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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
Mea
n sq
uare
vol
tage
per
uni
t ban
dwid
th
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
Manchester, Differential Manchester
Mea
n sq
uare
vol
tage
per
uni
t ban
dwid
th
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
Manchester, Differential Manchester
Mea
n sq
uare
vol
tage
per
uni
t ban
dwid
th
Normalized frequency (f/r)
2
Signal Power density, Watt/Hz
Frequency relative to data rate
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
Manchester, Differential Manchester
Mea
n sq
uare
vol
tage
per
uni
t ban
dwid
th
Normalized frequency (f/r)
2
Signal Power density, Watt/Hz
Frequency relative to data rate
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
a. Performance without noise: Bandwidth Efficiency BE
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
a. Performance without noise: Bandwidth Efficiency BE
Performance of D-A (Binary) Modulation Schemes
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
a. Performance without noise: Bandwidth Efficiency BE
Performance of D-A (Binary) Modulation Schemes
, 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:
Bandwidth Efficiency, BE:
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
Bandwidth Efficiency, BE:
)(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:
)](2cos[)( ttfAts cc
)()( 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:
)](2cos[)( ttfAts cc
)()( 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