Modulation and Coding Tradeoffs

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1 0 Modulation and Coding Tradeoffs

2 Contents 1 1. Design Goals 2. Error Probability Plane 3. Nyquist Minimum Bandwidth 4. Shannon Hartley Capacity Theorem 5. Bandwidth Efficiency Plane 6. Modulation and Coding Tradeoffs 7. Design Steps of Communication Systems 8. Bandwidth Efficient Modulation 9. Modulation and Coding for Bandlimited Channels

3 2 1. Design Goals

4 Design Goals 3 Maximum transmission bit rate R Minimum probability of bit error P B Minimum required power or E b /N 0 Minimum system bandwidth W Minimum system complexity Maximum system utilization For reliable service For a maximum number of users With minimum delay With maximum resistance to interference Tradeoff

5 Constraints and Limitations 4 Nyquist theoretical minimum bandwidth requirement Shannon 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 plane Bandwidth efficiency plane

6 5 2. Error Probability Plane

7 Bit Error Probability versus E b /N 0 6 Coherent MFSK: Coherent MPSK: Equi bandwidth Curves P B W E b /N 0 : fixed P B E W: fixed b/n 0 E b /N 0 W P B : fixed

8 7 3. Nyquist Minimum Bandwidth

9 Filtering Aspects of ISI 8 ISI for a typical baseband digital system: H t (f) H c (f) H r (f) Equivalent model: The tail of a pulse smears into adjacent symbol intervals. Transmitting filter Channel Receiving filter

10 System Transfer Function for Baseband Channel 9 System Transfer Function, H(f), and ideal Nyquist pulse, h(t) Sampling at the receiver at every T sec. If each pulse of a received sequence is of the form sinc(t/t), then the pulses can be detected without ISI.

11 Nyquist Minimum Bandwidth 10 The theoretical minimum bandwidth (Nyquist bandwidth) needed for the baseband transmission of R s symbols/s without ISI is R s /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/W measured in bits/s/hz A measurement of data throughput per hertz of bandwidth. Ex) MPSK system: For a fixed bandwidth W, as k increases, the allowable data rate R increases, and thus R/W increases at the expense of large E b /N 0.

12 11 4. Shannon Hartley Capacity Theorem C.E. Shannon, A Mathematical Theory of Communication, BSTJ, Vol.27, pp , 1948.

13 System Capacity 12 Shannon Hartley Capacity Theorem: (bits/s) C: System capacity for an AWGN channel W: Bandwidth S: Average received signal power (S = E b R) N: Average noise power (N = N 0 W) 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.

14 Normalized Channel Capacity/Bandwidth vs. SNR 13 Normalized channel capacity vs. SNR Normalized channel bandwidth vs. SNR C W = log2 1+ S N

15 Normalized Channel Capacity vs. E b /N 0 14 Suppose that the transmission bit rate, R = channel capacity, C N = N W 0 Eb STb S/ R S W = = = N N/ W N/ W N R 0 This is the case when R=C. E b S R = N N W 0 If R = C C S Eb C = log2 1+ = log2 1+ W N N0 W C/ W Eb C 2 = 1+ N W 0 Eb N 0 C/ W ( 2 1) W = C

16 Shannon Limit 15 Definition: The limit of E b /N 0 below which there can be no error free communication at any information rate Shannon s work provides a theoretical proof for the existence of codes that could improve the P B performance, or reduce the E b /N 0 required. Today, as much as 10 db improvement is realizable with turbo code. Mathematical limit: We use the identity: Let Then or As C/W 0 or x 0, we have or Again, this is the case when R=C.

17 Entropy, H 16 Definition: A metric for measuring the average amount of information per source output where p i is the probability of the i th output and. For a binary case: Maximum entropy when events occur equally likely.

18 Properties of Entropy 17 The 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.

19 Equivocation, H(X Y) (1/2) 18 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 message Y = received signal P(X,Y) = joint probability of X and Y P(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.

20 Equivocation, H(X Y) (2/2) 19 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 P B = Then the equivocation H(X Y) becomes ( ) H X Y = P( Y) P( X Y)log P( X Y) Y X = P(0) P(0 0)log 2P(0 0) + P(1 0)log 2P(1 0) P(1) P(0 1) log 2P(0 1) + P(11) log 2P(11) 1 = ( 1 B) log 2( 1 B) B log 2 B 2 P P + P P 1 PBlog2PB + ( 1 PB) log2( 1 PB) 2 = + ( 1 PB) log2( 1 PB) PBlog2PB 0.99log ( 0. 99) log ( 0.01) = 2 = bit/received symbol The channel introduces bit of uncertainty to each received symbol. 2 2

21 Effective Entropy, H eff 20 Effective entropy, H eff,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 P B = 0.01, the equivocation H(X Y) = Then the effective entropy becomes Thus, if the binary symbol rate is given by R = 1000 symbols/s, the effective information bit rate R eff can then be written as Note that in the extreme case, where P B = 0.5, we have H eff = 0 R eff = 0 The effective information bit rate is zero!

22 21 5. Bandwidth Efficiency Plane

: log2 M R= krs = ( log2 M) Rs = T For coherent MPSK: R log 2 M W

23 Bandwidth Efficiency Plane (R/W vs. E b /N 0 ) 22 Ideal performance Reflects how efficiently the bandwidth resource is utilized. P B R/W E b /N 0 : fixed Bandwidth efficiency plane at P B =10 5 : log2 M R= krs = ( log2 M) Rs = T For coherent MPSK: R log 2 M W = 1 W Ts For Noncoherent MFSK: W R W M T = s log2 M M s E b /N 0 R/W P B : fixed P B E b /N 0 R/W: fixed Equi error probability curves

24 23 6. Modulation and Coding Tradeoffs

25 Error Probability Plane vs. Bandwidth Efficiency Plane 24 Error Probability Plane Useful for examining power limited systems. Bandwidth Efficiency Plane Useful for examining bandwidth limited systems. G: gained, C: cost, F: fixed

26 25 7. Design Steps of Communication Systems

27 Before Designing Communication Systems 26 Need description of the channel: Received power Available bandwidth Noise statistics Other impairments, such as fading Need definition of the system requirements: Data rate Error performance Tradeoff between bandwidth power Given the channel description, we determine design choices that best match the channel and meet the performance requirements.

28 Requirements for MPSK and MFSK Signaling 27 Example: System rate, Nyquist minimum bandwidth, bandwidth efficiency, and required E b /N 0 at 9,600 bits/s: (3/8)

29 Example 1: Bandwidth Limited Uncoded System 28 Assumptions: AWGN radio channel Available bandwidth: W = 4,000 Hz Ratio of received signal power to noise power spectral density: P r /N 0 = 53 db Hz Required data rate: R = 9,600 bits/s Required bit error probability: P B = 10 5 Design steps: Received E b /N 0 : Use the relationship between P r /N 0 and E b /N 0 : or 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 E b /N 0 =13.2 db is greater than the required E b /N 0 =13.0 db).

30 Example 2: Power Limited Uncoded System 29 Assumptions: AWGN radio channel Available bandwidth: W = 45,000 Hz Ratio of received signal power to noise power spectral density: P r /N 0 = 48 db Hz Required data rate: R = 9,600 bits/s Required bit error probability: P B = 10 5 Design steps: Received E b /N 0 : Use the relationship between P r /N 0 and E b /N 0 : Determine modulation scheme: Choose MFSK signaling (since W > R, and the received E b /N 0 is relatively small for the required P B =10 5 ). Determine M: Choose M = 16 (since the minimum bandwidth required for M=16 is 38,400 Hz, and that for M=32 is 61,440 Hz). Check if we need an error correction coding: Not necessary (since the received E b /N 0 =8.2 db is greater than the required E b /N 0 =8.1 db).

31 Example 3: Bandwidth Limited and Power Limited Coded System Assumptions: AWGN radio channel Available bandwidth: W = 4,000 Hz Ratio of received signal power to noise power spectral density: P r /N 0 = 53 db Hz Required data rate: R = 9,600 bits/s Required bit error probability: P B = 10 9 Design steps: Received E b /N 0 : 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: Yes, it is necessary (since the received E b /N 0 is insufficient for the required P B ). 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. 30

32 31 8. Bandwidth Efficient Modulation

33 Bandwidth Efficient Modulation 32 Want to maximize bandwidth efficiency so that the spectral congestion problem is ameliorated. A constant envelope modulation is additionally required in order to relax Extraneous sidebands produced by system nonlinearities Adjacent channel interference problem Co channel interference problem with other communication systems Examples of the constant envelope modulation: Offset QPSK (OQPSK) Minimum shift keying (MSK)

34 QPSK (1/2) 33 Original data streams: ( ) The pulse streams are divided into an in phase and quadrature streams: d I (t) and d Q (t) each have half the bit rate of d k (t). Orthogonal realization of a QPSK waveform:

35 QPSK (2/2) 34 Modulation scheme: The two BPSK signals can be detected separately. Zero crossing takes place during transition. (The envelope of s(t) is not constant.)

36 Offset (or Staggered) QPSK (OQPSK) 35 Alignment of d I (t) and d Q (t) is offset by T. No zero crossing. (Less out of band interference comparing to QPSK)

Minimum Shift Keying (MSK) (1/4) 36 Motivation: QPSK OQPSK MSK Suppressing out of band interference Removing discontinuous phase transitions MSK can be viewed as either A special case of CPFSK

37 Minimum Shift Keying (MSK) (1/4) 36 Motivation: QPSK OQPSK MSK Suppressing out of band interference Removing discontinuous phase transitions MSK can be viewed as either A special case of CPFSK (continuous phase FSK), or A special case of OQPSK with sinusoidal symbol weighting MSK waveform: A phase constant for the k th bit interval If d k = 1, the transmitted frequency is (f 0 + 1/4T), and If d k = 1, the transmitted frequency is (f 0 1/4T). The tone spacing in MSK is 1/2T which is one half (i.e., minimum) that employed for noncoherently demodulated orthogonal FSK.

38 Minimum Shift Keying (MSK) (2/4) 37 Phase constant: During each T, the value of x k is constant, i.e., x k = 0 or π. The value for x k 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 a k = ±1 b k = ±1 The a k term can change value Sinusoidal only at symbol the zero crossing weightingof cos(πt/2t). The b k 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

39 Minimum Shift Keying (MSK) (3/4) 38 MSK as a special case of OQPSK: Modified I bit stream I bit stream times carrier Modified Q bit stream Q bit stream times carrier MSK waveform

40 Minimum Shift Keying (MSK) (3/4) 39 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 (f 0 + 1/4T) and (f 0 1/4T). The minimum tone separation is 1/2T which is equal to half the bit rate.

41 Power Spectral Densities for QPSK, OQPSK, and MSK 40 Average power of s(t), P=1 watt

42 Quadrature Amplitude Modulation (QAM) (1/3) 41 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 (n x, n y ), each with zero mean and variance N. Signal space for M=16 Canonical QAM modulator QAM channel model

43 Quadrature Amplitude Modulation (QAM) (2/3) 42 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 2 k equally likely equispaced amplitudes. By convention, the signal points are grouped about the center of the space at amplitudes. Bit error probability, P B, for QAM: Q(x): Gaussian Q function L: the number of amplitude levels for each signal coordinate. Gray code is assumed.

44 Quadrature Amplitude Modulation (QAM) (3/3) 43 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 exchange between R/W and E b /N 0 is possible than in MPSK.

Required spectral efficiency: M = 16 Since 16 ary QAM requires a lower E b /N 0 than 16

45 Waveform Design Example of QAM 44 Given the design parameters, Data rate: R = 144 Mbits/s Allowable DSB bandwidth = 36 MHz Q1: Which modulation scheme would you choose? Required spectral efficiency: M = 16 Since 16 ary QAM requires a lower E b /N 0 than 16 ary PSK, we choose 16 ary QAM. Q2: If the available E b /N 0 = 20, what would be the resulting P B? L = 4

46 45 9. Modulation and Coding for Band limited Channels

47 Three Major Research Areas 46 Optimum signal constellation boundaries: Want to choose a closely packed signal subset from any regular array or lattice of candidate points. Higher density lattice structures Want to add improvement to the signal subset choice by starting with the densest possible lattice for the space. Trellis coded modulation Want to use combined modulation and coding techniques for obtaining coding gain for band limited channel.

48 Commercial Telephone Modems (1/2) 47 A typical telephone channel: High SNR of approximately 30 db Bandwidth of approximately 3 KHz Evolution of leased line telephone modems:

49 Commercial Telephone Modems (1/2) 48 Evolution of dial line Telephone modem standards:

50 Signal Constellation Boundaries (1/2) 49 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 2 k 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

51 Signal Constellation Boundaries (2/2) 50 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.

52 Higher Dimensional Signal Constellation 51 Signaling 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

53 High Density Lattice Structure 52 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:

54 Combined Gain: N Sphere Mapping and Dense Lattice 53 The combined energy savings from N sphere mapping and dense lattices vs. N cube mapping and rectangular lattices:

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