Substitude S/N into Shannon's theorem, we get the maximum bps as 3k*log2^(1+S/N)=3log2^101kbps=19.97kbps.
However, the maximum data rate, assuming the channel is noise free, is only 6kbps, according to Nyquist rate. Therefore, the final answer should be 6kbps.
Notice that the M or k in bps uses base 10 intead of base 2.
Media | Bandwidth (Hz) |
Twisted pair (T3) | 16M |
Twisted pair (T5) | 100M |
Twisted pair (T6) | 250M |
Twisted pair (T7) | 600M |
Coax | 1G |
Fiber optics | 30T |
Note that this problem is designed to give you a sense how the media performs. You don't need to memorize those numbers.
B | ||
C | A | C |
B | C | B |
GSM uses 124 channels with at most 8 users sharing a single channel, which allows GSM to support up to 124*8=992 users simultaneously.
S.A = (-1+1-3+1-1-3+1+1).(-1-1-1+1+1-1+1+1) = (1-1+3+1-1+3+1+1)/8 = 1
S.B = (-1+1-3+1-1-3+1+1).(-1-1+1-1+1+1+1-1) = (1-1-3-1-1-3+1-1)/8 = -1
S.C = (-1+1-3+1-1-3+1+1).(-1+1-1+1+1+1-1-1) = (1+1+3+1-1-3-1-1)/8 = 0
S.D = (-1+1-3+1-1-3+1+1).(-1+1-1-1-1-1+1-1) = (1+1+3-1+1+3+1-1)/8 = 1
Therefore, A transmits 1, B transmits 0, C doesn't transmit, D transmits 1.