Duplicate Error Detection Problem. The sender sends an 8-bit word { x 1 , x 2 ,
ID: 3771347 • Letter: D
Question
Duplicate Error Detection Problem.
The sender sends an 8-bit word { x1, x2, … , x8 } in duplicate. The receiver receives the following two words meant to be identical copies of the word sent. Namely the receiver receives Y = { y1, y2, … , y8 } and Z = { z1, z2, … , z8 } . The receiver creates two parity calculator circuits. The first one, PC1, receives 8 input bits, namely the word Y, the second one, PC2, receives 8 input bits, namely the word Z. The output of PC1, is py , the parity of the word Y and the output of PC2, is pz , the parity of the word Z. That is:
py = y1 y2 y3 y4 y5 y6 y7 y8
pz = z1 z2 z3 z4 z5 z6 z7 z8
Then the receiver creates a final circuit for each pair of input bits of the Y and Z words. Namely for example one of these eight final circuits takes as input bits y1 , py , z1 , pz . The output bit w1 is to actually equal the original bit sent x1 .
Using the Boolean operators, +, * , , and create the output expression w1 involving the four input bits y1 , py , z1 , pz .
w1 = ___________________________________________________________________________
As partial support for your answer consider the following example.
Y = 1 0 0 0 1 1 0 0 Z = 0 0 0 1 1 1 0 1
Calculate py = _______________________________ calculate pz = _______________________________
In this example since y1 = 1 and z1 = 0 , then using your expression for w1 above
the value of w1 = _________________________________ which must have been the original value of x1 .
Explanation / Answer
consider the value of Y and z as follows.
Y = 1 0 0 0 1 1 0 0 Z = 0 0 0 1 1 1 0 1
In order to calculate py we need to place the parity bits of 4 to the given input value of Y .
Parity bit are placed based on powers of 2 calculation i.e
20 =1 that implies P1
21 =2 that implies P2
22 =4 that implies P4
23 =8 that implies P8
Therefore py = P1 P2 D1 P4 D2 D3 D4 P8 D5 D6 D7 D8
1 0 0 0 1 1 0 0
calcualation of P1 = XOR ( 3,5,7,9,11 bits)
=XOR ( 1,0,0,1,0)
= 0
calcualation of P2 = XOR ( 3,6,7,10,11 bits)
=XOR ( 1,0,0,1,0)
= 0
calcualation of P4 = XOR ( 5,6,7,12 bits)
=XOR ( 0,0,0,0)
= 0
calcualation of P8 = XOR ( 9,10,11,12 bits)
=XOR ( 1,1,0,0)
= 0
Now subsitutte the Parity bits
Therefore py = P1 P2 D1 P4 D2 D3 D4 P8 D5 D6 D7 D8
0 0 1 0 0 0 0 0 1 1 0 0
Parity calculator circuits PC1 = XOR ( 1 ,calcualation of P1 )
= 0
Parity calculator circuits PC2 = XOR ( 2 ,calcualation of P2 )
= 0
Parity calculator circuits PC4 = XOR ( 4 ,calcualation of P4 )
= 0
Parity calculator circuits PC8 = XOR ( 8 ,calcualation of P8 )
= 0
Since the above Parity calculcation circuit we found all the 4 bits of PC1 are 0 ,0 , 0 , 0 indicates that no error has occured
consider the value of Z as follows.
Z = 0 0 0 1 1 1 0 1
In order to calculate pz we need to place the parity bits of 4 to the given input value of Z .
Parity bit are placed based on powers of 2 calculation i.e
20 =1 that implies P1
21 =2 that implies P2
22 =4 that implies P4
23 =8 that implies P8
Therefore py = P1 P2 D1 P4 D2 D3 D4 P4 D5 D6 D7 D8
0 0 0 1 1 1 0 1
calcualation of P1 = XOR ( 3,5,7,9,11 bits)
=XOR ( 0,0,1,1,0)
= 0
calcualation of P2 = XOR ( 3,6,7,10,11 bits)
=XOR ( 0,0,1,1,0)
= 0
calcualation of P4 = XOR ( 5,6,7,12 bits)
=XOR ( 0,0,1,1)
= 0
calcualation of P8 = XOR ( 9,10,11,12 bits)
=XOR ( 1,1,0,1)
= 1
Now subsitutte the Parity bits
Therefore pz = P1 P2 D1 P4 D2 D3 P8 D4 D5 D6 D7 D8
0 0 0 0 0 0 1 1 1 1 0 1
Parity calculator circuits PC1 = XOR ( 1 ,calcualation of P1 )
= 0
Parity calculator circuits PC2 = XOR ( 2 ,calcualation of P2 )
= 0
Parity calculator circuits PC4 = XOR ( 4 ,calcualation of P4 )
= 0
Parity calculator circuits PC8 = XOR ( 8 ,calcualation of P8 )
= 0
Since the above Parity calculcation circuit we found all the 4 bits of PC2 are 0 ,0 , 0 , 0 indicates that no error has occured
since y1 = 1 and z1 = 0 then W1 is given by P1 P2 D1 P4 D2 D3 D4 P4 D5 D6 D7 D8 ,now replace the 4 paity bits with y1 , py , z1 , pz .
Therefore W1 = Y1 Py D1 Z1 D2 D3 D4 Pz D5 D6 D7 D8
1 0 0 0 0 0 0 0 0 0 0 0
let us have an error bit 1 make 1 to 0
then C1 = XOR ( 1 ,calcualation of P1 )
= 1
C2 = XOR ( 0 ,calcualation of P2 )
= 0
C4 = XOR ( 0 ,calcualation of P4 )
= 0
C8 = XOR ( 0 ,calcualation of P8 )
= 0
Hence the output C1C2C4C8 is 1 0 0 0 hence the error is at 1st bit
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