[Cryptography] Any Electrical Engineers here who know about noise filtering?

[Natanael]

Den 2 jan 2016 20:58 skrev "Henry Baker" <hbaker1 at pipeline.com>:
>
> Here's my problem:
>
> I'm trying to characterize a 1x pad.
>
> A 1x pad *adds* (modulo, but that shouldn't matter)
> uniformly distributed "noise" (the "key") to the
> "message" signal.
>
> Classical filtering theory says that given a
> noise spectrum, one can compute an optimal
> filter to remove as much noise from the signal
> as possible.
>
> I'd like to go through this mathematical
> exercise with modular addition and the
> noise spectrum one might want for a key
> distribution to show that such an
> optimal filter *for this case* is the
> identity -- i.e., even the optimal
> filter can do *nothing* to remove any
> "noise" -- i.e., the 1x pad random
> signal.
>
> I seem to recall ideas such as "Weiner
> filters", and the like.

Electrical noise (tempest) or signal noise in the OTP encrypted ciphertext?
The former can be solved with paired wires & transistors where the state
isn't 1 or 0, but where it is represented by which one in each pair is 1
and 0 respectively.

For the latter (which I guess you really mean), the pad provably brings up
the noise to a mathematically identical level for all bits, there is no
signal leak. *It isn't addictive noise* like with regular signals such as
electromagnetism or audio. You can't subtract anything based on any filter,
because everything is indistinguishable from random. The whole point is
that OTP eliminates correlations.

As for the proofs, Wikipedia has all the links you need, including fancy
math by academics: https://en.wikipedia.org/wiki/One-time_pad

In short, the fact that you start with a set of switches (bits) in unknown
starting positions and that  you flip a fully random set of switches means
that you can not tell what the starting points was. Like spinning a bottle
and having somebody guess the starting point after it stopped. All
possibilities have equal probability. For every plaintext bit, you do not
know if the bit was flipped by the corresponding key bit or if it wasn't.
For all plaintext bits, the chance is 50/50 that the key bit is identical,
50/50 that it wasn't.
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