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Algebra...

===== 1.archi.pdf =====

----- slide 34, example:

first problem/question:

"S/N=signal noise ratio (not expressed in dB)" what is the meaning? We shouldn't use dBs?

second problem/question:

C_max = B*log_2(1+S/N) = 3000*log(1+30) = (~) 3000*log(2^5) = 3000*5 = 15kb/s
Am I wrong?

---- slide 39, stop&go solution:

left-top:
packet P1 sent
T=L/b

Thanks.
Posted by Raphaël Tagliani on Tuesday 20 February 2007 at 17:50
Comments
----- slide 34, example:

first problem/question:

"S/N=signal noise ratio (not expressed in dB)" what is the meaning? We shouldn't use dBs?

A: the meaning is that you should use the ratio expressed in dB, but simply the ration of the

second problem/question:

C_max = B*log_2(1+S/N) = 3000*log(1+30) = (~) 3000*log(2^5) = 3000*5 = 15kb/s
Am I wrong?

A: yes, instead of 15, find and use corresponding "simple" ration of powers.

PS: the theorem is nicely defined in wiki pages:
http://en.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theorem :
...
C is the channel capacity in bits per second;
B is the bandwidth of the channel in hertz;
S is the total signal power over the bandwidth and
N is the total noise power over the bandwidth.
S/N is the signal-to-noise ratio of the communication signal to the Gaussian noise interference expressed as a straight power ratio (not as decibels).
Posted by Slavisa Sarafijanovic on Tuesday 27 February 2007 at 13:20
---- slide 39, stop&go solution:

left-top:
packet P1 sent
T=L/b

A: you are right.
Posted by Slavisa Sarafijanovic on Tuesday 27 February 2007 at 13:23
S/N[dB] = 10 log (S/N) or
S/N = 10^(S/N[dB] /10)

(Logarithm with base = 10)

=> 30 dB = 1000

It gives Cmax = 29,902 kb/s
Posted by Paule Franck on Thursday 1 March 2007 at 10:01