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?

Question

Computers & Technology

Catherine Gallagher

9 May, 09:44

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?

+1

Answers (1)

Know the Answer?

August Hardin · Answer 1

a) Converting each string to bit form:

73:

The string is 7 and 3

7 in bit form is : 111

3 in bit form is: 11

So

73: 0_111_0011

Similarly

F4:

F is 15 and its bit form is: 1111

4 in bit form: 100

So

F4: 1_111_0100

E5:

E is 14 and its bit form is: 1110

5 in bit form: 101

So

E5: 1_110_0101

76:

7 in bit form: 111

6 in bit form: 110

So

76: 0_111_0110

E5:

E is 14 and its bit form is: 1110

5 in bit form: 101

So

E5: 1_110_0101

4A:

4 in bit form: 100

A is 10

A in bit form: 1010

So

4A: 0_100_1010

EF:

E in bit form is: 1110

F in bit form is: 1111

So

EF: 1_110_1111

62:

6 in bit form: 110

2 in bit form: 10

So

62: 0_110_0010

73:

The string is 7 and 3

7 in bit form is : 111

3 in bit form is: 11

So

73: 0_111_0011

for 0_1110011: the decimal is: 115 which is equal to s

for 1_1110100: the decimal is: 116 which is equal to t

for 1_1100101: the decimal is: 101 which is equal to e

for 0_1110110: the decimal is: 118 which is equal to v

for 1_1100101: the decimal is: 101 which is equal to e

for 0_1001010: the decimal is: 74 which is equal to j

for 1_1101111: the decimal is: 111 which is equal to o

for 0_1100010: the decimal is: 98 which is equal to b

for 0_1110011: the decimal is: 115 which is equal to s

So the decoded character is: stevejobs

b) The parity used is odd.

for 0_1110011: There are 5 number of 1s and parity is 0 so the parity is odd.

for 1_1110100: There are 4 number of 1s and parity is 1 so the parity is odd.

So here we are counting the number of 1s and then checking if the parity is even or odd.

Similarly

for 1_1100101: parity is odd

for 0_1110110: parity is odd

for 1_1100101: parity is odd

for 0_1001010: parity is odd

for 1_1101111: parity is odd

for 0_1100010: parity is odd

for 0_1110011: parity is odd

Hence the parity used is odd.