Arbitrary non-Correlated Symbols


  When summation interval is obvious, we'll omit limits at summ sign: 
  j can mean j=0,n-1

  L          = Maximum number of chips in symbol.
  Tc         = Duration of one chip.
  C=Tc*L.  C = Maximum possible duration of one symbol.
  Signal (message)  x= j=0,n-1 xj        (1)
      xj(t)=Saj(t-cj)                                                (j) 
           - is a signal of symbol aj started (centered) at time cj 
      j - (time) index of symbol in a message,   
      aj is in Alphabet of symbols.
      For example N=2M for OOK.
      M = number of bits comprising each symbol.
      Sb(t) - "waveform,shape" of a symbol b.
      Sb is not neccesary a step function or reachtngular pulse.
      Rectangular pulse, Sb(t) = Abp(t-tb)                {P}
      will be consider later.
      Signals Sb do not necessary have the same time length L.
      For example, for DAPPM sequences 01 and 002.
      Hence, we don't assume that cj=Tc*L*j yet.      
  In (1), a is a random function aj of index j.
  In article [1], k denotes index such kTc=ta+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[x2(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/TI0,T  x(t)x(t-tau)dt         (R')

  Now it is time to involve Fourier transformation.
  Because of x has period T,  Rw=|xw|2,       (R'')
      where circular frequncy w=2 PI f = 2 PI m/ T,
      x=m=-oo,oo xweiwt.        (F)
  Like original signal x, its Fourier transform will be a summ of sinals via nodes j:

  xw=j xw,j.
  We need to find a way to exlcude "randomness" brought by random function Ss.
  We will take average:
  This is an artificial step, because Rw is already averaged:
  but this step make our derivations easy.
  We have:
        Rw=(1/I)jj'  <xwjx*wj'>

  Constant length symbols
  Consider messages where each symbol takes equal number of chips L.
  Because symbols follow independently:    
       <xwjx*wj'> = 
                <xwj><x*wj'>  for j=/=j',     (N)
                <xwjx*wj>=D     for j=j'.          (D)
  Expanding (j), we reduce average for S to average W via alphabet:
       <xwj> =  <Swj> = e-iwjC W.
       W=(1/N) a=1,N Saw.

  Average for j=j' does not contain message structure factor and is merely in form SS*:
      D=<xwjx*wj> =(1/N)bSbS*b.
  Our expression for R takes a form:
       Rw=(1/I)( Jw W'w + nDw),
       and factor:
       Jw = j=/=j' e-iwC(j-j') =
                     -n + jj' e-iwC(j-j') =
                     -n + FF*,
                     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,
                     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, Fw=n dw,0, where 
                     d is Kroneckers delta.
  Finally, for constant length symbols:
  We will often omit delta-member wich still will be correct for w=/=0:
  Now, let us to consider rectangular bar pulse {P}:
       Sbw = Abwpwe-iwdbTc 
       d is an offset of a pulse for symbol b relatively to t=0.
       pw=(1/T)0,Tc e-iwtdt=
          =(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)  
          f=m/T - frequency
          g=fG  - unitless parameter
          q=wG/2 = 2(pi)m/TG/2 = (pi)g
       We need only Pw = |pw|2 =  1/(nnLL) sinc2(g)
  For OOK, PPM, DAPPM(when neglecting effect of symbol offset shift?):
      D  = 1/(nnLL)PQ.
      W' = 1/(nnLL)PKH.
           Q=1/N    bAb2
           K=1/(NN) (s=1,A As)2  
           H=       dd'=0,L-1 eiw(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 R0, we replace KH with delta-member, -n/4 and obtain R0=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 + 2d=1,L-1 (L-d)cos(2(pi)gd)
      I   = 1/L
      R   = 1/(nLL) P (1-KH)
      To see R0, we replace KH with delta-member, -nKH=-nL and obtain R0=1/L.
      Does reader think this is correct?

      To make I=A2/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 Properties

Unit 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)]
                                               = 2MA2M/(2M+A) [1Tabl.2]
    PPM:    M * 2M / 2M  = M