Quiz 2.

I am extremely disappointed in the results of quiz 2.
I had told the class I would ask minor modifications of some problems 
in quiz 1. I did. Some students have not looked at all at quiz 1 after
making a mess of it.

Next time (probably in 7 - 10 days) I will AGAIN  include questions
that are minor modifications of problems from quizes 1 and 2.

So:

Study previous quizes. If you can not do them, ask questions.

Quiz 2:

1A:  PSTN = Public Switched Telephone Network.
1B:  PCM = Pulse Code Modulation.

2A: Designers of the telephone system with digital transport wanted to
    represent frequences to somewhat over 3000 Hz reasonably accurately.
    They rounded this off to 1 - 4000 Hz. The Nyquist result says that
    if you want to represent frequencies up to H Hz you must sample 
    2H times persecond. That is where the sampling rate of 8000 
    samples/sec comes from.
2B. 8000 samples/sec at 56,000 bits/sec gives 7 bits/sample.
    7 bits allows 2^(7) = 128 levels.
    A CIS (or COE, or IT) student should know the difference between 
    ``levels'' and ``bits''. 
    k bits allow 2^(k) levels.

3. The signal to noise ratio is 20 dB (decibels). 
   That means that  10*(log(S/N)) = 20 (log base 10). 
   Hence log(S/N) = 2, (S/N) = 100.
   The frequency range is H = (132 - 126)MHz = 6 MHz.
   By Shannon's result, the highest theoretically possible 
   data rate is H*log(1 + S/N) bits/sec  (log base 2).
   That is: 6,000,000*(log(101)) = 6,000,000*((ln(101)/(ln(2))) = 
   6,000,000*4.615/(.6931) = 6,000,000*6.6582 = 39,949,268.9 bits/sec.
   Let's call it 40 Mbit/sec.

   By now you must have understood I want you to remember Nyquist and 
   Shannon at least until Christmas.

4. I need to make a drawing. It will be put into a .pdf file in this 
   directory.

   If at time t the source sends frame(n) which makes it, but it 
   (or its ACK) gets delayed, so that at time (t+TO) there is a
   time-out at the source and Frame(n) gets re-transmitted, and the
   re-transmitted frame (and its ACK) also make it, 
   then the source twice gets ACK(n) and twice sends Frame(n+1), 
   and the destination twice sends ACK(n+1), etc, ``forever'', 
   or until some other accident occurs. 
   (The next ``accident'' is likely to make this worse, but you could
   be lucky and the next accident brings the system back to normal).

Teun Ott.