Question

4-   Two Systems 80 Km apart are communicating through a fibre Optic (note: signal speed is 2.1x10⁸ m/s). The data is in the form of 60 KB packets with transmission rate 622 M bps. Calculate the following:
a-   Data transmission time per packet.
b-   Propagation delay.
c-   bit duration.
d-   Bit length,
e-   Number of bits in the cable between the sender and receiver.

Answer 1

a)

Transmission Time per packet = Packet Size / Bandwidth = 60 KB / 622 Mbps = 0.790 milliseconds

b)

Propagation Delay = Distance / Speed = 80 km / 2.1 x 10^8 = 0.380 milliseconds

c)

Bit duration is the time taken to transmit 1 bit

Bit duration = 1 / 622 x 10^6 = 1.6 nanoseconds

d)

Bit length = Bit duration x Propagation Speed = 1.6 x 10^-9 x 2.1 x 10^8 = 0.336 meters

e)

Length of 1 bit = 0.336 meters

Length of the cable = 800000 m

Number of bits in the cable = 80000 / 0.336 m = 238095 bits