Question

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

Propogation Distance = 80 km = 80 x 1000m = 80,000m

Propogation Speed = 2.1 x 10^8 m/s

Transmission Length=60KB = 60 x 2^10 x 8 bits = 491520 bits

Transmission Speed/Rate = 622Mbps = 622 x 2^20 bits/sec = 652214272 bits/sec

(a) Data Transmission Time per packet = Packet Size / Bit Rate= 491520 bits / 652214272 bits/sec = 0.000753 sec

(b) Propagation Delay = Propogation Distance / Propogation Speed = 80,000m / 2.1 x 10^8 m/s = 38.095 msec

(c) Bit duration = Transmission length / Transmission Speed = 491520 bits / 652214272 bits/sec = 0.000753 sec

(d) Bit Length = Propogation Speed x Bit duration= 2.1 x 10^8 m/s x 0.000753 sec = 158130 m

(e) Number of bits in the cable between the sender and receiver = Propagation Length / Bit Length = 80,000m / 158130 m = 0.5059 bits