What is the result of AND’ing the binary value with the mask below? 01100111 bin

Question

What is the result of AND’ing the binary value with the mask below?

01100111 binary value

11110000 mask

What is the result of OR’ing the binary value with the mask below?

10110101 binary value

10101010 mask

What is the result of OR’ing the binary value with the mask below?

01100111 binary value

00001010 mask

What is the result of AND’ing the binary value with the mask below?

01100111 binary value

00001111 mask

Assume even parity is being used when transmitting 8-bit bytes. What is the value of the parity bit (either 0 or 1) for the byte below, and explain how you arrived at your answer?

01100011

Assume odd parity is being used when transmitting 8-bit bytes. What is the value of the parity bit (either 0 or 1) for the byte below, and explain how you arrived at your answer?

01100011

Explain Hamming Code, and calculate the 4-bit Hamming Code for the 8-bit word: 00111001.

Explanation / Answer

What is the result of AND’ing the binary value with the mask below?

01100111 binary value

11110000 mask

Ans: 01100000

What is the result of OR’ing the binary value with the mask below?

10110101 binary value

10101010 mask

Ans: 10111111

What is the result of OR’ing the binary value with the mask below?

01100111 binary value

00001010 mask

Ans: 01101111

What is the result of AND’ing the binary value with the mask below?

01100111 binary value

00001111 mask

Ans: 00000111

Assume even parity is being used when transmitting 8-bit bytes. What is the value of the parity bit (either 0 or 1) for the byte below, and explain how you arrived at your answer?

01100011

Ans: As per even parity rule, if the number of 1 present any binary number is even the parity bit is 0 and if number of 1 present in a binary number is odd then the parity bit is 1. The count of 1s in binary number 01100011 is 4 which is even. So, the parity bit would be 0. We need to add the parity bit at the end of binary number. So the binary number with parity bit is 011000110. Now, we can discard leading 0s, hence the 8 bit binary number with parity bit would be 11000110

Assume odd parity is being used when transmitting 8-bit bytes. What is the value of the parity bit (either 0 or 1) for the byte below, and explain how you arrived at your answer?

01100011

Ans: As per odd parity rule, if the number of 1 present any binary number is even the parity bit is 1 and if number of 1 present in a binary number is odd then the parity bit is 0. The count of 1s in binary number 01100011 is 4 which is even. So, the parity bit would be 1. We need to add the parity bit at the end of binary number. So the binary number with parity bit is 011000111. Now, we can discard leading 0s, hence the 8 bit binary number with parity bit would be 11000110

Explain Hamming Code, and calculate the 4-bit Hamming Code for the 8-bit word: 00111001.

Here is an excellent article about Hamming code. It's contains a nice graphical explanation of hamming code. Requesting you to read the article.

http://slopjong.de/2011/06/20/how-does-channel-coding-work-using-hamming-codes/

A byte of data: 00111001

Create the data word, leaving spaces for the parity bits: ? ? 0 ? 0 1 1 ? 1 0 0 1

Let's name the parity bits positions for better understanding: A B 0 C 0 1 1 D 1 0 0 1

For Position A, Start from A and traverse to the right, consider 1 bit then skip 1 bit. So we have:: ?, 0, 0, 1, 1, 0

We need to count the number of 1s encountered in considered bit which is 2. It's a even number. So, need to write 0 at position A. So now we have: 0 ? 0 ? 0 1 1 ? 1 0 0 1

For Position B, Start from B and traverse to the right, consider 2 bit then skip 2 bit. So we have: ?, 0, 0, 1, 1, 0

We need to count the number of 1s encountered in considered bit which is 2. It's a even number. So, need to write 0 at position B. So now we have: 0 0 0 ? 0 1 1 ? 1 0 0 1

For Position C, Start from C and traverse to the right, consider 4 bit then skip 4 bit. So we have: ?, 0, 1, 1, 1

We need to count the number of 1s encountered in considered bit which is 3. It's a odd number. So, need to write 1 at position C. So now we have: 0 0 0 1 0 1 1 ? 1 0 0 1

For Position D, Start from D and traverse to the right, consider 4 bit then skip 4 bit. So we have: ?, 1, 0, 0, 1

We need to count the number of 1s encountered in considered bit which is 2. It's a even number. So, need to write 0 at position C. So now we have: 0 0 0 1 0 1 1 0 1 0 0 1

4 bit humming code is : 0010

The data with humming code is: 000101101001

Hope it helps. :-)

Please feel free to comment you're having difficulties to understand any of above, I'll explain further.

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