08/05/2023
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De Morgan's Theorem and XOR Gates.
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State and prove De Morgan theorem. Design x-or gates using nand gates
De Morgan's Theorem is a fundamental law of logic that relates to the negation of logical expressions. It states that the negation of a conjunction (AND) is equivalent to the disjunction (OR) of the negations of the terms, and that the negation of a disjunction (OR) is equivalent to the conjunction (AND) of the negations of the terms. In other words:
~(p ∧ q) ≡ ~p ∨ ~q (Negation of conjunction)
~(p ∨ q) ≡ ~p ∧ ~q (Negation of disjunction)
Proof of De Morgan's Theorem for Negation of Conjunction:
To prove ~(p ∧ q) ≡ ~p ∨ ~q, we can use a truth table as follows:
p q p ∧ q ~p ~q ~p ∨ ~q ~(p ∧ q)
0 0 0 1 1 1 1
0 1 0 1 0 1 1
1 0 0 0 1 1 1
1 1 1 0 0 0 0
From the truth table, we can see that the values of ~(p ∧ q) and ~p ∨ ~q are the same for all possible combinations of p and q. Hence, we can conclude that ~(p ∧ q) ≡ ~p ∨ ~q is true.
Proof of De Morgan's Theorem for Negation of Disjunction:
To prove ~(p ∨ q) ≡ ~p ∧ ~q, we can use a truth table as follows:
p q p ∨ q ~p ~q ~p ∧ ~q ~(p ∨ q)
0 0 0 1 1 1 1
0 1 1 1 0 0 0
1 0 1 0 1 0 0
1 1 1 0 0 0 0
From the truth table, we can see that the values of ~(p ∨ q) and ~p ∧ ~q are the same for all possible combinations of p and q. Hence, we can conclude that ~(p ∨ q) ≡ ~p ∧ ~q is true.
Designing XOR Gates using NAND Gates:
The XOR gate is a fundamental logic gate that outputs a 1 (TRUE) when the number of inputs that are 1 is odd, and outputs a 0 (FALSE) otherwise. It can be implemented using a combination of NAND gates as follows:
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Copy code
_________
--|NAND gate|--(XOR output)
| |
--|NAND gate|--(XOR input A)
| |
--|NAND gate|--(XOR input B)
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The truth table for the XOR gate is as follows:
A B XOR
0 0 0
0 1 1
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Define Garay Code and convert (11101)Garay code to binary
Garay Code, also known as the balanced ternary code, is a non-standard ternary code in which the digits can take on the values -1, 0, or 1. It was invented by Frank Gray in 1953 and is used in certain applications such as analog-to-digital converters and digital signal processing.
To convert (11101)Garay code to binary, we can use the following steps:
Write the Garay code as a sequence of binary values by adding 1 to the first digit and taking the difference between adjacent digits:
Garay code: 1 1 1 0 1
Binary values: 1 0 0 1 0
Starting with the first binary digit, add the values of the binary digits until the end to get the decimal value of the binary number:
Binary: 1 0 0 1 0
Value: 1 2 4 5 0
The decimal value of the binary number is 12, so the binary representation of the Garay code (11101) is (1100)2.
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Constuct Hamming code for BCD0110. Use even parity
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To construct a Hamming code for BCD0110 with even parity, we can follow these steps:
Write the binary representation of the BCD code: 0110
Determine the number of parity bits needed by finding the smallest number, p, that satisfies the following inequality: 2^p >= n + p + 1, where n is the number of data bits. In this case, n = 4, so we need p parity bits such that 2^p >= 4 + p + 1. The smallest p that satisfies this inequality is p = 3, so we need 3 parity bits.
Insert the parity bits at the positions that correspond to powers of 2: 1, 2, 4, 8, 16, etc. In this case, we need to insert the parity bits in positions 1, 2, and 4.
Calculate the parity bits based on the values of the data bits in the corresponding positions. For even parity, the parity bit is set to 1 if the sum of the data bits and the parity bit is even, and 0 otherwise.
The resulting Hamming code for BCD0110 with even parity is: 0111010
Here is a breakdown of the code:
Position 1 (parity bit): calculated based on data bits in positions 3, 5, 7
Data bit in position 3: 1, so add 1 to the parity bit
Data bit in position 5: 0, so no change to the parity bit
Data bit in position 7: 1, so add 1 to the parity bit
Parity bit value: 1 (even)
Position 2 (parity bit): calculated based on data bits in positions 3, 6, 7
Data bit in position 3: 1, so add 1 to the parity bit
Data bit in position 6: 0, so no change to the parity bit
Data bit in position 7: 1, so add 1 to the parity bit
Parity bit value: 1 (even)
Position 3 (data bit): 1
Position 4 (parity bit): calculated based on data bits in positions 5, 6, 7
Data bit in position 5: 0, so no change to the parity bit
Data bit in position 6: 0, so no change to the parity bit
Data bit in position 7: 1, so add 1 to the parity bit
Parity bit value: 0 (even)
Position 5 (data bit): 0
Position 6 (data bit): 1
Position 7 (parity bit): calculated based on data bits in positions 1, 2, 4, 5, 6, 7
Data bit in position 1 (parity): 1, so add 1 to the parity bit
Data bit in position 2 (parity): 1, so add 1 to the parity bit
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