Understanding the XOR Expression \(Y = P \oplus Q \oplus R\)

The given expression is \(Y = P \oplus Q \oplus R\). Here, \(P\), \(Q\), and \(R\) are input variables, which in digital logic, can only take binary values: 0 or 1. The symbol \(\oplus\) represents the Exclusive OR (XOR) operation.

The XOR operation is a fundamental digital logic operation. For two inputs, the XOR output is 1 if the inputs are different, and 0 if they are the same.

• \(0 \oplus 0 = 0\)
• \(0 \oplus 1 = 1\)
• \(1 \oplus 0 = 1\)
• \(1 \oplus 1 = 0\)

A key property of the XOR operation, especially when extended to multiple inputs (like \(P \oplus Q \oplus R\)), is its relationship with the number of '1' inputs. A multi-input XOR output is 1 if there is an odd number of 1s among the inputs, and 0 if there is an even number of 1s among the inputs.

Analyzing the Output Y for Logic 0

For the expression \(Y = P \oplus Q \oplus R\), the output \(Y\) will be logic 0 if and only if an even number of the input variables (\(P\), \(Q\), \(R\)) are 1. Let's verify this by looking at all possible input combinations for \(P\), \(Q\), and \(R\).

The table clearly shows that the output \(Y\) is logic 0 when the number of 1s among the inputs \(P\), \(Q\), and \(R\) is either 0 or 2. In both these cases, the number of 1s is even.

Evaluating the Provided Options

Let's check which of the given options correctly describes the condition for \(Y\) being logic 0 based on our analysis:

• (A) an odd number of input variables are 1: From our analysis, an odd number of 1s results in \(Y = 1\). This option does not lead to \(Y = 0\).
• (B) an even number of input variables are 1: From our analysis and the truth table, an even number of 1s (0 or 2) results in \(Y = 0\). This option correctly states the condition for \(Y = 0\).
• (C) an odd number of input variables are 0: For three inputs, if the number of 0s is odd (1 or 3), then the number of 1s must be even (2 or 0, since total inputs is 3). As we've established, an even number of 1s results in \(Y=0\). So, this condition also results in \(Y=0\). However, option (B) directly corresponds to the defining property of the XOR output being 0 (even number of 1s).
• (D) an even number of input variable are 0: For three inputs, if the number of 0s is even (0 or 2), then the number of 1s must be odd (3 or 1). An odd number of 1s results in \(Y = 1\). This option does not lead to \(Y = 0\).
• (E) an odd number of input variable between 0 and 1: This phrasing is incorrect for binary inputs which are strictly 0 or 1. It is not a meaningful condition in this context.

Conclusion

Based on the truth table and the property of the XOR operation, the output \(Y = P \oplus Q \oplus R\) is logic 0 if and only if an even number of the input variables \(P\), \(Q\), and \(R\) are 1. This condition is precisely stated in option (B).

Revision Table: Condition for Y = 0

Additional Information: XOR, Parity, and Error Detection

The XOR operation is closely related to the concept of parity. Parity is used in digital communications and data storage to detect errors.

• Even Parity: In an even parity system, a parity bit is added to a group of data bits such that the total number of 1s (including the parity bit) is always even. A multi-input XOR gate can generate the parity bit. The output of the XOR of the data bits is the parity bit needed to make the total number of 1s even.
• Using XOR for Parity Check: If you transmit data bits along with an even parity bit, the receiver can XOR all the bits (data bits + parity bit) together. If no error occurred during transmission, the total number of 1s received should be even, and the XOR sum will be 0. If a single bit error occurs (a 0 flips to a 1 or a 1 flips to a 0), the total number of 1s becomes odd, and the XOR sum will be 1, indicating an error.

In our problem, \(Y = P \oplus Q \oplus R\) essentially performs an even parity check on the three inputs \(P\), \(Q\), and \(R\). The output \(Y\) being 0 signifies that the inputs \(P, Q, R\) have even parity (an even number of 1s).