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College of Information & Communications

The world goes wireless! Prepared by Sung Ho Cho

Hanyang University

Modulation and Coding Tradeoffs

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Contents1. Design Goals 2. Error Probability Plane3. Nyquist Minimum Bandwidth4. Shannon‐Hartley Capacity Theorem5. Bandwidth‐Efficiency Plane6. Modulation and Coding Tradeoffs7. Design Steps of Communication Systems8. Bandwidth‐Efficient Modulation9. Modulation and Coding for Bandlimited Channels

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1. Design Goals

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Design Goals

Maximum transmission bit rate RMinimum probability of bit error PBMinimum required power or Eb/N0

Minimum system bandwidth WMinimum system complexityMaximum system utilization

For reliable serviceFor a maximum number of usersWith minimum delayWith maximum resistance to interference

Tradeoff

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Constraints and Limitations

Nyquist theoretical minimum bandwidth requirementShannon‐Hartley capacity theorem (and Shannon limit)Government regulations (e.g., frequency allocations)Technological limitations (e.g., state‐of‐the‐art components)Other system requirements (e.g., satellite orbits)

Two major performance planes:Error probability planeBandwidth efficiency plane

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2. Error Probability Plane

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Bit Error Probability versus Eb/N0Coherent MFSK: Coherent MPSK:

PB↔WEb/N0: fixed

PB↔ EW: fixed

b/N0

Eb/N0 ↔WPB: fixed

Equi‐bandwidth Curves

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3. Nyquist Minimum Bandwidth

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Filtering Aspects of ISI

ISI for a typical baseband digital system:

Equivalent model:

Transmitting filter Channel Receiving filter

The tail of a pulse smears into adjacent symbol intervals.

Hr(f)Ht(f) Hc(f)

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System Transfer Function for Baseband Channel

System Transfer Function, H(f), and ideal Nyquist pulse, h(t)

If each pulse of a received sequence is of the form sinc(t/T), then the pulses can be detected without ISI.

Sampling at the receiver at every T sec.

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Nyquist Minimum Bandwidth

The theoretical minimum bandwidth (Nyquist bandwidth) needed for the baseband transmission of Rs symbols/s without ISI is Rs/2 Hz.

The maximum possible symbol transmission rate per hertz is 2 symbols/s/Hz.

In practice, the Nyquist minimum bandwidth is expanded by about 10% to 40% due to the constraints of real filters.

Typical baseband digital communication throughput is reduced from the ideal 2 symbols/s/Hz down to the range of about 1.8 to 1.4 symbols/s/Hz.

Bandwidth Efficiency: R/Wmeasured in bits/s/HzA measurement of data throughput per hertz of bandwidth.Ex) MPSK system:

For a fixed bandwidth W, as k increases, the allowable data rate Rincreases, and thus R/W increases at the expense of large Eb/N0.

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4. Shannon‐Hartley Capacity Theorem

C.E. Shannon,”A Mathematical Theory of Communication,”

BSTJ, Vol.27, pp.379‐423, 1948.

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System Capacity

Shannon‐Hartley Capacity Theorem:

C: System capacity for an AWGN channelW: BandwidthS: Average received signal power (S = EbR)N: Average noise power (N = N0W)

It is theoretically possible to transmit information over an AWGN channel at any rate R≤C, with an arbitrarily small error probability by using a sufficiently complicated coding scheme.

Shannon’s work shows that the values of S, N, and W set a limit on transmission rate, not an error probability.

(bits/s)

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Normalized Channel Capacity/Bandwidth vs. SNR

Normalized channel capacity vs. SNR Normalized channel bandwidth vs. SNR

2log 1C SW N

⎛ ⎞= +⎜ ⎟⎝ ⎠

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Normalized Channel Capacity vs. Eb/N0Suppose that the transmission bit rate, R = channel capacity, C

This is the case when R=C.

0N N W=

0

// /

b bE S T S R S WN N W N W N R

⎛ ⎞= = = ⎜ ⎟⎝ ⎠

0

bES RN N W

⎛ ⎞= ⎜ ⎟⎝ ⎠

2 20

log 1 log 1 bEC S CW N N W

⎛ ⎞⎛ ⎞ ⎛ ⎞= + = +⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

/

02 1C W bE C

N W⎛ ⎞= + ⎜ ⎟⎝ ⎠

( )/

02 1C WbE W

N C= −

If R = C

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Shannon LimitDefinition:

The limit of Eb/N0 below which there can be no error‐free communication at any information rateShannon’s work provides a theoretical proof for the existence of codes that could improve the PB performance, or reduce the Eb/N0 required.Today, as much as 10 dB improvement is realizable with turbo code.

Mathematical limit:

We use the identity:

Let

Then

As C/W→0 or x→0, we have

or

or

Again, this is the case when R=C.

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Entropy, H

Definition:A metric for measuring the average amount of information per source output

For a binary case:

where pi is the probability of the i‐th output and .

Maximum entropy when events occur equally likely.

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Properties of EntropyThe unit for H is average bit per event.

It is a measure of information content, and should not be confused with the term bit, meaning “binary digit.”

The term “entropy” has the same uncertainty connotation as it does in certain formulations of statistical mechanics.

The uncertainty in the event becomes maximum when the information source has equally likely possibilities.If we know the result before the event happens, then the result conveys no additional information.

Example: English alphabet If 26 letters occur with equal likelihood:

But, if

“There is a way of encoding the English language

with a fewer number of bits per character on the average.”

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Equivocation, H(X|Y) (1/2)Shannon uses a correction factor, call equivocation, to characterize the uncertainty in the received signal.Equivocation is defined as the conditional entropy of the message X, given Y, such that

where X = transmitted source messageY = received signalP(X,Y) = joint probability of X and YP(X|Y) = conditional probability of X given Y

For an error‐free channel, H(X|Y) = 0, because having received Y, there is no uncertainty about the message X, (i.e., P(X|Y) = 1). For a channel with a nonzero probability of symbol error, H(X|Y) > 0, because the channel introduces uncertainty.

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Equivocation, H(X|Y) (2/2)Example: For a binary sequence, X,

Let the a priori source probabilities be P(X=1) = P(X=0) = ½.Let the bit error probability at the receiver become PB = 0.01.

Then the equivocation H(X|Y) becomes

( )

( ) ( )

( ) ( )

( ) ( )

2

2 2

2 2

2 2

2 2

2 2

2

( ) ( ) log ( )

(0) (0 0)log (0 0) (1 0)log (1 0)

(1) (0 1)log (0 1) (11)log (11)

1 1 log 1 log2

1 log 1 log 12

1 log 1 log

0.99log 0.

Y X

B B B B

B B B B

B B B B

H X Y P Y P X Y P X Y

P P P P P

P P P P P

P P P P

P P P P

P P P P

= −

= − +⎡ ⎤⎣ ⎦− +⎡ ⎤⎣ ⎦

= − − − +⎡ ⎤⎣ ⎦

− + − −⎡ ⎤⎣ ⎦

= − − − +⎡ ⎤⎣ ⎦= −

∑ ∑

( ) ( )299 0.01log 0.01

0.081 bit/received symbol

+⎡ ⎤⎣ ⎦=

“The channel introduces 0.081 bit of uncertainty to each received symbol.”

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Effective Entropy, Heff

Effective entropy, Heff, is used to characterize the average effective information content, defined by

Example: For a system transmitting equally likely binary symbols,The entropy H(X) = 1.If the symbols are received with PB = 0.01, the equivocation H(X|Y) =0.081.Then the effective entropy becomes

Thus, if the binary symbol rate is given by R = 1000 symbols/s, the effective information bit rate Reff can then be written as

Note that in the extreme case, where PB = 0.5, we have

Heff = 0

Reff = 0 “The effective information bit rate is zero!”

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5. Bandwidth‐Efficiency Plane

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PB↔ R/WEb/N0: fixed

PB↔ Eb/N0R/W: fixed

Eb/N0 ↔ R/WPB: fixed

Equi‐error probability curves

Ideal performance

Bandwidth‐Efficiency Plane (R/W vs. Eb/N0)

( ) 22

loglogs ss

MR k R M RT

= = =

2logR MW

=

1

s

WT

≈

2log MRW M

=

s

MWT

≈

Reflects how efficiently the bandwidth resource is utilized.

Bandwidth‐efficiency plane at PB=10‐5:

For coherent MPSK:

For Noncoherent MFSK:

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6. Modulation and Coding Tradeoffs

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Error‐Probability Plane vs. Bandwidth‐Efficiency Plane

Error Probability Plane

G: gained, C: cost, F: fixed

Bandwidth‐Efficiency Plane“Useful for examining power‐limited systems.” “Useful for examining bandwidth‐limited systems.”

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7. Design Steps of Communication Systems

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Before Designing Communication Systems…

Need description of the channel:Received powerAvailable bandwidthNoise statisticsOther impairments, such as fading

Need definition of the system requirements:Data rateError performanceTradeoff between bandwidth ↔ power

Given the channel description, we determine design choices that best match the channel and meet the performance requirements.

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Requirements for MPSK and MFSK Signaling

Example:System rate, Nyquist minimum bandwidth, bandwidth efficiency, and required Eb/N0 at 9,600 bits/s:

(3/8)

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Example 1: Bandwidth‐Limited Uncoded SystemAssumptions:

AWGN radio channelAvailable bandwidth: W = 4,000 HzRatio of received signal power to noise power spectral density: Pr/N0 = 53 dB‐HzRequired data rate: R = 9,600 bits/sRequired bit error probability: PB = 10‐5

Design steps:Received Eb/N0:

Use the relationship between Pr/N0 and Eb/N0:

Determine modulation scheme:Choose MPSK signaling (since W < R).

Determine M: Choose M = 8 (since the minimum bandwidth required for M=4 is 4,800 Hz, and that for M=8 is 3,200 Hz).

Check if we need an error‐correction coding:Not necessary (since the received Eb/N0=13.2 dB is greater than the required Eb/N0=13.0 dB).

or

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Example 2: Power‐Limited Uncoded SystemAssumptions:

AWGN radio channelAvailable bandwidth: W = 45,000 HzRatio of received signal power to noise power spectral density: Pr/N0 = 48 dB‐HzRequired data rate: R = 9,600 bits/sRequired bit error probability: PB = 10‐5


Use the relationship between Pr/N0 and Eb/N0:

Determine modulation scheme:Choose MFSK signaling (since W > R, and the received Eb/N0 is relatively small for the required PB=10‐5).


Check if we need an error‐correction coding:Not necessary (since the received Eb/N0=8.2 dB is greater than the required Eb/N0=8.1 dB).

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Example 3: Bandwidth‐Limited and Power‐Limited Coded System

Assumptions:AWGN radio channelAvailable bandwidth: W = 4,000 HzRatio of received signal power to noise power spectral density: Pr/N0 = 53 dB‐HzRequired data rate: R = 9,600 bits/sRequired bit error probability: PB = 10‐9


Determine modulation scheme:Choose MPSK signaling (since W < R).


Check if we need an error‐correction coding:Yes, it is necessary (since the received Eb/N0 is insufficient for the required PB).

Optimum code selection:The output bit error probability of the combined modulation/coding system must meet the system error requirement.The rate of the code must not expand the required transmission bandwidth beyond the available channel bandwidth.The code should be as simple as possible.

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8. Bandwidth‐Efficient Modulation

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Bandwidth‐Efficient Modulation

Want to maximize bandwidth efficiency so that the spectral congestion problem is ameliorated.

A constant envelope modulation is additionally required in order to relaxExtraneous sidebands produced by system nonlinearitiesAdjacent channel interference problemCo‐channel interference problem with other communication systemsExamples of the constant envelope modulation:

Offset QPSK (OQPSK)Minimum shift keying (MSK)

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QPSK (1/2)Original data streams:

The pulse streams are divided into an in‐phase and quadrature streams:

dI(t) and dQ(t) each have half the bit rate of dk(t).

Orthogonal realization of a QPSK waveform:

( )

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QPSK (2/2)

Modulation scheme:

The two BPSK signals can be detected separately.

Zero‐crossing takes place during transition. (The envelope of s(t) is not constant.)

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Offset (or Staggered) QPSK (OQPSK)

Alignment of dI(t) and dQ(t) is offset by T.

No zero‐crossing. (Less out‐of‐band interference comparing to QPSK)

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Minimum Shift Keying (MSK) (1/4)Motivation:

MSK can be viewed as eitherA special case of CPFSK (continuous phase FSK), orA special case of OQPSK with sinusoidal symbol weighting

MSK waveform:

If dk = 1, the transmitted frequency is (f0 + 1/4T), and If dk = ‐1, the transmitted frequency is (f0 ‐ 1/4T).The tone spacing in MSK is 1/2Twhich is one‐half (i.e., minimum) that employed for noncoherently demodulated orthogonal FSK.

A phase constant for the k‐th bit interval

QPSK OQPSK MSK

Suppressingout‐of‐bandinterference

Removingdiscontinuous

phase transitions

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Minimum Shift Keying (MSK) (2/4)Phase constant:

During each T, the value of xk is constant, i.e., xk = 0 or π.The value for xk is determined by the requirement that the phase of the waveform be continuous at t=kT (i.e., continuous phase constraint).

Based on the continuous phase constraint, the MSK waveform can be rewritten as

The ak term can change value only at the zero‐crossing of cos(πt/2T).The bk term can change value only at the zero‐crossing of sin(πt/2T).Thus, the symbol weighting in either the I‐ or Q‐channel is a half‐cycle sinusoidal pulse during 2T seconds with alternating sign.

When viewed as a special case of OQPSK, the MSK waveform becomes

Sinusoidal symbol weighting

ak = ±1bk = ±1

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Minimum Shift Keying (MSK) (3/4)

MSK as a special case of OQPSK:

Modified I bit stream

I bit stream times carrier

Q bit stream times carrier

Modified Q bit stream

MSK waveform

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Minimum Shift Keying (MSK) (3/4)

Properties of MSK:The waveform s(t) has constant envelope.There is phase continuity in the RF carrier at the bit transition.The waveform s(t) can be regarded as an FSK waveform with signaling frequencies (f0 + 1/4T) and (f0 ‐ 1/4T).The minimum tone separation is 1/2Twhich is equal to half the bit rate.

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Power Spectral Densities for QPSK, OQPSK, and MSK

Average power of s(t), P=1 watt

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Quadrature Amplitude Modulation (QAM) (1/3)

QAM can be viewed as a combination of ASK and PSK.

A simple channel model for QAM:The signal point coordinate (x, y) are transmitted over separate channel, and independently perturbed by Gaussian noise variables (nx, ny), each with zero mean and variance N.

Signal space for M=16 Canonical QAM modulator QAM channel model

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The simplest QAM signaling:Use one‐dimensional PAM independently for each signal coordinate.Send k bits/dimension over a Gaussian channel, each signal point coordinate takes on one of 2k equally likely equispaced amplitudes.By convention, the signal points are grouped about the center of the space at amplitudes .

Bit error probability, PB, for QAM:

Q(x): Gaussian Q‐functionL: the number of amplitude levels for each signal coordinate.Gray code is assumed.

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Bandwidth‐power tradeoff of QAM:

Bandwidth efficiency:

QAM represents a method of reducing the bandwidth required for the transmission of digital data.

In QAM, a much more efficient exchangebetween R/W and Eb/N0 is possible than in MPSK.

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Waveform Design Example of QAM

Given the design parameters,Data rate: R = 144 Mbits/sAllowable DSB bandwidth = 36 MHz

Q1: Which modulation scheme would you choose?

Required spectral efficiency:

Since 16‐ary QAM requires a lower Eb/N0than 16‐ary PSK, we choose 16‐ary QAM.

Q2: If the available Eb/N0 = 20, what would be the resulting PB?

L = 4

M = 16

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9. Modulation and Coding for Band‐limited Channels

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Three Major Research Areas

Optimum signal constellation boundaries:Want to choose a closely packed signal subset from any regular array or lattice of candidate points.

Higher‐density lattice structuresWant to add improvement to the signal subset choice by starting with the densest possible lattice for the space.

Trellis‐coded modulationWant to use combined modulation and coding techniques for obtaining coding gain for band‐limited channel.

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Commercial Telephone Modems (1/2)

A typical telephone channel:High SNR of approximately 30 dBBandwidth of approximately 3 KHz

Evolution of leased‐line telephone modems:

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Commercial Telephone Modems (1/2)

Evolution of dial‐line Telephone modem standards:

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Signal Constellation Boundaries (1/2)

A large number of possible QAM signal constellations have been examined in a search for a designs that result in the best error performance for a given average signal‐to‐noise ratio.

Campopiano‐Glazer constellation construction rule:From an infinite array of points closely packed in a regular array or lattice, select a closely packed subset of 2k points as a signal constellation in a way that the average or peak power of the set is minimized for a given error probability.

M = 4

M = 8

M = 16

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Signal Constellation Boundaries (2/2)In a two‐dimensional signal space, the optimum boundary surrounding an array of points tends toward a circle.

Examples of 64‐ary (k=6) and 128‐ary (k=7) signal sets from a rectangular array.

The cross‐shaped boundaries are a compromise to the optimum circle.

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Higher‐Dimensional Signal ConstellationSignaling in a two‐dimensional space can provide the same error performance with less average (or peak) power than signaling in a one‐dimensional PAM space.

We choose points on a two‐dimensional lattice from within a circular rather than a rectangular boundary.

Further energy savings are possible by going to a higher number, N dimensions, and choosing points on an N‐dimensional lattice from within an N‐sphere rather than an N‐cube.

Such reduction in required energy for a given error performance is referred to as a shaping gain.

Energy savings from N‐sphere mapping vs. N‐cube mapping (i.e., shaping gain):

∞ 1.53

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High‐Density Lattice Structure

We want the densest possible lattice in the space.In a two‐dimensional signal space, the densest lattice is the hexagonal lattice.Examples of hexagonal packing:

Energy savings from the dense lattices vs. the rectangular lattice:

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Combined Gain: N‐Sphere Mapping and Dense Lattice

The combined energy savings from N‐sphere mapping and dense lattices vs. N‐cube mapping and rectangular lattices:

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