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Tuesday, 3 February 2015

CS601 Data Communication Assignment No.3 Solution Semester: FALL 2014 Due Date: 10 Feb, 2015

15:30    5 comments
Assignment No. 3
Semester: Fall 2014
Data Communication - CS601

Total Marks: 10

Due Date: 10/02/2015

Objective:

To have the understanding of Transmission Impairments and Multiplexing.

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                                                                                                       Assignment

Question. 1

Six channels, each with a 75-kHz bandwidth, are to be multiplexed together. What is the minimum bandwidth of the link if there is a need for a guard band of 10 kHz between the channels to prevent interference? Give your answer in following table with correct values.


Calculations
Bandwidth with proper Unit




Question. 2

A signal travels through an amplifier and its power is increased 100 times. Calculate the amplification of the signal. Give your answer in following table with correct values and proper unit.


Calculation
Amplification with proper unit








Wish You All The Best
J

5 comments:

  1. Crazy Romantic Love3 February 2015 at 17:06

    Students having CS601 Must Reply Here With Their Suggestions or Ideas regarding Assignment No.3.

    ReplyDelete
  2. Unknown6 February 2015 at 16:35

    This comment has been removed by the author.

    ReplyDelete
  3. Unknown6 February 2015 at 16:39

    Q1.
    For Six channels, we need at least five guard bands (Channels must be separated by strips of unused BW (guard bands) to prevent signals from Overlapping). This means that the required bandwidth is at least given as:


    Total Frequency of one channels=75KHz
    Guard Band Freq. b/w channels=10KHz
    Bandwidth=Freq. of one channel*Total Channel+Guard Band Freq.*Total Guard Bands
    Bandwidth=500KHz


    Calculations Bandwidth with proper Unit
    75*6+5*10 500KHz
    BW=450+50
    500KHz

    ReplyDelete
  4. Unknown6 February 2015 at 16:42

    Q2
    A signal travel through an amplifier and its power increases 100 times. This mean that P2 = 100P1
    In this case, the amplification (gain of power) can be calculated as
    10 log10 (P2/P1) = 10 log 10(100P1/P1)
    =10 log10 (100)
    = 10 (2)
    = 20dB








    Calculation Amplification with proper unit
    10 log 10(100P1/P1) 20dB
    =10 log10 (100)
    = 10 (2)
    = 20dB


    ReplyDelete
  5. Unknown8 February 2015 at 22:06

    ist question kis lecture sy ha

    ReplyDelete

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