Signal Encoding Techniques

1
COE 341 Data Computer Communications
(T061)Dr. Radwan E. Abdel-Aal

• Chapter 5
• Signal Encoding Techniques

2
Where are we
Chapter 7 Data Link Flow and Error control
Data Link
Chapter 8 Improved utilization Multiplexing
Physical Layer
Chapter 6 Data Communication Synchronization,
Error detection and correction
Chapter 4 Transmission Media
Transmission Medium
Chapter 5 Encoding From data to signals
Chapter 3 Signals and their transmission over
media, Impairments
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
4
3
2
1
5
Encoding Techniques

• 1. Digital data as digital signal
• 2. 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
x(t)
g(t)
g(t)
Encoder
Decoder
Analog Data
Analog Data
Digital Signal Transmission
Modulation
Shift in frequency
Shift back in frequency
m(t)
s(t)
Modulator
Demodulator
m(t)
Analog Data
Analog Data
Analog Signal Transmission
fc
fc
Higher frequency
Baseband
Link
Source
Destination
Baseband
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 same 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

• Not always Tb Ts !!
• Multi-symbol transmission (M
  4, 8, ) Tb lt Ts
• Return to zero (RZ) codes Ts lt Tb

• Mark and Space
• Binary 1 and Binary 0 respectively
• Duration of a bit (Tb)
• Time taken for transmitter to emit
• a data 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

10
Example Two different coding methods

• Data rate 1/1ms
• 1 M bps
• (in both cases)
• Signaling Rate for NRZI 1/1ms
• 1 M bauds
• Signaling Rate for Manchester 1/0.5ms
• 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
elements

• Schemes 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.

2
16
Aspects of comparison between schemes

• Clocking
• Synchronizing 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)

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 0s (so few edges are available for
  synchronization)
• Used for magnetic recording
• Not used much for signal transmission

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
? Power Spectral Density, Watt/Hz
1.5
B8ZS,HDB3
NRZ-L, NRZI
1
AMI, Pseudoternary
0.5
Mean square voltage per unit bandwidth
Manchester, Differential Manchester
0
1
0
0.5
1.5
2
Frequency relative to data rate (binary data)
-0.5
Normalized frequency (f/R)
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 duration 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
  Mapping data directly to signal levels

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 the 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

RX

• Effect 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 (for any data sequence!)
• Lower bandwidth than NRZ
• No loss of sync with a long string of 1z
  (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
27
Multilevel Spectrum
28
The Multilevel Binary Group Advantages
WK 9

• 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 1z
  (but zeros still a
  problem- Will try to solve it later)
• Alteration of pulse polarity also useful for
  error detection Next slide

29
Bipolar-AMI and Pseudoternary
1. All Single Pulse Errors- Detected
3. Double Pulse Error- Undetected
Adding
Canceling
2. Double Pulse Error- Detected
30
Disadvantages of Multilevel Binary
N Log2 (M)
No. of bits sent during each signal element
No. of signal levels used

• 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 gt 2 !
• We use 3 signal levels ? Enough to represent
  log23 1.58 bits gt 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 (bit error rate)

31
Performance with noise NRZ Vs AMI
Multi-Level Binary (AMI)
NRZ
A
A
In both cases signal level is 2A pk2pk
Noise level needed to cause an error
0
-A
-A

• For the same error rate AMI requires higher SNR
  noise (lower noise)
  
• ? i.e. higher Eb/N0
• (for same B and R)
• (hence the 3 dBs difference
• between the two curves)
• For the same SNR (same Eb/N0 )
• AMI has higher error rate
• i.e. AMI has poorer performance with noise

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

Examples of Self-Clocking Codes
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
Data Representation Points
34
Differential Manchester Encoding

• Mandatory midbit transition for clocking
• Data represented by transition or no transition
  at bit start
• ? Transition (either direction) represents 0
• (Invert on zeros)
• ? No transition represents 1

Any error detection capabilities??
Data Representation Points
35
Biphase Group Spectrum
Note higher frequency content
36
Biphase Pros and Cons

• Pros
• Guaranteed mid bit transitions
• Synchronization facility (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
• Modulation (signaling) rate as high as 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)

3 bits TXed
Data
Signal
k1
6 signal transitions 6 signal elements
Ts
Ts
Signal
k2
k 6/3 2

• k No. of signal elements/bit
• No. of signal transitions (both ways) No.
  of bits
  transmitted, n
• (over a given period of n Tbs)

38
Comparison of k for various encoding schemesat
various data bit sequences
k2
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
Use plot to verify values of k in Table 5.3 on
previous slide
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
  producing signal 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 produce 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 (violations alternate in
  polarity- no net DC added)
• 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-

• See how the insertion satisfies the 5
  requirements
• Detectable at RX
• Difficult for noise to generate
• Introduces transitions
• Does not introduce DC (alternate violations)
• Error detection capability

V Violation B Bipolar (Valid)
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-
• With insertions separated by n 1 pulses The
  new insertion is determined by the following
  rules (Table 5.4)
• If n is even, with last pulse p ( or -) ? p00p
• If n is odd, with last pulse p ( or -) ? 000p

45
HDB3
V Violation B Bipolar (Valid)
-000-00
1s
Even number of 1s after last substitution, with
the last pulse () ? p00p ? -00-
Odd number of 1s after last substitution, with
the last pulse (-) ? 000p ? 000-
p
p
46
B8ZS, HDB3 Spectrum
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 or more property of a carrier sine wave
• Amplitude shift keying (ASK)
• Frequency shift keying (FSK)
• Phase shift keying (PSK)

48
Modulation Techniques
Digital Data
Digital Signal
Analog Signals
FSK
Phase shift angles ?
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

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

Dfc
Dfc
fc
f1
f2
51
FSK
fc
f1 fc- Dfc
f2
f1
Spectrum spread due to chopping
Df
Df
f2 fc Dfc
52
FSK for digital data on Voice Grade Lines
Full Duplex Communication (in the 2 directions
simultaneously)
Amplitude
Spectrum of signal in one direction
Two Spectra overlap (Some Interference)
Frequency(Hz)
3400
300
1270
2025
Bell Systems 108 Series modem
fc ? for left and right
1070
2225
f1, f2 ?
f1, f2
Dfc ? for left and right
53
Multiple FSK (MFSK)
To improve BW utilization (efficiency) we send
one of multiple signal symbols (frequencies)
every signal element ? More than 1 bit at a time

• 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)

54
Multiple FSK (MFSK)
(Half the frequency separation)
(Dfc before)
i.e. different frequencies

• 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
  

Important Parameters
The closer the two frequencies are, the larger
Ts needed to discriminate between them
Ts
55
Multiple FSK (MFSK)
Frequency
Data sent
Correction!
kBauds
signaling
2fd50kHz
Bandwidth M (2fd) 8 x 50
400 kHz (lt 2 fc, so OK)
Min Ts 1/ (2fd) 1/50 KHz
20 ms
250kHz
75kHz
425kHz
fc
f1
f8
Max signaling rate 1/Ts 2fd
50 kBauds Max Data rate
Max Signaling rate x L 50 KHz x 3 150 Kbps
56
Multiple FSK (MFSK)
M 4 L Log2 (M) 2
11
10
01
00
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

-3?/4
1
-1
1
-1
Bit pair transmitted
-?/4
60
Quadrature PSK (QPSK) Implementation
Quadrature Component
In phase branch (I)
In-phase Component
n 1, 3, 5, 7
I and Q are derived from the 2 bits transmitted
1
Quadrature (90?) branch (Q)
-1
1
-1

• n I Q
• 1 1
• 3 -1 1
• -1 -1
• 7 1 -1

1 ? 1, 0 ? -1
Q
I
61
Quadrature PSK (QPSK) Implementation
Bits are taken 2 at a time .
Assign bit to I or Q?
1
-1
1
-1
I 1, Q -1

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

62
Quadrature Amplitude Modulation (QAM)
Constellation

• 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
• M 8, L 3
• Up to M256 is possible
• Large bandwidth savings
• But some susceptibility to
• noise
• QAM used on asymmetric
• digital subscriber line
• (ADSL) and some wireless
• systems

M8, 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 required is only 9600/4 2400 bauds
  (required BW is low i.e. can use on 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
a. Performance without noise

• Here, bandwidth requirement is the main concern
  (should be minimized)
• Bandwidth determined by baud rate

Modulated Signal
Modulation ? Filtering ? Transmission
r Filtering Coefficient
m(t)
x(t)
s(t)
Modulator
Filter (r)
Transmitted Signal
(Shift Keyer)
To TX
digital Data
Filter Truncates BW
Modulated Analog Signal
Filtered, band-limited signal
0 lt r lt1
fc
Carrier Signal
Larger r gives larger Transmission BT
Signaling rate D bauds
Data rate R bps
e.g.
(Limited Transmitted bandwidth, BT Hz), e.g.
(Wide, ? bandwidth)
65
Performance of Modulation Schemes

• Performance without noise Transmission
  Bandwidth (BT) Requirement

• 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 (or spectral) Efficiency, BE
  as
• Although it is Efficiency, BE can be greater
  than 1

66
Performance of Digital-Digital (Binary)
Modulation Schemes
a. Performance without noise Bandwidth
Efficiency BE

• Example NRZ and NRZI
• Transmitted BW is similarly truncated
• Transmission Bandwidth is given
• approximately by
• D R only for binary, therefore
• and therefore BE is

Larger r
r 0 BT 0.5R
r 1 BT R
67
Performance of D-A (Binary) Modulation Schemes
a. Performance without noise Bandwidth
Efficiency BE

• For BASK and BPSK
• BT directly related to the (signaling,
  modulation, baud) rate, D
• where r is the filtering coefficient 0lt r lt1
• With binary encoding (not multilevel), D R, so
• Bandwidth Efficiency, BE

68
Performance of D-A (Binary) Modulation Schemes
a. Performance without noise Bandwidth
Efficiency BE

• For BFSK
• Frequency of signal is changed by Df, about fc
  (i.e. 2 Df)
• BT is a function of both Df and the (signaling)
  modulation rate, D
• With binary encoding (not multilevel), D R, so
• Therefore BE is

69
Performance of D-A (Binary) Modulation Schemes
a. Performance without noise Bandwidth
Efficiency BE

• For BFSK, contd.
• Two extreme cases
• Df gtgt R (when fc is large)
• Df ltlt R (when fc is small)

, similar to that for BASK, BPSK
70
Performance of D-A (Multi-level) Modulation
Schemes
a. Performance without noise Bandwidth
Efficiency BE

• For MPSK M phases, L bits/signal element
• BT directly related to the (signaling) modulation
  rate, D
• where r is the filtering coefficient 0lt r lt1
• With M-level encoding,
  , so
• Bandwidth Efficiency, BE

Same as for BPSK
For multilevel, L 2 and r 1, so BE 1
So, BE is directly proportional with L
71
Performance of D-A (Multi-level) Modulation
Schemes
a. Performance without noise Bandwidth
Efficiency BE

• For MFSK M Frequencies, L bits/signal element
• At maximum signaling rate D 2fd
• Bandwidth Efficiency, BE

(Equation 5.11 in textbook)
72
Bandwidth Efficiency (BE) Data

• BE R/BT

Filtering Coefficient, r
73
Performance of D-A Modulation Schemes
b. Performance with noise ASK, FSK, PSK, QPSK

• Bit error rate (BER) Plotted Vs Eb/N0 (dBs)
• Curves to the left give better performance
• Lower S/N required for same Error rate
• Lower Error rate obtained 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!

74
Performance of D-A Modulation Schemes
b. Performance with noise MFSK, MPSK
Larger M ? Poorer error performance
Larger M ? Better error performance!
Orthogonal FSK
As expected
75
Eb/N0 in terms of the bandwidth efficiency
(BE)(for binary transmission)
BT is the Transmission Bandwidth
76
Example

• What is the bandwidth efficiency (BE) for ASK and
  PSK, for a bit error rate (BER) of 10-7 on a
  channel with a SNR of 12dB ?

• For ASK (binary) At BER 10-7, Eb/N0 14.3
  dBs
• Substituting in

• BEASK,FSK -14.3 12 -2.3 dBs ? R/BT 10-.23
  0.6
• However, for PSK ? Eb/N0 11.3 dBs)
• BEPSK R/BT 1.2 (doubled 3dB higher-
  improvement)

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

All supported by
Digital Signal (NRZ-L)
Codec
Or
Digital Mode of Transmission

• Will study two Types of Codec
• Pulse Code Modulation (PCM)
• Delta Modulation (DM)

Code Converter
Digital Signal (Manchester)
Or
Analog signals Carry data!
(Shift Keyer)
Analog Signal (ASK)
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Two basic tasks to be performed by a digitizer
Analog is continuous in both time and
amplitude Must discretize it in both!
Digital Out
Digitizer (Codec)
Analog In

• 1. Sampling in time

• 2. Quantization in amplitude

L bits (sent serially)
Number of quantization levels 2L, where L is
the number of bits allowed for the digital output
2. Quantization To a finite number of levels in
amplitude
PAM Samples
Signal values Between samples Are ignored- lost!
Digitizing the PAM Samples ? PCM
1. Sampling at discrete points in time
Maximum sampling interval allowed 1/(2fmax)
Where fmax is the maximum frequency in the analog
signal
2fmax is the minimum acceptable sampling rate
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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
• Requires sampling at a rate of at least
• 8000 sample per second

80
Quantization using 4 bits
Analog signal is band-limited, with bandwidth (0
to B Hz)
24 16 signal levels, numbered 0 to 15
Quantization
Signal Amplitude, Volts Vmax 16 V
PAM Sample
Quantization Error ?½ LSB
Level numbers starting from 0
Transmitted Serial Code representing the value
of the PAM Samples
1 LSB
Sampling rate 2B sample/s
Data Rate 2B x 4 bps
Each PAM sample is assigned the number of the
nearest quantization level and the corres.
digital code is transmitted
Must finish sending the n bits of the code within
the sampling interval .before the next sample
starts!
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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 power
  factor of 4)
• 256 quantization levels n 8 bits, SNR ? 50 dBs
  
• ? Quality comparable with analog transmission
• Voice 2 x 4000 8000 samples per second, with
  of 8 bits per sample, this is a data rate of
  80008 64 kbps

82
PCM Example

• Suppose we want to encode an analog signal that
  has voltage levels 0 5 V 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
  5/41.25 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
Non linear Encoding
Linear Encoding
Nonlinear Encoding
Quantization error is fixed- same for both
weaker and stronger signals
Weaker signals have smaller quantization errors
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
• 7000 gt 2 x 3000 ? OK
• 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 4.69
• Always 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
PCM How costly in terms of BW?

• For good voice reproduction
• PCM ? 128 levels (7 bits per sample)
• Voice bandwidth (baseband) 4 KHZ
• Data rate should be 2 x 4000 x 7 56 kbps
• Analysis of Bandwidth requirement
• PCM digital transmission requires 56 kbps
• Using Nyquist channel capacity criterion, this
  data rate requires a bandwidth of ? 28 KHz
• (B C/2 R/2 56/2 28)
• Base bandwidth of voice signal 4 KHz
• 28 KHz 7 x 4 KHz
• i.e. PCM digital encoding requires a Nyquist
  bandwidth which is 7 times the bandwidth of the
  baseband signal!
• ( n Bbaseband) ? PCM is costly in bandwidth,
  especially with lager n!

88
PCM Performance, Contd.

• 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 Compression and more efficient coding
  can be used to overcome the problem of the larger
  data rates (and BW) required by digital encoding

89
Delta Modulation A cheaper alternative to PCM

• An attempt to reduce complexity (and large R) for
  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 goes up (slope ive)
• Generate a 0 if staircase goes 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
Quantization
Lower for larger d
Larger for larger d
Sampling
1010 ...Alternating slope ? Signal is level
Digital O/P (Only 1 bit/sample!)
1 ive slope ? Signal increasing
0 - ive slope ? Signal decreasing
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
gt
lt
Generated Staircase
Transmitted bit sequence
Generated Staircase
At Source
Staircase Generator
Reconstructed Staircase
Generated Staircase
Received bit sequence
To filtering Analog Waveform Reconstruction
At Destination
93
Delta Modulation Important Design Parameters

• Two important parameters in DM scheme
• Size of amplitude step (d) 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 d
• When waveform changes slowly, quantizing noise
  increases with a larger d (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
WK 11

• 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)

95
Types of Analog Modulation
Carrier
Modulating Signal
X(t)
Signal to be Transmitted, x(t)
Modulated Signals
Amplitude Modulation (AM)
A ? x(t)

• Angle Modulation
• Phase, PM

f ? x(t)
A sin (wt) f(t) wt
2. Frequency, FM
f ? x(t)
Effect of modulation on signal power?
Effect of modulation on signal BW?
96
Amplitude Modulation (AM)

• Simplest form of modulation
• Accos 2pfct is the carrier,
• and x(t) Amcos 2pfmt is the input modulating
  signal
• Modulated signal expressed as
• na is the modulation index (0 lt 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)

Amplitude of modulated wave
Portion of the modulating signal
Units of na?
97
DSBTC Amplitude Modulation - Example

• Given the amplitude-modulating signal x(t)Amcos
  2pfmt , 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 lt1, envelope is exact reproduction of the
  modulating signal,
• So it can be recovered at receiver by a low pass
  filter
• With na gt1, envelope crosses the time axis and
  information is lost

Ac
Am/2
Am/2
Two Sidebands contain modulating signal power
fc
fm
fm
So, keep na ? 1
98
DSBTC Amplitude Modulation - Examples
MatLab Simulations
Different vertical scales for the 3 plots
Modulating Signal fm ?
Am ?
Carrier fc ?
Ac ?
Envelope
Modulated Signal
na ?
na 0.5/1 0.5
(10.5cos2pit) (1nacos2pit)
99
DSBTC Amplitude Modulation - Example
Maximum modulation allowed (na 1)
na 1/1 1
100
DSBTC Amplitude Modulation - Example
Beyond maximum modulation allowed (na gt 1)
na 2/1 2 (gt1) (not allowed)
101
Spectrum of DSBTC signal
Modulating signal has a single frequency, fm
fc 40 Hz fm 1 Hz Ac 2 V Am 2 V na
Am/Ac 1
2
1
102
Spectrum of an DSBTC signal
Let modulating signal have a bandwidth 0-B
Hz (Baseband)

• Spectrum of AM signal is original
• carrier plus spectrum of original
• signal translated on both sides of fc
• Portion of spectrum f gt fc is
• upper sideband
• Portion of spectrum f lt 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

Bandwidth of Modulated Signal
Note orientation of the two sidebands
Note Modulating signal amplitude does not affect
bandwidth of modulated signal . But affects its
power (next slide)
103
DSBTC Amplitude Modulation
DSBTC

• Total transmitted power Pt in modulated s(t) is
  given by
• Pc is transmitted carrier power
• na should be maximized (but lt1) to allow
  transmission of more power in sideband 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 at TX to pass
  only one of the sidebands the carrier, saves on
  BW ( B)
• SSBSC single sideband suppressed carrier, uses a
  filter to select only one of the sidebands
  (without the carrier), 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.

Note Modulating signal amplitude affects power
of modulated signal
Am na Ac
104
DSBSC Double Sideband Suppressed Carrier -
Example

• Signal is expressed as

Suppressed Carrier
105
Angle Modulation
What parameters can I change to change the
angle of the modulated signal?

• Includes
• Frequency modulation (FM) and
• Phase modulation (PM)
• Modulated signal is given by
• Phase modulation (PM) (the direct way)
• Instantaneous phase is proportional to the
  modulating signal
• np is the phase modulation index
• Frequency modulation (FM) (the indirect way)
• Instantaneous angular frequency deviations from
  wc is proportional to the modulating signal,
• and we have
• So make the derivative of f proportional to
  modulating signal
• nf is the frequency modulation index

Total Angle
Units of np? Units of nf?
106
Angle Modulation

• The total phase angle of s(t) at any instant is
  2pfctf(t)
• Instantaneous phase deviation from that of the
  carrier is f(t)
• Phase Modulation (PM)
• f(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, f(t) is made proportional to x(t)
  ? f(t) nf x(t)
• So, instantaneous frequency deviations from the
  carrier frequency are proportional to x(t).
  ? df (t) (nf/2p)
  x(t)

df (t)
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 2pfmt
• We know that s(t)
• For PM, f (t) is given by
• Then s(t) is
• Instantaneous frequency of s(t) is

np is Radians/Volt
Peak frequency deviation for the PM signal
Note Frequency variations in s(t) phase-lead
x(t) amplitude variations by 90?
108
Frequency Modulation FM

• From equations opposite,
• Peak frequency deviation DF 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 DF ? increases 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 (angle modulation in
  general), Am affects the BW but not the power
  budget
• While in Amplitude Modulation, Am affects the
  power budget but not the bandwidth

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 2pfmt
• The FM modulated signal s(t) is
• For FM, f(t) is given by
• Then f(t) is
• We have
• Substituting for DF we get

nf is (Radians/s)/Volt
But frequency varies as f, i.e. as sin not as
cos !!
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 fcfm, fc2fm,
  (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
For AM

• Carsons Rule of thumb
• Since ? is gt 0, both FM and PM require a larger
  bandwidth than AM (2B)
• For FM, BT 2DF 2B

DF is the peak frequency deviation