## 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_________

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

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**

36 MHz | |

360 kHz | |

3.96 MHz | |

396 kHz |

Question 2 Explanation:

Let f

Bandwidth = f

_{h}is the highest frequency and f_{i}is the lowest frequency.Bandwidth = f

_{h}– f_{i}= 4000 - 40 KHz = 3960 KHz = 3.96 MHzQuestion 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:**

109 | |

216 | |

218 | |

219 |

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).

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.