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# CS601 Assignment No. 4 Semester Fall 2012 Data Communication Due Date: 29-01-2013 Please share your ideas here

Assignment No. 4
Semester Fall 2012
Data Communication - CS601

Total Marks: 30

Due Date: 29-01-2013

Objective: Understanding the T & E lines and Error detection method.

Instructions:

Please read the following instructions carefully before solving & submitting assignment:

Assignment should be in your own wordings not copied from net, handouts or books.

It should be clear that your assignment will not get any credit (marks) if:

• The assignment is submitted after due date.
• The submitted assignment does not open or file is corrupt.
• The assignment is copied (from other student or copied from handouts or internet).

For any query about the assignment, contact at cs601@vu.edu.pk

GOOD LUCK

Q. 1. Telephone Companies use a hierarchy of Digital Signals Service (DS) which have different levels as DS-0, DS-1, DS-2, DS-3 and DS-4. All these levels have different data rates and are implemented through different types of T lines (from T-1 to T-4) or E lines (from E-1 to E-4) on the basis of data rate in each level.

Now you are required to fill  the columns of the given table according to following  instructions:

1. Calculate the corresponding data rate against each number of given voice channels.
2. Write the best suitable type of line (form T-1 to T-4 or from E-1 to E-4 line) with total number of T or E lines corresponding to calculated data rate. [15 Marks]

Note: Type of line (form T-1 to T-4 or from E-1 to E-4 line) and total number of T or E lines must be selected on the basis of calculated data rate having minimum wastage of capacity of the line. Only optimal solution is required for these columns.

 No. of Voice Channels Data Rate in Mbps Type of T/E line used Total No. of T/E lines 50 90 140 195 600

Q. 2.  Using Modulo-2 Division and Polynomial x4 + x3 + x + 1, perform the complete process of Cyclic Redundancy Check (CRC) on the given message by writing all steps for both sender and receiver end.

Assume that the message has been received by the receiver without any error. [15 Marks]

Message:

1010111001100010

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### Replies to This Discussion

tariq bhai plz tel how we choose t or E line .& overhead also be added in data rate.plz tel

Nice and keep it up

Solution:

 No. of Voice Channels Data Rate in Mbps Type of T/E line used Total No. of T/E lines DS0 64Kbps 1/24 of T-1 1 Channel DS1 1.544Mbps 1 T-1 24 Channels DS1C 3.152 Mbps 2 T-1 48 Channels DS2 6.312 Mbps 4 T-1 96 Channels DS3 44.736 Mbps 28 T-1 672 Channels DS3C 89.472 Mbps 56 T-1 1344 Channels DS4 274.176 Mbps 168 T-1 4032 Channels

Q. 2.  Using Modulo-2 Division and Polynomial x4 + x3 + x + 1, perform the complete process of Cyclic Redundancy Check (CRC) on the given message by writing all steps for both sender and receiver end.

Assume that the message has been received by the receiver without any error. [15 Marks]

Message:

1010111001100010

in 2nd q remainder is 1111 .it will rejected at receiver how we make it error free at ath receiver end...

Malik bhai ye sol wrong hai

What would the solution for question # 2

Ha its total CRC checker do it (2Q)

I'm do it through it.....

Crc checker result may come zero.crc generator result is 1111what i am saying

right?

can anyone explain it in simple wordssss .............  still data rate he clear hoa hai

message.

1 1 0 1 0 0 0 1 1 0 0 0 1 1 0 1

1 1 0 1 1          1 0 1 0 1 1 1 0 0 1 1 0 0 0 1 0 1 1 1 1

1 1 0 1 1

1 1 1 0 1

1 1 0 1 1

0 1 1 0 1

0 0 0 0 0

1 1 0 1 0

1 1 0 1 1

0 0 0 1 0

0 0 0 0 0

0 0 1 0 1

0 0 0 0 0

0 1 0 1 1

0 0 0 0 0

1 0 1 1 0

1 1 0 1 1

1 1 0 1 0

1 1 0 1 1

0 0 0 1 0

0 0 0 0 0

0 0 1 0 1

0 0 0 0 0

0 1 0 1 0

0 0 0 0 0

1 0 1 0 1

1 1 0 1 1

1 1 1 0 1

1 1 0 1 1

0 1 1 0 1

0 0 0 0 0

1 1 0 1 1

1 1 0 1 1

0 0 0 0 Reminder

 No. of Voice Channel Data Rates in Mbps Type of T/E line used Total no. T/E line 50 50*64kbps/1024=3.125 Mbps T-1 Line 2 line of T-1 (1.544*2=3.088Mbps) with minimum wastage of 0.037 Mbps 90 90*64kbps/1024=5.625 Mbps T-1 Line 3 line of T-1 (1.544*3=4.632 Mbps) with minimum wastage of 0.993 Mbps 140 140*64kbps/1024=8.75 Mbps E-2 Line 1 line of E-1 (8.44 Mbps) with minimum wastage of 0.31 Mbps 195 195*64kbps/1024=12.187 Mbps T-1 , E-1 , E-2 Lines 1 line of Each T-1 , E-1 , E-2  (1.544+2.048+8.44=12.032 Mbps)  with minimum wastage of 0.155 Mbps 600 600*64kbps/1024=37.5 Mbps T-1 ,E-1 , E-3 Lines 1 line of Each E-1 , E-3 (1.544+2.048+34.368=37.96  Mbps)  with minimum wastage of 0.46 Mbps

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