coe 341: data & computer communications (t081) dr. marwan abu-amara

COE 341: Data & Computer Communications (T081)Dr. Marwan Abu-Amara

Chapter 3: Data Transmission

COE 341 – Dr. Marwan Abu-Amara 2

Agenda Concepts & Terminology Decibels and Signal Strength Fourier Analysis Analog & Digital Data Transmission Transmission Impairments Channel Capacity


Terminology (1) Transmitter Receiver Medium

Guided medium e.g. twisted pair, optical fiber

Unguided medium e.g. air, water, vacuum


Terminology (2) Direct link

No intermediate devices Point-to-point

Direct link Only 2 devices share link

Multi-point More than two devices share the link


Terminology (3) Simplex

One direction e.g. Television

Half duplex Either direction, but only one way at a time

e.g. police radio

Full duplex Both directions at the same time

e.g. telephone


Frequency, Spectrum and Bandwidth Time domain concepts

Analog signal Varies in a smooth way over time

Digital signal Maintains a constant level then changes to another

constant level Periodic signal

Pattern repeated over time Aperiodic signal

Pattern not repeated over time


Analogue & Digital Signals


PeriodicSignals

S (t+nT) = S (t);Where:t is timeT is the waveform periodn is an integer

T

Temporal Period


Sine Wave – s(t) = A sin(2ft +) Peak Amplitude (A)

maximum strength of signal unit: volts

Frequency (f) rate of change of signal unit: Hertz (Hz) or cycles per second Period = time for one repetition (T) = 1/f

Phase () relative position in time unit: radians

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


Varying Sine Wavess(t) = A sin(2ft +)


Wavelength () Distance occupied by one cycle Distance between two points of corresponding

phase in two consecutive cycles Assuming signal velocity v

= vT f = v For an electromagnetic wave,

v = speed of light in the medium In free space, v = c = 3*108 m/sec


Frequency Domain Concepts

Signal usually made up of many frequencies Components are sine waves Can be shown (Fourier analysis) that any

signal is made up of component sine waves Can plot frequency domain functions


Addition of FrequencyComponents(T=1/f)

Fundamental Frequency


FrequencyDomainRepresentations


Spectrum & Bandwidth Spectrum

range of frequencies contained in signal Absolute bandwidth

width of spectrum Effective bandwidth

Often just bandwidth Narrow band of frequencies containing most of

the energy DC Component

Component of zero frequency


Signal with a DC Component

t

1V DC Level

1V DC Component

t


Bandwidth for these signals:fmin fmax Absolute

BWEffective

BW

1f 3f 2f 2f

0 3f 3f 3f

0 1/x ?


Bandwidth and Data Rate Any transmission system supports only a limited band of

frequencies for satisfactory transmission “system” includes: TX, RX, and Medium Limitation is dictated by considerations of cost, number

of channels, etc. This limited bandwidth degrades the transmitted signals,

making it difficult to interpret them at RX For a given bandwidth: Higher data rates More

degradation This limits the data rate that can be used with given

signal and noise levels, receiver type, and error performance

More about this later!!!


Bandwidth Requirements

)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 =

…

Larger BW

needed for better representation

Mo

re d

iffic

ult

rece

ptio

n w

ith m

ore

lim

ited

BW

1,3

1,3,5

1,3,5,7

1,3,5,7 ,9,…

1

2

3

4Fourier Series for a Square Wave


Decibels and Signal Strength Decibel is a measure of ratio between two

signal levels NdB = number of decibels

P1 = input power level

P2 = output power level

Example: A signal with power level of 10mW inserted onto a

transmission line Measured power some distance away is 5mW Loss expressed as NdB =10log(5/10)=10(-0.3)=-3 dB

1

210log10

P

PNdB


Decibels and Signal Strength Decibel is a measure of relative, not absolute, difference

A loss from 1000 mW to 500 mW is a loss of 3dB A loss of 3 dB halves the power A gain of 3 dB doubles the power

Example: Input to transmission system at power level of 4 mW First element is transmission line with a 12 dB loss Second element is amplifier with 35 dB gain Third element is transmission line with 10 dB loss Output power P2

(-12+35-10)=13 dB = 10 log (P2 / 4mW)

P2 = 4 x 101.3 mW = 79.8 mW


Relationship Between Decibel Values and Powers of 10 Power Power

RatioRatiodBdB Power Power

RatioRatiodBdB

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


Decibel-Watt (dBW) Absolute level of power in decibels Value of 1 W is a reference defined to be 0 dBW

Example: Power of 1000 W is 30 dBW Power of 1 mW is –30 dBW

W

PowerPower W

dBW 1log10 10


Decibel & Difference in Voltage Decibel is used to measure difference in

voltage. Power P=V2/R

Decibel-millivolt (dBmV) is an absolute unit with 0 dBmV equivalent to 1mV. Used in cable TV and broadband LAN

1

22

1

22

1

2 log20/

/log10log10

V

V

RV

RV

P

PNdB

mV

VoltageVoltage mV

dBmV 1log20


Fourier AnalysisSignals

Periodic (fo) Aperiodic

Discrete Continuous Discrete Continuous

DFS FS

DTFT

FT

DFT

Infinite time Finite time

FT : Fourier TransformDFT : Discrete Fourier TransformDTFT : Discrete Time Fourier TransformFS : Fourier SeriesDFS : Discrete Fourier Series


Fourier Series (Appendix B) Any periodic signal of period T (f0 = 1/T) can be

represented as sum of sinusoids, known as Fourier Series

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

fundamental frequency


Fourier Series Amplitude-phase representation

1

00 )2cos(

2)(

nnn tnfC

Ctx

00 AC 22nnn BAC

n

nn A

B1tan


Fourier Series Representation of Periodic Signals - Example

1

-1

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

T

0111212)(2)(2

2)(

2 1

2/1

2/1

0

1

0

2

00

0 dtdtdttxdttxdttxT

AT

x(t)

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


Fourier Series Representation of Periodic Signals - Example

1

0

0

2/

0

0

0

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

)2cos()(2

dttnftxdttnftxT

dttnftxT

ATT

n

2sin

4)2cos(2)2cos(2

1

2/1

0

2/1

0

0

n

ndttnfdttnf

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

2/

0

0

2/

0

0 )2sin()(2

)2sin()(2 TT

dttnftxT

dttnftxT

Replacing t by –tin the first integralsin(-2nf t)=- sin(2nf t)


Fourier Series Representation of Periodic Signals - ExampleSince x(– t)=x(t) as x(t) is an even function, then

Bn = 0 for n=1, 2, 3, …

1

000 )2sin()2cos(

2)(

nnn tnfBtnfA

Atx

tnn

ntx

n

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

Cosine is an even function


Another Example

1

-1

1-1 2

T

-2

x1(t)

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

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

4 1 1 1 1 1 1 11( ) cos cos 3 cos 5 cos 7

2 3 2 5 2 7 2x t t t t t

x(t)


Another Example

4 1 3 1 5 1 71( ) cos cos 3 cos 5 cos 7

2 3 2 5 2 7 2x t t t t t

4 1 1 11( ) in in3 sin 5 in7

3 5 7x t s t s t t s t

tt sin2

cos

tt 3sin

2

33 cos

tt 5sin2

55 cos

tt 7sin

2

77 cos

Sine is an odd function

Where:


Fourier Transform For a periodic signal, spectrum consists of

discrete frequency components at fundamental frequency & its harmonics.

For an aperiodic signal, spectrum consists of a continuum of frequencies (non-discrete components). Spectrum can be defined by Fourier Transform For a signal x(t) with spectrum X(f), the following

relations hold

dfefXtx ftj 2 )()(

dtetxfX ftj 2 )()(


Fourier Transform Example

x(t)A

22

dtetxfX ftj 2 )()(

2/

2/

22/

2/

2

2 )(

ftjftj efj

AdteAfX


Fourier Transform Example

2/2

)2/2sin(

2

2

2

2

22

2 2/22/2

f

ff

f

A

j

ee

f

A fjfj

f

fA

f

fAfX

)sin(

2/2

)2/2sin()(

j

ee jj

2sin

2cos

jj ee

A

1/

f

Study the effect of the pulse width

Sin (x) / x“sinc” function


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

The Extreme Cases


Power Spectral Density & Bandwidth

Absolute bandwidth of any time-limited signal is infinite

However, most of the signal power will be concentrated in a finite band of frequencies

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

Power spectral density (PSD) describes the distribution of the power content of a signal as a function of frequency


Signal Power A function x(t) specifies a signal in terms of

either voltage or current Assuming R = 1

Instantaneous signal power = V2 = i2 = |x(t)|2

Instantaneous power of a signal is related to average power of a time-limited signal, and is defined as

For a periodic signal, the averaging is taken over one period to give the total signal power

22

1

1( )

2 1

t

t

x t dtt t

2

0

1( )Total

T

P x t dtT


Power Spectral Density & Bandwidth For a periodic signal, power spectral density

is

where (f) is

Cn is as defined before on slide 27, and f0 being the fundamental frequency

2

0( ) ( )nn

PSD f C f nf

1 =00 0( ) f

ff


Power Spectral Density & Bandwidth For a continuous valued function S(f), power

contained in a band of frequencies f1 < f < f2

For a periodic waveform, the power through the first j harmonics is

2

1

)(2f

f

dffSP

j

nnCCP

1

220 2

125.0


Power Spectral Density & Bandwidth - Example Consider the following signal

The signal power is

7in

7

1 5sin

5

1 3in

3

1 in)( tsttststx

watt586.0 49

1

25

1

9

11

2

1

Power


Fourier Analysis Example Consider the half-wave rectified cosine signal from

Figure B.1 on page 793:1. Write a mathematical expression for s(t)

2. Compute the Fourier series for s(t)

3. Write 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 a value of n such that Fourier series for s(t) contains 95% of the total power in the original signal

6. Determine the corresponding effective bandwidth for the signal


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

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


Example (Cont.)2. Fourier Analysis:

1 )2/sin( where, 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

T

T

T

T

o

T

T

T

dttxT

A0

0 )(2

f0 = (1/T)


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)



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



/ 4 / 4

1

/ 4 / 4

/ 42

/ 4

/ 4

/ 4

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

2 cos (2 )

sin(4 )2 2 sin( ) sin( )

2 4 2 4 8

2

T T

n o o o

T T

T

o

T

T

o

o oT

AA s t f t dt f t f t dt

T T

Af t dt

T

f tA t A T

T f T f

A

Note: cos2 = ½(1 + cos 2)



/ 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



/ 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



2

2

1

(1 )

22,4,6,...

o 1

(1 )

2

( ) cos(2 ) sin(2 )2

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

2 1

2C ,

20 , is odd and 1

2 ( 1) , 2, 4,

( 1)

n

n

on o n o

n

o on

n

n

As t A nf t B nf t

A A Af t nf t

n

A AC

C n n

AC n

n

6,...

00 AC 22nnn BAC


Example (Cont.)3. Power Spectral Density function (PSD):

Or more accurately:

220

1

1

4 2 nn

CPSD C

220

1

1( ) ( )

4 2 n on

CPSD f C f nf


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


Example (Cont.)4. Total Power:

3 / 4 / 422 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)


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



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



6. Effective bandwidth with 95% of total power:

Beff = fmax – fmin

= 2f0 – 0 = 2f0

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

0 f0 2f0 3f0

…

f

Beff


Data Rate and Bandwidth

Any transmission system has a limited band of frequencies

This limits the data rate that can be carried Example on pages 74 – 76 of textbook


Bandwidth and Data Rates

(Bsys = 2B) = (Bsig = 4f)

Bsys = (Bsig = 4f)

f 3f

f 3f 5f

Period T = 1/f

1 0 1 0

Data Element,Signal Element

T/2

Data rate = 1/(T/2) = 2/T bits per sec = 2fGiven a bandwidth B,Data rate = 2f = 2(B/4) = B/2

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

1. Double the transmission system bandwidth, with the same receiver and error rate (same received waveform)

f 3f 5f

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

2. Same transmission system bandwidth, B, with a better receiver, higher S/N, or by tolerating more error (poorer received waveform)

(Bsys = B) = (Bsys = 2f)

Bandwidth: B,Data rate = 2f = 2(B/2) = B

1 1 1 10 0 0 0

1 1 1 10 0 0 0

B

2B

B


Bandwidth and Data Rates Increasing the data rate (bps) with the same BW

means working with inferior waveforms at the receiver, which may require: Better signal to noise ratio at RX (larger signal relative to

noise): Shorter link spans Use of more repeaters/amplifiers Better shielding of cables to reduce noise, etc.

More sensitive (& costly!) receiver Dealing with higher error rates

Tolerating them Adding more efficient means for error detection and correction-

this also increases overhead!.


Bandwidth and Data Rates


Analog and Digital Data Transmission Data

Entities that convey meaning Signal

Electric or electromagnetic representations of data Transmission

Communication of data by propagation and processing of signals


Analog and Digital Data Transmission Data

Can be either Analog data or Digital data Signal

Can use either Analog signal or Digital signal to convey the data

Transmission Can use either Analog transmission or Digital

transmission to carry the signal


Analog and Digital Data Analog

Continuous values within some interval e.g. sound, video

Digital Discrete values e.g. text, integers


Acoustic Spectrum (Analog)


Analog and Digital Signals Means by which data are propagated Analog

Continuously variable Various media

wire, fiber optic, space Speech bandwidth 100Hz to 7kHz Telephone bandwidth 300Hz to 3400Hz Video bandwidth 4MHz

Digital Use two DC components


Advantages & Disadvantages of Digital Signals Advantages:

Cheaper Less susceptible to noise

Disadvantages: Greater attenuation

Pulses become rounded and smaller Leads to loss of information


Attenuation of Digital Signals


Components of Speech Frequency range (of hearing) 20Hz-20kHz

Speech 100Hz-7kHz Easily converted into electromagnetic signal

for transmission Sound frequencies with varying volume

converted into electromagnetic frequencies with varying voltage

Limit frequency range for voice channel 300-3400Hz


Conversion of Voice Input into Analog Signal


Video Components USA - 483 lines scanned per frame at 30 frames

(scans) per second 525 lines but 42 lost during vertical retrace

So 525 lines x 30 frames (scans) = 15750 lines per second 63.5s per line

11s for retrace, so 52.5 s per video line Max frequency if line alternates black and white Horizontal resolution is about 450 lines giving 225

cycles of wave in 52.5 s Max frequency (for black and white video) is 4.2MHz


Binary Digital Data

From computer terminals etc. Two dc components Bandwidth depends on data rate


Conversion of PC Input to Digital Signal


Data and Signals

Usually use digital signals for digital data and analog signals for analog data

Can use analog signal to carry digital data Modem

Can use digital signal to carry analog data Compact Disc audio


Analog Signals Carrying Analog and Digital Data


Digital Signals Carrying Analog and Digital Data


Four Data/Signal Combinations Signal

Analog Digital

Data

Analog

- Same spectrum as data (base band): e.g. Conventional Telephony

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

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

(pulse code modulation)

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

V.90 standard

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


Analog Transmission

Analog signal transmitted without regard to content

Analog signal may be analog or digital data Attenuated over distance Use amplifiers to boost signal Also amplifies noise


Digital Transmission Concerned with content Integrity endangered by noise, attenuation

etc. Repeaters used

Repeater receives signal Extracts bit pattern Retransmits

Attenuation is overcome by a repeater by reconstructing the signal

Noise is not amplified


Four Signal/Transmission Mode Combinations

Transmission mode

Analog- Uses amplifiers- Not concerned with what data the signal represents

- Noise is cumulative

Digital- Uses repeaters- Assumes signal represents digital data, recovers it and represents it as a new outbound signal

- This way, noise is not cumulative

Signal

Analog

OK

Makes sense only if the analog signal represents digital data

Digital Avoid OK


Advantages of Digital Transmission Digital technology

Low cost LSI/VLSI technology Data integrity

Longer distances over lower quality lines Capacity utilization

High bandwidth links economical High degree of multiplexing easier with digital techniques

Security & Privacy Encryption

Integration Can treat analog and digital data similarly


Transmission Impairments Signal received may differ from signal

transmitted Analog signal - degradation of signal

quality Digital signal - bit errors Caused by

Attenuation and attenuation distortion Delay distortion Noise


Attenuation Signal strength falls off with distance Depends on medium (guided vs. unguided) Attenuation affects received signal strength

received signal strength must be enough to be detected received signal strength must be sufficiently higher than noise to

be received without error signal strength can be achieved by using amplifiers or repeaters

Attenuation is an increasing function of frequency Different frequency components of a signal get attenuated

differently Signal distortion Particularly significant with analog signals

for digital signals, strength of signal falls of rapidly with frequency Can overcome signal distortion using equalizers


Delay Distortion Only in guided media Propagation velocity varies with frequency

Highest at 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)

Badly affects digital data due to bit spill-over (intersymbol interference) major limitation to max bit rate over a transmission channel

Can overcome delay distortion using equalizers


Noise Additional unwanted signals inserted between

transmitter and receiver The most limiting factor in communication

systems Noise categories:

Thermal Intermodulation Crosstalk Impulse


Thermal (White) Noise Due to thermal agitation of electrons Uniformly distributed across the bandwidth Power of thermal noise present in a bandwidth B

(Hz) is given by

T is absolute temperature in kelvin and k is Boltzmann’s constant (k = 1.3810-23 J/K)

0 (watts)

= 228.6 10log 10log (dBw)

N kTB N B

T B

= =

- + +

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


Intermodulation Occurs when signals at different frequencies

share same transmission medium Produces signals that are the sum and/or the

difference of original frequencies sharing the medium f1, f2 (f1+f2) and (f1-f2) Caused by nonlinearities in the medium and

equipment, e.g. due to overdrive and saturation of amplifiers

Resulting frequency components (i.e. f1+f2 and f1-f2) may fall within valid signal bands, thus causing interference


Crosstalk & Impulse Crosstalk

A signal from one channel picked up by another channel e.g. Coupling between twisted pairs, antenna pick up,

leakage between adjacent channels in FDM, etc.

Impulse Irregular pulses or spikes Short duration High amplitude e.g. External electromagnetic interference Minor effect on analog signals but major effect on digital

signals, particularly at high data rates


Channel Capacity Channel capacity: Maximum data rate usable under given

communication conditions How BW, signal level, noise and other impairments, and the amount

of error tolerated limit the channel capacity? Max data rate

= Function (BW, Signal wrt noise, Error rate allowed) Max data rate: Max rate at which data can be communicated, 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) Error rate: bits received in error/total bits transmitted. Equal to the bit

error probability


1. Nyquist Bandwidth: (Noise-free, Error-free) Idealized, theoretical Assumes a noise-free, error-free channel Nyquist: If rate of signal transmission is 2B then a signal with

frequencies no greater than B is sufficient to carry that signalling rate

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

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

For the same B, data rate can be increased by sending one of M different signal levels (symbols): as a signal level now represents log2M bits

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

Signals/s bits/signal


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

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

Telephone Channel: B = 3400-300 = 3100 Hz With a binary signal (M = 2):

C = 2B log2 2 = 2B = 6200 bps With a quandary signal (M = 4):

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

Practical limit: larger M makes it difficult for the receiver to operate, particular with noise

0

1

00

01

10

11


2. Shannon Capacity Formula: (Noisy, Error-Free)

Assumes error-free operation with noise Data rate, noise, error: A given noise burst affects more bits at

higher data rates, which increases the error rate So, maximum error-free data rate increases with reduced noise Signal to noise ratio SNR = signal / noise levels

SNRdB= 10 log10 (SNR) dBs

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 errors C/B: Spectral (bandwidth) efficiency, BE, (bps/Hz) (>1) Larger BEs mean better utilizing of a given B for transmitting data

fast.

Caution! Log2 Not Log10

Caution! Ratio- Not log


Shannon Capacity Formula: Comments

Formula says: for data rates calculated C, it is theoretically possible to find an encoding scheme that gives error-free transmission.

But it does not say how… It is a theoretical approach based on thermal (white) noise

only However, in practice, we also have impulse noise and

attenuation and delay distortions So, maximum error-free data rates obtained in practice are

lower than the C predicted by this theoretical formula However, maximum error-free data rates can be used to

compare practical systems: The higher that rate the better the system is


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!

High SNR obtained through excessive amplification may introduce nonlinearities: distortion and intermediation noise!

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


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 = 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


3. Eb/N0 (Signal Energy per Bit/Noise Power density per Hz) (Noise and Error Together) Handling both noise and error together

Eb/N0: A standard quality measure for digital communication system performance

Eb/N0 Can be independently related to the error rate Expresses SNR in a manner related to the data rate, R Eb = Signal energy per bit (Joules)

= Signal power (Watts) x bit interval Tb (second)

= S x (1/R) = S/R N0 = Noise power (watts) in 1 Hz = kT

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


Eb/N0 Example 1:

Given: Eb/No = 8.4 dB (minimum) is needed to achieve a bit error rate of 10-4

Given: The effective noise temperature is 290oK (room temperature) Data rate is 2400 bps

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

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


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

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

Which encoding scheme is better: A or B?

Get Eb/N0 to achieve a desired error rate, then determine other parameters from formula, e.g. S, SNR, R, etc. (Design)

Error performance of a given system (Analysis)

Effect of S, R, T on error performance

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


Eb/N0 (Cont.) From Shannon’s formula:

C = B log2(1+SNR)We have:

From the Eb/N0 formula:

With R = C, substituting for SNR we get:

Relates achievable spectral efficiency C/B (bps/Hz) to Eb/N0


Eb/N0 (Cont.) Example 2

Find the minimum Eb/N0 required to achieve a spectral efficiency (C/B) of 6 bps/Hz:

Substituting in the equation above:

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

coe 341: data & computer communications (t081) dr. marwan abu-amara

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