4.2-21 4.2.4 Orthogonal, Biorthogonal and Simplex Signals • In PAM, QAM and PSK, we had only one basis function. For orthogonal, biorthogonal and simplex signals, however, we use more than one orthogonal basis function, so N-dimensional Examples: the Fourier basis; time-translated pulses; the Walsh-Hadamard basis. o We’ve become used to SER getting worse quickly as we add bits to the symbol – but with these orthogonal signals it actually gets better. o The drawback is bandwidth occupancy; the number of dimensions in bandwidth W and symbol time Ts is N ≈ 2W Ts o So we use these sets when the power budget is tight, but there’s plenty of bandwidth. Orthogonal Signals • With orthogonal signals, we select only one of the orthogonal basis functions for transmission: The number of signals M equals the number of dimensions N. pulse position modulation (PPM). • Energy and distance: o The signals are equidistant.4. as can be seen from the sketch or from ⎡ ⎤ ⎢ ⎥ Es ⎥ ← location i ⎢ ⎥ si − s j = ⎢ ⎢ ⎥ ⎢ − Es ⎥ ← location j ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ so si − s j = 2 Es = 2log 2 ( M ) Eb = 2kEb = d min . ∀i. and choice of Walsh-Hadamard functions (note that with Fourier basis. it’s FSK.2-22 • Examples of orthogonal signals are frequency-shift keying (FSK). j . not OFDM). so receiver is o Error probability is same for all signals. adding more bits increases the minimum distance – strong contrast to PAM. o But adding each bit doubles the number of signals M.4. equiprobable signals. Equal energy. If s1 was sent. PSK. c=⎢ ⎥ ⎢ ⎢ ⎥ ⎢ nM ⎥ ⎣ ⎦ .2-23 • Effect of adding bits: o For a fixed energy per bit. QAM. which equals the number of dimensions N – and that doubles the bandwidth! • Error analysis for orthogonal signals. then the correlation vector is ⎡ Es + n1 ⎤ ⎢ ⎥ n2 ⎥ with real components. o The unconditional probability of symbol error (SER) is Pes = 1 − Pcs . The probability of a correct symbol decision.2-24 o Assume s1 was sent.4. is Pcs (c1 ) = P ⎡( n2 < c1 ) ∧ ( n3 < c1 ) ∧ … ∧ ( nM < c1 ) ⎤ ⎣ ⎦ ⎛ ⎛ c1 ⎞ ⎞ = ⎜1 − Q ⎜ ⎟⎟ ⎜ N 2 ⎟⎟ ⎜ 0 ⎝ ⎠⎠ ⎝ M −1 o Then the unconditional probability of correct symbol detection is ⎛ ⎛ c1 ⎞ ⎞ Pcs = ⎜ 1 − Q ⎜ ⎟⎟ ⎜ N 2 ⎟⎟ ⎜ 0 ⎝ ⎠⎠ −∞ ⎝ ∫ ∫ ∞ ∞ ∞ M −1 pc1 ( c1 ) dc1 M −1 ⎛ ⎛ c1 ⎞ ⎞ = ⎜1 − Q ⎜ ⎟⎟ ⎜ N 2 ⎟⎟ ⎜ 0 ⎝ ⎠⎠ −∞ ⎝ = 1 2π ⎛ 1 2 1 c1 − Es exp ⎜ − 2 N0 N0 2 ⎝ ( ) 2 ⎞ ⎟ dc1 ⎠ −∞ ∫ (1 − Q ( u ) ) M −1 1 ⎛ 1 exp ⎜ − u − 2γ s 2π ⎝ 2 ( ) 2 ⎞ ⎟ du ⎠ Needs a numerical evaluation. conditioned on the value of the received c1 . 1 prob of symbol error 0. so ⎝i⎠ Ps k ⎛ k ⎞ Ps k ⎛ k − 1⎞ Peb = k ∑i ⎜ ⎟ = ∑k ⎜ ⎟ 2 − 1 i =1 ⎝ i ⎠ 2k − 1 i =1 ⎝ i − 1 ⎠ 2k −1 k Ps ≈ Ps =k k 2 2 −1 About half the bits are in error in the average symbol error. 1 .01 1 . gamma_b (dB) . M −1 2 −1 ⎛k ⎞ o Can get i bit errors in ⎜ ⎟ equiprobable ways.10 1 .10 3 M=2 M=4 M=8 M = 16 The SER improves as we add bits! 4 1 .10 But the bandwidth doubles with each additional bit.10 5 6 2 0 2 4 6 8 10 12 14 16 SNR per bit. • The SER for orthogonal signals is striking: Symbol Error Prob. o All errors equally likely. Orthogonal Signals 1 0.2-25 • Now for the bit error rate.4. at Pes P = k es . No point in Gray coding. So fall back to a union bound.2-26 • Why does SER drop as M increases? o All points are separated by the same d = 2log 2 ( M ) Eb . assume s1 was transmitted. Since d 2 increases linearly with k = log 2 ( M ) . M as the event that r is closer to s m than to s1 .… .4. o Which effect wins? Can’t get much analytical traction from the integral expression two pages back. and M increases exponentially with k. m = 2. the number of bits per symbol. . the pairwise error probability (between two specific points) decreases exponentially with k. o Define Em . o Offsetting this is the number M-1 of neighbours of any point: M-1 ways of getting it wrong. • Union bound analysis: o Without loss of generality. 2-27 o Note that Em is a pairwise error.…. .4. EM and they are not mutually exclusive. It does not imply that the receiver’s decision is m. so the SER is bounded by the sum of the pairwise event probabilities1 Pes = P [ E ] = P [ E2 ∪ E3 ∪ … ∪ EM ] ≤ ⎛ d 2 ⎞ = ( M − 1)Q ⎜ ⎟ = ( M − 1) Q ⎜ N 2⎟ 0 ⎝ ⎠ 1 ≤ 2 e − k γb 2 k 1 m=2 ∑ P [ Em ] log 2 ( M ) γ b M ( ) 2 <e ⎛γ ⎞ − k ⎜ b − ln(2) ⎟ 2 ⎝ ⎠ A general expansion is M M M P [ A1 ∪ A2 ∪ … ∪ AM ] = ∑ P [ A i ] − ∑∑ P ⎡ Ai ∩ A j ⎤ ⎣ ⎦ i =1 i =i j =1 j ≠i + ∑∑∑ P ⎡ Ai ∩ A j ∩ Ak ⎤ − ⎣ ⎦ i =1 j =1 k =1 j ≠i k ≠ j k ≠i M M M etc. since r could lie closer to two or more points than to s1 : o The overall error event E = E2 ∪ E3 ∪ … ∪ EM . In fact. we have events E2 . ” Thresholds are characteristic of coded systems.10 1 . SNR per bit.10 1 .10 1 . • Remember that this analysis is based on bounds… SER and Bound. bounds are dashed. o Conversely.1 prob of symbol error 0.10 3 4 5 6 True SERs are solid lines. gamma_b (dB) .01 1 . Orthogonal Signals 10 1 0. if γ b < 2ln ( 2 ) . o Too bad the dimensionality. M=2 M=2 M=4 M=4 M=8 M=8 M = 16 M = 16 2 0 2 4 6 8 10 12 14 16 Upper bound is useful to show decreasing SER.2-28 o From the bound. but is too loose for a good approximation.386 . increasing k causes the upper bound on SER to rise exponentially (SER itself saturates at 1).4. then loading more bits onto a symbol causes the upper bound on SER to drop exponentially to zero. if γ b > 2ln ( 2 ) = 1. • This scheme is called “block orthogonal coding. the number of correlators and the bandwidth also increase exponentially with k. • Energy and distance: All signal points are equidistant from si si − s k = 2 Es = d min (the same as orthogonal signals) except one – the reflection through the origin – which is farther away s i − − s i = 2 Es . Or keep the same M and cut the bandwidth in half.4.2-29 Biorthogonal Signals • If you have coherent detection. why waste the other side of the axis? Double the number of signal points (add one bit) with ± Es . Now M = 2 N . with no increase in bandwidth. o So the union bound on SER is Ps ≤ ( M − 2) Q ( γ ) + Q ( 2γ ) ≤ ( M − 2) Q ( γ ) second term redundant (Sec'n 4. A signal is equidistant from all other signals but its own complement.5. since it has half the number of dimensions. for the same M and same γ b .2-30 • Symbol error rate: The probability of error is messy.4) = ( M − 2) Q ( log ( M ) γ ) s s s 2 b which is slightly less than orthogonal signaling. o And the bit error probability is Pb ≈ log 2 ( M ) Ps 2 . but the union bound is easy.4. The major benefit is that biorthogonal needs only half the bandwidth of orthogonal. .2-31 Simplex Signals • The orthogonal signals can be seen as a mean values. M o The mean does not contribute to distinguishability of signals – so why not subtract it? ⎡ −1 M ⎤ ⎢ −1 M ⎥ ⎢ ⎥ ⎢ ⎥ s′ = Es ⎢ ⎥ m 1 − 1 M ⎥ ← location m ⎢ ⎢ ⎥ ⎢ ⎥ ⎣ −1 M ⎦ s′ = s m − s m where Es is the original signal energy.4. shared by all. and signal-dependent increments: 1 The mean signal (the centroid) is s = M m =1 ∑ sm . After rotation into tidier coordinates: • They still have equal energy (though no longer orthogonal). That energy is ′ Es = s′ m 2 1 2 ⎛ = Es ⎜ M 2 + 1 − M ⎝ M 1 ⎞ ⎞ ⎛ ⎟ = E ⎜1 − ⎟ ⎠ ⎝ M⎠ The energy saving is small for larger M. .2-32 o Removing the mean lowers the dimensionality by one.4. • The SER is easy. 4. s′ ) 1 Re [ρmn ] = m n = M = − 1 s′ ⋅ s′ M −1 m n 1− M − for m ≠ n A uniform negative correlation.2-33 • The correlation among signals is: 1 ( s′ . it is BPSK on each frequency at once – a simple OFDM o With the time translate basis. • Vertices of a hypercube is straightforward: two-level PAM (binary antipodal) on each of the N dimensions: o With the Fourier basis. Translation doesn’t change the error rate.2. we will now allow more than one dimension to be used in a symbol.5 Vertices of a Hypercube and Generalizations • More signals defined on a multidimensional space. it is a classical NRZ transmission: . Unlike orthogonal.4. so the SER is that of orthogonal signals with a M ( M − 1) SNR boost. 2-34 o Signal vectors: ⎡ sm1 ⎤ ⎢s ⎥ sm = ⎢ m2 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ smN ⎦ with smn = ± Es N for m = 1.… M and M = 2 N . it looks like this o Every point has the same distance from the origin.4. o In space. hence the same energy sm 2 = Es = log 2 ( M ) Eb = N Eb . . keeping Eb fixed? • It’s easy to generalize from binary to PAM. PSK. sm − sn = 2 d H Es = 2 d H Eb N o What are the effects on d min and bandwidth if we increase the number of bits. on each of the dimensions: o With the Fourier basis. etc. it is OFDM o With time translates. QAM. it is the usual serial transmission. for d H disagreements (Hamming distance).2-35 o Minimum distance occurs for differences in a single coordinate: ⎡1⎤ ⎢1⎥ Es ⎢ ⎥ ⎢1⎥ N ⎢⎥ ⎢1⎥ ⎢1⎥ ⎣⎦ ⎡1⎤ ⎢1⎥ Es ⎢ ⎥ ⎢ −1⎥ N ⎢ ⎥ ⎢1⎥ ⎢1⎥ ⎣ ⎦ sm = sn = d min = s m − s n = 2 Es = 2 Eb N More generally.4. 4.2-36 • Error analysis: All points have same energy Es = N Eb Euclidean distance for Hamming distance h is d = 2 h Eb o If labeling is done independently on different dimensions (see sketch). just ) binary antipodal . then the independence of the noise causes it to decompose to N independent detectors. So the following structure Ps ≤ N Q ( 2γ b = log 2 ( M )Q ( ) 2γ b ) Pb depends on labels becomes Ps is meaningless Pb = Q ( 2γ b . like QAM. This is a binary block code. o Advantage: greater Euclidean distance o Disadvantage: lower data rate .4. with points selected for greater minimum distance. These signals form a finite lattice. Again.2-37 • Generalization: user multilevel signals (PAM) in each dimension. if labeling is independent by dimension. Not too exciting. then it’s just independent and parallel use. • Generalization: use a subset of the cube vertices. Yet.
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