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Assignment No. 5 
Semester Spring 2012
Data Communication - CS601

 

Total Marks: 20

 

Due Date: 00-07-2012

 

Objective: To understand the concept related to the error detection mechanism.

 

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).
  • Student ID is not mentioned in the assignment File or name of file is other than student ID.

 

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

 

GOOD LUCK

 

 

 

Q. 1. Suppose a sender sent a word “assignment” for which the receiver received the following binary:

 “01100001011010010111001101101001011001110110111001101110011001010110111001110100”

 

By applying the complete process of LRC using even parity, you are required to check that, is the same word received by the receiver or not? Also write the word received by the receiver.

 

Binary Code

Alphabet

Binary Code

Alphabet

Binary Code

Alphabet

Binary Code

Alphabet

0110 0001

a

0110 1000

h

0110 1111

o

0111 0110

v

0110 0010

b

0110 1001

i

0111 0000

p

0111 0111

w

0110 0011

c

0110 1010

j

0111 0001

q

0111 1000

x

0110 0100

d

0110 1011

k

0111 0010

r

0111 1001

y

0110 0101

e

0110 1100

l

0111 0011

s

111 1010

z

0110 0110

f

0110 1101

m

0111 0100

t

 

 

0110 0111

g

0110 1110

n

0111 0101

u

 

 

Table 1

 

 

 

 

HINT: First convert the word “assignment” into binary equivalent with the help of the given table 1.

The LRC calculated by the sender is not appended with the given binary. It is also your task to calculate the LRC by yourself.

 

 

 

Note:

 

  • Assignment should be in your own wordings not exactly copied from Internet, handouts or books.

 

  • Your answer should be “to the point”.

 

Views: 6583

Replies to This Discussion

heading waghera ap na bhi den to its ok, just find binary stream of word given and finds its LRC then find alphabet equivalents of received binary with the help of given table and then its LRC and you are done..

aqsa check ur lrc sirf yea gahalt hea sis.change it..........

LRC is 100% correct..

jo word receive hua hay agar wo different hay kisi corruption ki waja say to phir sending binary LRC same nahi ho sakti received wali say..

because received binary stream kuch or hay or jo binary stream send ki gai hay wo kuch or hay..

to phir in dono ka LRC bhi different hi niklay ga..

ap nay assignment me b yehi likhna hay k sent word or received word same nahi hay or in ka LRC bhi same nahi hay..

hope all things will be cleared to you now..

LRC OF word assignment is

                               00010011

and LRC of received word is

00001010

sis can you tell is it right or wrong

mera bhi yehi aya hai

yes it is absolutely right..

its correct abdullah

Nice job done dear...

Well done....

received bits calculate karnay k liye ap apni assignment file  check karen, wahan received binary stream di hui hay, ap us k equivalent alphabets table me say find kar k likh len..

Please see the attached file for the solution ...... thnx!

Attachments:

S no

word

Binary equivalent

1

A

0

1

1

0

0

0

0

1

2

S

0

1

1

1

0

0

1

1

3

S

0

1

1

1

0

0

1

1

4

I

0

1

1

0

1

0

0

1

5

G

0

1

1

0

0

1

1

1

6

N

0

1

1

0

1

1

1

0

7

M

0

1

1

0

1

1

0

1

8

E

0

1

1

0

0

1

0

1

9

N

0

1

1

0

1

1

1

0

10

T

0

1

1

1

0

1

0

0

 

LRC

(Using even Parity)

0

0

0

1

0

0

1

1

 

 

reciever receive the word aisignnent”

 

S no

word

Binary equivalent

1

A

0

1

1

0

0

0

0

1

2

I

0

1

1

0

1

0

0

1

3

S

0

1

1

1

0

0

1

1

4

I

0

1

1

0

1

0

0

1

5

G

0

1

1

0

0

1

1

1

6

N

0

1

1

0

1

1

1

0

7

N

0

1

1

0

1

1

1

0

8

E

0

1

1

0

0

1

0

1

9

N

0

1

1

0

1

1

1

0

10

T

0

1

1

1

0

1

0

0

 

LRC

0

0

0

1

0

0

1

1

 

 

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