Re: Q6.11 solution

Ed Casas (edc@ece.ubc.ca) Thu, 2 Mar 2000 17:42:17 -0800 

Date: Thu, 2 Mar 2000 17:42:17 -0800
From: Ed Casas <edc@ece.ubc.ca>
Subject: Re: Q6.11 solution



On Tue, Feb 29, 2000 at 06:28:54PM -0800, Maria Lei [4681] wrote:
> >From the solution given for Q6.11, P(X) is the Q function.  However, 
> it was indicated that P(x) does not involve the Q function earlier
> in the message board.  Could anyone please explain?

I'll try.  The first sentence in 6.11 is misleading. Also,
stating that y is the BER is wrong (the person who wrote the
solutions was more polite and simply said "it would be more
suitable to call y the complement of the system reliability.")

The value of y specified in this question is not a BER (either
instantaneous or average) -- instead it is the probability that
the instantaneous BER will be above a certain value.

The reason why this quantity is of interest is that in many cases
we are more interested in answering the question "what fraction
of the time is the system usable" than the question "what is the
long-term fraction of bits that are wrong".  Usually a customer
is interested in whether the system is usable and that will be
determined by the short-term error rate rather than a long-term
average.  A similar argument applies when we specify the frame or
packet error rate of a system instead of the raw bit error rate
(typically a frame with any number of errors must be discarded).

When I answered your question, I didn't read the question
thoroughly enough to realize the mistake.

The answer to the question "what is the Eb/No .. to sustain a
1E-3 error rate" is certainly not the value given in the
solutions.  Instead, the answer given in the solutions is the
Eb/No required to make probability of the BER being below 1E-3
equal to 1E-3 (for some reason the person who wrote up the
solutions just used 1E-3 for both y and x).

Thanks for pointing this out.

-- 
Ed Casas   edc@ece.ubc.ca   http://casas.ece.ubc.ca   +1 604 822-2592