Arbitrary non-Correlated SymbolsNotations: When summation interval is obvious, we'll omit limits at summ sign:_{j}can mean_{j=0,n-1}Definitions: L = Maximum number of chips in symbol. T_{c}= Duration of one chip. C=T_{c}*L. C = Maximum possible duration of one symbol. Signal (message) x=_{j=0,n-1}x_{j}(1) where: x_{j}(t)=S_{aj}(t-c_{j}) (j) - is a signal of symbol a_{j}started (centered) at time c_{j}j - (time) index of symbol in a message, a_{j}is in Alphabet of symbols. Alphabet={0,1,...N-1}. For example N=2^{M}for OOK. M = number of bits comprising each symbol. S_{b}(t) - "waveform,shape" of a symbol b. Note: S_{b}is not neccesary a step function or reachtngular pulse. Rectangular pulse, S_{b}(t) = A_{b}p(t-t_{b}) {P} will be consider later. Signals S_{b}do not necessary have the same time length L. For example, for DAPPM sequences 01 and 002. Hence, we don't assume that c_{j}=T_{c}*L*j yet. In (1), a is a random function a_{j}of index j. In article [1], k denotes index such kT_{c}=t_{a}+jC for equilength symbol messages.Our first goal is to reseach spectrum of power.Autocorrelation function is R(tau)= E[x(0)x(0-tau)]/I. (R) I - average power, I=E[x^{2}(0)]. We are not interested in research of specific sequences of symbols. We are reseaching properties of communication channel. So, for our convenience, we take a sequence of uniformly distributed non-correlated symbols. Intuitively, because of set of symbols is finite, there exists enough big T such the interval [0,T] will be sufficient sample to bear atocorrelation properties for tau << T. For our convenience, we extend function x(t) periodically below 0 and beyond T. x is defined on all |R-axis. We are ready to calculate expectation in following form: R(tau)=1/TI_{0,T}x(t)x(t-tau)dt (R') Now it is time to involve Fourier transformation. Because of x has period T, R_{w}=|x_{w}|^{2}, (R'') where circular frequncy w=2 PI f = 2 PI m/ T, x=_{m=-oo,oo}x_{w}e^{iwt}. (F) Like original signal x, its Fourier transform will be a summ of sinals via nodes j: x_{w}=_{j}x_{w,j}. We need to find a way to exlcude "randomness" brought by random function S_{s}. We will take average: <R_{w}>=<x_{w}x_{w}^{*}> This is an artificial step, because R_{w}is already averaged: R_{w}=<R_{w}> but this step make our derivations easy. We have: R_{w}=(1/I)_{jj'}<x_{wj}x^{*}_{wj'}>Constant length symbolsConsider messages where each symbol takes equal number of chips L. Because symbols follow independently: <x_{wj}x^{*}_{wj'}> = <x_{wj}><x^{*}_{wj'}> for j=/=j', (N) <x_{wj}x^{*}_{wj}>=D for j=j'. (D) Expanding (j), we reduce average for S to average W via alphabet: <x_{wj}> = <S_{wj}> = e^{-iwjC}W. W=(1/N)_{a=1,N}S_{aw}. Average for j=j' does not contain message structure factor and is merely in form SS*: D=<x_{wj}x^{*}_{wj}> =(1/N)_{b}S_{b}S*_{b}. Our expression for R takes a form: R_{w}=(1/I)( J_{w}W'_{w}+ nD_{w}), where: W'_{w}=W_{w}W_{w}^{*}, and factor: J_{w}=_{j=/=j'}e^{-iwC(j-j')}= -n +_{jj'}e^{-iwC(j-j')}= -n + FF*, where: F=_{j}e^{-iwCj}= (using geometric-progression formula) =(1-e^{-iwCn})(1-e^{-iwC})=0 because of Cn=T and wCn=2 PI m. However, F=0 is correct for w=/=0. Expanding exponents till first order in wCn, we have: (1-(1-iwCn))/(1-(1-iwC))=n. Because of F is continuous in 0, F_{w}|_{w=0}=n. The only interest we have in knowing values at w=0 is to see that final formulas for R will have correct average = <R>. In other words, F_{w}=n d_{w,0}, where d is Kroneckers delta. Finally, for constant length symbols: R_{w}=n/I(D-(1-nd_{w,0})W'); We will often omit delta-member wich still will be correct for w=/=0: R_{w}=n/I(D-W'); Now, let us to consider rectangular bar pulse {P}: S_{bw}= A_{bw}p_{w}e^{-iwdbTc}d is an offset of a pulse for symbol b relatively to t=0. p_{w}=(1/T)_{0,Tc}e^{-iwt}dt= =(1-e^{-iwG})/(iwT) = e^{-iwG/2}(2i sin(iwG/2))/(iwGnL) = = e^{-iwG/2}/nL sin((pi)g)/((pi)g) = = e^{-iwG/2}/nL sinc(g) G=T_{c}f=m/T - frequency g=fG - unitless parameter q=wG/2 = 2(pi)m/TG/2 = (pi)g We need only P_{w}= |p_{w}|^{2}= 1/(nnLL) sinc^{2}(g) For OOK, PPM, DAPPM(when neglecting effect of symbol offset shift?): D = 1/(nnLL)PQ. W' = 1/(nnLL)PKH. Q=1/N_{b}A_{b}^{2}K=1/(NN) (_{s=1,A}A_{s})^{2}H=_{dd'=0,L-1}e^{iw(d-d')G}. R=1/(nLLNI) P(Q-KH) For OOK, For OOK, we considered Alphabet as having 2 symbols and each chip will carry exactly one symbol 0 or 1: C=G. N=2. L=1. A=1. I=1/2. Q = 1/2 K = 1/4 H = 1 R = 1/n P(1/2-KH) = 1/(4n) P To see R_{0}, we replace KH with delta-member, -n/4 and obtain R_{0}=1/4. Does reader think this is correct? Without normalizing by I, R=1/(8n) P. What if to take P/8 as a reference? For PPM: N = L Q = (1+1+....+1)/N=1 K = 1/N H =_{dd'=0,L-1}cos(2(pi)g(d-d') = L + 2_{d=1,L-1}(L-d)cos(2(pi)gd) I = 1/L R = 1/(nLL) P (1-KH) To see R_{0}, we replace KH with delta-member, -nKH=-nL and obtain R_{0}=1/L. Does reader think this is correct? To make I=A^{2}/L the same levlel as for OOK, 1/2, we need to increase amplitude in sqrt(L/2) times which increases non-normalized R by L/2 which gives 1/(nLL2) P(...). We have to compare OOK with 1/(2LL) P(1-KH), or P against 4/(LL).Coding Scheme PropertiesUnit of measure denoted in [...]. (Really strange definition:) Transimission Capacity = (Number of Bits in full alphabet)/(Average number of chips in symbol). [Bit/Chip]. DAPPM: M *(Number of symbols) / ((L+1)/2) = 2MAL/(L+1). [1(8)] = 2MA2^{M}/(2^{M}+A) [1Tabl.2] PPM: M * 2^{M}/ 2^{M}= M