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yan
16 days ago
6

The following is a string of ASCII characters whose bit patterns have been converted into hexadecimal for compactness: 73 F4 E5

76 E5 4A EF 62 73.
Of the eight bits in each pair of digits, the leftmost is a parity bit. The remaining bits are the ASCII code.

(a) Convert the string to bit form and decode the ASCII.

(b) Determine the parity used: odd or even?
Computers and Technology
1 answer:
Rzqust [894]16 days ago
5 0

Answer:

a) Transforming each character into its binary equivalent:

73:

The digits are 7 and 3

7 in binary is: 111

3 in binary is: 11

Thus

73: 0_111_0011

Likewise

F4:

F stands for 15 and its binary is: 1111

4 in binary: 100

Consequently

F4:  1_111_0100

E5:

E corresponds to 14 and its binary form is: 1110

5 in binary: 101

Therefore

E5:  1_110_0101

76:

7 in binary: 111

6 in binary: 110

Hence

76:  0_111_0110

E5:

E has a binary representation of: 1110

5 in binary: 101

Consequently

E5:  1_110_0101

4A:

4 in binary: 100

A represents 10

A in binary form: 1010

Therefore

4A:  0_100_1010

EF:

E in binary is: 1110

F in binary is: 1111

Thus

EF: 1_110_1111

62:

6 in binary form: 110

2 in binary form: 10

Therefore

62:  0_110_0010

73:

The digits are 7 and 3

7 in binary is: 111

3 in binary is: 11

Thus

73: 0_111_0011

for 0_1110011: the decimal equivalent is: 115 which translates to s

for 1_1110100:  the decimal equivalent is: 116 which translates to t

for 1_1100101:  the decimal equivalent is: 101 which translates to e

for 0_1110110: the decimal equivalent is:  118 which translates to v

for 1_1100101:  the decimal equivalent is: 101 which translates to e

for 0_1001010: the decimal equivalent is:  74 which translates to j

for 1_1101111: the decimal equivalent is: 111 which translates to o

for 0_1100010:  the decimal equivalent is: 98 which translates to b

for 0_1110011: the decimal equivalent is:  115 which translates to s

Thus the decoded sequence is:  stevejobs

b) The parity being utilized is odd.

for 0_1110011:  There are 5 instances of 1s and the parity is 0 indicating it is odd.

for 1_1110100: There are 4 instances of 1s and the parity is 1 indicating it is odd.

Thus, we count the number of 1s and then verify if the parity is odd or even.

Likewise

for 1_1100101:  the parity is odd

for 0_1110110: the parity is odd

for 1_1100101:  the parity is odd

for 0_1001010: the parity is odd

for 1_1101111: the parity is odd

for 0_1100010: the parity is odd

for 0_1110011: the parity is odd

Therefore, the parity being used is odd.

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