power formulas for finite pulse trains REVD

Power Formulas for Finite Pulse Trains PhL 8.2.13 Here is where I realized that WK also works for finite pulse trains. I added all this into FT, but it is split into two places inside FT. Question #1: What does WK say for a finite pulse train? In FT I keep assuming infinite pulse trains at critical points, but I would like the finite N formulas as well. Let's start on the R"(z) side of the docket. I define rs ≡ limN→∞ [!Syntax Error, I yn yn+s ] ≡ <yn yn+s> 1 . (32.16) But for finite N and pulse train of length 2N+1, which has only y-N .... y0....yN we get rs = !Syntax Error, I yn yn+s where terms vanish that maybe I can make explicit this way rs = !Syntax Error, I yn yn+s Θ(-N ≤ n+s ≤ N) = !Syntax Error, I yn yn+s Θ(-N-n ≤ s ≤ N-n) For n = N we find Θ(-N-N ≤ s ≤ N-N) = Θ(-2N ≤ s ≤ 0) so rs vanishes when s < -2N. and similarly on the high side, we I think we can say rs is non-vanishing only in the range -2N to +2N. Now we go on to the next step: compute R"(z). We can still use ∞ endpoints with the understanding that rs vanishes out of range either way. And the same for yn etc. So copy and paste to get R"(z) ≡ !Syntax Error, I rs z-s = !Syntax Error, I{ !Syntax Error, I yn yn+s } z-s = !Syntax Error, I !Syntax Error, I [yn zn ] [yn+s z-(n+s)] = !Syntax Error, I !Syntax Error, I [yn zn ] [ym z-m] m ≡ n+s = [ !Syntax Error, I yn zn] [!Syntax Error, Iym z-m ] = Y"(z)* Y"(z) So the result continues to be valid for our finite pulse train where T = (2N+1)T1 and Y"(z) = !Syntax Error, I yn z-n Conclusion: for finite pulse train (-N,N), we can still claim WK which says R"(z) = !Syntax Error, I rs z-s = (2N+1)-1 Y"(z)* Y"(z) Y"(z) = !Syntax Error, I yn z-n Question #2: Is the power formula valid for a finite pulse train? Look at summary box (34.4). First we have P(ω) ≡ = which is certainly valid for any pulse train, finite or not. Then we know that X(ω) = (1/T1)Xpulse(ω) Y'ω) = Xpulse(ω) Y"(z) so everything "goes through" for a finite pulse train, in particular |X(ω)|2 = |Xpulse(ω)|2 | Y"(z) |2 and P(ω) = Ppulse(ω) | Y"(z) |2 where now the ratio is (2N+1)-1. Then we apply the above WK to get P(ω) = Ppulse(ω) R"(z) Conclusion: This power formula IS VALID for a finite pulse train.