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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:

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

Views: 9893

Attachments:

### Replies to This Discussion

Solution 1

No. of Voice Channels Data Rate in Mbps Type of T/E line used Total No. of T/E lines
50 3.2Mbps T1 /E1 3 lines of T1 or 2 lines of E1
90 5.76Mbps T2 /E1 1 Line of T2 or 3 lines of E1
140 8.96Mbps T2+T1/E1+E2 I line of T2 + 2 lines of T1 or 1 line E1 + 1 lines E2
195 12.48Mbps T2+T1 /E-1 1 line of T1 + 2 lines of T2 or 7 lines of E1
600 38.4Mbps T3/E2+E3 1 line of T3 or 1 line of E2 + 1 line of E3

Solution 2

Our message reaches with out alteration because our reminder is nonzero

The message =1010111001100010

Polynomial (Devisor) x4 + x3 + x + 1

Binary (Devisor)

x4 + x3 + x +1

+x2
1 1 0 1 1

Therefore Devisor = 11011

Whole strings (original message + 0000) is 10101110011000100000

Sender Side

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 0 0 0 0
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 0
1 1 0 1 1
1 1 1 1 0
1 1 0 1 1
0 1 0 1 0
0 0 0 0 0
1 0 1 0 0
1 1 0 1 1
1 1 1 1 CRC

Message + CRC
1 0 1 0 1 1 1 0 0 1 1 0 0 0 1 0 +1 1 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 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

cs601 assignmnt4 solution fall 2012

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:

Calculate the corresponding data rate against each number of given voice channels.
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

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

Question no 2
Polynomial x4 + x3 + x + 1,
by the roule of CRc x4=1,x3=1 or x2 missing hy than the result is o or last par x=1 or 1 hy than duble 11 aey ga. so the deviser is 11011 ho ga
es k bad hum es diver ko Question maindeye gaey 1010111001100010 es data se devide karen gy es main 4 zerro add kar k than the valu become 10101110011000100000
at the end u will get 1111 this is ur sender

Solution 1

No. of Voice Channels Data Rate in Mbps Type of T/E line used Total No. of T/E lines
50 3.2Mbps T1 /E1 3 lines of T1 or 2 lines of E1
90 5.76Mbps T2 /E1 1 Line of T2 or 3 lines of E1
140 8.96Mbps T2+T1/E1+E2 I line of T2 + 2 lines of T1 or 1 line E1 + 1 lines E2
195 12.48Mbps T2+T1 /E-1 1 line of T1 + 2 lines of T2 or 7 lines of E1
600 38.4Mbps T3/E2+E3 1 line of T3 or 1 line of E2 + 1 line of E3

question 1 is wrong  ( also check channels are > or =  to give channels   and min wastage capacity

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