1- Imagine that you have trained your dog, to carry a box of three 8mm tapes. These
tapes each contain 7 gigabytes. The dog can travel to your side, wherever you may
be, at 18 km/hour. For what range of distances does The dog have a higher data rate
than a transmission line whose data rate (excluding overhead) is 150 Mbps ?
Ans .
1- 3 tapes * 7 gigabytes=21 gigabytes
21 gigabytes*8=168 gigabits
18 km/hour=18km/(60*60)= 0.005 km/sec
Distance=speed*time
Time to travel distance x = x/0.005 km/sec= 200 x sec
Data rate=data/time = 168gbps/200x
data rate = 168/200x Gbps =168*1000/200x Mbps= 840/x Mbps
distance related to 150 Mpbs=840/x Mbps
150=840/x  x=5.6 km
So, For data rate of 150 Mbps distance < 5.6 km
*/
According to Shannon's Channel Capacity Equation:
R = W*log2(1 + C/N) = W*log2(1+ SNR)
Where,
R = Maximum Data rate (symbol rate)
W = Bw = Nyquist Bandwidth = samples/sec = 1/Ts
C = Carrier Power
N = Total Noise Power
SNR = Signal to Noise Ratio
2. Given the following information, find the minimum bandwidth required for the
path:
FDM Multiplexing
Five devices, each requiring 4000 Hz.
200 Hz guard band for each device.
Solution:
No. of devices = 5.
No. of guard bands required between these is 4.
Hence total bandwidth = (4000 x 5) + (200 x 4)
= 20.8 KHz.