Transmission-and-Propagation-Delay

Question 1
Consider a 100 Mbps link between an earth station (sender) and a satellite (receiver) at an altitude of 2100 km. The signal propagates at a speed of 3x10 8 m/s. The time taken (in milliseconds, rounded off to two decimal places) for the receiver to completely receive a packet of 1000 bytes transmitted by the sender is_________
A
7.08 ms
Question 1 Explanation: 
The time required for the receiver to receive the packet is transmission time + propagation time i.e. Tt +Tp
Tt = L / B => 1000 x 8 bits / 10^8 bps = 0.08 ms
Tp = D / V => 2100 x 1000 m / 3 x10^8 ms = 7 ms
Therefore, Total time = 7.08 ms
Question 2
Optical fiber uses reflection to guide light through a channel, in which angle of incidence is ________ the critical angle.
A
equal to
B
less than
C
greater than
D
less than or equal to
Question 2 Explanation: 
→ Optical fiber uses reflection to guide light through a channel, in which angle of incidence is greater than the critical angle.
→ Total internal reflection is the phenomenon which occurs when a propagated wave strikes a medium boundary at an angle larger than a particular critical angle with respect to the normal to the surface.
→ If the refractive index is lower on the other side of the boundary and the incident angle is greater than the critical angle, the wave cannot pass through and is entirely reflected.
Question 3
Time taken by a packet to travel from client to server and then back to client is called
A
STT
B
RTT
C
PTT
D
Total time
Question 3 Explanation: 
→Round-trip time (RTT), also called round-trip delay, is the time required for a signal pulse or packet to travel from a specific source to a specific destination and back again.
→In this context, the source is the computer initiating the signal and the destination is a remote computer or system that receives the signal and retransmits it
There are 3 questions to complete.

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