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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: 9853

Attachments:

### Replies to This Discussion

 No. of Voice Channels Data Rate in Mbps Type of T/E line used Total No. of T/E lines 50 64Kbps 1/24 of T-1 1 Channel 90 1.544Mbps 1 T-1 24 Channels 140 3.152 Mbps 2 T-1 48 Channels 195 6.312 Mbps 4 T-1 96 Channels 600 44.736 Mbps 28 T-1 672 Channels

CS601_Assignment#04_Solution

Attachments:

1010111001100010

______ 1001100111

11011 ) 1010111001100010

11011

111111001100010

11011

1001001100010

11011

100101100010

11011

10011100010

11011

1000100010

11011

101000010

11011

11110010

11011

101010

00000 (because the left most bit is zero, So all zero from original)

101010

11011

11100

11011

100 Cyclic Redundancy Check (CRC) is 100

x2 Not included in polynomial so it will be“0”,

11011

its Divisor

bhai jo reviser decide karta hain hum us sa 1 kam 0 hum denominator man add nai karta kya???

 No. of Voice Channels Data Rate in Mbps Type of T/E line used Total No. of T/E lines 50 3.208 Mbps T-1 2 90 5.768 Mbpc T-2 6 140 9.128 Mbps T-3 12 195 12.648 Mbps T-1 16 600 38.568 Mbps T-1 25

T/E Carrier and Packetized Communications
T/E Carrier
T/E carrier systems use digital multiplexing to transmit voice and data signals. Maxim Integrated's solutions for T/E communications include line interface units (LIUs), framers, and single-chip transceivers (SCTs).
Carrier Ethernet
Carrier Ethernet is a standardized carrier-class service and network for communication among business, academic, and government local area networks (LANs), providing flexible bandwidth and scalability. Maxim's Ethernet-over-PDH (EoPDH)/SDH mapping devices extend Ethernet LAN segments for transmission over PDH/SDH data streams.

 No. of Voice Channels Data Rate in Mbps Type of T/E line used Total No. of T/E lines 50 3.208 Mbps T-1 2 90 5.768 Mbpc T-2 6 140 9.128 Mbps T-3 12 195 12.648 Mbps T-1 16 600 38.568 Mbps T-1 25

 No. of Voice Channels Data Rate in Mbps Type of T/E line used Total No. of T/E lines 50 3.208 Mbps T-1 2 90 5.768 Mbpc T-2 6 140 9.128 Mbps T-3 12 195 12.648 Mbps T-1 16 600 38.568 Mbps T-1 25

Q,2 exact sol :)

Attachments:

question no:01 ka exact solution knsa hai hai

full solution kon sa hai

can any one help in questio n 1 solution...?

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