Question

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

Answer 1

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(T_t ) = L/B

=   

= seconds

Propagation delay(T_p):

Time taken for one bit to travel from sender to receiver

Propagation delay(T_p)=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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