Human speech: 100Hz-7kHz Telephone voice channel: Spectrum is further limited to 300-

3400Hz (why?) Mechanical sound waves (data) are easily converted into

electromagnetic signal for processing and transmission: Mechanical waves (Sound) of varying pitch and loudness (Data)

is represented as:

Electromagnetic signals of different frequencies and amplitudes (Signal)

77

Analog Example: 1. The Acoustic Spectrum

Frequency Range Dyn

amic

Ran

ge o

f S

igna

l Pow

er

Source Data

Hearing Spectrum

Log Scale

SPEECH

Dynamic range of the human ear can be as high as 120 dBs!

-70

dBs

78

Conventional Telephony: Analog data – Analog Signal

Telephone mouthpiece converts mechanical voice analog data into electromagnetic analog electrical signal

Signal travels on telephone lines At receiver, speaker re-converts received electrical

signal to voice

79

Analog Signal Example 2. Video Data Electrical signal proportional to the brightness of

image spot on a raster-scanned phosphor screen

Interlaced Scan

Line Scan

Frame Scan

52.5 s (Active)11 s

80

Bandwidth of a Black & White Video Signal USA Specification: 525 lines per frame scanned at the rate

of 30 frames per second 525 lines = 483 active scan lines + 42 lost during vertical retrace

So 525 lines x 30 frames/second = 15750 lines per second Line scan interval = 1/15750 = 63.5s 11s go for horizontal retrace, so 52.5 s for active video per line

Effective vertical resolution = 0.7 x 483 = 338 lines Horizontal resolution = 338 x aspect ratio

= 338 x (4/3) = 450 dots Max frequency is when black and white dots alternate 450 picture dots correspond to 225 cycles in 52.5 s

Time period = 52.5/225 s fmax = 1/Period = 4.2 MHz

fmin (DC) = 0 Bandwidth = fmax - fmin = 4.2 MHz

81

Digital Signals

Advantages: Cheaper and easier to generate: No extra processing

needed Less susceptible to noise

(The threshold effect)

Disadvantages: When noise is above threshold Total data reversal

(Bit error) (1 0, 0 1) Greater attenuation

Line capacitances make pulses rounded and smaller in amplitude, leading to loss of information

More so at higher data rates and longer distances So, use at low data rates over short distances

82

Attenuation of Digital Signals

Effect of line capacitances

Worse at higher data rates (narrower pulses)

Pulse shapingDue to line capacitances:Worse over longer distances

1 1 1 1 . . .

0 0 0 0 . . .

83

Digital Binary Signal Example: Between keyboard and computer Two bipolar dc levels (+ and – : Why?) Bandwidth required depends on the signal

frequency, which depends on: The data rate (bps) and The actual data sequence transmitted

_

-Data rate = ?

- Maximum f = ?

- Minimum f = ? Signal

Data

1

-

Data element

84

Data and Signal combinations We have seen above: (data and signal of same type)

Analog signals carrying analog data: Telephony, Video Digital signals carrying digital data: Keyboard to PC

Simple- one only needs a transducer/transceiver But we may also have: (data and signal of different types)

Analog signal representing digital data: Data over telephone wires (using a modem)

Digital signal representing analog data: CD Audio, PCM (pulse code modulation) (using a codec) More complex- We Need a converter

So, all the four data-signal combinations are possible!

85

Analog Signals can carry Analog Data or Digital Data

(Base band)

(Converter)

We need a converter when the signal type is different from the data type

(Transducer)i.e. in its original form

86

Digital Signals can carry Analog Data or Digital Data

Coder-Decoder

Transmitter-Receiver

(Converter)

We need a converter when the signal type is different from the data type

Digitized Analog Samples

e.g. using PCM(Pulse Code Modulation)

87

Four Data/Signal Combinations Signal

Analog Digital

Data

Analog

Two ways: Signal has:- Same spectrum as data (base band): e.g. Telephony to exchange

- Different spectrum (through modulation): e.g. AM Radio, FDM

Use a (converter): codec, e.g. for PCM

(pulse code modulation)

Digital Use a (converter): modem e.g. the V.90

standard

- Simple two signal levels: e.g. NRZ code- Special Encoding: e.g. Manchester code (Chapter 5)

88

Two Modes of Transmitting Signals: 1. The Analog Mode (associated with FDM)

Treats the signal as “analog” regardless of what it represents (Not interested in the data content of signal)

Following attenuation over distance, signal level is boosted using “amplifiers”

Unfortunately, this also amplifies in-band noise With cascaded amplifiers (i.e. one after the other at locations

along the link), effect on noise and distortion is cumulative, i.e. they get amplified again and again

Effect of noise and distortion on analog systems may be tolerated, e.g. with telephony you can still manage to get it! (Humans are good at filling-in gaps!)

But digital systems are more sensitive to the effects of excessive noise and distortion unacceptable errors

So… Do not transmit digital signals the analog way!

89

Two Modes of Transmitting Signals: 2. The Digital Mode (Associated with TDM)

Concerned with the data content of the signal It assumes that the signal carries digital data Uses “repeaters” (not amplifiers), which:

Receive the signal Extract the data bit stream from it Retransmit a fresh, strong signal representing the

extracted bit stream This way:

Effect of attenuation is overcome Noise and distortion are not cumulative

90

Four Signal/Transmission Mode Combinations Transmission mode

Analog- Uses amplifiers- Not concerned with what data the signal represents- Noise and distortion are cumulative-- Associated with FDM

Digital- Uses repeaters- Assumes signal represents digital data, recovers this data and represents it as a new outbound signal- This way, noise and distortion are not cumulative- Associated with TDM

Signal

Analog

OK

Makes sense only if the analog signal represents digital data! (Ask yourself: What data is the repeater going to extract?!)

Digital Avoid OK

FDM: Frequency Division MultiplexingTDM: Time Division Multiplexing

Which transmission mode is more versatile

and useful for integrating different signal types?

91

Advantages of Digital Mode of Transmission Use of digital technology

Lower cost, smaller size, and high speed VLSI technology Higher data integrity (reliability) as noise effects are not cumulative (fresh

signal restoration en-route) Cover longer distances, at higher data rate, at low error rates, over lower

quality lines: Easier to implement multiplexing for improved utilization of link capacity

High bandwidth links are now economical (Fiber, Satellite…) To utilize them efficiently we need to do a lot of multiplexing This is done more efficiently using digital (TDM) rather than analog (FDM)

(Chapter 8) Encryption for data security

and confidentiality is digital Easier to integrate different data types

Convert analog data to digital signals…and use one system to handle all voice, video, and data, e.g. one network for all types of traffic

Time Division Multiplexing

Frequency Division

Multiplexing

92

Transmission Impairments Signal received is often a degraded form of the signal

transmitted Why? What happens en-route?... Impairments:

Attenuation: Limits the bandwidth of the received signal In-band signals arrive weaker Attenuation distortion (Attenuation is not uniform over bandwidth)

Delay Delay distortion Noise and interference (including crosstalk)

Effect: On analog data - Some degradation in signal quality On digital data – Fatal bit errors (total bit reversals)

93

Attenuation Signal strength falls off with distance traveled Nature of loss in signal power depends on medium:

Guided (Wires, etc.): Exponential drop is signal power with distance: Pd = P0 e-d

10 ln (Pd/P0) = -d

10 log (Pd/P0) = -’d Loss: ’ dBs per km (’ depends on medium type e.g. fiber, twisted pair, cable)

Unguided (Open space): Inverse square law spread with distance: P P0 /d2

Loss: 6 dBs for each distance doubling Absorption, scattering May also depend on weather, e.g. rain, sunspots,

Signal power after traveling distance d

94

Effects of Attenuation Received signal strength must be:

Sufficiently Large enough to be detected Sufficiently higher than noise to be interpreted correctly

(without error) To overcome these problems:

Use amplifiers (analog transmission mode) or repeaters (digital transmission mode) en-route

Amplifier gains should not be too large as this may cause signal distortion due to saturation (nonlinearities)

Problem with networks: distance actually traveled (hence attenuation) will depend on actual route taken through the network!

95

Attenuation Distortion

Attenuation usually increases with frequency This causes bandwidth limitation (understood) Moreover, over the transmitted bandwidth itself:

Different frequency components of the signal get attenuated differently Signal distortion

Affects analog signals more To overcome this problem:

Use Equalizers that reverse the effect of frequency-dependent attenuation distortion: Passive: e.g. loading coils in telephone circuits Active: Amplifier gain designed specifically for this purpose

96

Attenuation DistortionEqualizationTo Reduce Attenuation Distortion

Q. What is the signal ?

97

Delay Distortion

Happens only on guided media Wave propagation velocity varies with frequency:

Highest at the center frequency (minimum delay) Lower at both ends of the bandwidth (larger delay)

Effect: Different frequency components of the signal arrive at slightly different times! (Dispersion in time)

Affects digital data more: due to bit spill-over (timing is more critical here than for analog data)

Again, equalization can help overcome the problem

98

Delay DistortionEqualizationTo Reduce Delay Distortion

Without Equalizer

With Equalizer

99

Noise (1) Definition: Any additional unwanted signal inserted

between transmitter and receiver The most limiting factor in communication systems Noise Types:

Thermal Noise Inter-modulation Noise Crosstalk Noise Impulse Noise

100

Noise (2) Thermal (White) Noise

Due to thermal agitation of electrons

(Increases with temperature) Uniformly distributed over frequency (White noise)

Difficult to eliminate

(exists even in the same bandwidth as your signal!) Effect is more significant on weak received signals,

e.g. from satellites

PSD

f

101

Thermal Noise, Contd. Thermal noise power density in 1 Hz of bandwidth, N0

(Constant, Independent of frequency):

k Boltzmann’s constant = 1.3810-23 J/K T temperature in degrees Kelvin (= 273 + t C)

Thermal noise power in a bandwidth of B Hz:

)Hz/W(kTN 0

10 log k

Example: at t = 21 C (T = 294 K) and for a bandwidth of 10 MHz:

N = -228.6 + 10 log 294 + 10 log 107

= - 133.9 dBW

f

PSD

N0

1 Hz

B

Can you see some disadvantage now in having a larger

BW?

102

Noise (3) Inter-modulation Noise

Signals having the sum and difference (frequency mixing) of original frequencies sharing a transmission system

(e.g. in FDM systems)

f1, f2 (f1+f2) and (f1-f2) Caused by nonlinearities in the medium and equipment,

e.g. due to overdrive and saturation of amplifiers Danger: Resulting new frequency components may fall

within valid signal bands, thus causing interferenceA cos 1 + B cos 2 Linear System

K(A cos 1 + B cos 2)

A’ cos 1 + B’ cos 2

A cos 1 + B cos 2 Non-Linear System

K(A cos 1 + B cos 2) + K(A cos 1 + B cos 2)2

A’ cos 1 + B’ cos 2+ f(21)+f(22)+f(1-2)+f(1+2)

Inter-modulation componentsInput

Ou

tpu

t

New spurious components can fall within genuine signal bands causing interference

WK 6

Input

103

Noise (4)

Crosstalk Noise A signal from one channel picked up by another channel in

close proximity Examples:

Physical proximity: coupling between adjacent twisted pair channels

Shield cables properly Directional proximity: antenna pick up from other directions

Use directional antennas Spectral proximity: leakage between adjacent channels in

frequency division multiplexing (FDM) systems

Use guard bands between adjacent channels

104

Noise (5)

Impulse Noise Pulses (spikes) of irregular shape and high amplitude lasting

short durations Causes: External electromagnetic interference due to

switching large currents, car ignition, lightning, … Minor effect on analog signals (e.g. crackling noise in voice

channels) Major effect on digital signals- Bit reversal error! More damage at higher data rates

(a noise pulse of a given width can destroy a larger block of bits)

105

Effect of Impulse Noise on a Digital Signal

Q: What is the effect of the same noise at 10 times the data rate?

Impulse

+

=

RX

106

Channel Capacity Channel capacity: Maximum data rate usable under a

given set of communication conditions How channel BW (B), signal level, noise and impairments,

and the amount of data error that can be tolerated limit the channel capacity?

In general, Max possible data rate, C, on a given channel = Function (B, Signal wrt noise, Bit error rate allowed) Max data rate: Max rate at which data can be communicated on

the channel, bits per second (bps) Bandwidth: BW of the transmitted signal as constrained by the

transmission system, cycles per second (Hz) Signal relative to Noise, SNR = signal power/noise power ratio

(Higher SNR better communication conditions higher C) Bit error rate (BER) allowed: in (bits received in error)/(total bits

transmitted). Equal to the bit error probability. e.g. Higher allowed higher usable data rates higher C

107

Channel Capacity, C: So, in general: C bps = F(B, SNR, BER) Three Formulations under different assumptions:

Assumptions Formulation

Ideal: Noise-free, Error-free: C = F(B) Nyquist

Noisy, Error-free: C = F(B, SNR) Shannon

Practical: Noisy, Error: C = F(B, SNR, BER) Eb/N0 Vs Error Rate

Realistic

Idealistic

108

Bandwidth (or Spectral) Efficiency (BE):

Measures how well we are utilizing a given bandwidth to send data at a high rate….

Can be greater than 1 (not like engineering efficiencies) The larger the better

HzbpsBBandwidth

CCapacityChannelBE / ,

109

1. Nyquist Channel capacity: (Noise-free, Error-free) Idealized, theoretical Assumes a noise-free error-free channel Nyquist showed that (without noise, without errors): If rate of signal transmission

is 2B then a signal with frequency components up to B Hz is sufficient to carry that signalling rate

In other words: Given bandwidth B, highest signalling rate possible is 2B signal elements/s

How much data rate does this represent? (depends on how many bits are represented by each signal element!) Given a binary signal (1,0), data rate is same as signal rate

Data rate supported by a BW of B Hz is 2B bps C = 2B For the same B, data rate can be increased by sending one of M

different signals (symbols): as each signal level now represents log2M bits

Generalized Nyquist Channel Capacity, C = 2B log2M bits/s (bps)

Bandwidth efficiency = C/B = 2 log2M (bits/s)/Hz : Dimensionless quantity Signals/s

bits/signal

110

Nyquist Bandwidth: Example C = 2B log2M bits/s

C = Nyquist Channel Capacity B = Bandwidth M = Number of discrete signal levels (symbols) used

Data on telephone Channel: B = 3400-300 = 3100 Hz

With a binary signal (M = 2 symbols, e.g. 2 amplitudes):

C = 2B log2 2 = 2B x 1 = 6200 bps With a quadnary signal (M = 4 symbols):

C = 2B log2 4 = 2B x 2 = 4B = 12,400 bps

Channel capacity increased, but disadvantage: Larger number of signal levels (M) makes it more difficult for the receiver to determine data correctly in the presence of noise

0

1

00

01

10

11

2 bits/Symboli.e. 2 bits /signal element

Signal Element

111

2. Shannon Capacity Formula: (Noisy, Error-Free) Highest error-free data rate in the presence of noise

Signal to noise ratio SNR = signal / noise levelsSNRdB

= 10 log10 (SNR ratio) Errors are less likely with lower noise (larger SNR ratios).

This allows higher error-free data rates i.e. larger Shannon channel capacities

Shannon Capacity C = B log2(1+SNR):

Highest data rate transmitted error-free with a given noise level For a given BW, the larger the SNR the higher the data

rate I can use without introducing errors C/B: Spectral (bandwidth) efficiency, BE, (bps/Hz) (>1) Larger BEs mean better utilization of a given bandwidth

B for transmitting data fast.

Caution! Log2 Not Log10

Caution! Ratio- Not dBs

112

Shannon Capacity Formula: Comments Formula says: for data rates calculated C, it is

theoretically possible to find an encoding scheme that achieves error-free transmission at the given SNR… But it does not say how!Also:

It is a theoretical approach based on thermal (white) noise only. But in practice, we also have impulse noise, attenuation and delay distortions, etc… So, maximum error-free data rates measured in practice

are expected to be lower than the C predicted by the Shannon formula due to the greater noise

However, maximum error-free data rates can be used to compare practical systems: The higher that rate the better the system…

113

Shannon Capacity Formula: Comments Contd. Formula suggests that changes in B and SNR can

be done arbitrarily and independently… but

In practice, this may not be the case! Higher SNR obtained through excessive

amplification may also introduce nonlinearities increased distortion and inter-modulation noise … which reduces SNR!

High Bandwidth B opens the system up for more thermal noise (kTB), and therefore reduces SNR!

114

Shannon Capacity Formula: Example

Spectrum of communication channel extends from 3 MHz to 4 MHz SNR = 24dB Then B = 4MHz – 3MHz = 1MHz

SNRdB = 24dB = 10 log10 (SNR)

SNR (ratio) = log-110 (24/10) = 1024/10 = 251

Using Shannon’s formula: C = B log2 (1+ SNR)

C = 106 * log2(1+251) ~ 106 * 8 = 8 Mbps Based on Nyquist’s formula, determine M that gives the above

channel capacity:

C = 2B log2 M

8 * 106 = 2 * (106) * log2 M

4 = log2 M

M = 16

115

3. Eb/N0 Vs Error Rate Formulation (Noise and Error are both specified Together) Handling both noise and a quantified error rate simultaneously

We introduce Eb/N0: A standard quality measure of three channel parameters (B, SNR, R) and can also be independently related to the error rateR is the data rate. Max value of R is the channel capacity C

It expresses SNR in a manner related to the data rate, R Eb = Signal energy in one bit interval (Joules)

= Signal power (Watts) x bit interval Tb (second) = S x (1/R) = S/R

N0 = Noise power (watts) in 1 Hz = kT. Two formulations:

0 0

/b bE ST S R S

N N kT kTR

0 0

/b T TE B BS R SSNR

N N N R R

= SNR/BE

Tb = 1/R

116

Eb/N0 (Cont.) Bit error rate for digital data is a

decreasing function of Eb/N0 for a given signal encoding scheme

Analysis: For a given system (SNR, B, R) (Eb/N0), determine error rate BER

Design: Given a desired error rate BER, get Eb/N0 to achieve it, then determine other parameters from formula, e.g. S, SNR, R, etc.

Effect of S, R, T on error performance

Which encoding scheme is better: A or B?

0

10log 10log 10log

10log 228.6 10log

bdBW

dB

dBW

ES R k T

N

S R dBW T

Lower E

rror Rate: larger E

b/N0

A B

0 0

/b T TE B BS R SSNR

N N N R R

BetterEncoding

= SNR

BE

Max R = C, BE = C/B

BER vs Eb/N0 curve for a given encoding scheme

117

Example:

Given: The effective noise temperature, T, is 290oK The data rate, R, is 2400 bps Would like to operate with a bit error rate of 10-4 (e.g. 1 error in 104 bits)

What is the minimum signal level required for the received signal?

From curve, a minimum Eb/No needed to achieve a bit error rate of 10-4 = 8.4 dB

8.4 = S(dBW) – 10 log 2400 + 228.6 dBW – 10 log290 = S(dBW) – (10)(3.38) + 228.6 – (10)(2.46)

S = -161.8 dBW

0

10log 10log 10log

10 log 228.6 10log

bdBW

dB

dBW

ES R k T

N

S R dBW T

Design or Analysis?

118

Eb/N0 in terms of BE, assuming Shannon channel capacity

From Shannon’s formula:

C = B log2(1+SNR)

We have:

From the Eb/N0 formula:

C/B (bps/Hz) is the spectral (bandwidth) efficiency BE based on Shannon channel capacity

)12()12( / BEBCSNR

)12(1

0

BEb

BEBE

SNR

N

E

119

Example

Find the minimum Eb/N0 required to achieve a Shannon bandwidth efficiency (BE=CShannon/B) of 6 bps/Hz:

Substituting in the equation above:

Eb/N0 = (1/6) (26 - 1) = 10.5 = 10.21 dB

)12(1

0

BEb

BEN

E