[rec.ham-radio] Spread Spectrum

guest@hyper.UUCP (guest) (03/02/88)

I think I can demonstrate the whole concept of spread spectrum
stealthiness with the following simple BASIC program:

Input "Value to be sent";V
Input "Number of bands used";B
Vx=V/B
R=0
 For J=1 to B
  R=R+Vx+(1-RND(0)*2)
 Next J
Print "Value received =";R
Print "Signal power in each band =";Vx
End

As you can see, the signal V has its energy distributed at B points with
strength Vx.  Vx is superimposed on the pre-existing random noise (here
confined to a range of + and - 1) at each point J in B.

As long as Vx is not too far below +and- 1, and there are sufficient bands,
the random noise tends to cancel, leaving the original signal R, which
is correlated here by simple addition.

There is no need, however, to have the bands evenly spaced. In fact the
spacing of the bands should be a changing function. Any other stealth
xmitter would, in all probability, be using a different band spacing
function at any given time. Thus there is no interference because there
is no correlation.  Correlation is the key to reception, interference and
detection.

So if the signal (in its discrete components) looks like noise, tastes like
noise and feels like noise (to an uninformed receiver) then it must be
noise. You can claim you have detected it, but I think not.


John Logajan            umn-cs!hyper!ns!logajan
Network Systems Corp
7600 Boone Ave
Brooklyn Park, MN  55428

paulf@umunhum.STANFORD.EDU (Paul Flaherty) (03/03/88)

I missed your reply to my first message, but it's clear that you don't have
a fundamental grasp of Shannon's Law;  Spread Spectrum raises the noise 
floor across the communication passband, and that increase in noise can
be detected.

The BASIC simuation that you've proposed is invalid for a number of reasons,
not limited to:

	1. The BASIC RND function is uniform; to model a radio channel,
	one needs Gaussian noise, plus impulse noise, plus Rayleigh fading
	on the signal.

	2. No statistical test is proposed to separate the spreading code
	and the noise; however, even in the case that you have given,
	it can be shown statistically that the resultant is not uniformly
	distributed, and therefore a signal exists on top of the noise.

Certainly, spread spectrum is less detectable than a narrowband signal, but
to claim that it is next to impossible to detect is foolish.




-=Paul Flaherty, N9FZX	    |
Computer Systems Laboratory |  "panic: getfs: bad magic"
Stanford University	    |
->paulf@shasta.Stanford.EDU |		--4.3bsd Unix