Solution#1
Transmission Time = Number of bits / Bit Rate
= 1 / (10 x 106) sec
= 0.1 µ sec.
Solution#2
Case(a) :
Propagation delay, Tp = 100 / 2 x 108
= 0.5 µ sec
Transmission Delay, Tx= 1000 / 10 x 103 sec
= 0.1 sec
a= 0.5 x 10-6 / 0.1
= 5 x 10-6
Case(b) :
Tp = 10 x 103 / 2 x 108
= 50 µ sec
Tx= 1000 / 1 x 16 sec
= 1 msec
a= 50 x 10-6 / 0.001
= 50 x 10-3
Case(c) :
Tp = 50000 x 103 / 2 x 108
= 0.25 sec
Tx= 1000 / 10 x 16 sec
= 0.1 msec
a= 0.25 / 0.0001
= 2.5 x 103
For the given data block and for the given medium propagation time increases with the distance of transmission. The transmission delay depends on the transmission rate of the medium. So for coaxial cable transmission delay is reduced and it is further reduced for the satellite link because of the still higher data rate. But for propagation delay it goes on increasing as the distance is increasing.
In cases (a) and (b) Tp is less than Tx and therefore a < 1 but for satellite link because of the large distance between source and destination Tp > Tx. Therefore for the satellite link at a time many bits are in travel between source and destination.
Number of bits in transit between source and destination = propagation delay x
transmission rate
= 0.25 x 10 x 106
= 2.5 x 106 bits
Solution#3
A noiseless channel can carry an arbitrarily large amount of information, no matter how often it is sampled. Just send a lot of data per sample. For the 4 kHz channel, make 8000 samples/sec. If each sample is 16 bit, the channel can send 128 kbps. If each sample is 1024 bits, the channel can send 8.2 Mbps. The keyword here is "noiseless". With a normal 4 kHz channel, the Shannon limit would not allow this.
Solution#4
Signal to noise ratio = 20 dB = 100
Shannon limit = 3 x 103 log2 [1 + 100]
= 19.5 kbps
Nyquist limit = 2 x 3 x 103 log22
= 6 kbps
The bottleneck is therefore a Nyquist limit, giving a maximun channel capacity of 6 kbps.