Question

In: Computer Science

Two Systems 80 Km apart are communicating through a direct Metallic cable (note: signal speed is...

Two Systems 80 Km apart are communicating through a direct Metallic cable (note: signal speed is 2 x 10^8 m/s). The data is in the form of 60 KB packets with transmission rate 1M bps. Calculate the following:

  1. Data transmission time per packet.
  2. Propagation delay.
  3. bit duration.
  4. Bit length.
  5. Number of bits in the cable between the sender and receiver

Solutions

Expert Solution


Solution

Given distance between two systems=80 Km = 80 * 103 = 8000 metres

Bandwidth(B) =1Mbps =106 bits /second

Length of the Packet(L) =60 KB=60 * 210 Bytes = 60 * 210 * 8= 48 * 211 bits

Speed = 2 *108 metres/second

Transmission time:

Time taken to put the packet on the transmission link

Transmisson time(Tt ) = L/B

=   

= seconds

Propagation delay(Tp):

Time taken for one bit to travel from sender to receiver

Propagation delay(Tp)=distance/speed

=8000/2*108

=4 * 10-5  seconds

Bit Rate:

The number of bits per second sent or received.

   

  

Bit duration:

Bit duration

seconds

Bit Length :

Bit Length = Propagation speed * Bit duration

= 2 * 108 * 0.5 * 10-6

=200 metres

Bit Rate:

No of bits in the cable between sender and receiver:

is nothing but bitrate .the no of bits in the cable per second

The number of bits per second sent or received.





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