d5 and d6 mystery REVD

d5 and d6 mystery PhL 8.4.13 This matter is resolved. See red notes below The mystery is that for a large N finite pulse train, I seem to get different spectra depending on which of the two methods I use. The ensemble method gives δ6 as I show in both old and new Section 35. But what happens in the autocorrelation method? I get this paradox: R"(z) must be positive by its definition, but my calculation gives R"(z) of both signs. First, we know this, P(ω) = Ppulse(ω) | Y"(z) |2 = Ppulse(ω) !Syntax Error, I !Syntax Error, I ym* yn eiω(m-n)T (34.14) R"(z) = | Y"(z) |2 = !Syntax Error, I rs z-s which seems to say R"(z) must be positive definite. But what is rs ? s=0 r0 = <yn2>1 = μ2 + σ2 s ≠0 rs = <ymym+s>1 = μ2 = !Syntax Error, I yn yn+s no limit (32.16) (35.15) This last expression is the problem! The expression is "correct" as stated if you understand that the quantity yn yn+s has implicit θ function restrictions of this form (taken from Section 35 start) ymym+s = ymym+s θ(m ≤ N)θ(m ≥-N) θ(m+s ≤ N)θ(m+s ≥ -N) = ymym+s θ(m ≤ N)θ(m ≥-N) θ(s ≤ N-m)θ(s ≥ -N-m) (35.1a) This is that all-important gray parallelogram support region idea. These θ's restrict the n sum so in reality it is this rs = !Syntax Error, I ymym+s so there is s-dependence of rs now from the sum limits! When you then go to the ensemble level, this is <rs> = f(s) where f(s) = ymym+s stationary Then when independence is added f(s) = . // stationarity and independence assumed (35.15) you get this <rs> = μ2 s ≠ 0 <r0> = 1 (σ2+μ2) s ≠ 0 The first result is very different from saying rs = μ2! You then end up with the δ6 result, not the δ5 result. This was a fairly subtle issue and it had me confused for several days. I think the extent of rs is now limited to -2N to +2N so if would be zero outside that range. Then is the same, but it cuts off at ±2N. Then we should have R"(z) = !Syntax Error, I rs z-s = !Syntax Error, I{ μ2 } z-s + { σ2+μ2 } = μ2 !Syntax Error, I z-s + σ2 I then call upon !Syntax Error, I eink = 2π { } ≡ 2π δ5(k,N) . -∞ < k < ∞ (13.3) which is going to give a δ5 result, not a δ6 result! Then z = eiωT from (24.1) we have z-s = e-iωTs, so setting k = ωT1 we get !Syntax Error, I z-s = 2π δ5(ωT1,2N) and so R"(z) = σ2 + μ2 2π δ5(ωT1,2N) Now δ5 looks like this Fig A.6 and similarly from a new plot I made Then if σ2 is very small and μ2 large, which must be possible, we have R"(z) = σ2 + μ2 2π δ5(ωT1,2N) and this goes negative about half the time! But I said R"(z) had to be positive all the time! I think something is wrong about rs. Go back to R"(z) = !Syntax Error, I rs z-s If you know that rs is symmetric in s, easy to show that R"(z) = !Syntax Error, Irs cos(ωT1s) but this can have either sign! So just being symmetric in s is not enough to make R"(z) be positive! It has to be some other detail of rs. rs =!Syntax Error, I ymym+s = < ymym+s >1 Isn't there some tapering off at the ends? But not really. The above formula is exact, what is not exact is this assumption < ymym+s >1 = < ym >12 = μ2 Except for extremely large N, there is no justification to write this! But for extremely large N, we will find that δ5 and δ6 give the same result and the negativity of δ5 becomes less and less So I guess that is the resolution of this mystery. (1) for finite N, we cannot say < ymym+s >1 = < ym >12 even though we can say < ymym+s > = < ym >2 for independent Comment: If we blindly repeat the finite N calculation using the autocorrelation method with the same rs shown in *** , we obtain R"(z) = !Syntax Error, I rs z-s = !Syntax Error, I{ μ2 } z-s + { σ2+μ2 } = μ2 !Syntax Error, I z-s + σ2 = σ2 + μ2 2π δ5(ωT1,2N) which does give the correct result as N→∞. But for finite N, as Fig A.6 shows, if σ2 is small and μ2 large, then this function R"(z) goes negative in places near the peaks, which conflicts with R"(z) = | Y"(z) |2 having to be positive definite ( δ6 never goes negative). This paradox is explained by realizing that for finite N, if the ensemble random variables are independent, we might have for s≠ 0 that < ymym+s > = < ym >2, but the identifications of (35.9) are only approximate for finite N so < ymym+s >1 = < ym >12 is only approximately true, allowing the error just noted. OK , this matter is now resolved and I entered a the above as Comment 2 at the end of the Section 35 section on the finite calculation.