A First Course in Digital Communications

A First Course in Digital Communications Ha H. Nguyen and E. Shwedyk February 9 A First Course in Digital Communications 1/46

Introduction There are benefits to be gained when M-ary (M = 4 signaling methods are used rather than straightforward binary signaling. In general, M-ary communication is used when one needs to design a communication system that is bandwidth efficient. Unlike QPSK and its variations, the gain in bandwidth is accomplished at the expense of error performance. To use M-ary modulation, the bit stream is blocked into groups of λ bits the number of bit patterns is M = λ. The symbol transmission rate is r s = 1/T s = 1/(λT b = r b /λ symbols/sec there is a bandwidth saving of 1/λ compared to binary modulation. Shall consider M-ary ASK, PSK, QAM (quadrature amplitude modulation and FSK. A First Course in Digital Communications /46

Optimum Receiver for M-ary Signaling i m (t s i r(t mˆ i w(t w(t is zero-mean white Gaussian noise with power spectral density of N (watts/hz. Receiver needs to make the decision on the transmitted signal based on the received signal r(t = s i (t + w(t. The determination of the optimum receiver (with minimum error proceeds in a manner analogous to that for the binary case. A First Course in Digital Communications 3/46

Represent M signals by an orthonormal basis set, {φ n (t} N n=1, N M: s i (t = s i1 φ 1 (t + s i φ (t + + s in φ N (t, s ik = Ts s i (tφ k (tdt. Expand the received signal r(t into the series r(t = s i (t + w(t = r 1 φ 1 (t + r φ (t + + r N φ N (t + r N+1 φ N+1 (t + For k > N, the coefficients r k can be discarded. Need to partition the N-dimensional space formed by r = (r 1,r,...,r N into M regions so that the message error probability is minimized. A First Course in Digital Communications 4/46

N dimensional observation space r = ( r, r,, r 1 M R Choose s ( t or m 1 1 1 R M M Choose s ( t or m M R Choose s ( t or m The optimum receiver is also the minimum-distance receiver: Choose m i if N k=1 (r k s ik < N k=1 (r k s jk ; j = 1,,...,M; j i. A First Course in Digital Communications 5/46

M-ary Coherent Amplitude-Shift Keying (M-ASK s i (t = V i cos(πf c t, t T s T s = [(i 1 ]φ 1 (t, φ 1 (t = cos(πf c t, t T s, T s i = 1,,...,M. s ( s ( s ( s (t s k M ( s M (t 1 t t 3 t (k 1 (M (M 1 1 t φ ( t 1 t = kt s s i (t r(t kt s r1 ( dt! " ( k 1 Ts mˆ i w(t φ 1 ( t N WGN, strength watts/hz A First Course in Digital Communications 6/46

Minimum-Distance Decision Rule for M-ASK ( s k (t, if k 3 < r1 < ( k 1, k =,3,...,M 1 Choose s 1 (t, if r 1 < s M (t, if r 1 > ( M 3. f ( r s ( 1 k t Choose s1 ( t (k 1 Choose s M ( t Choose s k ( t # r 1 A First Course in Digital Communications 7/46

Error Performance of M-ASK f ( r s ( 1 k t P[error] = M P[s i (t]p[error s i (t]. i=1 P[error s i (t] = Q ( / N, i =,3,...,M 1. P[error s i (t] = Q ( / N, i = 1,M. (M 1 ( P[error] = M Q / N. A First Course in Digital Communications 8/46 r 1

Modified M-ASK Constellation The maximum and average transmitted energies can be reduced, without any sacrifice in error probability, by changing the signal set to one which includes the negative version of each signal. s i (t = (i 1 M }{{} V i '( '* & E s = E b = 3 M i=1 E i M T s cos(πf c t, t T s, i = 1,,..., M. = 4M E s log M = (M 1 1 log M = 3 % φ 1 ( t φ ( t 1 M (i 1 M = (M 1. 1 i=1 (1 log ME b M 1 A First Course in Digital Communications 9/46

Probability of Symbol Error for M-ASK ( (M 1 P[error] = M Q 6E s (M = 1N ( (M 1 M Q 6log M M 1 ( P[bit error] = 1 (M 1 P[symbol error] = λ M log M Q 6log M M 1 1 1 E b. N E b (with Gray mapping N P[symbol error] 1 1 3 1 4 1 5 1 6 M= (W=1/T b M=4 (W=1/T b M=8 (W=1/3T b M=16 (W=1/4T b 1 7 5 1 15 5 E /N (db b W is obtained by using the WT s = 1 rule-of-thumb. Here 1/T b is the bit rate (bits/s. A First Course in Digital Communications 1/46

Example of -ASK (BPSK and 4-ASK Signals 1 Baseband information signal 1 Tb Tb 3Tb 4Tb 5Tb 6Tb 7Tb 8Tb 9Tb 1Tb BPSK Signalling Tb Tb 3Tb 4Tb 5Tb 6Tb 7Tb 8Tb 9Tb 1Tb 4 ASK Signalling Tb 4Tb 6Tb 8Tb 1Tb A First Course in Digital Communications 11/46

M-ary Phase-Shift Keying (M-PSK [ s i (t = V cos πf c t ] (i 1π, t T s, M i = 1,,...,M; f c = k/t s, k integer; E s = V T s / joules [ ] [ ] (i 1π (i 1π s i (t = V cos cos(πf c t+v sin sin(πf c t. M M φ 1 (t = V cos(πf ct, φ (t = V sin(πf ct. Es Es s i1 = [ ] (i 1π E s cos, s i = E s sin M M [ (i 1π The signals lie on a circle of radius E s, and are spaced every π/m radians around the circle. ]. A First Course in Digital Communications 1/46

Signal Space Plot of 8-PSK [ s i (t = V cos πf c t ] (i 1π, t T s, M i = 1,,...,M; f c = k/t s, k integer; E s = V T s / joules φ ( t s3( t 11 1 s 4 ( t s( t 1 E s 11 s 5( t π 4 s1( t φ 1 ( t 111 s6( t s8( t 1 s7( t 11 A First Course in Digital Communications 13/46

Signal Space Plot of General M-PSK [ s i (t = V cos πf c t ] (i 1π, t T s, M i = 1,,...,M; f c = k/t s, k integer; E s = V T s / joules φ ( t s ( t E s π M π M s ( t 1 φ 1 ( t s M (t A First Course in Digital Communications 14/46

Optimum Receiver for M-PSK r(t φ 1 ( t T, s T- s ( dt ( dt t = T s t = T s r 1 r Compute ( r1 si 1 + ( r+ si for i = 1,,, M and choose the smallest mˆ i r φ ( t s ( t Region Choose s ( t P[error] = P[error s 1 (t] = 1 r 1,r Region 1 f(r 1, r s 1 (tdr 1 dr. E s π M s ( t 1 Region1 Choose s ( t 1 r 1 A First Course in Digital Communications 15/46

Lower Bound of P[error] of M-PSK r Region Choose s ( t r s ( t R 1 s ( t E s π M s ( t 1 Region1 Choose s ( t 1 r 1 π M s 1 ( t ( E s, Es sin ( π M r 1 P[error s 1 (t] > P[r 1,r fall in R 1 s 1 (t], or { ( π } P[error s 1 (t] > Q sin Es /N. M A First Course in Digital Communications 16/46

Upper Bound of P[error] of M-PSK r r π M s 1 ( t Region r Choose s 1 (, t R 1 s ( t ( E s Es sin ( π M s ( t E s π M s ( t 1 Region1 Choose s ( t 1 r 1 r π M ( E s, s ( t 1 Es r 1 sin ( π M R s M (t P[error] < P[r 1,r fall in R 1 s 1 (t] + P[r 1,r fall in R s 1 (t], or ( ( π P[error] < Q sin Es /N, M A First Course in Digital Communications 17/46

Symbol Error Probability of M-PSK 1 1 M=3 1 M=16 1 3 P[symbol error] 1 4 1 5 Exact M=4 M=8 1 6 1 7 Lower bound Upper bound M= 5 1 15 5 E /N (db b With a Gray mapping, the bit ( error probability is approximated as: P[bit error] M-PSK 1 log M Q λsin ( π Eb M N. A First Course in Digital Communications 18/46

Comparison of BPSK and M-PSK ( ( π P[error] M-PSK Q λsin Eb, where E s = λe b. M N P[error] BPSK = Q( E b /N. λ M M-ary BW/Binary BW λsin (π/m M-ary Energy/Binary Energy 3 8 1/3.44 3.6 db 4 16 1/4.15 8. db 5 3 1/5.5 13. db 6 64 1/6.144 17. db A First Course in Digital Communications 19/46

M-ary Quadrature Amplitude Modulation (M-QAM M-QAM constellations are two-dimensional and they involve inphase (I and quadrature (Q carriers: φ I (t = cos(πf c t, t T s, T s φ Q (t = sin(πf c t, t T s, T s The ith transmitted M-QAM signal is: s i (t = V I,i T s cos(πf c t + V Q,i T s sin(πf c t, t T s i = 1,,...,M = E i T s cos(πf c t θ i V I,i and V Q,i are the information-bearing discrete amplitudes of the two quadrature carriers, E i = VI,i + V Q,i and θ i = tan 1 (V Q,i /V I,i. A First Course in Digital Communications /46

Rectangle Rectangle Triangle M = 4 M = 8 (1,3 (1,7 (4,4 A First Course in Digital Communications 1/46

R 1 Rectangle Triangle (1,5,1 R M = 16 R 1 R 1 R R Hexagon (8,8 (4,1 A First Course in Digital Communications /46

A Simple Comparison of M-QAM Constellations With the same minimum distance of all the constellations, a more efficient signal constellation is the one that has smaller average transmitted energy. 3456789:4./1 ; < =>?@AB CDEDF M = 8 E s for the rectangular, triangular, (1,7 and (4,4 constellations are found to be 1.5, 1.15, 1.16 and 1.183, respectively. A First Course in Digital Communications 3/46

Rectangular M-QAM φ ( t Q M = 64 M =16 M = 3 M = 8 φ I ( t M = 4 The signal components take value from the set of discrete values {(i 1 M /}, i = 1,,..., M. A First Course in Digital Communications 4/46

Modulation of Rectangular M-QAM Each group of λ = log M bits can be divided into λ I inphase bits and λ Q quadrature bits, where λ I + λ Q = λ. Inphase bits and quadrature bits modulate the inphase and quadrature carriers independently. Inphase bits Select VI, i GHI JKLMNOLHPHQ Infor. bits s (t Inphase ASK i cos(πf ct T s XYZ WN[VSL Quadrature bits Select VQ, i Quadrature ASK RST UQSVWJNMMHQ sin(πf ct T A First Course in Digital Communications 5/46 s

Demodulation of Rectangular M-QAM Due to the orthogonality of the inphase and quadrature signals, inphase and quadrature bits can be independently detected at the receiver. T s ( dt t = T s Inphase ASK decision r ( t = si ( t + w( t φi ( t = cos( π fct T s T s ( dt t = T s Quadrature ASK decision Multiplexer Decision φq ( t = sin( π fct T s (b Receiver The most practical rectangular QAM constellation is one which λ I = λ Q = λ/, i.e., M is a perfect square and the rectangle is a square. A First Course in Digital Communications 6/46

Symbol Error Probability of M-QAM For square constellations: P[error] = 1 P[correct] = 1 ( 1 P M [error], ( ( P M [error] = 1 1 3E s Q, M (M 1N where E s /N is the average SNR per symbol. For general rectangular constellations: [ ( ] 3E s P[error] 1 1 Q (M 1N ( 3λE b 4Q (M 1N where E b /N is the average SNR per bit. A First Course in Digital Communications 7/46

1 1 1 M=64 M=56 P[symbol error] 1 3 1 4 1 5 M=4 M=16 1 6 1 7 M= Exact performance Upper bound 5 1 15 5 E b /N (db A First Course in Digital Communications 8/46

Performance Comparison of M-PSK and M-QAM For M-PSK, approximate P[error] Q( Es N sin π M For M-QAM, use the upper bound 4Q( 3λEb (M 1N. Comparing the arguments of Q( for the two modulations gives: 3/(M 1 κ M = sin (π/m. M 1log 1 κ M 8 1.65 db 16 4. db 3 7. db 64 9.95 db 56 15.9 db 14 1.93 db. A First Course in Digital Communications 9/46

Performance Comparison of M-ASK, M-PSK, M-QAM 1 1 1 P[symbol error] 1 3 1 4 4 QAM or QPSK M=4, 8, 16, 3 1 5 1 6 M ASK M PSK M QAM 5 1 15 E /N (db b A First Course in Digital Communications 3/46

M-ary Coherent Frequency-Shift Keying (M-FSK s i (t = { V cos(πfi t, t T s, elsewhere, i = 1,,..., M, where f i are chosen to have orthogonal signals over [, T s ]. ( 1 (k ± i T f i = s, (coherently orthogonal (, i =, 1,,... 1 (k ± i, (noncoherently orthogonal T s φ ( t s ( t E s E s s ( t 1 φ 1 ( t s ( t 3 E s φ ( t 3 A First Course in Digital Communications 31/46

Minimum-Distance Receiver of M-FSK Choose m i if M M (r k s ik < (r k s jk k=1 k=1 j = 1,,...,M; j i, r(t s ( t φ 1 = 1 ( t E s Tf s Tg s ( dt h ( dt Choose m i if r i > r j, j = 1,,...,M; j i. t = T s t = T s r 1 h r M \]^^_` a]` bcde`_a Decision sm ( t φm ( t = E s A First Course in Digital Communications 3/46

Symbol Error Probability of M-FSK P[correct s 1 (t] = = r 1= P[error] = P[error s 1 (t] = 1 P[correct s 1 (t]. P[(r < r 1 and and (r M < r 1 s 1 (t sent]. P[(r < r 1 and and (r M < r 1 {r 1 = r 1, s 1 (t}]f(r 1 s 1 (tdr. [(r < r 1 and and(r M < r 1 {r 1 = r 1, s 1 (t}] = P[r j < r 1 {r 1 = r 1, s 1 (t}] = P[correct] = r 1= [ r1 1 πn exp λ= r1 M P[(r j < r 1 {r 1 = r 1, s 1 (t}] j= 1 πn exp 1 πn exp { (r 1 E s N { λ { λ N } dr 1. N } dλ. } dλ] M 1 A First Course in Digital Communications 33/46

1 1 1 P[symbol error] 1 3 1 4 1 5 M=64 M=3 M=16 M=4 M= M=8 1 6 P[error] = 1 1 π 1 7 4 6 8 1 1 14 16 E b /N (db ( [ 1 π y M 1 e dx] x / exp 1 y log ME b N dy. A First Course in Digital Communications 34/46

Bit Error Probability of M-FSK Due to the symmetry of M-FSK constellation, all mappings from sequences of λ bits to signal points yield the same bit error probability. For equally likely signals, all the conditional error events are equiprobable and occur with probability Pr[symbol error]/(m 1 = Pr[symbol error]/( λ 1. There are ( λ k ways in which k bits out of λ may be in error The average number of bit errors per λ-bit symbol is λ ( λ Pr[symbol error] k k λ 1 k=1 = λ λ 1 λ Pr[symbol error]. 1 The probability of bit error is simply the above quantity divided by λ: Pr[bit error] = λ 1 λ Pr[symbol error]. 1 A First Course in Digital Communications 35/46

Union Bound on the Symbol Error Probability of M-FSK P[error] = P[(r 1 < r or (r 1 < r 3 or,,or (r 1 < r M s 1 (t]. Since the events are not mutually exclusive, the error probability is bounded by: P[error] < P[(r 1 < r s 1 (t]+ P[(r 1 < r 3 s 1 (t] + + P[(r 1 < r M s 1 (t]. ( Es But P[(r 1 < r j s 1 (t] = Q /N, j = 3,4,...,M. Then ( ( P[error] < (M 1Q Es /N < MQ Es /N < Me Es/(N. { where the bound Q(x < exp x } has been used. A First Course in Digital Communications 36/46

An Upper Bound on Q(x Q(x = 1 x 1 π exp } { λ dλ < exp } { x Q(x and its simple upper bound 1 5 1 1 Q(x exp( x / 1 15 1 3 4 5 6 7 8 x A First Course in Digital Communications 37/46

Interpretations of P[error] < Me E s/(n 1 Let M = λ = e λln and E s = λe b. Then P[error] < e λ ln e λe b/(n = e λ(e b/n ln /. As λ, or equivalently, as M, the probability of error approaches zero exponentially, provided that E b N > ln = 1.39 = 1.4 db. Since E s = λe b = V T s /, then P[error] < e λ ln e V T s/(4n = e Ts[ r b ln+v /(4N ] If r b ln + V /(4N >, or r b < V 4N ln the probability or error tends to zero as T s or M becomes larger and larger. A First Course in Digital Communications 38/46

Comparison of M-ary Signaling Techniques A compact and meaningful comparison is based on the bit rate-to bandwidth ratio, r b /W (bandwidth efficiency versus the SNR per bit, E b /N (power efficiency required to achieve a given P[error]. M-ASK with single-sideband (SSB transmission, W = 1/(T s and ( rb = log W M (bits/s/hz. SSB-ASK M-PSK (M > must have double sidebands, W = 1/T s and ( rb = log W M, (bits/s/hz, PSK (Rectangular QAM has twice the rate of ASK, but must have double sidebands QAM and SSB-ASK have the same bandwidth efficiency. For M-FSK with the minimum frequency separation of 1/(T s, W = M T s = M (λ/r b = M log M r b, and ( rb = log M W FSK M. A First Course in Digital Communications 39/46

USSB Transmission of BPSK Signal cos( π f t c m( t + s ( t USSB LTI Filter + h( t H( f H ( f = j sgn( f mˆ ( t sin( π f t ( ˆm(t = m(t h(t = m(t 1 = 1 πt π c m(t λ dλ λ A First Course in Digital Communications 4/46

Example of USSB-BPSK Transmitted Signal V (a BPSK signal V Tb Tb 3Tb 4Tb 5Tb 6Tb t (b USSB BPSK signal V V Tb Tb 3Tb 4Tb 5Tb 6Tb 7Tb 8Tb t A First Course in Digital Communications 41/46

Power-Bandwidth Plane (At P[error] = 1 5 r b /W (bits/s/hz 1 5 3 1.5.3 M=4 M= M=8 M=16 M=3 M=16 M=8 M=64 M=16 M=3 M=64 Bandwidth limited region: r b /W>1 Power limited region: r b /W<1. M=64 PSK QAM and ASK (SSB.1 FSK 1 5 1.6 5 1 15 5 3 SNR per bit, E b /N (db A First Course in Digital Communications 4/46

Two Statements Consider information transmission over an additive white Gaussian noise (AWGN channel. The average transmitted signal power is P av, the noise power spectral density is N / and the bandwidth is W. Two statements are: 1 For each transmission rate r b, there is a corresponding limit on the probability of bit error one can achieve. For some appropriate signalling rate r b, there is no limit on the probability of bit error one can achieve, i.e., one can achieve error-free transmission. Which statement sounds reasonable to you? A First Course in Digital Communications 43/46

Shannon s Channel Capacity ( C = W log 1 + P av, WN where W is bandwidth in Hz, P av is the average power and N / is the two-sided power spectral density of the noise. Shannon proved that it is theoretically possible to transmit information at any rate r b, where r b C, with an arbitrarily small error probability by using a sufficiently complicated modulation scheme. For r b > C, it is not possible to achieve an arbitrarily small error probability. Shannon s work showed that the values of P av, N and W set a limit on transmission rate, not on error probability! The normalized channel capacity C/W (bits/s/hz is: C W = log ( 1 + P av WN = log ( 1 + C W A First Course in Digital Communications 44/46 E b N.

Shannon s Capacity Curve E b N = C/W 1 C/W r b /W (bits/s/hz 1 5 3 1.5.3 Shannon limit Channel capacity limit, C/W M=4 M= M=8 M=16 M=3 M=16 M=8 M=64 M=16 M=3 M=64 Bandwidth limited region: r b /W>1 Power limited region: r b /W<1. M=64 PSK QAM and ASK (SSB.1 FSK 1 5 1.6 5 1 15 5 3 SNR per bit, E /N (db b A First Course in Digital Communications 45/46

Spectrum Efficiency of DVB-S Standard 5 4.5 Example : 8 Mbit/s in 36MHz at 9.5dB with 8PSK Spectrum Efficiency 4 3.5 3.5 1.5 1.5 Example 1: 5Mbit/s in 36MHz at 4dB with QPSK -4-4 6 8 1 1 14 16 SNR (db More information: www.dvb.org A First Course in Digital Communications 46/46