The Poisson Channel at Low Input Powers

THE POISSON CHANNEL A T LO W INPUT PO WERS Amos Lapidoth, Ligong W ang ETH Zurich Zurich, Switzerland { lapidoth, wang } @isi. ee.ethz.ch J effr e y H. Shapir o MIT Cambridge, MA, USA jhs@mit.ed u V in odh V enkatesan IBM Research Zurich, Switzerland ven@zurich .ibm.com ABSTRA CT The asymptotic capacity at low input powers of an average- power limited or an a verage- and peak-p ower limited discrete- time Poisson channel is considered . For a Poisson chan- nel whose dark cu rrent is zero o r decays to zero line arly with its average input p ower E , capacity scales like E log 1 E for small E . For a Poisson channel whose dark curren t is a n onzero constan t, capacity scales, to within a constant, like E log log 1 E for small E . Index T erms — Asymptotic, Capac ity , Low SNR, Pois- son channel. 1. INTR ODUCTION W e consider the d iscrete-time memo ryless Poisson ch annel whose input x is in the set R + 0 of no nnegative reals and whose output y is in the set Z + 0 of nonnegative integers. Condition al on the in put X = x , the output Y has a Poisson distribution of mean ( λ + x ) where λ ≥ 0 is called the d ark cu rr en t . W e denote the Poisson distribution of mean ξ by P ξ ( · ) so P ξ ( y ) = e − ξ ξ y y ! , y ∈ Z + 0 . W ith this n otation the channel law W ( ·|· ) is given b y W ( y | x ) = P λ + x ( y ) , x ∈ R + 0 , y ∈ Z + 0 . (1) This chann el is often used to model pulse-amplitud e m od- ulated optical commu nication with a direct-detection rece i ver [1]. Here the input x is p ropor tional to the p rodu ct of the transmitted ligh t intensity by the p ulse du ration; the output y models the number of p hotons ar riving at the recei ver during the pulse dur ation; and λ mo dels the average n umber of extra- neous counts that appear in y in additio n to those associated with the illumination x . The average-power constraint 1 on the input is E [ X ] ≤ E , (2) 1 The wo rd “po wer” here has the meaning “a vera ge number of pho tons transmitt ed per channel use. ” If we denote by P the standard “po wer” in physics, namely , ener gy per unit ti me (in w att s), then the nota tion of “ po wer” in this pa per i s rea lly ηP T / ~ ω , where η is the detecto r’ s quantum effic ienc y , T is t he pul se duration (i n sec), and ~ ω is the phot on ene rgy (in joules) at the operati ng frequency ω (in rad/ sec). where E > 0 is the maximum allo w ed a verage power . The peak-p ower constr aint on the input is that with prob- ability one X ≤ A . (3) When no pe ak-power con straint is impo sed, we write A = ∞ . No analytic expression f or the capacity of the Poisson channel is known. In [1] Shamai showed that capacity- achieving inp ut distributions are d iscrete who se num bers o f mass points dep end on E a nd A . In [2, 3] Lapidoth and Moser d erived the asy mptotic cap acity o f the Po isson chan- nel in the regime where both the average and peak powers tend to infinity with their ratio fixed. In th e pr esent paper, we seek the asymptotic ca pacity of the Poisson chan nel when the average in put p ower tend s to zero. The peak- power co nstraint, when considered, is h eld constant a nd hence d oes not ten d to zero with th e average power . W e consider tw o dif ferent cases for the d ark c ur- rent λ . The fir st case is wh en the da rk curr ent tends to zero propo rtionally with the average power . This corresp onds to the wide-band regime where the pulse du ration tends to zero. The seco nd case is whe n the dark cur rent is constant. This correspo nds to the regime where the transmitter is weak. Our lower bounds on channel capacity ar e all based on binary inputs. In some cases we show that this is asymptoti- cally o ptimal. Our upper b ounds are derived using the duality expression (see [4] and ref erences ther ein). An ef ficient w ay to co mpute asympto tic capacities a t low average input pow- ers is to comp ute the capacity per un it cost [ 5]. Howe ver, we shall see that, apart from one case (Equation (7)), the capacity per unit cost d oes n ot exist, namely , the capacity tend s to zero more slo wly than linearly with the av erage power . Among the re sults in this paper, the special case of zero dark current has been derived indepen dently in [6, 7 ]. The rest of the paper is arranged as follows: in Section 2 we state the re sults o f th is p aper; in Section 3 we p rove the lower bounds; and in Sec tion 4 we sketch the pr oofs for th e upper boun ds. 2. RESUL TS Let C ( λ, E , A ) denote the cap acity of the Poisson c hannel with dark current λ under Constraints (2) and (3) C ( λ, E , A ) = sup I ( X ; Y ) where th e supremu m is over all input distributions satisfy- ing (2) and (3). When λ is pro portion al to E , the a symptotic cap acity o f the Poisson ch annel as E ↓ 0 is given in the following propo- sition. Note that th is also include s the case wh ere th e dark current is the constant zero. Proposition 1 (Dark C urrent Proportional to E ) . F or any c ≥ 0 and A ∈ (0 , ∞ ] , lim E ↓ 0 C ( c E , E , A ) E log 1 E = 1 . Recall th at, for any α, β > 0 , the sum o f two indepen- dent rand om variables with th e Po isson distributions P α ( · ) and P β ( · ) has the Po isson distribution P α + β ( · ) . Thus, we can p roduc e any Poisson channel with nonzer o dark curren t by adding no ise to a Poisson c hannel with zero dark current. Consequently , C (0 , E , A ) ≥ C ( c E , E , A ) , c, E , A > 0 . Thus, to prove Pr oposition 1, we only need to show the fol- lowing two boun ds: lim E ↓ 0 C ( c E , E , A ) E log 1 E ≥ 1 , c > 0 , A ∈ (0 , ∞ ] , (4) lim E ↓ 0 C (0 , E , A ) E log 1 E ≤ 1 , A ∈ (0 , ∞ ] . (5) W e shall prove (4) in Section 3.1 an d shall sketch a proof for (5) in Section 4.1. Remark 1. The bound (5) can also be derived b y noting that the cap acity of the P oisson channel with zer o dark c urr ent under an average-power co nstraint only is upper -bound ed by the capa city of the pur e-loss bosonic channel, and by using the explicit formu la [8] C bosonic ( E ) = (1 + E ) log (1 + E ) − E log E (6) of the latter . Remark 2. Because the pur e-loss boson ic chann el with co - her ent inp ut states a nd d ir ec t detection reduces to a P oisson channel, the lower b ound (4) and the achievability of its left- hand side u sing binary signaling combine with (6) to show that the a symptotic (quan tum-r eceiver) capa city of the pu r e- loss bosonic channel is achievable with bin ary modulation (on-o ff ke ying) and dir ect detection . For a Poisson channel with constant n onzero dar k current, we hav e the following result. Proposition 2 (Constant Nonzero D ark Current) . F or any λ > 0 , lim E ↓ 0 C ( λ, E , A ) E =  1 + λ A  log  1 + A λ  − 1 , A < ∞ , (7) and 1 2 ≤ lim E ↓ 0 C ( λ, E , ∞ ) E lo g log 1 E ≤ lim E ↓ 0 C ( λ, E , ∞ ) E lo g log 1 E ≤ 2 . (8) The proof of (7) is a simple application of the formu la for capacity per unit cost [5 , Theor em 2]. The pr oof of the lower bound in (8) is in Section 3.2 ; and a sketch of the proof of the upper boun d in (8) is in Section 4. 2. 3. THE LO WER BOUNDS The achiev ab ility r esults in th is section are obtained by choos- ing bina ry inpu t distributions and th en comp uting the mutua l informa tions. W e denote by Q b the binar y d istribution X = ( 0 , w .p. (1 − p ) , ζ , w .p. p, (9) where ζ > 0 , p ∈ (0 , 1) . I f we choose the parameters ζ and p in such a way that Constraints (2) and (3) are satisfied, then C ( λ, E , A ) ≥ I ( Q b , W ) . (10) 3.1. Dark Current Pr oportional to E In this sub section we shall der i ve Inequality (4). T o th is end, we co mpute th e mutual in formatio n I ( Q b , W ) for inp ut d is- tribution Q b giv en by (9): I ( Q b , W ) = H ( Y ) − H ( Y | X ) = − ∞ X y =0  (1 − p ) P λ ( y ) + p P λ + ζ ( y )  · log  (1 − p ) P λ ( y ) + p P λ + ζ ( y )  + (1 − p ) ∞ X y =0 P λ ( y ) log P λ ( y ) + p ∞ X y =0 P λ + ζ ( y ) log P λ + ζ ( y ) = I 0 ( λ, ζ , p ) + I 1 ( λ, ζ , p ) , (11) where in the last equality we defined I 0 ( λ, ζ , p ) , −  (1 − p ) e − λ + pe − ( λ + ζ )  · log  (1 − p ) e − λ + pe − ( λ + ζ )  − (1 − p ) λe − λ − p ( λ + ζ ) e − ( λ + ζ ) , I 1 ( λ, ζ , p ) , − ∞ X y =1  (1 − p ) P λ ( y ) + p P λ + ζ ( y )  · log  (1 − p ) P λ ( y ) + p P λ + ζ ( y )  + (1 − p ) ∞ X y =1 P λ ( y ) log P λ ( y ) + p ∞ X y =1 P λ + ζ ( y ) log P λ + ζ ( y ) . Note that in the ab ove deco mposition we to ok out the terms correspo nding to y = 0 in all three summations to form I 0 ( λ, ζ , p ) and collected the remainin g terms in I 1 ( λ, ζ , p ) . W e lower-bound I 0 ( λ, ζ , p ) as I 0 ( λ, ζ , p ) ≥ 0 − (1 − p ) λe − λ − p ( λ + ζ ) e − ( λ + ζ ) ≥ − λ − p ( λ + ζ ) . (12) W e lower-bound I 1 ( λ, ζ , p ) as I 1 ( λ, ζ , p ) = − ∞ X y =1  (1 − p ) P λ ( y ) + p P λ + ζ ( y )  · log  (1 − p ) P λ ( y ) P λ + ζ ( y ) + p  + (1 − p ) ∞ X y =1 P λ ( y ) log P λ ( y ) P λ + ζ ( y ) = − ∞ X y =1  (1 − p ) P λ ( y ) + p P λ + ζ ( y )  ·  log p + log  1 + 1 − p p P λ ( y ) P λ + ζ ( y )  + (1 − p ) ∞ X y =1 P λ ( y ) log e − λ λ y y ! e − ( λ + ζ ) ( λ + ζ ) y y ! = − ∞ X y =1  (1 − p ) P λ ( y ) + p P λ + ζ ( y )  · log p + log  1 + 1 − p p P λ ( y ) P λ + ζ ( y )  | {z } ≤ 1 − p p P λ ( y ) P λ + ζ ( y ) ! + (1 − p ) ζ ∞ X y =1 P λ ( y ) | {z } =1 − e − λ +(1 − p ) log λ λ + ζ ∞ X y =1 P λ ( y ) y | {z } = λ ≥ − ∞ X y =1  (1 − p ) P λ ( y ) + p P λ + ζ ( y )  | {z } =(1 − p )(1 − e − λ )+ p (1 − e − ( λ + ζ ) ) log p − ∞ X y =1  (1 − p ) P λ ( y ) + p P λ + ζ ( y )  1 − p p P λ ( y ) P λ + ζ ( y ) + (1 − p )(1 − e − λ ) ζ − (1 − p ) λ lo g  1 + ζ λ  =  (1 − p )(1 − e − λ ) + p (1 − e − ( λ + ζ ) )  log 1 p − (1 − p ) 2 p ∞ X y =1 ( P λ ( y )) 2 P λ + ζ ( y ) | {z } = e ζ 2 λ + ζ P λ 2 λ + ζ ( y ) − (1 − p ) ∞ X y =1 P λ ( y ) | {z } =1 − e − λ + (1 − p )(1 − e − λ ) ζ − (1 − p ) λ lo g  1 + ζ λ  =  (1 − p )(1 − e − λ ) + p (1 − e − ( λ + ζ ) | {z } ≤ e − ζ )  log 1 p − (1 − p ) 2 p | {z } ≤ 1 p e ζ 2 λ + ζ | {z } ≤ e ζ  1 − e − λ 2 λ + ζ  | {z } ≤ λ 2 λ + ζ ≤ λ 2 ζ − (1 − p ) | {z } ≤ 1 (1 − e − λ ) | {z } ≤ λ − (1 − p ) | {z } ≤ 1 (1 − e − λ ) | {z } ≤ λ ζ − (1 − p ) λ log  1 + ζ λ  ≥ (1 − p )(1 − e − λ ) log 1 p + p (1 − e − ζ ) log 1 p − 1 p λ 2 ζ e ζ − λ − λζ − (1 − p ) λ log  1 + ζ λ  . (13) Choose any ζ ∈ (0 , A ] and , for sm all enoug h E , let p = E /ζ . Then the d istribution (9) satisfies b oth Constrain ts (2) and (3). Let λ = c E . Using (1 2) we ca n b ound the as ymptotic behavior o f I 0 ( λ, ζ , p ) as lim E ↓ 0 I 0  c E , ζ , E ζ  E log 1 E ≥ − lim E ↓ 0 c E E log 1 E − lim E ↓ 0 E ζ ( c E + ζ ) E log 1 E = 0 . (14) Similarly , u sing (1 3) w e can boun d the asym ptotic behavior of I 1 ( λ, ζ , p ) as lim E ↓ 0 I 1  c E , ζ , E ζ  E log 1 E ≥ 1 − e − ζ ζ . (15) Combining (10), (11), (14), and (15) we obtain lim E ↓ 0 C ( c E , E , A ) E log 1 E ≥ 1 − e − ζ ζ , for all ζ ∈ (0 , A ] . (16) W e can make the righ t-hand side (RHS) o f (16) arbitrar ily close to 1 by cho osing ar bitrarily small po siti ve values for ζ . Thus we obtain (4). 3.2. Constant Nonzero Da rk Current In this sub section we shall prove the first i nequality in (8 ). T o this en d, we lower-bound on the mutual infor mation I ( Q b , W ) for the input distribution (9) as fo llows: I ( Q b , W ) = H ( Y ) − H ( Y | X ) = − ∞ X y =0 ((1 − p ) P λ ( y ) + p P λ + ζ ( y )) · log ((1 − p ) P λ ( y ) + p P λ + ζ ( y )) + (1 − p ) ∞ X y =0 P λ ( y ) log P λ ( y ) + p ∞ X y =0 P λ + ζ ( y ) log P λ + ζ ( y ) = − p ∞ X y =0 P λ + ζ ( y ) log  (1 − p ) P λ ( y ) P λ + ζ ( y ) + p  − (1 − p ) ∞ X y =0 P λ ( y ) log  (1 − p ) + p P λ + ζ ( y ) P λ ( y )  = − p ∞ X y =0 P λ + ζ ( y ) · log P λ ( y ) P λ + ζ ( y ) + lo g  (1 − p ) + p P λ + ζ ( y ) P λ ( y )  ! − (1 − p ) ∞ X y =0 P λ ( y ) log  (1 − p ) + p P λ + ζ ( y ) P λ ( y )  = p ∞ X y =0 P λ + ζ ( y ) log P λ + ζ ( y ) P λ ( y ) − ∞ X y =0 (1 − p ) P λ ( y ) | {z } ≥ 0 + p P λ + ζ ( y ) | {z } ≥ 0 ! log  (1 − p ) + p P λ + ζ ( y ) P λ ( y )  | {z } ≤ log “ 1+ p P λ + ζ ( y ) P λ ( y ) ” ≤ p P λ + ζ ( y ) P λ ( y ) ≥ p ∞ X y =0 P λ + ζ ( y ) log P λ + ζ ( y ) P λ ( y ) − ∞ X y =0  (1 − p ) P λ ( y ) + p P λ + ζ ( y )  p P λ + ζ ( y ) P λ ( y ) = p ∞ X y =0 P λ + ζ ( y ) log e − ( ζ + λ ) ( ζ + λ ) y y ! e − λ λ y y ! ! − (1 − p ) p ∞ X y =0 P λ ( y ) P λ + ζ ( y ) P λ ( y ) − p 2 ∞ X y =0 P λ + ζ ( y ) P λ + ζ ( y ) P λ ( y ) = p ∞ X y =0 P λ + ζ ( y ) log  e − ζ  1 + ζ λ  y  − (1 − p ) p ∞ X y =0 P λ + ζ ( y ) | {z } =1 − p 2 ∞ X y =0  e − ( ζ + λ ) ( ζ + λ ) y y !  2 e − λ λ y y ! = p ∞ X y =0 P λ + ζ ( y )  − ζ + y log  1 + ζ λ  − (1 − p ) p − p 2   ∞ X y =0 e − ( λ +2 ζ )  λ + 2 ζ + ζ 2 λ  y y ! e − ζ 2 λ   | {z } = P ∞ y =0 P λ +2 ζ + ζ 2 λ ( y )=1 e ζ 2 λ = − pζ ∞ X y =0 P λ + ζ ( y ) | {z } =1 + p ∞ X y =0 P λ + ζ ( y ) y | {z } =( ζ + λ ) log  1 + ζ λ  − p + p 2 − p 2 e ζ 2 λ = p ( ζ + λ ) log  1 + ζ λ  − pζ − p − p 2  e ζ 2 λ − 1  . (17) For small eno ugh E , we ch oose ζ = q λ log 1 E and p = E ζ = E √ λ log 1 E . By using (10) and (17) an d letting E ten d to zero we establish the lower bound in (8). 4. THE UPPER BOUNDS In this section we shall sketch the proofs of the upper bounds on the asymp totic cap acities of the Poisson channel. W e shall use the du ality b ound [4] wh ich states that, for any distribu- tion R ( · ) on the ou tput, the channel capacity satisfi es C ≤ sup E  D  W ( ·| X ) k R ( · )  , (18) where the supr emum is taken over all allowed inpu t distribu- tions. W e shall d escribe the choices o f R ( · ) that lead to our upper boun ds, but we shall omit the details. 4.1. Dark Current Pr oportional to E In this sub section we shall sketch the p roof for (5). T o this end, as in [3], we introd uce the P oisson chan nel with con tin- uous output whose inp ut x is the same as the original Poisson channel, and wh ose o utput is ˜ y ∈ R + 0 . The con ditional den- sity ˜ W ( ·|· ) is ˜ W ( ˜ y | x ) = P λ + x ( ⌊ ˜ y ⌋ ) . (19) W e deno te the capacity of ( 19) u nder Constraints ( 2) and (3) by ˜ C ( λ, E , A ) . I t is sho wn in [2] that C ( λ, E , A ) = ˜ C ( λ, E , A ) . Thus, to prove (5), it suffices to prove lim E ↓ 0 ˜ C (0 , E , A ) E log 1 E ≤ 1 , A ∈ (0 , ∞ ] . (20) T o this end, we choose the distribution ˜ R ( · ) on ˜ Y to be o f density f ˜ R ( ˜ y ) =    (1 − p ) , 0 ≤ ˜ y < 1 p · ˜ y ν − 1 e − ˜ y β β ν Γ( ν, 1 β ) , ˜ y ≥ 1 , where β > 0 is arbitrary , ν ∈ (0 , 1 ] and p ∈ (0 , 1 ) will be spe cified later, and Γ( · , · ) deno tes the Inco mplete Gamma Function giv en by Γ( a, ξ ) = Z ∞ ξ t a − 1 e − t t . , ∀ a, ξ ≥ 0 . Applying (1 8) on ˜ C (0 , E , A ) with the a bove choice of f ˜ R ( · ) in the place of R ( · ) and with the choice ν = 1 2 yields that, f or ev ery p ∈ (0 , 1) and β > 0 C (0 , E , A ) ≤ E log 1 p + lo g 1 1 − p + E β + E max ( 0 , 1 2 log β + log Γ( 1 2 , 1 β ) √ π + 1 2 β !) . Choosing p = E 1+ E in the above inequality and letting E tend to zero yield (5). 4.2. Constant Nonzero Da rk Current In this sub section we sha ll sketch the proo f of the upper bound in (8). W e choo se the distrib ution R ( · ) on the output Y to b e R ( y ) = ( e − λ λ y y ! , y ∈ { 0 , 1 , . . . , N − 1 } δ (1 − p ) p y − N , y ∈ { N , N + 1 , . . . } , where N ∈ Z + and p ∈ (0 , 1) are con stants to be spec- ified later , an d δ is a n ormalizin g factor given by δ , P ∞ y = N e − λ λ y y ! . W e next apply (18) to upper-boun d C ( λ, E , A ) . Calculation (with repeated applications o f the Chernoff bound ) yields C ( λ, E , A ) ≤  N lo g N + 1 12 N + 1 2 log(2 π N ) + log 1 1 − p  ·  E N − √ N − λ + ex p ( N + N log λ − N log N )  +  1 + log 1 p + lo g λ  ·  E + λ E N − √ N − λ + λ · e N − 1 − λ +( N − 1) log λ − ( N − 1) log( N − 1)  + E ·  1 + λ N − λ  · max  0 , log 1 λ  + E · N log N λ N − λ . 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