Question

Please Answer these following Question very urgent "file attch"

Answer 1

Answer:

1. ∑ numbers=57

Overall ∑ is 57 hence.

And each number is less than 15, and hence they can be represented in 4 bits

However, we cannot write 57 in 4 bits.

However, 111001 is 57 in binary, and it is as well greater than 4 bits.

Hence, we are adding the two extreme left bits to right 4 bits as mentioned below.

Left are 11, right bits are 1001

1001

   11

1110 is the checksum

=14 and this is 14 bits.

2. The degree of given polynomial is 3, and hence,

The divisor is X3+1 =1001

Original data word=10110

Since the size of CRC is 3 bits, we should place 3 zeros to the right of the original data

And thus we get,  

10110000

         10100

1001√10110000

         1001

         00100000

         1001

         --------------

         0100   CRC

We send our data 10110 its reminder (CRC) will be added with it like 10110100

Receiver will receive 10110100 with same divisor 1001

         10100

1001√10110100

         1001

---------------------------

         01001

           1001

--------------------------------

           000

And since the last 3 bits are zero, and hence its error free.

3.

At 7 and 5 we have 1 and similarly in 2nd at 8, 4 and 0

  Bits    Polynomial

1010000        X7+X5+0

10001001 X8+X4+1

Explanation:

Please check answer.