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# CS601 Assignment No 03 Solution & Discussion Due Date: 10-02-2015

CS601 Assignment No 03 Solution & Discussion Due Date: 10-02-2015

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

Views: 12405

### Replies to This Discussion

thnx bro

excellent

Good work

thank you zain for sharing this.plz tell us the source of this example...plz plz plz

You can copy the question in google, it will show you lot of sources e.g.

See attached file - Slide No. 14

Attachments:

kese nikala yh solution ??

correct

Dear Students Don’t wait for solution post your problems here and discuss ... after discussion a perfect solution will come in a result. So, Start it now, replies here give your comments according to your knowledge and understandings....

Dear guys please tell me the correct formulas of both questions.

Guys tell me that this answer is right which i have solved.

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 Channels = 6

Total Frequency of one channel = 75 KHz

Guard Band frequency b/w channels = 10 KHz

Bandwidth = Freq. of one channel × Total Channels + Guard band freq. × Total guard bands

Bandwidth = 75 × 6 + 5 × 10

Bandwidth = 450 + 50

Bandwidth = 500 KHz           Ans.

 Calculations Bandwidth with proper Unit Minimum Bandwith 500 KHz

Excellent!

Question. 2

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

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