ECE 461 Fall 2006September 20, 2006Digital Modulation• After possible source and error control encoding, we have a sequence {mn} of message symbolsto be transmitted on the channel. The message symbols are assumed to come from a finitealphabet, say {0, 1, . . . , M − 1}. In the simplest case of binary signaling, M = 2. Each symbolin the sequence is assigned to one of M waveforms {s0(t), . . . , sM−1(t)}.• Memoryless modulation versus modulation with memory. If the symbol to waveform mapping isfixed from one interval to the next, i.e., m 7→ sm(t), then the modulation is memoryless. If themapping from symbol to waveform in the n-th symbol interval depends on previously transmittedsymbols (or waveforms) then the modulation is said to have memory.• For memoryless modulation, to send the sequence {mn} of symbols at the rate of 1/Tssymbolsper second, we transmit the signals(t) =Xnsmn(t − nTs) . (1)• Linear versus nonlinear modulation. A digital modulation scheme is said to be linear if wecan write the mapping from the sequence of symbols {mn} to the transmitted signal s(t) asconcatenation of a mapping from the sequence {mn} to a complex sequence {cn}, followed by alinear mapping from {cn} to s(t). Otherwise the modulation is nonlinear.Linear Memoryless Modulation• In this case, the mapping from symbols to waveforms can be written in complex baseband as:sm(t) =pEmejθmg(t) , m = 1, 2, . . . , M , (2)where g(t) is a real-valued, unit energy, pulse shaping waveform.• The signal sm(t) can be represented by a point in the complex plane, i.e., the signal spacecorresponding to a symbol interval is a 1-d (complex) space with basis function g(t).PSfrag replacementsIIQQsm√Emθms1s2s3s4sMRepresentation of sm(t) Signal constellationcV.V. Veeravalli, 2006 1• In real passband,˜sm(t) = Re[√2sm(t) ej2πfct] =p2Emcos(2πfct + θm)g(t) . (3)• As we saw in class, the signal energy is the same in both the real passband and complex basebanddomains and equals Em.• The average symbol energy for the constellation is given byEs=1MMXm=1Em. (4)• The average bit energy for the constellation (assuming that M = 2ν, for some integer ν) is givenbyEb=Eslog2M=Esν. (5)• The distance between signals skand smis dk,m= ksk−smk, and the minimum distance is givenbydmin= mink,mdk,m. (6)• A measure of goodness of the constellation is the ratioζ =d2minEb. (7)Note that ζ is independent of scaling of the constellation.• Some commonly used signal constellations are:◦ Pulse Amplitude Modulation (PAM). Information only in amplitude:θm= 0 andpEm= (2m − 1 − M)d2, m = 1, 2, . . . , M . (8)We can compute ζ as a function of M. For example, ζ = 4 for M = 2.◦ Phase Modulation or Phase Shift Keying (PSK). Information only in phase:θm=2πmMand Em= E , m = 1, 2, . . . , M . (9)We showed in class thatdmin=q2E1 − cos2πM=⇒ ζ = 2 log2M1 − cos2πMFor QPSK, ζ = 4 (as in BPSK).◦ Quadrature Amplitude Modulation (QAM). Information in phase and amplitude. We candesign constellations to maximize ζ for a given M. Rectangular constellations are convenientfor demodulation. For rectangular 16-QAM, ζ = 1.6.cV.V. Veeravalli, 2006 2Orthogonal Memoryless Modulation• Here the signal set is given bysm(t) =√E gm(t), m = 1, 2, . . . , M (10)where {gm(t)} are (possibly complex) unit energy signals, i.e., kgm(t)k = 1.• The correlation between signals sk(t) and sm(t) is given by:ρkm=hsk(t), sm(t)iE= hgk(t), gm(t)i (11)• There are two kinds of orthogonality:◦ Orthogonality only in the real component of the correlation, i.e. Re{ρkm} = 0, for k 6= m.This form of orthogonality is enough for coherent demodulation.◦ Complete orthogonality, i.e., ρkm= 0, for k 6= m. This is required for noncoherent demodula-tion.• Examples of orthogonal signal sets◦ Separation in time:gm(t) = g (t − (m − 1)Ts/M) (12)where g(t) is such that hg(t −kTs/M), g(t − mTs/M)i = δkm. For example, g(t) = pTs/M(t), arectangular pulse of width Ts/M.This signal set is completely orthogonal. We can also create a signal set of twice the size whichsatisfies orthogonality only in the real component of the correlation by adding {jgm(t)} to theabove signal set as we saw in class.◦ Separation in frequency:gm(t) = ej2π(m−1)∆ftpTs(t) (13)It is easy to show thatρkm= sinc[Ts(k − m)∆f] ejπTs(k−m)∆f(14)and thatRe{ρkm} = sinc[2Ts(k − m)∆f] . (15)Thus the smallest value of ∆fsuch that ρkm= 0, for k 6= m, is 1/Ts, and such that Re{ρkm} =0, for k 6= m, is 1/2Ts.◦ Separation in time and frequency: One way to do this is to pick {gm(t)} to be the Walsh-Hadamard functions on [0, Ts] as we saw in class.cV.V. Veeravalli, 2006