6.441-23 Channel side information, wide-band channels

LECTURE 23 Last time: Finite-state channels • Lower capacity and upper capacities • Indecomposable channels • Markov channels • Lecture outline Spreading over fading channels • Channel model • Upper bound to capacity • Interpretation • Spreading over fading channels Channel decorrelates in time T Channel decorrelates in frequency W Recall Markov channels: difficulty arises when we do not know the channel Gilbert-Eliot channel: hypothesis testing for what channel state is Consider several channels in parallel in frequeenc (recall that if channels are known, we can water-fill) Channel model Block fading in bandwidth and in time Over each coherence bandwidth of size W ,the channel experiences Rayleigh flat fadingAll the channels over distinct coherence bandwiddth are independent, yielding a block-fading model in frequency We transmit over µ coherence bandwidths The energy of the propagation coefficient F [i]j over coherence bandwidth i at samplle time j is σF For input X[i]j at sample time j (we sample at the Nyquist rate W ), the corresponding output is Y [i]j = F [i]jX[i]j + N[i]j, where the N[i]js are samples of WGN bandlimited to a bandwidth of W , with energy normalizze to 1 Channel model The time variations are block-fading in natuur The propagation coefficient of the channel remains constant for T symbols (the coherrenc interval), then changes to a value independent of previous values Thus, F [i](j+1)TW is a constant vector and jT W +1 the F [i](j+1)TW are mutually independent jT W +1 for j =1, 2,.... Signal constraints: For the signals over each coherence band₭ width, the second moment is upper bounded by E[X2] ≤Eµ The amplitude is upper bounded by γ �E[X2]• E Upper bound to capacity Capacity is C �W, E,σ2 , T, µ, γ� F C �W, E, σF 2 , T, µ, γ� = lim max µ⎛1 k1 k→∞ pX �⎞�� I �X[j](i+1)TW ; Y [j](i+1)TW ⎝k T iTW +1 iT W +1 ⎠ i=1 j=1 (1) where the fourth central moment of X[j]i is upper bounded by γ 2 and its average energy µconstraint is Eµ . Since we have no sender channel side informmatio and all the bandwidth slices are independent, we may use the fact that mutuua information is concave in the input distribbutio to determine that selecting all the inputs to be IID maximizes the RHS of (1). Upper bound to capacityWe first rewrite the mutual information term:1 �X[j](i+1)TW ; Y [j](i+1)TW � TI iTW +1 iT W +1 1 �Y [j](i+1)TW � = h T iTW +1 1 �Y [j](i+1)TW X[j](i+1)TW �− Th iTW +1 |iT W +1 (2)We may upper bound the first term of (2):1 �Y [j](i+1)TW �Th iTW +11 �� � ln (2πe)TW Λ Y [j](i+1)TW2T≤ iT W +1 Gaussian distribution bound⎛ TW≤ 21 T ln ⎝(2πe)TW ��σF 2 σ2 ⎠X[i]+1�⎞ i=1 from Hadamard’s inequality TW= W ln (2πe)+ 1 � Fσ2ln �σ2 X[i]+1� 22Ti=1 WW � σ2 � ≤ 2 ln (2πe)+ 2 ln Fµ E + 1 (3) from concavity of ln and our average energy constraint. Upper bound to capacity We now proceed to minimize the second term of (1) X[j](i+1)TW Y [j](i+1)TWConditioned on iT W +1 , iT W +1 is Gaussian, since F [j](i+1)TW is GaussianiT W +1 and N(i+1)TW Gaussian and independentiT W +1 of F T 1 h �Y [j](i+1)TW |X[j](i+1)TW � iT W +1 iT W +1 1 ��� ��= EX ln (2πe)T Λ Y [j](i+1)TW2T iTW +1(4) Λ(i+1)TW has kth diagonal term σF 2 x[k]2+ Y [j]iT W +1 1 and off-diagonal (k, j) term equal to x(k)x(j)σF 2, conditioned on X[j](i+1)TW = x =[x(1),...,x(TW )]iT W +1 The eigenvalues λj of ΛY are 1 for j = 1 ...TW − 1 and ||x|| 2 σ2 +1 for j = TW .F Upper bound to capacityHence, we may rewrite (3) as 1 �(i+1)TW (i+1)TW � Th YiTW +1 |XiT W +1 1 � �� (i+1)TW 2 �� = EX ln xiT W +1 σF 2 +1 2TW+ ln(2πe) (5)2 We seek to minimize the RHS of (5) subjeec to the second moment constraint holdiin with equality and the subject to the peak amplitude constraint The distribution for X which minimizes the RHS of (5) subject to our constraints can be found using the concavity of the ln functiio The distribution is such that the only valuue which Xcan take are 0 and γ with|| 22 √µE probabilities 1 −Eand Eγ2, respectively. γ2 Upper bound to capacityThus, we may lower bound (5) by 1 �(i+1)TW (i+1)TW � Th Y |XiT W +1 iT W +1 2Eln � TW γ2 σF 2 +1 � + W ln(2πe)≥ 2Tγ2 µE 2 (6) Combining (6), (3) and (1) yields C �W, E, σF 2 , T, µ, γ� µW ln � σF 2 E +1 � ≤ 2 µ µE2 ln � TW γ2 σF 2 +1 � (7)−2Tγ2 µE Upper bound to capacityWhat is the limit for µ infinite? ln(1 + x) � x for small x First term goes to σ2 F W E2 Second term also goes to σ2 F W E2 Graphical interpretation InterpretationOver any channel (slice of of bandwidth of size W ), we do not have enough energy to measure the channel satisfactorily Necessary assumptions: energy scales per bandwidth slice over • the whole bandwidth peak energy per bandwidth slice over • the whole bandwidth InterpretationWe may relax the assumption of the peak bandwidth Assume second moment (variance) scales as µ 1 and fourth moment (kurtosis) scales 1as 2µThe mutual information goes to 0 as µ → ∞ We may also relax the assumption regardiin the channel block-fading in time andfrequency as long as we have decorrelationMIT OpenCourseWarehttp://ocw.mit.edu 6.441 Information Theory Spring 2010 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.