## 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
Computer-Networks       Transmission-and-Propagation-Delay       GATE 2022       Video-Explanation
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
What is the bandwidth of the signal that ranges from 40 kHz to 4 MHz
 A 36 MHz B 360 kHz C 3.96 MHz D 396 kHz
Computer-Networks       Transmission-and-Propagation-Delay       ISRO CS 2008
Question 2 Explanation:
Let fh is the highest frequency and fi is the lowest frequency.
Bandwidth = fh – fi = 4000 - 40 KHz = 3960 KHz = 3.96 MHz
 Question 3
Assume that each character code consists of 8 bits. The number of characters that can be transmitted per second through an asynchronous serial line at 2400 baud rate, and with two stop bits, is:
 A 109 B 216 C 218 D 219
Computer-Networks       Transmission-and-Propagation-Delay       ISRO CS 2008
Question 3 Explanation:
Here, the baud value represents the serial port is capable of transferring a maximum number of bits per second.
Given 2400 baud value,
Sending maximum of 2400 bits per second with serial port .
Total Data To send = 1 bit(start) + 8 bits(char size) + 2 bits(Stop) = 11 bits.
Number of 8-bit characters that can be transmitted per second = 2400/11 = 218.18
The effective number of characters transmitted = 218 So, the correct option is (C).
There are 3 questions to complete.

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