Chapter 5: Signal Encoding Techniques COE 341: Data & Computer Communications (T061) Dr. Radwan E. Abdel-Aal

Chapter 5:

Signal Encoding Techniques

COE 341: Data & Computer Communications (T061)Dr. Radwan E. Abdel-Aal

2

Where are we:

Physical Layer

Transmission Medium

Data Link

Chapter 4: Transmission Media

Chapter 3: Signals and their transmission over

media, Impairments

Chapter 5: Encoding: From data to signals

Chapter 7: Data Link: Flow and Error control

Chapter 6: Data Communication: Synchronization,

Error detection and correction

Chapter 8: Improved utilization: Multiplexing

3

Agenda Overview:

Implementation of the 4 encoding combinations introduced in chapter 3

Encoding Digital Data as Digital Signals Encoding Digital Data as Analog Signals Encoding Analog Data as Digital Signals Encoding Analog Data as Analog Signals

4

Four Data/Signal Combinations Signal

Analog Digital

Data

Analog

- Same spectrum as data (base band): e.g. Telephony- Different spectrum (modulation of a carrier): e.g. AM, FM, PM

Use a (converter): codec, e.g. PCM (pulse code modulation)

Digital Use a (converter): modem e.g. ASK, FSK, PSK

-Two signal levels: e.g. NRZ-More complex encoding: e.g. Manchester

3

2 1

4

5

Encoding Techniques1. Digital data as digital signal2. Digital data as analog signal: Converter (Modem)3. Analog data as digital signal: Converter (Codec)4. Analog data as analog signal In general:

When the outcome is a digital signal we use an Encoding process

When the outcome is an analog signal we use a Modulation process But we call the modulation of analog signal by digital data

shift-keying

6

Encoding:

Encoder Decoder

Modulator Demodulator

digitalor

AnalogData

DigitalSignalTransmission

g(t)

m(t)

fc

x(t)

t

s(f)

ffc

g(t)

m(t)

x(t)

s(t)

digitalor

AnalogData

AnalogData

AnalogSignalTransmission

AnalogData

Modulation:

fc

Source DestinationLink

m(f)

0

m(f)

0

Shift in frequency

7

Encoding and Modulation: Remarks Encoding is simpler and less expensive than modulation

Encoding into digital signals allows use of modern digital transmission and switching equipment Basis for Time Division Multiplexing (TDM)

Modulation shifts baseband signals to a higher region in the frequency spectrum (needs fcs at both ends) Basis for Frequency Division Multiplexing (FDM)

Optical fibers and unguided media and can carry only analog signals

8

Terminology Unipolar Signals

Binary data represented by signals of the same polarity, e.g. 0: +5 V, 1: +10 V DC content

Bipolar (Polar) Signals Binary data represented by signals of opposite polarity,

e.g. 0: +5 V, 1: -5 V ideally Zero DC content

9

Terminology, Contd.Data rate and Signaling rate

Mark and Space Binary 1 and Binary 0 respectively

Duration of a bit (Tb) Time taken for transmitter to emit a bit

Data rate, R ( = 1/Tb) Rate of data transmission Measured in bits per second (bps)

Duration of a Signal Element (Ts) Minimum duration of a signal pulse

Modulation (signaling) rate, D (1/Ts) Rate at which the signal level changes with time Measured in bauds = signal elements per second

Not always Tb = Ts !!

- Multi-symbol transmission (M = 4, 8, …): Tb < Ts

- Return to zero (RZ) codes: Ts < Tb

10

Example: Two different coding methods

Data rate = 1/1s

= 1 M bps

Signaling Rate for NRZI: = 1/1s

= 1 M bauds

Signaling Rate for Manchester: = 1/0.5s

= 2 M bauds

Tb

Ts

Ts

11

Interpretation of the Received Signal

12

Interpreting Received Signals Requirements at RX: Determine timing of bits – Bit start and end (When to look)

Need Synchronization (Chapter 6)

Detect signal levels at mid-bit points Compare signal level with a threshold level to decide on data

Factors affecting successful signal interpretation

(Affect bit error rate) Bandwidth Signal to noise ratio Data rate Also Encoding/Modulation scheme, e.g. binary or multi-level

13

1. Digital Data, Digital Signal Digital signal

Voltage/current pulses having a few discrete levels (2 levels for binary)

Each pulse is a signal element Binary data is encoded into those signal elements

14

Encoding SchemesEncoding: Mapping data to signal elementsSchemes for encoding digital data as digital signals

The Nonreturn to Zero (NRZ) Group: Nonreturn to Zero-Level (NRZ-L) Nonreturn to Zero Inverted (NRZI)

The Multi-level Binary Group: Bipolar-AMI (Alternate Mark Invert) Pseudoternary

The Bi-Phase (RZ) Group: Manchester Differential Manchester

Scrambling Group: B8ZS (Bipolar with 8-Zeros Substitution) HDB3 (High Density Bipolar 3-Zeros)

15

Why so Many Encoding Schemes? Aspects of comparison between schemes: Signal Spectrum: Desirable Features Small high frequency content: Reduces effective bandwidth No dc component: Allows ac transformer/capacitor coupling,

required sometimes for electrical isolation Concentrate

signal power in

the middle of

the bandwidth:

Avoids problems

at BW edges, e.g.

delay distortion.

2Normalized frequency (f/r)

0 0.5 1 1.5

1

1

3

24

Power Spectral Density, Watt/Hz

16

ClockingSynchronizing RX to TX can be achieved using: An external clock,

or better: A built-in synchronizing mechanism in the signal itself!

(so, a code with many signal transitions is better)

Error detection Mostly handled by higher layers, e.g. data link control But error detection capabilities built into the signal

encoding scheme would help! Advantage: Implemented much faster (in hardware)

Aspects of comparison between schemes:

17

Comparison of Encoding Schemes, contd. Performance with interference and noise

Some encoding schemes perform better than others:

e.g. with differential encoding: data is encoded as signal transition/no signal transition, and data detection at RX is less affected by noise

Cost and complexity Some codes require signaling at a rate greater than the

data rate (e.g. RZ)

At higher signaling rates this requires higher bandwidth, faster circuits, etc. (larger costs)

18

NRZ GroupPros and Cons:

Pros Easy to implement Modest bandwidth requirements

Cons Large DC component Poor TX-RX synchronization: e.g. No signal transitions for long strings of all 0’s

(so few edges are available for synchronization) Used for magnetic recording Not used much for signal transmission

0 0.5 1 1.5

1

1

3

2 4

19

The RZ Solution

Advantages of RZ: Lower DC content (signal spends more time around 0V) Guarantees an edge per bit (Better TX-RX synchronization)

Disadvantages of RZ: Higher frequency content More difficult to implement

20

NRZ Spectrum

-0.5

0

0.5

1

1.5

0 0.5 1 1.5

NRZ-L,NRZI

B8ZS,HDB3

AMI, Pseudoternary

Manchester, Differential Manchester

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Normalized frequency (f/R)

2

Power Spectral Density, Watt/Hz

Frequency relative to data rate (binary data)

21

NRZ-L: Non return to Zero-Level

Two different signal voltages for the 0 and 1 data bits Voltage level is constant (no return to zero, so no signal

transition) for the full during of the data bit interval e.g. 0 V for zero and a positive voltage for one More often, negative voltage for one data value and

positive for the other (bipolar signal) (Why?) An example of absolute encoding:

Encoding data directly as a signal level

22

NRZI: Nonreturn to Zero Invert

Still constant voltage level for bit duration of (hence NRZ) But data is encoded as presence or absence of signal

transition at beginning of bit time: Transition (low to high or high to low): Denotes binary 1 No transition: Denotes binary 0

This is an example of differential encoding: Encoding data as a change/no change in signal level

23

Differential Encoding Data is represented by signal transitions rather

than signal levels Advantages;

With noise, signal transitions (or lack of them) are detected more easily than signal levels Better noise immunity

In complex transmission layouts, it is easy to accidentally lose sense of polarity

+_

RXEffect of swapping terminals on:- NRZ-L - NRZI

24

The Multilevel Binary Group Uses more than two signal levels (3 in this case) Signal is multi-level but data is still binary! Bipolar-AMI (Alternate Mark (1) Inversion)

0 data is represented by no line signal 1 data represented by positive or negative pulse The “1” pulses alternate in polarity (why? 2 reasons!) Advantages:

No net dc component Lower bandwidth than NRZ No loss of sync with a long string of 1’z

(but zeros still a problem- Will try to solve it later) Alteration of pulse polarity also useful for error detection

25

Pseudoternary

Opposite of Bipolar-AMI: 1 represented by no line signal 0 represented by alternating positive and negative

pulses Could be called Bipolar-ASI: (Why?) No advantage or disadvantage over bipolar-AMI

26

Bipolar-AMI and Pseudoternary

CancelingAdding

27

Multilevel Spectrum

-0.5

0

0.5

1

1.5

0 0.5 1 1.5

NRZ-L,NRZI

B8ZS,HDB3

AMI, Pseudoternary


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Normalized frequency (f/r)

2

Signal Power density, Watt/Hz

Frequency relative to data rate

28

The Multilevel Binary Group: Advanatges No net dc component Spectrum centered at the middle of the BW Lower bandwidth than NRZ No loss of sync with a long string of 1’z

(but zeros still a problem- Will try to solve it later) Alteration of pulse polarity also useful for error detection: Next slide

WK 9

29

Bipolar-AMI and Pseudoternary All Single Pulse Errors-

Detected

CancelingAdding

Double Pulse Error-Undetected

Double Pulse Error-Detected

30

Disadvantages of Multilevel Binary Coding scheme not as efficient as NRZ:

We send only one bit at a time (1 or 0 data) Only M = 21 = 2 signal levels should be enough, but we are sending 3 levels > 2

We use 3 signal levels Enough to represent log23 = 1.58 bits > 1 bit

Receiver Design and Noise Performance Now receiver must distinguish between three signal

levels (+A, -A, 0) Need better receiver design Requires approximately 3dB higher SNR for the

same probability of bit error (error rate)

N = Log2 (M)

No. of bitssent during eachsignal element

No. of signal levels used

31

Performance with noise: NRZ Vs AMINRZ Multi-Level Binary (AMI)

+A

-A

+A

-A

0

For the same error rate: AMI requires higher SNR noise (lower noise)

i.e. double the Eb/N0

(for same B and R)

(hence the 3 dBs differencebetween the two curves)

For the same SNR (same Eb/N0 ) AMI has higher error rate

i.e. AMI has poorer performance with noise

Noise level needed to cause an error

In both cases signal level is 2A pk2pk

32

The Biphase Group (2 signal phases per bit) Manchester Transition in middle of each bit period Transition serves both as a clock edge and data representation

Low to high represents 1 High to low represents 0

Used by the IEEE 802.3 specification for Ethernet LAN (short distances)

Differential Manchester Dedicated mid-bit transition used only for clocking Data representation is at start of bit:

No transition at start of a bit period represents 1 Transition at start of a bit period represents 0

(Invert on 0’s – opposite of NRZI) An example of differential encoding Used by IEEE 802.5 specification for Token Ring LAN

33

Manchester Encoding• Mandatory transition in middle of each bit period• Low to high represents 1• High to low represents 0• Transitions at start of bit only where required

Any error detection

capabilities??

Note: This is not differential

34

Differential Manchester Encoding• Mandatory midbit transition for clocking• Transition (either direction) at bit start represents 0 (Invert on zeros)• No transition at bit start represents 1

Any error detection

capabilities??

35

Biphase Group Spectrum

-0.5

0

0.5

1

1.5

0 0.5 1 1.5

NRZ-L,NRZI

B8ZS,HDB3

AMI, Pseudoternary


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2



Note higher frequency content

36

Biphase Pros and Cons Pros

Guaranteed mid bit transitions Synchronization facility (Example of self clocking codes)

Ideally no dc component (using bipolar signals) Error detection

Detecting absence of expected (mandatory) transitions

Cons At least one transition per bit time and possibly two

Maximum modulation (signaling) rate is twice that of NRZ So, requires more bandwidth Therefore, used over shorter distances (in LANs)

37

Data rate & Modulation (signaling) rate Data rate, R = 1/Tb bps

Signaling Rate, D = 1/Ts bauds

If we use k signal elements per bit, then:

Signaling (modulation) rate, D = Data rate, R (bit/s

x k (signal elements/bit)

Signal elements/s (bauds)

Tb

Ts

k = No. of signal elements/bit

= No. of signal transitions ÷ No. of bits transmitted

(over a given period of n Tbs)

k=1

k=2

Data

Signal

Signal

3 bits TXed

Ts Ts

6 signal transitions= 6 signal elements

k = 2

38

Comparison of k for various encoding schemes

k=2

e.g., here k = 1.5 i.e. baud rate D is 1.5

x data rate R

39

Digital data, Digital signal Encoding

Bipolar -AMI

Pseudoternary

0 1 0 0 1 1 0 0 0 1 1

NRZ

NRZI

Manchester

DifferentialManchester

40

Scrambling Group: B8ZS, HDB3Modifications on Bipolar Multilevel codes Use bit scrambling to replace data bit sequences that would otherwise produce a constant signal voltage, with a more appropriate bit sequence containing changes

Helps overcome constant DC problems with Multilevel Binary codes (poor synch)

So, a “filling” (replacement) bit sequence is inserted where necessary

Criteria for a “Filling sequence” Should produce enough transitions for synchronization Must be recognized by receiver for replacement with original data Not likely to be generated by noise

(difficult for noise/interference to produces it) Should occupy the same bit length as original data

(so no extra overhead in the data rate)

41

Scrambling Group: B8ZS, HDB3

Advantages: No long sequences of zero level line signal No dc component No reduction in useful data rate (No extra data sent) Built-in error detection capability

42

B8ZS Bipolar With 8 Zeros Substitution Improvement on bipolar-AMI If an octet of 8 zeros and the last pulse preceding was

positive (+):Transmitter encodes the 8 zeros as 000+-0-+(how many level changes does this introduce?)

If an octet of 8 zeros and last voltage pulse preceding was negative (-): Transmitter encodes as 000-+0+- (shown in Fig. 5.6)

Each insertion has two intentional violations of the basic AMI code rule:+000+-0-+

-000-+0+- A strange event unlikely to be caused by noise Receiver should detect it and interpret as an octet of 8 zeros

(original data) No additional data sent No penalty on genuine data rate

43

B8ZS

-000-+0+-

V: Violation

B: Bipolar (Valid)

See how the insertion satisfies the 5 requirements:-Detectable at RX-Difficult for noise to generate-Introduces transitions-Does not introduce DC-Error detection capability

44

HDB3 High Density Bipolar 3 Zeros Also based on bipolar-AMI 4th zero always replaced with an intentional code violation String of four zeros replaced with either:

1 pulse -000- or +000+ (violation with preceding pulse) or 2 pulses -+00+ or +-00- (internal violation within the insertion)

What determines whether 1 or 2 pulses? Successive insertion violations must alternate in polarity (why?):

-00000000 -000-+00+ or +00000000 +000+-00- If insertions are separated by ‘1’ pulses: The new insertion is

determined by the following rules (Table 5.4) Even number of 1s, with last pulse p (+ or -) p00p

Odd number of 1s, with last pulse p (+ or -) 000p

45

HDB3V: Violation

B: Bipolar (Valid)

-000-+00+

Even number of 1s after last substitution, with the last pulse (+) p00p -00-

1s

Odd number of 1s after last substitution, with the last pulse (-) 000p 000- p

p

46

B8ZS, HDB3 Spectrum

-0.5

0

0.5

1

1.5

0 0.5 1 1.5

NRZ-L,NRZI

B8ZS,HDB3

AMI, Pseudoternary


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47

2. Digital Data, Analog Signal Encoding e.g. over public telephone system

300Hz to 3400Hz Use modem (modulator-demodulator)

Modulation (here called shift keying) manipulates one property of a carrier sine wave: Amplitude shift keying (ASK) Frequency shift keying (FSK) Phase shift keying (PSK)

48

Modulation Techniques

Phase shift angles = ?

Digital Data

Digital Signal

Analog Signals

FSK

PSK

49

Amplitude Shift Keying (ASK) Values represented by different amplitudes of the

carrier sine wave Usually, one amplitude is zero

i.e. presence and absence of carrier

e.g. switching the light sent through a fiber on and off Susceptible to noise and sudden changes in gain Up to 1200bps on voice grade lines Used over optical fiber

cos(2 ) binary 1( )

0 binary 0cA f t

s t

50

Frequency Shift Keying (FSK) Most common form is binary FSK (BFSK)

The two binary data values represented by two different frequencies (near and on both sides of a central carrier frequency fc)

Less susceptible to noise than ASK(Same as with FM Radio: Frequency can be detected correctly in the presence of noise better than amplitude)

Applications: Up to 1200bps on voice grade lines Also used at High frequency radio (3-30 MHz) And at even higher frequencies on LANs using coaxial cables

0binary )2cos(

1binary )2cos()(

2

1

tfA

tfAts

fcf1 f2

fc fc

51

FSK

f1

Carrier 2

Datasignal

Carrier 1

vd(t)

v1(t), f1

v2(t), f2

vFSK(t)

Signalpower

Frequency

frequency spectrum

f2fc

f f

f1 = fc- fc

f2 = fc+ fcSpectrum spread due to chopping

52

FSK for digital data on Voice Grade Lines

300

1070 2225

1270 2025 3400

Amplitude

Frequency(Hz)

Spectrum of signalin one direction

f1, f2 f1, f2

Bell Systems108 Series modem

Full Duplex Communication

(in the 2 directions simultaneously)

fc = ? for left and right

Two Spectra overlap(Some Interference)

fc = ? for left and right

53

Multiple FSK (MFSK)

More than two frequencies used An example of multi-level coding (M levels) Each signalling element conveys more than one bit (L

bits, L = log2 M) This increases bandwidth efficient

(high BE = C/B values) (Higher data rates for the same signalling rate)

But in general, multi-level coding is more prone to error due to noise(Unless you do something about it, e.g. orthogonally)

To improve BW utilization (efficiency) we send one of multiple signal symbols (frequencies) every signal element More than 1 bit at a time

54

Multiple FSK (MFSK)

- Frequency separation = 2 fd

- Bandwidth Required = M (2fd)

- Minimum Ts (signal element duration) = 1/(2fd) Max signaling rate D = 1/Ts = 2fd

Max data rate R = D L = 2fd L

(Half the frequency separation)

Important Parameters

fc before)

i.e. different frequencies

Ts

55

Multiple FSK (MFSK)

250kHz

fc

75kHz

2fd=50kHz

f1

425kHz

f8

Bandwidth = M (2fd) = 8 x 50 = 400 kHz

(< 2 fc, so OK)

Min Ts = 1/ (2fd) = 1/50 KHz = 20 s

Max signaling rate = 1/Ts = 2fd

= 50 kBaudsMax Data rate = Max Signaling rate x L = 50 KHz x 3 = 150 Kbps

signaling kBauds

Data sent:

Correction!

Frequency:

56

Multiple FSK (MFSK)M = 4L = Log2 (M) = 2

00

111001

b

57

Phase Shift Keying (PSK) Phase of carrier signal is shifted to represent data Binary PSK: Absolute

Two phases (spaced at 180) represent the two binary digits

Where d(t) = +1 for ‘1’ data and -1 for ‘0’ data

58

Differential PSK (DPSK) Phase shifted relative to the previous signal element,

rather than some reference signal:

0: Do not reverse phase 1: Reverse phase (as with NRZI, invert on 1)) (A form of differential encoding) Advantage: - No need for a reference oscillator at RX to determine absolute phase

59

Multi-level PSK (MPSK) 4 different phases spaced at /2 (90o) Multilevel signaling, so:

More efficient use of bandwidth (i.e higher data rate for the same signaling rate)

Each signal element represents log2 4 = 2 bits

1-1

1

-1

-3/4

-/4

Bit pair transmitted

60

Quadrature PSK (QPSK) Implementation )2(sin

2

1 )2(cos

2

1 )

42cos(1 tftfntf ccc

)2(sin Q(t) 2

1 )2(cos I(t)

2

1 )( tftfts cc

In phase branch (I)

Quadrature (90) branch (Q)

n = 1, 3, 5, 7

1-1

1

-1

I Q1 1, 0 -1

I and Q are derivedfrom the 2 bits transmitted

61

Quadrature PSK (QPSK) Implementation

)2(sin Q(t) 2

1 )2(cos I(t)

2

1 )( tftfts cc

1 0 1 1 0 0 0 1 1 1

+1

-1+1

-1

- Started with how many phases?- 4 for the price of 2?- Expect error performance similar to BPSK…!

Assign bit to I or Q?

Bits are taken 2 at a time ….

I = 1, Q = -1

62

Quadrature Amplitude Modulation (QAM) An extension of the QPSK just described

Combines both ASK and PSK For example, ASK with 2 levels and

PSK with 4 levels give 4 x 2 i.e. 8-QAM Up to M=256 is possible Large bandwidth savings But some susceptibility to

noise QAM used on asymmetric

digital subscriber line

(ADSL) and some wireless

systems

Constellation

M=8, L = 3

63

True Multilevel PSK (MPSK) Can use more phase angles and more than one

amplitude For example, 9600 bps modems use 12 phase

angles, four of which have 2 amplitudes Gives 16 different signal elements M = 16 and

L = log2 (16) = 4 bits Every signal element carries 4 bits

(Data sent 4 bits at a time) Baud rate D is only 9600/4 = 2400 bauds

(required BW is low … OK for a voice grade lines!)

Complex signal encoding allows high data rates to be sent on voice grade lines having a limited bandwidth

64

Performance of D-A Modulation Schemes Here, bandwidth requirement is the

main concern

(should be minimized)

a. Performance without noise:

Modulator Filter(r)

fc

m(t) s(t)

digitalData Modulated

AnalogSignal

To TX

Modulation Filtering Transmission

CarrierSignal

Filtered, band-limited

signal

x(t) f

s(f)

fc

(Wide bandwidth)

(Limited Transmitted bandwidth, BT Hz), e.g.

0 < r <1

Larger r gives larger

Transmission BT

ModulatedSignal

TransmittedSignal

Data rate R bps

r = Filtering Coefficient

Signaling rate D bauds

DrBT )1(

FilterTruncates BW

65

Performance of D-A Modulation Schemes

We would like to optimize the use of available bandwidth i.e. send data at a high rate with the minimum bandwidth

possible Define the Bandwidth Efficiency, BE as

Although it is ‘Efficiency’, BE can be greater than 1

a. Performance without noise: Transmission Bandwidth (BT) Requirement

TB

RBE

66

For BASK and BPSK : BT directly related to the (signaling, modulation, baud) rate, D

where r is the filtering coefficient; 0< r <1 With binary encoding (not multilevel), D = R, so:

Bandwidth Efficiency, BE:

RrBT )1(

DrBT )1(

a. Performance without noise: Bandwidth Efficiency BE

rB

RBE

TBPSKBASK

1

1,

Performance of D-A (Binary) Modulation Schemes

67

For: NRZ and NRZI Transmission Bandwidth is given approximately by;

D = R for binary, therefore:

and therefore BE is:

)1( 5.0 rDBT


Performance of Digital-Digital (Binary) Modulation Schemes

)1( 5.0 rRBT

rB

RBE

TBPSKBASK

1

2,

r = 0:

BT = 0.5R

Larger r

r = 1:

BT = R

68

For BFSK : Frequency of signal is changed by ± f, about fc (i.e. 2 f)

BT is a function of both f and the (signaling) modulation rate, D:

With binary encoding (not multilevel), D = R, so:

Therefore BE is:

RrfBT )1(2

DrfBT )1(2

12 fffff cc

)1(2

1

)1(2 rR

fRrf

R

B

RBE

TBFSK



69

For BFSK, contd.:

Two extreme cases: f >> R (when fc is large):

f << R (when fc is small):

rallforB

RBE

TfBFSK ;0large ,

fall forrB

RBE

TBFSK

;

1

1small f ,

)1(2

1

)1(2 rR

fRrf

R

B

RBE

TBFSK



, similar to that for BASK, BPSK

70

For MPSK: M phases, L bits/signal element

BT directly related to the (signaling) modulation rate, D

where r is the filtering coefficient; 0< r <1 With M-level encoding, , so:


L

RrBT

)1(

DrBT )1(

r

L

B

RBE

TMPSK

1

Performance of D-A (Multi-level) Modulation Schemesa. Performance without noise: Bandwidth

Efficiency BE

)(log2 M

R

L

RD

L ≥ 2 and r ≤ 1, so BE ≥ 1

)(log2 ML

Same as for BPSK

71

For MFSK: M Frequencies, L bits/signal element

At maximum signaling rate: D = 2fd


)(log2 ML

Mr

L

B

RBE

TMFSK )1(

Performance of D-A (Multi-level) Modulation Schemesa. Performance without noise: Bandwidth

Efficiency BE

RMLog

Mr

L

MRrMDrBT )

)1((

)1()1(

2

)]2()[1(

)1(

fdMr

widthSpectrumrBT

(Equation 5.11 in textbook)

72

Bandwidth Efficiency (BE) DataBE = R/BT

Digital-Analog Modulation Scheme

r = 0 r = 0.5 r = 1

BASK 1.0 0.67 0.5

BFSK (wideband f >> R) 0 0 0

BFSK (narrowband f << R) 1.0 0.67 0.5

BPSK 1.0 0.67 0.5

MPSK: M=4 (L=2) 2.0 1.33 1.0

MPSK: M=8 (L=3) 3.0 2.00 1.5

MPSK: M=16 (L=4) 4.0 2.67 2.0

MPSK: M=32 (L=5) 5.0 3.33 2.5

Filtering Coefficient, r

73

Bit error rate (BER) Plotted Vs Eb/N0 (dBs)

Curves to the left give better performance: Lower S/N for same Error rate Lower Error rate for same SNR

Why QPSK and PSK give the same performance? 2 phase levels (+1,-1) in both

cases Remember QPSK gave 4 phase

levels for the price of 2!

b. Performance with noise: ASK, FSK, PSK, QPSK

Performance of D-A Modulation Schemes

74

b. Performance with noise: MFSK, MPSKPerformance of D-A Modulation Schemes

Larger M Poorer error performanceLarger M Better error performance!

Orthogonal FSK As expected

75

Eb/N0 in terms of the bandwidth efficiency (BE)(for binary transmission)

dBdB

dBTdBdB

b

TTT

bb

dBTdB

BEN

S

B

R

N

S

N

E

BRNS

BNRS

BN

ST

N

E

B

RBE

0

0

BT is the Transmission Bandwidth

76

Example What is the bandwidth efficiency (BE) for FSK, ASK, PSK,

for a bit error rate (BER) of 10-7 on a channel with a SNR

of 12dB ?

dBdBb BESNR

N

E

0

For ASK and FSK (binary): At

BER = 10-7, Eb/N0 = 14.3 dBs Substituting in:

BEASK,FSK = R/BT = 0.6

Similarly for PSK (with Eb/N0 = 11.3 dBs):

BEPSK = R/BT = 1.2 (doubled: 3dB higher)

77

3. Analog Data, Digital Signal Digitization

Conversion of analog data into signals suitable for the digital mode of transmission/storage The digital data can be transmitted digitally as is (e.g. NRZ-L) Or converted to a more appropriate digital code, e.g.

Manchester Or even converted to analog signal for transmission, e.g. ASK

CodeConverter

Digital Signal(NRZ-L)

Analog Signal(ASK)

Digital Signal(Manchester)

Digital Mode ofTransmission

Will study two Types of Codec:

- Pulse Code Modulation (PCM)- Delta Modulation (DM)

CodecOr:

Or:

(Shift Keyer)

78

Two basic tasks to be performed by a digitizer:

Samplingat discrete points in time

QuantizationTo a finite number of levelsin amplitude

• Sampling in time• Quantization in amplitude

Maximum sampling interval allowed = 1/(2fmax); Where fmax is the maximum frequency in the analog signal

Number of quantization levels = 2L, where L is the number of bits allowed for the digital output

Digitizer(Codec)

Analog In Digital Out

L bits (sent serially)

PAM Samples

Digitizing the PAM Samples PCM

“Analog” is continuous in both time and amplitude… Must discretize it in both

79

Sampling Nyquist Sampling Theorem:

If a signal is sampled at regular intervals at a rate higher than twice the highest signal frequency fmax, the samples contain all the information in the original signal

Original signal may be reconstructed from these samples using an ideal low-pass filter

Example: Voice data limited to 4000Hz Require sampling at a rate of at least 8000 sample

per second

80

Quantization using 4 bits

PAM SampleSignal Amplitude, Volts

Vmax = 16 V

Transmitted Serial Code representing the PAM Samples:

24 = 16 signal levels, numbered 0 to 15

Each PAM sample is assigned the number of the nearest quantization level and its digital code is transmitted

Sampling rate: 2B sample/s

Analog signal is band-limited, with bandwidth (0 to B Hz)

Quantization Error = ½ LSB

Must finish sending the n bits of the code within the sampling interval ….before the next sample starts!

Level numberstarting from 0

1 LSB

Qu

anti

zati

on

Data Rate: 2B x 4 bps

81

Pulse Code Modulation (PCM) Start with the analog sampled pulses (Pulse Amplitude Modulation, PAM)

Assign each sample a digital value (= number of the closest quantization level)

n = 4 bit system gives M = 16 levels (M = 2n) Quantization error or noise

Larger for small M (number of levels) Approximations mean it is impossible to recover the original signal exactly SNR for quantization error using n bits is

Each additional bit used for quantization increases SNR by about 6 dB (a voltage factor of 2 = a power factor of 4)

8 bit quantization uses 256 levels Quality comparable with analog transmission

Voice: 2 x 4000 = 8000 samples per second, with of 8 bits each gives a data rate of 8000*8 = 64 kbps

dBndBSNR n 76.102.6 76.12log20 10

82

PCM Example Suppose we want to encode an analog signal that

has voltage levels 0-5v using 2-bit PCM (n = 2 bits) (M = 22 = 4 levels)

We divide the max voltage level into four intervals, so the size of each interval is 5/4=1.25 V Level intervals: 0-1.25, 1.25-2.5, 2.5-3.75, 3.75-5

We select the quantization levels at the middle of each level interval i.e. selected levels are: 0.625, 1.875, 3.125, 4.375 This guarantees a maximum quantization error

of ½ (5V /4) = 0.625 (=1/2 LSB) and quantization SNR = 6 x 2 + 1.76 = 13.76 dB

83

Problem with Linear (Uniform) Encoding Absolute quantization error for each sample is the

same regardless of signal level Signals with lower amplitudes are relatively more

distorted One Solution: make quantization levels not evenly

spaced (denser for low amplitudes) i.e. higher number of quantization steps for lower

amplitudes and smaller number for larger ones Reduces overall signal distortion This is Nonlinear Encoding

84

Effect of Nonlinear Coding

Linear Encoding Non linear EncodingNonlinear Encoding

Weaker signals have smaller quantization errors

Quantization error is fixed- same for both weaker and stronger signals

85

Companding: An analog solution to the problem

Effect of nonlinear coding can also be reduced by companding the analog signal before a linear digital encoding Compressing-expanding At TX: More gain for weak

signals than for strong signals- before encoding

At RX: Reverse operation (de-companding?) How would the de-companding

curve look like?

No Companding(Linear encoding)

86

Example (Problem 5-20) Consider an audio signal with spectral

components in the range of 300 to 3000 Hz. Assuming a sampling rate of 7000 samples per second will be used to generate the PCM signal. To obtain a quantization SNR of 30 dB, what is the

number of uniform quantization levels needed? (SNR)dB = 6.02 n + 1.76 = 30 dB

n = (30 – 1.76)/6.02 = 4.69Always round off to the next higher integer n = 5 bits 25 = 32 quantization levels

What is the data rate required? R = 7000 samples/sec 5 bits/sample = 35 Kbps

87

CODEC - Performance Good voice reproduction

PCM - 128 levels (7 bits) Voice bandwidth (baseband) = 4 KHZ Data rate should be 2 x 4000 x 7 = 56 kbps for PCM

Analysis of Bandwidth requirement: PCM digital transmission requires 56 kbps for 4 KHz

analog signal Using Nyquist channel capacity, this data rate requires

approximately a bandwidth of 28 KHz (B = C/2 = R/2 = 56/2 = 28)

i.e. PCM digital encoding requires a Nyquist bandwidth which is 7 times the bandwidth of the baseband signal!

(= n Bbaseband)

88

CODEC – Performance, Contd. A common PCM scheme for color TV uses 10-bit codes

For bandwidth=4.6 MHz 92 Mbps (i.e. 2*4.6*10) Requires a bandwidth 10 Bbaseband (= n Bbaseband)

Nevertheless, digital encoding continues to grow in popularity, because they allow: Use of repeaters: No cumulative noise Time-division multiplexing (TDM) without the inter

modulation noise of the alternative analog scheme (FDM) Use of the more efficient digital switching techniques in

networks Solution: More efficient coding can be used to overcome the

problem of the larger BW required by digital encoding

89

Delta Modulation: A cheaper alternative to PCM An attempt to reduce complexity (and large R) of PCM

Analog input is approximated by a staircase function Move up or down one fixed amplitude increment () at each

sample interval to track changes in the analog waveform A single bit stream is produced to approximate the

derivative of the analog signal rather than its amplitude Generate a 1 if staircase is to go up (slope + ive) Generate a 0 if staircase is to go down (slope - ive)

Transmit this sequence of 1,0 data (1-bit per sample) Receiver uses this bit stream to reconstruct the

staircase waveform and approximate the original analog waveform

90

Delta Modulation - example

Digital O/P(Only 1 bit/sample!)

Smaller for larger

Larger for larger

1: + ive slope: Signal increasing

0: - ive slope: Signal decreasing

1010 ...Alternating slope: Signal is level

Sampling

Qu

anti

zati

on

91

Delta Modulation - Implementation At mid sampling interval, compare the analog input

to current value of the approximating staircase function If input exceeds staircase function, transmit a 1 and

increment staircase by for the next sample Otherwise generate a 0 and decrement staircase by

for the next sample Output of the DM is a binary bit sequence to be

used for generating the staircase function at RX Reconstruct staircase function at receiving end and

smooth by a low pass filter to reconstruct an approximation of the analog signal

92

Delta Modulation - Implementation +

_

Staircase Generator

Generated Staircase

Reconstructed Staircase

To filtering &Analog WaveformReconstruction

><

Received bitsequence

Transmitted bitsequence

Generated Staircase

Generated Staircase

At Source

At Destination

93

Delta Modulation: Important Design Parameters Two important parameters in DM scheme Size of amplitude step () assigned to each binary digit

Must be chosen to produce a balance between two types of errors or noise (conflicting requirements) When waveform changes rapidly, slope overload noise increases with a

smaller When waveform changes slowly, quantizing noise increases with a

larger (the usual quantization error) Sampling rate, increasing it:

Improves the accuracy of the scheme But increases the data rate requirement

Main advantages of DM: Lower data rate required (1 bit samples!) Simple to implementation

Disadvantage: Larger quantization errors (lower SNR) compared to PCM

94

4. Analog Data, Analog Signals

Modulation Combining an input signal m(t) and a carrier at frequency fc

to produce signal s(t) with bandwidth centered at fc

We had to use a form of modulation (shift keying) to represent digital data as analog signals.

But why modulate signals that are already analog? Higher frequency may be needed for effective transmission

For unguided transmission: impossible to send low frequency baseband signals, e.g. speech, as required antennas would have dimensions in kilometers!

Allows implementing frequency division multiplexing (FDM)

WK 11

95

Types of Analog Modulation

Angle Modulation:

1. Phase, PM

2. Frequency, FM

Amplitude Modulation (AM)

)()(

tdt

td

Carrier

A sin (t)tt

Signal to beTransmitted,x(t)

ModulatingSignal

ModulatedSignals

Effect of modulation on power? Effect of modulation on BW?

A x(t)

f x(t)

x(t)

96

Amplitude Modulation (AM) Simplest form of modulation Accos 2fct is the carrier,

and x(t)= Amcos 2fmt is the input modulating signal Modulated signal expressed as:

na is the modulation index (0 < na 1):

Added ‘1’ is a DC component to prevent loss of information - there will always be a carrier

Scheme is known as double sideband transmitted carrier (DSBTC)

tfAtfnts ccma 2cos]2cos1[)(

Amplitude of modulated wave

c

ma A

An

Portion of the modulating signal

Units of na?

97

DSBTC Amplitude Modulation - Example Given the amplitude-modulating signal x(t)=Amcos 2fmt , find s(t):

Resulting signal has three components: One at the original carrier frequency fc A pair of additional components (side bands),

each spaced fm Hz from the carrier Envelope of resulting signal

With na <1, envelope is exact reproduction of the modulating signal,So it can be recovered at receiver

With na >1, envelope crosses the time axis and information is lost

tffA

tffA

tfA

tffAn

tffAn

tfA

tftfnAts

mcm

mcm

cc

mcca

mcca

cc

cmac

)(2cos2

)(2cos2

2cos

)(2cos2

)(2cos2

2cos

2cos]2cos1[)(

c

ma A

An Am/2Am/2

Ac

fc fmfm

Two Sidebandscontainmodulatingsignal power

So, keep na 1

98

DSBTC Amplitude Modulation - Examples

na = 0.5/1 = 0.5

MatLab Simulations

Envelope

ModulatingSignalfm = ?

Carrierfc = ?

ModulatedSignal

=(1+0.5cos2*pi*t)=(1+nacos2*pi*t)

Am = ?

Ac = ?

Note different vertical scales

na = ?

99

DSBTC Amplitude Modulation - Example

na = 1/1 = 1

Maximum modulation allowed (na = 1)

100

DSBTC Amplitude Modulation - Example

na = 2/1 = 2 (>1)(not allowed)

Beyond maximum modulation allowed (na > 1)

101

Spectrum of DSBTC signal:Modulating signal hasa singlefrequency, fm

tffA

tffA

tfA

tffAn

tffAn

tfA

tftfnAts

mcm

mcm

cc

mcca

mcca

cc

cmac

)(2cos2

)(2cos2

2cos

)(2cos2

)(2cos2

2cos

2cos]2cos1[)(

Am/2Am/2

Ac

fc fmfm

Ac = 2 Vna = 1 = Am/AcAm = Ac = 2V

2

1

102

Spectrum of an DSBTC signal Spectrum of AM signal is: original carrier plus spectrum of original

signal translated on both sides of fc Portion of spectrum f > fc is

upper sideband Portion of spectrum f < fc is

lower sideband Bandwidth Requirement: 2B Example: voice signal 300-3000Hz

With fc 60 KHz Upper sideband is 60.3-63 KHz Lower sideband is 57-59.7 KHz

Bandwidth Requirement: 2 fmmax

Let modulating signal havea bandwidth 0-B Hz(Baseband)

Note orientation of the two sidebands

Bandwidthof ModulatedSignal

Note: Modulating signal amplitude does not affect bandwidth of modulated signal

103

DSBTC Amplitude Modulation Total transmitted power Pt in modulated s(t) is given by

Pc is transmitted power in carrier na should be maximized (but <1) to allow transmission of more power in

signals that carry information Modulated signal contains redundant information (duplicate side bands)

Only one of the sidebands is enough for restoring the modulating signal Possible ways to economize on transmitted power:

SSB: single sideband, uses a filter to select only one of the sidebands and the carrier, saves on BW (= B)

SSBSC: single sideband suppressed carrier, uses a filter to select only one of the sidebands, saves on BW (= B)

DSBSC: double sideband suppressed carrier, carrier is not transmitted, no saving on BW (= 2B)

Suppressing the carrier may not be OK in some applications, e.g. ASK, where the carrier can provide TX-RX synchronization.

21

2a

ct

nPP

Am/2Am/2

Ac

fc fmfm

Am = na Ac

Note: Modulating signal amplitude affects power of modulated signal

104

DSBSC: Double Sideband Suppressed Carrier - Example

Signal is expressed as tftxAts cc 2cos)]([)(

Suppressed Carrier

105

Angle Modulation Includes:

Frequency modulation (FM) and Phase modulation (PM)

Modulated signal is given by

Phase modulation (PM): (the direct way) Instantaneous phase proportional to the modulating signal: np is the phase modulation index

Frequency modulation (FM): (the indirect way) Instantaneous angular frequency deviations from c proportional

to the modulating signal, and we have: So make the derivative of proportional to modulating signal nf is the frequency modulation index

)()( txnt p

)()()( txntt f

)](2cos[)( ttfAts cc

Total Angle

What parameters can I change to change the angle of the modulated signal?

f(t) ’(t)

Units of np?Units of nf?

106

Angle Modulation The total phase angle of s(t) at any instant is [2fct+(t)] Instantaneous phase deviation from that of the carrier is (t) Phase Modulation (PM):

(t) = npx(t), instantaneous phase variations are directly proportional to m(t)

Frequency Modulation (FM): Instantaneous angular frequency, , can be defined as the rate

of change of total phase So, for the modulated signal, s(t)

In FM, ’(t) is made proportional to x(t). So, instantaneous frequency deviations from the carrier frequency are proportional to x(t).

)(2

1)(

)(2

)(2)(2)(

tftf

tf

tfttfdt

dt

ci

i

cci

)(ti

107

Phase Modulation (PM)- Example Derive an expression for a phase-modulated signal s(t) and its instantaneous frequency given: Ac= 5V, and the modulating signal

x(t) = 3 sin 2fmt We know that s(t):

For PM, (t) is given by:

Then s(t) is:

Instantaneous frequency of s(t) is:


)()( txnt p

]2sin32cos[5)( tfntfts mpc

tffnftffn

ftf mmpcmmp

ci

2cos32cos2

)2(3)(

Note: Frequency variations in s(t) phase-lead x(t) amplitude variations by 90

Peak frequency deviation for the PM signal

np is Radians/Volt

phase total2

1)(

dt

dtfi

108

Frequency Modulation: FM From equations opposite,

Peak frequency deviation F is given by:

Where Am is the peak value of the modulating signal x(t)

An increase in the amplitude Am of x(t) increases F, which increases the

bandwidth requirement BT

But average power level of the FM modulated signal is fixed at AC2/2, (does not

increase with Am)

i.e. in Frequency Modulation, Am affects the BW but not the power budget

While in Amplitude Modulation, Am affects the power budget but not the bandwidth

Hz2

1mf AnF

)2sin()(

)()(

)(2

1

tfAtxand

txntand

tff

mm

f

ci

109

Frequency Modulation - Example Derive an expression for a frequency-modulated signal s(t) with Ac= 5V, given the modulating signal

x(t) = 3 sin 2fmt The FM modulated signal s(t) is: For FM, ’(t) is given by:

Then (t) is:

We have:

Substituting for F we get:


)()( txnt f

tff

ndttfndttt m

m

fmf

2cos

2

3 2sin3)()(

Hz 2

3fnF

]2cos2cos[5)( tff

Ftfts m

mc

But frequency varies as ’,

i.e. as sin not as – cos !!

nf is (Radians/s)/Volt

110

Bandwidth Requirement

All AM, FM, and PM result in a modulated signal whose bandwidth is centered around fc

Let B be the bandwidth of the modulating signal (0-B Hz) AM gives only sums & differences of frequencies with fc,

and we have: BT = 2B for DSB systems Angle modulation includes a term of the form cos(…

+cos()) which is a nonlinear term producing a wide range of frequencies fc+fm, fc+2fm, … (the Bessel function)

i.e. Theoretically, an infinite bandwidth is required to transmit an FM or PM signal

111

Practical Bandwidth Requirement for Angle Modulation

Carson’s Rule of thumb

Since is > 0, both FM and PM require a larger bandwidth than AM

For FM, BT= 2F + 2B

BBT 2)1(

FMfor 2B

F

PMfor

B

AnAn

mf

mp

F is the peak frequency deviation

For AM: BBT 2

Documents

Chapter 5: Signal Encoding Techniques COE 341: Data & Computer Communications (T061) Dr. Radwan E. Abdel-Aal