1

Modulation and Multiplexing

Joe MontanaIT 488 - Fall 2003

2

Agenda

• Modulation Concept

• Analog Communication

• Digital Communication

• Digital Modulation Schemes• Error Detection and Correction

3

Modulation

4

Why Modulate Signals?If we transmit signal through electromagnetic waves, we need antennas to recover them at a remote point.At low frequencies (baseband), the wavelengths are very large.

Ex. Voice, at approx. 4 kHz, has a wavelength of 75 Km!!

If we “move” those signals to higher frequencies, we can get more manageable antennas. After receiving the signal, we need to “move” them back to the original frequency band (baseband) through demodulation.Therefore, you can see the modulation task as “giving wings” to the information message.

5

Modulation – Basic PrinciplesModulation is achieved by varying the amplitude, phase or frequency of a high frequency sinusoid.The initial high frequency sinusoid that will have a parameter modified is called the “Carrier”.The original message signal (baseband) is called the “Modulating” signal.The resulting bandpass signal is the “Modulated” signal, which is a combination of the carrier and the original message.

6

Modulation – Basic Principles

Modulating Signal V(t), at baseband(fB)

Carrier (fC)

Action on carrier’s amplitude, frequency or phase

Modulated Signal carrying the information of V(t), bandpass (fC)

fC

fC

7

MODULATION AND MULTIPLEXING - 1

MODULATIONTHIS IS THE WAY INFORMATION IS ENCAPSULATED FOR TRANSMISSION

MULTIPLEXINGTHIS IS THE WAY MORE THAN ONE LINK CAN BE CARRIED OVER A SINGLE COMMUNICATIONS CHANNEL

WE WILL BE LOOKING AT MODULATION INITIALLY, BUT WHERE DO MODULATION AND MULTIPLEXING FIT INTO A SYSTEM?

8

MODULATION AND MULTIPLEXING - 2

Fig. 5.1 in text: (A) At uplink earth station (B) At downlink earth station

9

MODULATION AND MULTIPLEXING - 3KEY POINTS

You have to multiplex before modulating on the transmit side (that is, you have to get all of the output signals together prior to modulating onto a carrier)You have to demodulate before demultiplexing on the receive side (that is, before you can separate - i.e. demultiplex - the incoming signals, you have to demodulate the carrier to obtain the transmitted information)

10

Analog Communications

11

ANALOG TELEPHONY - 1Baseband voice signal

300 - 3400 Hz (CCITT, now called ITU-T)300 - 3100 Hz (Bell)We will use the ITU-T definition

12

ANALOG TELEPHONY - 2

KEY POINTTHE NUMBER OF VOICE CHANNELS A SATELLITE TRANSPONDER CAN CARRY VARIES INVERSELY WITH THE AVERAGE POWER LEVEL PER CHANNEL

NOTE:A pessimistic choice (power level set too high) will lower capacity estimate; An optimistic choice (power level set too low) can reduce quality of signals

Simple Example

13

CHANNEL LOADING EXAMPLE - 1

A 25 W transponder is designed to carry 250 two-way telephone channels (giving 500 channels at RF).

Q1. How much power is available for each telephone channel?

Answer: Power per channel = (25) / (500) = 50 mW

Q2. If the amplifier requires to be backed off 3 dB to preserve linearity, what is the power available per telephone channel now?

Answer: Power per channel = (25/2) / (500) = 25 mW

Q2. What is the power per channel in the second case if 1000 RF channels are carried?

Answer: Power per channel = (25 mW) / 2 = 12.5 mW

14

SATELLITE ANALOG

Satellite transponders are bandwidth limitedA flexible scheme is therefore required for loading analog voice channelsearth stations may transmit in multiples of 12 voice channels (from 12 to 1872)

NOTE: There is very little analog (FM) voice traffic over satellites now. The bulk of the high capacity traffic is carried over optical fibers. The majority of voice capacity is in small digital carriers

called IDR (Intermediate Digital Rate)

15

FREQUENCY MODULATION - 1

DEFINITION“Frequency modulation results when the deviation, f, of the instantaneous frequency, f, from the carrier frequency fc is directly proportional to the instantaneous amplitude of the modulating voltage”.

LET’S LOOK AT THIS PICTORIALLY

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FREQUENCY MODULATION - 1

Output Frequency

Input voltage

Transfer characteristicVmax

Vmin

Range of Input Voltage, v(t)

Range of Output Frequency

f min f max

Instantaneous Input Voltage

Instantaneous Output Frequency

f

NOTE: In this example, fmin = the carrier frequency, fc

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FREQUENCY MODULATION - 2

Schematic representation of a sinusoidal modulating

signal, vp, on a carrier signal, frequency fc

NOTE: instantaneous frequency increases with increase in modulating voltage, and vice versa

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FREQUENCY MODULATION - 3

The Frequency Modulated output signal, , will be as follows:

(5.2)

tvFM

ttAtvtv cFM mod

mod

sincos

Maximum angular frequency deviation of the modulator

c = 2fc = carrier radian frequency

Maximum value of input modulating radian frequency

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CARSON’S RULE - 1Carson’s rule states that the transmission bandwidth, BT, is given by: mod2 fB f

Where B is the bandwidth of the modulating signal which, for a sinusoidal modulating signal, is the highest modulating frequency, fmod.

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CARSON’S RULE - 2A. Single-frequency sinusoid:

Approximate value for required bandwidth B:

(5.5)

Maximum frequency deviation

Modulating frequency

B. Real signal (practical case): Approximate value for required bandwidth B:

max2 fB f (5.6)

Maximum modulating frequency

mod2 fB f

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FM IMPROVEMENTFM modulation is relatively inefficient with the use of transmission spectrum

A small baseband bandwidth is converted into a large RF bandwidth

FM demodulation and detection converts the wide RF bandwidth occupied into a small baseband bandwidth occupied

Ratio of RF to baseband bandwidths gives an improvement in signal to noise ratio which leads to the so-called FM IMPROVEMENT

22

Digital Communications

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DIGITAL COMMUNICATIONS -1

Many signals originate in digital formdata from computersdata from digital fixed and mobile systemsdigitized information (e.g. voice)

World-wide network is moving towards all-digital systemComputers can only handle digital signals

§5.4 in Chapter 5 + updated material

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Why Digital Transmission?

RobustnessGenerally less susceptible to degradationsBut...when it does degrade tends to fail quickly

AdaptivenessCan easily combine a mix of signal information

• Data, voice, video, multiple user signals

Compatibility - with digital storage, etc.Security - not easily received except by recipient

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DIGITAL COMMUNICATIONS -2

At baseband, send V (volts) to represent a logical 1 and 0At RF - digitally modulate the carrier

ASK Amplitude Shift KeyingFSK Frequency Shift KeyingPSK Phase Shift Keying

Binary forms of these areOOK, BFSK, and BPSK, respectively

Let’s first look at basic Digital Communications from the book by COUCH (7th. Edition)

26

DIGITAL COMMUNICATIONS -3

NOTE: from Couch

27

DIGITAL COMMUNICATIONS - 4

From Couch, Fig. 3-15

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DIGITAL COMMUNICATIONS - 5

From Couch, Fig. 3-13

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DIGITAL COMMUNICATIONS - 5Analog-to-Digital recap; we have:

Sampled at 2 times highest frequencyStored the sampled valueCompared stored value with a quantized levelSelected the nearest quantized levelTurned the selected quantized level into a digital value using the selected number of bits

We now need to generate a line codeLine Codes are serial bit streams that are used to drive the digital modulator

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LINE CODES - 1Couch Fig.

3-15

Usually used in digital circuits

Always have net

zero voltage

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LINE CODES - 2

SELECTION OF LINE CODE BASED ON

NEED TO HAVE SYNCHRONIZATION (OR OTHERWISE)NEED TO HAVE A NET ZERO VOLTAGE (OR OTHERWISE)NEED TO PREVENT STRING OF SAME VOLTAGE LEVEL SIGNALSSPECTRAL EFFICIENCYSOME TYPICAL SPECTRA

32

TYPICAL SPECTRA

Couch Fig. 2-6

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PULSE SPECTRA

2

sin

b

bbf fT

fTTG

A random train of ones and zeroes has a spectrum (power spectral density) of

(5.40)

X = fTb, Tb = bit period, and f = frequency in Hz

Max value of Tb at f = 0

G(f) extends to f =

2

2sin

X

XTb

Filtering affects the pulse shape

34

EFFECT OF FILTERING - 1

Fig. 5.8 in text

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EFFECT OF FILTERING - 2

Rectangular pulses (i.e. infinite rise and fall times of the pulse edges) need an infinite bandwidth to retain the rectangular shapeCommunications systems are always band-limited, so

send a SHAPED PULSE

Attempt to MATCH the filter to the spectrum of the energy transmitted

Before FILTERS, let’s look at Inter-Symbol Interference

36

INTER-SYMBOL INTERFERENCE

Sending pulses through a band-limited channel causes “smearing” of the pulse in time“Smearing” causes the tail of one pulse to extend into the next (later) pulse periodParts of two pulses existing in the same pulse period causes Inter-Symbol Interference (ISI)ISI reduces the amplitude of the wanted pulse and reduces noise immunity

Example of ISI

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ISI - contd. - 1

Form Couch, Fig. 3-23

38

ISI - contd. - 2

To avoid ISI, you can SHAPE the pulse so that there is zero energy in adjacent pulses

Use NRZ; pulse lasts the full bit periodUse Polar Signaling (+V & -V); average value is zero if equal number of 1’s and 0’sCommunications links are usually AC coupled so you should avoid a DC voltage componentThen use a NYQUIST filterNyquist Filter???

39

NYQUIST FILTER - 1

Bit Period is Tb

Sampling of the signal is usually at intervals of Tb

Thus, if we could generate pulses that are at a one-time maximum at t = Tb and zero at each succeeding interval of Tb (i.e. t = 2Tb, 3Tb, ….. , NTb then we would have no ISIThis is called a NYQUIST filter

40

NYQUIST FILTER - 2

0 Tb 2Tb 3Tb 4Tb

Impulse at this point t

Sampling instant is

CRITICAL

41

NYQUIST FILTER - 3

Fig. 5.9 in text

NOTE: At each sampling interval, there is only one

pulse contribution - the others being

at zero level

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NYQUIST FILTER - 4

Arranging to sample at EXACTLY the right instant is the “Zero ISI” technique, first proposed by Nyquist in 1928Networks which produce the required time waveforms are called “Nyquist Filters”. None exist in practice, but you can get reasonably close

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NYQUIST FILTER - 5Noise into receiver must be held to a minimumPlace half of Nyquist filter at transmit end of link, half at receive end, so that the individual filter transfer function H(f) is given by

Vr(f)NYQUIST = H(f) H(f)

Filter is a “Square Root Raised Cosine Filter” NYQUISTr fVfH )(

H(f) matches pulse characteristic,

hence it is called a “matched filter”

Matched Filter

44

MATCHED FILTER - 1

6 dB

f f

Roll-off factor = = (f / f0 )

where f0 = 6 dB bandwidth

B = absolute bandwidth (here shown for = 0.5) and

B = f + f0

f1 = start of ‘roll-off’ of the filter characteristic

Bf1 f0

Fig. 5.10 in text

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MATCHED FILTER - 2A Raised Cosine Filter gives a Matched Filter responseThe “Roll-Off Factor”, , determines bandwidth of Raised Cosine Low Pass Filter (LPF)Gives zero ISI when the output is sampled at correct time, with sampling rate of Rb (i.e. at a sampling interval of Tb)

BUT how much bandwidth is required for a given transmission rate???

46

BANDWIDTH REQUIRED - 1Bandwidth required depends on whether the signal is at BASEBAND or at PASSBAND

Bandwidth needed to send baseband digital signal using a Nyquist LPF isBandwidth = (1/2)Rb(1 + )Bandwidth needed to send passband digital signal using a Nyquist Bandpass filter isbandwidth = Rb(1 + )

NOTE: It is the Symbol Rate that is key to bandwidth, not the Bit Rate

47

BANDWIDTH REQUIRED - 2

SYMBOL RATE is the number of digital symbols sent per secondBIT RATE is the number of digital bits sent per secondDifferent modulation schemes will “pack” different numbers of Bits in a single Symbol

BPSK has 1 bit per symbolQPSK has 2 bits per symbol

48

BANDWIDTH REQUIRED - 3OCCUPIED BANDWIDTH, B, for a signal is given by B = Rs ( 1 + ) where Rs is the symbol rate and is the filter roll-off factorNOISE BANDWIDTH, BN, for a channel will not be affected by the roll-off factor of filter. Thus BN = Rs

49

BANDWIDTH EXAMPLE - 1GIVEN:

Bit rate 512 kbit/sQPSK modulationFilter roll-off, , is = 0.3

FIND: Occupied Bandwidth, B, and Noise Bandwidth, BN

SOLUTION:Symbol Rate = Rs = (1/2)

(512 103) = 256 103

2 bits per symbol

Number of bits/s

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BANDWIDTH EXAMPLE - 2

Occupied Bandwidth, B, isB = Rs (1 + ) = 256 103 ( 1 + 0.3) = 332.8 kHz

Noise Bandwidth, BN, is BN = Rs = 256 kHz

Now what happens if you have FEC?Example with FEC

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BANDWIDTH EXAMPLE - 3

SAME Example, but 1/2-rate FEC is now used

SOLUTIONSymbol Rate, Rs = (1/2) (2) (512

103)= 512 103 symbols/s

Occupied Bandwidth, B, is B = Rs ( 1 + )

= 665.6 kHz

2 bits per symbol

Number of bits/s

1/2-rate FEC used

52

BANDWIDTH EXAMPLE - 3

Noise Bandwidth, BN, isBN = Rs = 512 103 = 512 kHz

Summary:High Modulation Index More Bandwidth EfficientFEC (Block or Convolutional) Increases bandwidth required

53

Digital Modulations

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Digital ModulationsIn digital communications, the modulating signal is a binary or M-ary data.The carrier is usually a sinusoidal wave.Change in Amplitude: Amplitude-Shift-Keying (ASK)Change in Frequency: Frequency-Shift-Keying (FSK)Change in Phase: Phase-Shift-Keying (PSK)Hybrid changes (more than one parameter). Ex. Phase and Amplitude change: Quadrature Amplitude Modulation (QAM)

55

Binary Modulations – Basic Types

These two have constant envelope (important for amplitude sensitive channels)

56

Coherent and Non-coherent DetectionCoherent Detection (most PSK, some FSK):

Exact replicas of the possible arriving signals are available at the receiver.This means knowledge of the phase reference (phased-locked).Detection by cross-correlating the received signal with each one of the replicas, and then making a decision based on comparisons with pre-selected thresholds.

Non-coherent Detection (some FSK, DPSK): Knowledge of the carrier’s wave phase not required. Less complexity.Inferior error performance.

57

Design Trade-offsPrimary resources:

Transmitted Power.Channel Bandwidth.

Design goals:Maximum data rate.Minimum probability of symbol error.Minimum transmitted power.Minimum channel bandiwdth.Maximum resistance to interfering signals.Minimum circuit complexity.

58

Coherent Binary PSK (BPSK)Two signals, one representing 0, the other 1.

Each of the two signals represents a single bit of information.Each signal persists for a single bit period (T) and then may be replaced by either state.Signal energy (ES) = Bit Energy (Eb), given by:

ts

tfAtfAts

tfAts

cc

c

1

2

1

2cos2cos

2cos

2

2TAEE bS

b

b

T

EA

2Therefore

59

Orthonormal basis representation

Gram-Schmidt Orthogonalization: basis of signals that are both ortogornal between them and normalized to have unit energy.

Allows representation of M energy signals {si(t)} as linear combinations of N orthonormal basis functions, where N<=M.Ex.: N=2

ji

jidttt j

T

i if 0

if 1)()(

0

Tt0 )2sin(2

)(

Tt0 )2cos(2

)(

2

1

tfT

t

tfT

t

c

c

60

BPSK representationLet’s consider the unidimensional base (N=1) where:

Let’s also rewrite the signal amplitudes as a function of their energy:

tfT

Ets

tfT

Ets

cb

b

cb

b

2cos2

2cos2

2

1

Tt0 )2cos(2

)(1 tfT

t cb

61

BPSK representationTherefore, we can write the signals s1(t) and s2(t) in terms of 1(t):

b12

b11

Tt0 )()(

Tt0 )()(

tEts

tEts

b

b

• Which can be graphically represented as:

62

BPSK Physical Implementation

cos(2fct)

t1

0

t

0 1 0 0 1 1 1 0-A

+A

63

Detection of BPSK

Actual BPSK signal isreceived with noiseWe assume AWGN inthis classOther noise properties are possible

AWGN is a good approximationOther noise models are more complex

Constellation becomes a distribution because of noise variations to signal

64

Recall Gaussian DistributionArea to the right of this

line represents Probability (x>x0)

xx0

2

-

2

1-Q)x(x Probabiliy b0b0

0

xerfc

x

Both Q(.) and erfc(.) functions are integrals widely tabled and available as functions in Excel and calculators

= mean=standard deviation

y

z dzeyerfc22

Where:

y

eyerfc

y2

Approximation for large positive values of y

65

Calculating Error Probability

Noise Spectral Density = N0 Noise Variance:

202 N

i

-A +A

AWGN on Signal

0

P(-A/+A) = P(+A/-A)

0

b

0

b

bbb

E

2

1

22

E

2

1

2

E

2

1EQExP/0)1P(/1)0P(

Nerfc

Nerfc

erfc

20N

BPSK error probability

66

Bit Error Rate (BER) for BPSKBER is therefore given by

o

b

N

E

o

b

o

b

e

NE

BER

N

EBER

2821.0

erfc2

1 Eb/No

(dB) BER0 0.082 0.044 0.0146 0.00278 2*10-4

10 4*10-6

10.543 10-6

Approximationvalid for Eb/Nogreater than ~4 dB

Note that these calculations are for synchronous detection

67

Ambiguity Resolution

We haven’t discussed yet how to tell which signal state is a 1 and which a 0Because of variations in the signal path, its impossible to tell a prioriTwo common approaches resolutions:

Unique WordDifferential Encoding

68

Unique Word Ambiguity Resolution

A specific, known unique word is sentThe unique word is sent at a known time in

the dataThe correct signal state is chosen as 1 to

correctly decode the unique wordUsually implemented with two detectors - the output of the correct one is simply usedCould lead to problems until a new UW is RX if a phase slip occurs

All bits after slip will be received wrong!

69

Differential Encoding Ambiguity Resolution

Data is not transmitted directlyEach bit is represented by:0 => phase shift of radians1 => no phase shift

in the carrierThis results in ~ doubling the BER since any error will tend to corrupt 2 bitsBER is then

o

b

N

E

o

bo

bBPSKDBPSK e

NEN

EBERBER

5642.0erfc2

Valid for BER<~0.01

70

Coherent Quaternary PSK (QPSK)Four signals are used to convey information

This leads to a constellation of:when shown as a phasorreferenced to the signal phase, Each of the two states representsa two bits of information

2/2cos

2cos

2/2cos

2cos

4

3

2

1

tfAts

tfAts

tfAts

tfAts

c

c

c

c

i

q

+A-A

+jA

-jA

Constant Modulus =>

71

QPSK Constellation Representation

In this case we use the following orthonormal basis:

Which gives, after application of some trigonometric identities, the following constellation representation:

Tt0 )2sin(2

)(

Tt0 )2cos(2

)(

2

1

tfT

t

tfT

t

c

c

72

QPSK Constellation

73

QPSK Waveform

74

QPSK Physical Implementation

cos(2fct)

90o

Demux

1/2 rate data

1/2 rate data

full ratedata, Rb

QPSK symbolsRs = Rb/2

Note that the QPSKsignal can be seen tobe two BPSK signalsin phase quadrature

75

Bit Error Rate (BER) for QPSKThe BER is still the probability of choosing the wrong signal state (symbol now)Because the signal is Gray coded (00 is next to 01 and 10 for instance but not 11) the BER for QPSK is that for BPSK:BER (after a lot of derivation) is given by:

o

b

N

E

o

bo

bQPSK e

N

EN

EBER

2821.0erfc

2

1Approximationvalid for Eb/Nogreater than ~4 dB

Note that Eb is here, not Es!

76

Frequency Shift KeyingTwo signals are used to convey information

In principle, the transmitted signal appears as 2 sinx/x functions at carrier frequenciesEach of the two states representsa single bit of informationEach state persists for a single bit period and then may be replaced either stateBER is: 2x BPSK BER for coherent

for non-coherent

222

111

2cos

2cos

tfAts

tfAtsConstant Modulus =>

o

b

N

E

eBER 2

2

1

77

Frequency Shift Keying

78

Other Modulations (cont.)

M-ary PSKPSK with 2n states where n>2Incr. spectral eff. - (More bits per Hertz)Degraded BER compared to BPSK or QPSK

QAM - Quadrature Amplitude ModulationNot constant envelopeAllows higher spectral eff.Degraded BER compared to BPSK or QPSK

79

M-ary PSK

80

M-ary QAM

81

Other Modulations

OQPSKQPSKOne of the bit streams delayed by Tb/2

Same BER performance as QPSK

MSKQPSK - also constant envelope, continuous phase FSK1/2-cycle sine symbol rather than rectangularSame BER performance as QPSK

82

Shannon Bound

1948 Shannon demonstrated that, with proper coding a channel capacity of

dB 6.1or 443.1

lim since

443.1lim

e therefor since

1log1log 22

o

b

o

bbb

Bb

o

bb

B

o

bb

o

bb

b

o

bo

bb

N

E

N

ERR

CapacityR

N

ERCapacity

N

E

BW

R

N

S

N

E

BW

R

E

N

R

BW

N

ER

N

SBWCapacity

Required channel qualityfor error free communications=>we’re doing much worse

83

Modulation Schemes Error Performance

84

M-ary PSK Error Performance

85

Operation Point Comparison

86

Error Detection and Correction

87

Coding position on a transmission system

88

Error Protection CodingThree types to discuss

Parity Bits (error detection only, really a subset of BC)Block Coding (eg. Reed-Solomon)Convolutional Coding (eg. Viterbi or Turbo)

All impose an overhead on channelAdditional information must be transmittedThis additional information is the redundant information of the error coding

Block codes develop less coding gain but are (much) easier to process (esp. at high data rates)Often advantageous to use both togetherGain depends on BER - must be careful hereCoding ~ necessary for non-lin. ch.s (discuss BER flare)

Forward ErrorCorrection codes

89

Parity Bits

The data is parsed into uniform k-bit words7 bits is a common data lengthAn extra bit is added to this to make an k+1 bit transmission wordThe value of the k+1th bit is determined by:

Even parity:Odd parity:

Doesn’t correct errors just detects, and only an odd number of errors (discuss why)

kk

kk

bitbitbitbit

bitbitbitbit

211

211

90

Block Codes - 1

The data is parsed into uniform k-bit blocksCoder adds n-k unique redundant bitsAn n-bit block is transmittedCoder is memoryless - only this block usedTransmitted data rate is then:Redundant bits used to correct errors

k

nRR bc

91

Block Codes - 2

Hamming, Golay, BCH, Reed-Solomon, maximal-length are different types of block codesImportant for this class

Depending on amount of redundancy added, block codes may be used to detect only or to actually correct bit errors.Block codes correct burst errors (ie. adjacent errors) as well as they do random errors. Not as powerful as convolutional

k

nRR bc

92

Ciclic Codes (block codes)

rRR bc1

93

Convolutional Codes - 1

Process as sliding window of dataUse constraint length of k (window length)Transmit at rate of where r is rate Fairly high coding gainTurbo codes are even higher (but harder)Do not handle burst errors well

rRR bc1

r=1/3 r=1/2 r=2/3 r=3/4Eb/Nouncoded

(dB)

BERk=7 k=8 k=5 k=6 k=7 k=6 k=8 k=6 k=9

6.8 10-3 4.2 4.4 3.3 3.5 3.8 2.9 3.1 2.6 2.69.6 10-5 5.7 5.9 4.3 4.6 5.1 4.2 4.6 3.6 4.2

11.3 10-7 6.2 6.5 4.9 5.3 5.8 4.7 5.2 3.9 4.8infinite 0 7.0 7.3 5.4 6.0 7.0 5.2 6.7 4.8 5.7

Coding Gain (dB) for various Viterbi codes

94

Convolutional Codes - 2

95

Trellis Coding - 1

96

Trellis Coding - 2

97

Interleaving and Code on Code

Problem: Noise often happens in burstsCan use interleaving - spreading adjacent bits of convolutional code over time to avoid having adjacent bits corruptedBut, we still have a quandary:

Block codes are robust against burstsConvolutional codes provide more gain

Solution: use both inner convolutional and outer block codes to get both effects

98

Summary of Useful Formulas

99

Summary of Digital Communications -1

Bw = Bandwidth in Hertz

= Roll-off factor (from 0 to 1)

Gc = Coding Gain (convert from dB to linear to use in formulas)

Ov = Channel Overhead (convert from % to fraction : 0 to1)

M = modulation size. (Ex: 2, 4, 16, 64)

Legend of variables mentioned in this section:

BER = Bit Error Rate

100

Summary of Digital Communications - 2

•Bits per Symbol: MLogBs 2

• Symbol Rate [symbol/second]: Ws BR

1

1

•Gross Bit Rate [bps]: WssG BMLogRBR

1

12

•Net Data Rate [bps]:

)1(1

1)1( 2 OvBMLogOvRR WGi

101

Summary of Digital Communications - 3

•Required Eb/No (assuming no coding) [adimensional]:(function of modulation scheme and required bit error rate – see table later)

BER)Scheme,n (Modulatiofunction Table 1

theoryfrom Req0

N

Eb

•Required Eb/No (using coding gain) [adimensional]:

theoryfrom Req0Re0

1

N

E

GN

E b

cq

b

•Required C/N [adimensional]:

W

G

q

b

q B

R

N

E

N

C*

Re0Re

102

Summary of Digital Communications - 4

 •Required Signal Strength [Watts]: Where k = Boltzman constant = 1.38e-23 J/Hz

TS = System Noise TemperatureT0 = ambient temperature (usually 290 K)F = System Noise figure in linear scale (not in dB)

FBkTN

CBkT

N

C

NN

CC

Wq

Wsq

qq

0ReRe

ReRe

103

BER Calculation as a Function of Modulation Scheme and Eb/No Available

• Equations given on next slide are used to calculate the bit error rate (BER) given the bit energy by spectral noise ratio (Eb/No) as input.

• These functions are used in their direct form for the bit error rate calculations. Excel and some scientific calculators provide the solution for the “erfc” function.

• The formulas provided can be inverted by numerical methods to obtain the Eb/No required as a function of the BER.

• Also possible to draw the graphic and obtain the “inverse” by graphical inspection.

104

BER Calculation as a Function of Modulation Scheme and Eb/No Available - 2

Modulation

Scheme

Coh-PSK BER = 0.5*ERFC(SQRT((Eb/No)))

Coh-DPSK BER = ERFC(SQRT((Eb/No)))-0.5*(ERFC(SQRT((Eb/No)))) 2̂

Coh-QPSK BER = ERFC(SQRT((Eb/No)))-0.25*(ERFC(SQRT((Eb/No)))) 2̂

Ncoh-QPSK(Dif) BER = ERFC(SQRT(2*(Eb/No))*SIN(PI()/4))

Coh-8-PSK BER = ERFC(SQRT(3*(Eb/No))*SIN(PI()/8))

Ncoh-8PSK(Dif) BER = ERFC(SQRT(2*3*(Eb/No))*SIN(PI()/(2*8)))BER = ((1-1/K)/(LOG(K)/LOG(2)))*ERFC(SQRT(3*(LOG(K)/LOG(2))/(K 2̂-1)*(Eb/No)))Where K = 4BER = ((1-1/K)/(LOG(K)/LOG(2)))*ERFC(SQRT(3*(LOG(K)/LOG(2))/(K 2̂-1)*(Eb/No)))Where K = 6BER = ((1-1/K)/(LOG(K)/LOG(2)))*ERFC(SQRT(3*(LOG(K)/LOG(2))/(K 2̂-1)*(Eb/No)))Where K = 8BER = ((1-1/K)/(LOG(K)/LOG(2)))*ERFC(SQRT(3*(LOG(K)/LOG(2))/(K 2̂-1)*(Eb/No)))Where K = 16

Coh-4FSK BER = 0.5*ERFC(SQRT((Eb/No)/2))

256-QAM

32-QAM

64-QAM

Theoretical BER Calculation

16-QAM