old section 2_5h REVD

Save old section 2.5 (h) PhL 3.26.05 I did this before I learned there was a much simpler expression for the MLS spectrum. So basically I wasted a lot of Maple time here, but fine. No need to review, just keep for a while. (h) Plots of MLS spectra for symbos in {0,1} For a square pulse shape of unit amplitude and width T1 we have (see FT Sec 9 with τ = T1 and A = 1) xpulse(t) = [ θ(t + T1/2) - θ(t - T1/2) ] // time domain FT (9.1) Xpulse(ω) = T1 sinc(ωT1/2) . // Fourier Integral Transform FT (9.2) Ppulse(ω) ≡ = (1/ω1) sinc2(ωT1/2) // ω1 ≡ 2π/T1 FT (34.4) Then (2.5.25) becomes P(ω) = sinc2(ωT1/2) !Syntax Error, Iδ(ω - ω1m/P) (1/4)(1 + 1/P) [1 + 2π δ6(ωT1, ) ] = (1/4)(1 + 1/P) !Syntax Error, Iδ(ω - m[ω1/P]) [sinc2(ωT1/2) + sinc2(ωT1/2) 2π δ6(ωT1, ) ] where 2π δ6(ωT1, ) =   The function 2π δ6 has zeros when PωT1/2 = Nπ, or when ω = N(ω1/P), for N = ±1, ±2, ... . . Thus, the second term above vanishes except for the m = 0 term in the sum over m, where 2π δ6 = P. Thus we can simplify to get P(ω) = (1/4)(1 + 1/P) { !Syntax Error, Iδ(ω - m[ω1/P]) sinc2(ωT1/2) + 1 } Let K ≡ (1/4)(1 + 1/P) and rewrite as (we do this so plots have compatible scales) P(ω) = !Syntax Error, Iδ(ω - ω1m/P) (1/K){Ksinc2(ωT1/2)} {K [1 + 2π δ6(ωT1, ) ]}. = !Syntax Error, Iδ(ω - ω1m/P) (1/K){f1(ω)} {f2(ω) ]} = !Syntax Error, Iδ(ω - ω1m/P)f3(ω) What does this line spectrum look like? The half width of the central peak of δ6(k,N) is Δk = π/N which in our case says T1Δω = 2π/(P-1) or Δω = ω1/(P-1). The half width of the sinc function is (Δω)' = 2π/T1 = ω1. Finally, the spacing between the lines of integer m is (Δω)" = ω1/P. Thus, Δω = ω1/(P-1) half width of central peak of δ6 (Δω)' = ω1 half width of sinc function (Δω)" = ω1/P spacing between lines for integer m values What this says is that there will be a strong central line for m = 0, but all other lines including m = 1 fall outside the central peak of δ6 and thus only the "1" term in [1 + 2π δ6] contributes to the spectrum. Eventually the sinc2 function kills off all the lines. The following Maple code is useful to display the spectrum. We first tell; Maple about the δ6 function Next we enter the functions which appear in P(ω) above, For plotting purposes, we set T1 = 2π to get ω1 = 1, and we enter a value for P, We next plot the three functions fi(ω) and a set of vertical lines at ωm = m(ω1/P) which are the delta function line frequencies, Here are the resulting plots The red curve is f3(ω) and this provides the "envelope" function for P(ω). The lines are at ωm = m(ω1/P) which in this case is ωm = m/7. The delta function amplitudes in P(ω) are the points of intersection of the vertical lines with the red curve. The line amplitudes for m = 0 to 10 are given by For larger values of P, the predominance of the central peak become stronger. For P = 31, Now the sinc curve is nearly flat in the region of interest. For P = 511 the spectrum settles in to the expected "white sequence" pattern, It is useful now to compare the above spectra to the spectrum of a "white sequence". This is shown in FT to be <P(ω)> = Ppulse(ω) { (β-α) + α!Syntax Error, I2π δ(ωT1- 2πm) } FT (35.11) where, from (2.5.34) for Case 1, we have α = 1/4, β = 1/2, and (β-α) = 1/4 so the above becomes <P(ω)> = Ppulse(ω) (1/4){1 + !Syntax Error, I2π δ(ωT1- 2πm) } Installing our same square wave pulse with Ppulse(ω) = (1/ω1) sinc2(ωT1/2) we find <P(ω)> = (1/ω1) sinc2(ωT1/2) (1/4) [ 1 + ω1!Syntax Error, Iδ(ω- mω1) ] = (1/ω1) (1/4) [sinc2(ωT1/2) + ω1!Syntax Error, Iδ(ω- mω1) sinc2(mπ) ] = (1/ω1) (1/4) [sinc2(ωT1/2) + ω1 δ(ω) ] = (1/4) [(1/ω1) sinc2(ωT1/2) + δ(ω) ] We now have an m = 0 line with amplitude (1/4) = .2500000 and the rest is a continuous spectrum which we now plot assuming, as in earlier plots, T1 = 2π and ω1 = 1.