COE 341: Data & Computer
Communications (T061)
Dr. Radwan E. Abdel-Aal
Chapter 5:
Signal Encoding Techniques
Where are we:
Chapter 7: Data Link:
Flow and Error control
Data Link
Chapter 8: Improved
utilization: Multiplexing
Physical Layer
Chapter 4:
Transmission Media
Transmission Medium
Chapter 3: Signals and
their transmission over
media, Impairments
Chapter 6: Data
Communication:
Synchronization,
Error detection and
correction
Chapter 5: Encoding:
From data to signals
2
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
3
Four Data/Signal Combinations
Signal
Analog
Digital
Same spectrum as
data (base band): e.g. Use a (converter):
Telephony
4
codec, e.g. PCM (pulse
Analog
- Different spectrum
code modulation)
(modulation of a
3
carrier): e.g. AM, FM,
Data
PM
-Two signal levels:
e.g. NRZ
Digital Use a (converter):
modem e.g. ASK,
-More complex
FSK, PSK
encoding: e.g.
2
1
4
Manchester
-
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
5
Encoding:
x(t)
x(t)
g(t)
digital
or
Analog
Data
Encoder
t
Decoder
Digital
Signal
Transmission
g(t
)
digital
or
Analog
Data
Modulation: s(f)
m(t)
Analog
Data
m(f)
s(t)
fc
m(t)
Demodulator
Modulator
fc
f
Analog
Signal
Transmission
Analog
Data
m(f)
fc
0
0
Source
Link
Destination
6
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


Modulation shifts baseband signals to a higher
region in the frequency spectrum (needs fcs)


Basis for Time Division Multiplexing (TDM)
Basis for Frequency Division Multiplexing (FDM)
Unguided media and optical fibers can carry only
analog signals
7
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
8
Terminology, Contd.

Mark and Space




Rate of data transmission
Measured in bits per second (bps)
- Multi-symbol
transmission
(M = 4, 8, …): Tb < Ts
- Return to zero (RZ)
codes: Ts < Tb
Duration of a Signal Element (Ts)


Time taken for transmitter to emit a bit
Data rate, R ( = 1/Tb)


Binary 1 and Binary 0 respectively
Duration of a bit (Tb)

Not always Tb = Ts !!
Minimum signal pulse duration
Modulation (signaling) rate, D (1/Ts)


Rate at which the signal level changes with time
Measured in bauds = signal elements per second
9
Example: Two different coding methods
Data rate = 1/1ms
= 1 M bps
Signaling Rate for NRZI:
= 1/1ms
= 1 M bauds
Tb
Ts
Ts
Signaling Rate for
Manchester: = 1/0.5ms
= 2 M bauds
10
Interpretation of the Received Signal
11
Interpreting Signals

Requirements at RX:




Determine timing of bits - start and end (When to look)
 Need Synchronization (Chapter 6)
Detect signal levels at mid-bit points
Threshold level for comparison with signal to decide on
data
Factors affecting successful signal interpretation
(Affect bit error rate)




Signal to noise ratio
Data rate
Bandwidth
Also Encoding/Modulation scheme
12
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
13
Encoding Schemes
Encoding: Mapping data to signal elements
Schemes for encoding digital data as digital signals

The Nonreturn to Zero (NRZ) Group:



The Multi-level Binary Group:



Bipolar-AMI (Alternate Mark Invert)
Pseudoternary
The Bi-Phase (RZ) Group:



Nonreturn to Zero-Level (NRZ-L)
Nonreturn to Zero Inverted (NRZI)
Manchester
Differential Manchester
Scrambling Group:


B8ZS (Bipolar with 8-Zeros Substitution)
HDB3 (High Density Bipolar 3-Zeros)
14
Why so Many Encoding Schemes?
Aspects for comparison:

Signal Spectrum: Desirable Features



Low high frequency content: Reduces effective bandwidth
No dc component: Allows ac transformer/capacitor coupling,
required sometimes for electrical isolation
Concentrate
3
signal power in
Power Spectral Density, Watt/Hz
1
the middle of
2
4
the bandwidth:
Avoids problems
1
at BW edges, e.g.
delay distortion.
0
0.5
1
1.5
Normalized frequency (f/r)
2
15
Comparison of Encoding Schemes, contd.

Clocking
Synchronizing RX to TX can be achieved using:
 An external clock,
or better:
 A built-in synchronizing mechanism based on the signal
itself! (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)
16
Comparison of Encoding Schemes, contd.

Performance with interference and noise


Some encoding schemes perform better than others:
e.g. when data is encoded as signal transition/no signal
transition, 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)
Higher signaling rates require higher bandwidth, faster
circuits, etc. (larger costs)
17
NRZ Group
NRZ pros and cons:

Pros



Cons




Easy to implement
Modest bandwidth requirements
3
1
2
4
1
Large DC component
0.5
1
1.5
0
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
18
The RZ Solution

Advantages of RZ:



Lower DC content
Guarantees an edge per bit (Better TX-RX synchronization)
Disadvantages of RZ:


Higher frequency content
More difficult to implement
19
Mean square voltage per unit bandwidth
NRZ Spectrum
Power Spectral Density,
Watt/Hz
1.5
1
NRZ-L,
NRZI
B8ZS,HDB3
AMI, Pseudoternary
0.5
Manchester,
Differential Manchester
0
-0.5
0
0.5
1
1.5
2
Frequency
relative to
data rate (binary data)
Normalized frequency (f/R)
20
NRZ-L: Non return to Zero-Level





Two different signal voltages for the 0 and 1 data bits
Voltage level constant (no return to zero, so no signal
transition) during the data bit interval
e.g. 0 V for zero and 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
21
NRZI: Nonreturn to Zero Invert




Nonreturn to zero, invert signal on 1’s data
Still constant voltage level for bit duration of (hence NRZ)
But data 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
An example of differential encoding: Encoding data as a
change in signal level
22
Differential Encoding


Data 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
Effect of swapping terminals on:
- NRZ-L
- NRZI
+
_
RX
23
The Multilevel Binary Group



Use 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
“1” pulses alternate in polarity (why? 2 reasons!)
No loss of sync with a long string of 1’z
(but zeros still a problem- Will try to solve it later)
Advantages:



No net dc component
Lower bandwidth than NRZ
Alteration of pulse polarity useful for error detection
24
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
25
Bipolar-AMI and Pseudoternary
Double Pulse ErrorUndetected
All Single Pulse ErrorsDetected
Adding
Canceling
Double Pulse ErrorDetected
26
Mean square voltage per unit bandwidth
Multilevel Spectrum
Signal Power density,
Watt/Hz
1.5
1
NRZ-L,
NRZI
B8ZS,HDB3
AMI, Pseudoternary
0.5
Manchester,
Differential Manchester
0
-0.5
0
0.5
1
1.5
2
Frequency
relative
to data rate
Normalized frequency (f/r)
27
Disadvantages
of Multilevel Binary

No. of bits
sent during each
signal element
No. of signal
levels used
Coding scheme not as efficient as NRZ:



N = Log2 (M)
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)
28
Performance with noise: NRZ Vs AMI
NRZ
Multi-Level Binary (AMI)
+A
Noise level
needed to cause an error
-A

+A
In both cases
0 signal level is
2A pk2pk
-A
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 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
29
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
30
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
31
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??
32
Mean square voltage per unit bandwidth
Biphase Group Spectrum
Signal Power density,
Watt/Hz
Note higher frequency content
1.5
1
NRZ-L,
NRZI
B8ZS,HDB3
AMI, Pseudoternary
0.5
Manchester,
Differential Manchester
0
-0.5
0
0.5
1
1.5
2
Frequency
relative
to data rate
Normalized frequency (f/r)
33
Biphase Pros and Cons

Pros

Guaranteed mid bit transitions



Ideally no dc component (using bipolar signals)
Error detection


Synchronization facility (Example of self clocking codes)
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)
34
Data rate & Modulation (signaling) rate


Data rate, R = 1/Tb bps
Signaling Rate, D = 1/Ts
bauds
3 bits TXed
Tb
Data
Signal
k=1
If we use k signal elements per
Ts
bit, then:
Signal
 Signaling (modulation) rate,
D = Data rate, R (bit/s
x k (signal elements/bit) k=2
Signal elements/s (bauds)

Ts Ts
k = No. of signal elements/bit
= No. of signal transitions ÷ No. of bits transmitted
(over a given period of n Tbs)
35
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
36
Digital data, Digital signal Encoding
0 1 0 0
1 1 0 0
0 1 1
NRZ
NRZI
Bipolar -AMI
Pseudoternary
Manchester
Differential
Manchester
37
Scrambling Group: B8ZS, HDB3
Modifications 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
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)
38
Scrambling Group: B8ZS, HDB3

Advantages:




No dc component
No long sequences of zero level line signal
No reduction in useful data rate (No extra data sent)
Built-in error detection capability
39
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-+
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 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
40
B8ZS
-000-+0+-
V: Violation
B: Bipolar (Valid)
41
HDB3



High Density Bipolar 3 Zeros
Also based on bipolar-AMI
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)
4th zero always replaced with a code violation
What determines whether 1 or 2 pulses?


Successive violations must alternate in polarity (why?):
-00000000  -000-+00+ or +00000000  +000+-00If 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
42
HDB3
V: Violation
B: Bipolar (Valid)
-000-+00+
1s
Odd number of 1s after last substitution,
with the last pulse (-)  000p  000p
Even number of 1s after last substitution,
with the last pulse (+)  p00p  -00p
43
Mean square voltage per unit bandwidth
B8ZS, HDB3 Spectrum
Signal Power density,
Watt/Hz
1.5
1
NRZ-L,
NRZI
B8ZS,HDB3
AMI, Pseudoternary
0.5
Manchester,
Differential Manchester
0
-0.5
0
0.5
1
1.5
2
Frequency
relative
to data rate
Normalized frequency (f/r)
44
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)
45
Modulation Techniques
Digital Data
Digital Signal
Analog Signals
FSK
Phase shift angles = ?
PSK
46
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
 A cos(2 f c t )
s (t )  
0




binary 1
binary 0
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
47
Frequency Shift Keying (FSK)


Most common form is binary FSK (BFSK)
The two binary data values represented by two different
frequencies (near and both sides of a central carrier frequency fc)
 A cos( 2f1t )
s(t )  
 A cos( 2f 2t )




binary 1
binary 0
Dfc Dfc
f1 fc f2
Less susceptible to noise than ASK
(Same as with FM Radio: Frequency can be better detected in the
presence of noise compared to 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
48
FSK
Data
signal
vd(t)
Carrier 1
v1(t), f1
Carrier 2
v2(t), f2
vFSK(t)
Signal
power
f1 = fc- Dfc
f2 = fc+ Dfc
frequency spectrum
f1
Df
fc
Frequency
Df
f2
Spectrum spread
due to chopping
49
FSK for digital data on Voice Grade Lines
Full Duplex
Communication
Amplitude
Spectrum of signal
in one direction
Two Spectra overlap
(Some Interference)
300
Bell Systems
108 Series modem
1270
1070
 f1, f2
2025
2225
f1, f2 
3400
Frequency(Hz)
fc = ? for left and right
Dfc = ? for left and right
50
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 represents more than one bit
(L bits, L = log2 M)
More 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)
51
Multiple FSK (MFSK)
(Half the frequency separation) (Dfc before)
i.e. different
frequencies
Important
Parameters
Ts
-
Frequency separation = 2 fd
-
Bandwidth Required = M (2fd)
-
Minimum Ts (signal element duration) = 1/(2fd)
 Max signaling rate D  Max data rate R = D L
52
Multiple FSK (MFSK)
Frequency: Data sent:
Correction!
Min Ts = 1/ (2fd)
= 1/50 KHz
= 20 ms
kBauds
signaling
2fd=50kHz
75kHz
250kHz
Bandwidth = M (2fd)
= 8 x 50
= 400 kHz
425kHz
(< 2 fc, so OK)
fc
f8
Max signaling rate = 1/Ts f1
= 2fd
= 50 kBauds
Max Data rate = Max Signaling rate x L = 50 KHz x 3 = 150 Kbps
53
Multiple FSK (MFSK)
M=4
L = Log2 (M) = 2
11
10
01
00
b
54
Phase Shift Keying (PSK)


Phase of carrier signal is shifted to represent data
Binary PSK

Two phases (spaced at 180) represent the two binary digits
Where d(t) = +1 for ‘1’ data and -1 for ‘0’ data
55
Differential PSK (DPSK)
Phase shifted relative to 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
56
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
-/4
57
Quadrature PSK (QPSK) Implementation

1
1
1cos( 2f ct  n )  
cos(2f ct ) 
sin (2f ct )
4
2
2
In phase branch (I)
n = 1, 3, 5, 7
1
Quadrature (90) branch (Q)
-1
1
s (t ) 
-1
I
Q
0  -1
1
1
I(t) cos(2f ct ) 
Q(t) sin (2f ct )
2
2
58
Quadrature PSK (QPSK) Implementation
Bits are taken 2 by 2 ….
Assign bit to I or Q?
1
0
1
1
0
0
0
1
1
1
+1
-1
+1
-1
I = 1, Q = -1
- Stared with how many phases?
- 4 for the price of 2?
- Expect error performance similar to
BPSK…!
s(t ) 
1
1
I(t) cos(2f ct ) 
Q(t) sin (2f ct )
2
2
59
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
60
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 OK for a voice grade line!)

Complex signal encoding allows high data
rates to be sent on voice grade lines having a
limited bandwidth
61
Performance of D-A Modulation Schemes
a. Performance without noise:
 Here, bandwidth requirement is the
main concern
(should be minimized)
Modulated
Signal
s(f)
Modulation 
Filtering  Transmission
r = Filtering Coefficient
m(t)
s(t)
Modulator
digital
Data
Carrier
Data rate Signal
R bps
fc
Filter
(r)
Modulated
Analog
Signal
0 < r <1
Modulation
rate D
bauds
Larger r
gives larger
BT
(Wide
bandwidth)
x(t)
To TX
fc
f
Transmitted
Signal
Filtered,
band-limited
signal
(Limited Transmitted
bandwidth, BT Hz), e.g. BT  (1  r )D
62
Performance of D-A Modulation Schemes
a.




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 Efficiency, BE as
R
BE 
BT
Although it is ‘Efficiency’, BE can be greater than 1
63
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
BT  (1  r ) D

where r is the filtering coefficient; 0< r <1
With binary encoding (not multilevel), D = R, so:
BT  (1  r ) R

Bandwidth Efficiency, BE:
BE BASK , BPSK
R
1


BT 1  r
64
Performance of Digital-Digital (Binary) Modulation Schemes

a. Performance without noise: Bandwidth Efficiency BE
For: NRZ and NRZI

Transmission Bandwidth is given approximately by;
BT  0.5D (1  r )

D = R, therefore:
BT  0.5R (1  r )

and therefore BE is:
Larger r
BE BASK , BPSK
R
2


BT 1  r
r = 0:
BT = 0.5R
r = 1:
BT = R
65
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)
Df  f 2  f c  f c  f1

BT is a function of both Df and the (signaling) modulation
rate, D:
BT  2Df  (1  r ) D

With binary encoding (not multilevel), D = R, so:
BT  2Df  (1  r ) R

Therefore BE is:
BE BFSK 
R
R
1


BT 2Df  (1  r ) R 2Df  (1  r )
R
66
Performance of D-A (Binary) Modulation Schemes
a. Performance without noise: Bandwidth Efficiency BE

For BFSK, contd.:
BE BFSK 


R
R
1


BT 2Df  (1  r ) R 2Df  (1  r )
R
Two extreme cases:
Df >> R (when fc is large):
BE BFSK,Df large 

R
 0; for all r
BT
Df << R (when fc is small):
BE BFSK, Df small 
R
1

; for all Df
BT 1  r
, similar to that for BASK, BPSK
67
Performance of D-A (Multi-level) Modulation Schemes
a. Performance without noise: Bandwidth Efficiency BE

For MPSK: M phases, L bits/signal element
L  log 2 (M )

BT directly related to the (signaling) modulation rate, D
BT  (1  r ) D


Same as for BPSK
where r is the filtering coefficient; 0< r <1
R
R , so:
With M-level encoding, D 

log 2 ( M ) L
(1  r ) R
BT 
L
Bandwidth Efficiency, BE:
BE MPSK
R
L


BT 1  r
L ≥ 2 and r ≤ 1, so BE ≥ 1
68
Performance of D-A (Multi-level) Modulation Schemes
a. Performance without noise: Bandwidth Efficiency BE

For MFSK: M Frequencies, L bits/signal element
L  log 2 (M )
BT  (1  r ) Spectrum width
 (1  r )[ M ( 2 fd )]

At maximum signaling rate: D = 2fd
BT  (1  r ) MD 

(1  r ) MR
(1  r ) M
(
)R
L
Log2 M
(Equation 5.11
in textbook)
Bandwidth Efficiency, BE:
BE MFSK 
R
L

BT (1  r ) M
69
Bandwidth Efficiency (BE) Data
BE = R/BT
Digital-Analog Modulation
Scheme
BASK
BFSK (wideband Df >> R)
BFSK (narrowband Df << R)
BPSK
MPSK: M=4 (L=2)
MPSK: M=8 (L=3)
MPSK: M=16 (L=4)
MPSK: M=32 (L=5)
Filtering Coefficient, r
r=0
r = 0.5
r=1
1.0
0
1.0
1.0
2.0
3.0
4.0
5.0
0.67
0
0.67
0.67
1.33
2.00
2.67
3.33
0.5
0
0.5
0.5
1.0
1.5
2.0
2.5
70
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 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!
71
Performance of D-A Modulation Schemes
b. Performance with noise: MFSK, MPSK
Larger M  Better error performance!
Orthogonal FSK
Larger M  Poorer error performance
As expected
72
Eb/N0 in terms of the bandwidth efficiency (BE)
(for binary transmission)
BE dB
R

BT
dB
S
S
Eb STb

 R  N
N
N
R
N0
BT
BT
BT
Eb
N0

dB
S
N
S

N
R

BT
dB
dB
 BE dB
dB
73
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 ?

For ASK and FSK (binary): At
BER = 10-7, Eb/N0 = 14.3 dBs

Substituting in:
Eb
 SNRdB  BE dB
N0

BEASK,FSK = R/BT = 0.6

Similarly for PSK (with Eb/N0 = 11.3 dBs):
BEPSK = R/BT = 1.2 (doubled: 3dB higher)
74
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
Codec
Or:
Will study two Types of Codec:
- Pulse Code Modulation (PCM)
- Delta Modulation (DM)
Digital Signal
(NRZ-L)
Code
Converter
Or:
(Shift Keyer)
Digital Signal
(Manchester)
Analog Signal
(ASK)
Digital Mode of
Transmission
75
Two basic tasks to be performed by a digitizer:
“Analog” is continuous in both time and amplitude… Must discretize it in both
• Sampling in time
• Quantization in amplitude
Analog In
Digitizer
(Codec)
Digital Out
L bits (sent serially)
Number of quantization levels = 2L, where L is the
number of bits allowed for the digital output
Quantization
To a finite number of levels
in amplitude
PAM Samples
Digitizing the PAM Samples
 PCM
Sampling
at discrete points in time
Maximum sampling interval allowed = 1/(2fmax); Where fmax is
the maximum frequency in the analog signal
76
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 8000 sample per
second
77
Quantization using 4 bits
Analog signal is band-limited,
with bandwidth (0 to B Hz)
Quantization
24 = 16 signal levels, numbered 0 to 15
PAM Sample
Level
number
starting
from 0
Signal Amplitude, Volts
Vmax = 16 V
Quantization
Error = ½ LSB
1 LSB
Transmitted Serial Code representing
the PAM Samples:
Sampling,
rate: 2B
Each PAM sample is assigned the number of the nearest
quantization level and its digital code is transmitted
78
Must finish sending the n bits of the code within the sampling interval ….before the next sample starts!
Pulse Code Modulation (PCM)




Start with the analog sampled pulses (Pulse Amplitude Modulation,
PAM)
Each sample assigned 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
SNR  20 log 10 2  1.76 dB  6.02n  1.76 dB
n




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 gives 256 levels
Quality comparable with analog transmission
8000 samples per second of 8 bits each gives a data rate of
8000*8 = 64 kbps
79
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
80
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
81
Effect of Nonlinear Coding
Linear Encoding
Quantization error is fixedsame for both weaker and stronger signals
Non
linear Encoding
Nonlinear
Encoding
Weaker signals have smaller
quantization errors
82
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
No Companding
signals than for strong
(Linear encoding)
signals- before encoding
At RX: Reverse
operation (de-companding?)

How would the de-companding
curve look like?
83
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 = 4.69
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
84
CODEC - Performance


Good voice reproduction
 PCM - 128 levels (7 bits)
 Voice bandwidth 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 bandwidth which
is 7 times the bandwidth of the baseband signal!
(= n Bbaseband)
85
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
More efficient coding can be used to overcome the problem
of the larger BW required by digital encoding
86
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


A single bit stream is produced to approximate the
derivative of the analog signal rather than its amplitude




Move up or down one fixed amplitude increment () at each
sample interval to track changes in the analog waveform
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
87
Quantization
Delta Modulation - example
Smaller for larger 
Sampling
Larger for larger 
101 ...Alternating slope:
Signal is level
Digital O/P
(Only 1
bit/sample!)
1: + ive slope:
0: - ive slope:
Signal increasing Signal decreasing
88
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
89
Delta Modulation - Implementation
+
_
Generated Staircase
>
<
Transmitted bit
sequence
Generated
Staircase
At Source
Staircase Generator
Reconstructed
Staircase
Received bit
sequence
At Destination
Generated
Staircase
To filtering &
Analog Waveform
Reconstruction
90
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
91
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 be kilometers in diameter!
Allows implementing frequency division multiplexing (FDM)
92
Types of Analog
Modulation
Signal to be
Transmitted,
x(t)
Carrier
Modulating
Signal
Modulated
Signals
Amplitude Modulation (AM)
Angle Modulation:
1. Phase, PM
d (t )

  (t )
dt
2. Frequency, FM
A sin (t)
(t)  t
Effect of modulation on power?
Effect of modulation on BW?
93
Amplitude Modulation (AM)




Simplest form of modulation
Accos 2fct is the carrier,
and x(t)= Amcos 2fmt is the input modulating signal
Modulated signal expressed as:
Amplitude of
modulated wave
s(t )  [1  na cos 2f mt ] Ac cos 2f ct

na is the modulation index (0 < na  1):
Am
na 
Ac


Portion of the
modulating signal
Units of na?
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)
94
DSBTC Amplitude Modulation - Example

Given the amplitude-modulating signal x(t)=Amcos 2fmt , find s(t):
s (t )  Ac [1  na cos 2f m t ] cos 2f c t

na Ac
na Ac
 Ac cos 2f c t 
cos 2 ( f c  f m )t 
cos 2 ( f c  f m )t
2
2
A
A
 Ac cos 2f c t  m cos 2 ( f c  f m )t  m cos 2 ( f c  f m )t
2
2
Resulting signal has three components:
A



At the original carrier frequency fc
Am
n

A pair of additional components (side bands), a
Ac
each spaced fm Hz from the carrier
Envelope of resulting signal


c
Am/2
Am/2
fm fc fm
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
So, keep na below 1
Sidebands
contain
modulating
signal
power
95
DSBTC Amplitude Modulation - Examples
MatLab Simulations
Modulating
Signal
Carrier
Note: Different vertical scales
Am = ?
Ac = ?
Envelope
Modulated
Signal
na = 0.5/1
= 0.5
na = ?
=(1+0.5cos2*pi*t)
=(1+nacos2*pi*t)
96
DSBTC Amplitude Modulation - Example
Maximum modulation allowed (na = 1)
na = 1/1
=1
97
DSBTC Amplitude Modulation - Example
Beyond maximum modulation allowed (na > 1)
na = 2/1
=2
(>1)
(not allowed)
98
Spectrum of DSBTC signal:
s (t )  Ac [1  na cos 2f mt ] cos 2f c t
na Ac
n A
cos 2 ( f c  f m )t  a c cos 2 ( f c  f m )t
2
2
A
A
 Ac cos 2f c t  m cos 2 ( f c  f m )t  m cos 2 ( f c  f m )t
2
2
 Ac cos 2f c t 
Modulating
signal has
a single
frequency, fm
Ac
Am/2
Am/2
fm fc fm
99
DSBTC Amplitude Modulation

Total transmitted power Pt in modulated s(t) is given by

na2 
Pt  Pc 
1  2 



Ac
Am/2
Am/2
fm fc fm
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.




100
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
Modulating
signal has
a bandwidth 0-B Hz
Note orientation of the
two sidebands
101
DSBSC: Double Sideband Suppressed Carrier Example

Signal is expressed as s(t )  Ac [ x(t )] cos 2f ct
Suppressed
Carrier
102
Angle Modulation

Includes:



Frequency modulation (FM) and
Phase modulation (PM)
Modulated signal is given by
What parameters
can I change to change
the angle of the
modulated signal?
Total Angle
s(t )  Ac cos[ 2f ct   (t )]

Phase modulation (PM): (the direct way)  (t )  n p x(t )



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,


(
t
)


(t )  n f x (t )
f(t)  ’(t)
and we have:
So make the derivative of  proportional to modulating signal
nf is the frequency modulation index
Units of np?
103
Units of nf?
Angle Modulation



The total phase angle of s(t) at any instant is [2fct+(t)]
Instantaneous phase deviation from carrier is (t)
Phase Modulation (PM):


(t) = npx(t), instantaneous phase variations are directly
proportional to m(t)
Frequency Modulation (FM):


Instantaneous angular frequency, i (t ) , can be defined as the rate
of change of total phase
So, for the modulated signal, s(t)
i (t )  2f i (t ) 
 fi  f c 

d
2f c t   (t )  2f c   (t )
dt
1
 (t )
2
In FM, ’(t) is proportional to x(t). So, instantaneous frequency
deviations from the carrier frequency is proportional to x(t).
104
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 2fmt
We know that s(t):

For PM,  (t) is given by:


s(t )  Ac cos[ 2f ct   (t )]
 (t )  n p x(t )
np is Radians/Volt
Then s(t) is:
s (t )  5 cos[ 2f c t  n p 3 sin 2f mt ]

Peak frequency deviation
for the PM signal
Instantaneous frequency of s(t) is:
3n p (2f m )
f i (t )  f c 
cos 2f mt  f c  3n p f m cos 2f mt
2
Note: Frequency variations in s(t) lead x(t) amplitude variations by 90
105
Frequency Modulation: FM

Peak frequency deviation DF is given by:
1
DF 
n f Am Hz
2





1
 (t )
2
and  (t )  n f x(t )
fi  fc 
and x(t )  Am sin( 2f mt )
Where Am is the peak value of the modulating signal x(t)
An increase in the amplitude Am of x(t) increases DF, 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
106
Frequency Modulation - Example



Derive an expression for a frequency-modulated signal s(t) with
Ac= 5V, given the modulating signal
x(t) = 3 sin 2fmt
The FM modulated signal s(t) is: s(t )  Ac cos[ 2f c t   (t )]
For FM, ’(t) is given by:
 (t )  n f x(t )

nf is (Radians/s)/Volt
Then (t) is:
 (t )    (t )dt   n f 3 sin 2f mt dt 

We have:

Substituting for DF we get:
3
DF 
n f Hz
2
 3n f
2f m
cos 2f mt
But frequency
varies as ’,
i.e. as sin not
as – cos !!
DF
s (t )  5 cos[ 2f c t 
cos 2f mt ]
fm
107
Bandwidth Requirement





All AM, FM, and PM result in a modulated signal whose
bandwidth is centered at 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
108
Practical Bandwidth Requirement for Angle Modulation

Carson’s Rule of thumb
BT  (   1)2B
n p Am

   DF n f Am


2B
B
for PM
for FM
DF is the peak frequency deviation


Since  is > 0, both FM and PM require a larger
bandwidth than AM
For FM, BT= 2DF + 2B
109