Signal Encoding Techniques - PowerPoint PPT Presentation

1 / 109
About This Presentation
Title:

Signal Encoding Techniques

Description:

Signal Encoding Techniques – PowerPoint PPT presentation

Number of Views:628
Avg rating:3.0/5.0
Slides: 110
Provided by: icsd
Category:

less

Transcript and Presenter's Notes

Title: 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
digital or Analog Data
digital or Analog Data
Digital Signal Transmission
Modulation
m(t)
s(t)
Modulator
Demodulator
m(t)
Analog Data
Analog Data
Analog Signal Transmission
fc
fc
Link
Source
Destination
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)
  • Basis for Frequency Division Multiplexing (FDM)
  • Unguided media and optical fibers 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.
  • 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 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
  • Not always Tb Ts !!
  • Multi-symbol transmission (M
    4, 8, ) Tb lt Ts
  • Return to zero (RZ) codes Ts lt Tb

10
Example Two different coding methods
  • Data rate 1/1ms
  • 1 M bps
  • 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 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

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 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
  • signal power in
  • the middle of
  • the bandwidth
  • Avoids problems
  • at BW edges, e.g.
  • delay distortion.

2
16
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)

17
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)

18
NRZ GroupNRZ pros 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
  • 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 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

22
NRZI Nonreturn to Zero Invert
  • Nonreturn to zero, invert signal on 1s 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

23
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

RX
  • Effect of swapping terminals on
  • NRZ-L
  • NRZI


_
24
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 1z
    (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

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
All Single Pulse Errors- Detected
Double Pulse Error- Undetected
Adding
Canceling
Double Pulse Error- Detected
27
Multilevel Spectrum
28
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 (error rate)

29
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. 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

30
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 0s
    opposite of NRZI)
  • An example of differential encoding
  • Used by IEEE 802.5 specification for Token Ring
    LAN

31
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
32
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??
33
Biphase Group Spectrum
Note higher frequency content
34
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)

35
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
Ts
Ts
Signal
k2
  • k No. of signal elements/bit
  • No. of signal transitions No. of bits
    transmitted
  • (over a given period of n Tbs)

36
Comparison of k for various encoding schemes
k2
e.g., here k 1.5 i.e. baud rate D is
1.5 x data rate R
37
Digital data, Digital signal Encoding
Bipolar -AMI
Pseudoternary
38
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
  • 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)

39
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

40
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

41
B8ZS
-000-0-
V Violation B Bipolar (Valid)
42
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 ? 000-00-
  • 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

43
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
44
B8ZS, HDB3 Spectrum
45
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)

46
Modulation Techniques
Digital Data
Digital Signal
Analog Signals
FSK
Phase shift angles ?
PSK
47
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

48
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)
  • 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

Dfc
Dfc
fc
f1
f2
49
FSK
Df
fc
Df
f1 fc- Dfc
f2
f1
Spectrum spread due to chopping
f2 fc Dfc
50
FSK for digital data on Voice Grade Lines
Full Duplex Communication
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
51
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)

52
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 ? Max data rate R D L

Important Parameters
Ts
53
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
54
Multiple FSK (MFSK)
M 4 L Log2 (M) 2
11
10
01
00
b
55
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
56
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

57
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
58
Quadrature PSK (QPSK) Implementation
In phase branch (I)
n 1, 3, 5, 7
1
Quadrature (90?) branch (Q)
-1
1
-1
0 ? -1
Q
I
59
Quadrature PSK (QPSK) Implementation
Bits are taken 2 by 2 .
Assign bit to I or Q?
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!

60
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
  • 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
61
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

62
Performance of D-A Modulation Schemes
a. Performance without noise
  • Here, bandwidth requirement is the main concern
  • (should be minimized)

Modulated Signal
Modulation ? Filtering ? Transmission
r Filtering Coefficient
m(t)
x(t)
s(t)
Modulator
Filter (r)
Transmitted Signal
To TX
digital Data
Modulated Analog Signal
Filtered, band-limited signal
0 lt r lt1
fc
Carrier Signal
Larger r gives larger BT
Modulation rate D bauds
Data rate R bps
(Limited Transmitted bandwidth, BT Hz), e.g.
(Wide bandwidth)
63
Performance of D-A 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 Efficiency, BE as
  • Although it is Efficiency, BE can be greater
    than 1

64
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

65
Performance of Digital-Digital (Binary)
Modulation Schemes
a. Performance without noise Bandwidth
Efficiency BE
  • For NRZ and NRZI
  • Transmission Bandwidth is given approximately by
  • D R, therefore
  • and therefore BE is

Larger r
r 0 BT 0.5R
r 1 BT R
66
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

67
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
68
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
L 2 and r 1, so BE 1
69
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)
70
Bandwidth Efficiency (BE) Data
  • BE R/BT

Filtering Coefficient, r
71
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!

72
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
73
Eb/N0 in terms of the bandwidth efficiency
(BE)(for binary transmission)
74
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
  • BEASK,FSK R/BT 0.6
  • Similarly for PSK (with Eb/N0 11.3 dBs)
  • BEPSK R/BT 1.2 (doubled 3dB higher)

75
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

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
(Shift Keyer)
Analog Signal (ASK)
76
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
  • Sampling in time
  • Quantization in amplitude

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
77
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

78
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 number starting from 0
Transmitted Serial Code representing the PAM
Samples
1 LSB
Sampling, rate 2B
Each PAM sample is assigned the number of the
nearest quantization level and its digital code
is transmitted
Must finish sending the n bits of the code within
the sampling interval .before the next sample
starts!
79
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
  • 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 80008 64 kbps

80
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
    5/41.25 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

81
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

82
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
83
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)
84
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

85
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)

86
CODEC Performance, Contd.
  • A common PCM scheme for color TV uses 10-bit
    codes
  • For bandwidth4.6 MHz ? 92 Mbps (i.e. 24.610)
  • 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

87
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

88
Delta Modulation - example
Quantization
Smaller for larger d
Larger for larger d
Sampling
101 ...Alternating slope Signal is level
Digital O/P (Only 1 bit/sample!)
1 ive slope Signal increasing
0 - ive slope Signal decreasing
89
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

90
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
91
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

92
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)

93
Types of Analog Modulation
Carrier
Modulating Signal
Signal to be Transmitted, x(t)
Modulated Signals
Amplitude Modulation (AM)
Angle Modulation 1. Phase, PM
2. Frequency, FM
A sin (wt) f(t) wt
Effect of modulation on power?
Effect of modulation on BW?
94
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?
95
DSBTC Amplitude Modulation - Example
  • Given the amplitude-modulating signal x(t)Amcos
    2pfmt , find s(t)
  • Resulting signal has three components
  • 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
  • With na gt1, envelope crosses the time axis and
    information is lost

Ac
Am/2
Am/2
Sidebands contain modulating signal power
fc
fm
fm
So, keep na below 1
96
DSBTC Amplitude Modulation - Examples
MatLab Simulations
Note Different vertical scales
Modulating Signal
Am ?
Carrier
Ac ?
Envelope
Modulated Signal
na ?
na 0.5/1 0.5
(10.5cos2pit) (1nacos2pit)
97
DSBTC Amplitude Modulation - Example
Maximum modulation allowed (na 1)
na 1/1 1
98
DSBTC Amplitude Modulation - Example
Beyond maximum modulation allowed (na gt 1)
na 2/1 2 (gt1) (not allowed)
99
Spectrum of DSBTC signal
Modulating signal has a single frequency, fm
100
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 lt1) 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.

101
Spectrum of an DSBTC signal
Modulating signal has a bandwidth 0-B Hz
  • 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

Note orientation of the two sidebands
102
DSBSC Double Sideband Suppressed Carrier -
Example
  • Signal is expressed as

Suppressed Carrier
103
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 proportional to the
    modulating signal
  • np is the phase modulation index
  • Frequency modulation (FM) (the indirect way)
  • Instantaneous angular frequency deviations from
    wc 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
df(t) ? f(t)
Units of np? Units of nf?
104
Angle Modulation
  • The total phase angle of s(t) at any instant is
    2pfctf(t)
  • Instantaneous phase deviation from 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 proportional to x(t). So,
    instantaneous frequency deviations from the
    carrier frequency is proportional to x(t).

105
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) lead x(t)
amplitude variations by 90?
106
Frequency Modulation FM
  • 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, 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

107
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 !!
108
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 fcfm, fc2fm,
    (the Bessel function)
  • i.e. Theoretically, an infinite bandwidth is
    required to transmit an FM or PM signal

109
Practical Bandwidth Requirement for Angle
Modulation
  • Carsons Rule of thumb
  • Since ? is gt 0, both FM and PM require a larger
    bandwidth than AM
  • For FM, BT 2DF 2B

DF is the peak frequency deviation
Write a Comment
User Comments (0)
About PowerShow.com