Detecting unipolar and repeating preamble on wired communications?

I am trying to implement a preamble detection.

A couple of things that,
I’m not using wireless communication but wired communication.
And also, I’m not using a bipolar signal, but a unipolar signal.

(Bipolar: a signal swings between +A volts (or amperes) and -A volts;
Ethernet, USB, Wi-Fi, … are using bipolar signals,
Unipolar: a signal swings between +A volts and 0 volts)

Using a unipolar signal, I am having a difficulties on determining
whether
a signal has a repetition.

A preamble has a typical form of 1, 0, 1, 0, 1, 0, 1, 0, (and data
payload
follows)
and it is mapped into a real signal and received at a receiver: +A+n, n,
+A+n, n, +A+n, n, +A+n, n, (and data payload follows)
(n is very small noise and fluctuation, n can be either negative or
positive)

I think an optical fiber communication might use unipolar signalling,
but
I’m not sure.

Is there anyone already implemented unipolar preamble detection
succesffully, or could anyone give me some hints on mathematical
approach
or something?

Regards,
Jeon.

Hi,

(Bipolar: a signal swings between +A volts (or amperes) and -A volts;
Ethernet, USB, Wi-Fi, … are using bipolar signals,
Unipolar: a signal swings between +A volts and 0 volts)

There is no difference … “0 volt” is just an arbitrary reference
point.

0 <-> +A swing is the same as
-A/2 <-> A/2 swing if you shift your reference …

Ethernet and USB are differential but that’s just to improve signal
integrity / noise immunity and such. At the receiver end, the
differential input buffer essentially compares both and see which one
is higher. (a bit simplified, but that’s the gist of it).

Is there anyone already implemented unipolar preamble detection
succesffully, or could anyone give me some hints on mathematical approach or
something?

If ‘A’ is known, just remove -A/2 from your input signal.
If it is unknown,you can use a very long time constant low pass to
find it. (assuming your signal doesn’t have long runs of 0 or 1).

Cheers,

Sylvain

Dear, Sylvain M.,

Thanks.I got a hint from your answer.

What about using moving average instead LPF?
I think moving average (MA) is much simpler than LPF if computational
cost
is considered.
But, another issue is, precision on estimation A and A/2. In that case,
LPF
would be better.

Anyway, I will do applying either LPF and MA on my flow graph. Thanks.

Regards,
Jeon.

2015-05-05 16:27 GMT+09:00 Sylvain M. [email protected]:

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