Question

If a signal passing through a gate is inhibited by sending a low into one of the inputs, and the output is HIGH, the gate is a(n):

• A. NOR
• B. AND
• C. OR
• D. NAND

Hint:

A useful way to remember NAND and NOR gate properties: \[\begin{array}{rl} \bullet & \text{NAND: Any '0' input gives a '1' output. (Think: Not-AND. AND needs all 1s for a 1 output; NAND is the opposite).} \\ \bullet & \text{NOR: Any '1' input gives a '0' output. (Think: Not-OR. OR needs any 1 for a 1 output; NOR is the opposite).} \\ \end{array}\] The problem describes the NAND gate's property.

Answer 1

The Correct Option is D

Let's break down the description of the gate's behavior. 

1. "Inhibited by sending a low into one of the inputs": This means if any input is LOW (logic 0), the output is forced to a specific, fixed state, regardless of the other inputs. 

2. "the output is HIGH": This fixed state is HIGH (logic 1). 

So, we are looking for a logic gate where if any input is 0, the output is always 

1. Let's test the options for a 2-input gate with inputs A and B. We'll set B=0 and see what happens to the output. \[\begin{array}{rl} \bullet & \text{NOR: Output = \( (A+B)' \). If B=0, Output = \( (A+0)' = A' \). The output depends on A, so it's not inhibited to a fixed state.} \\ \bullet & \text{AND: Output = \( A \cdot B \). If B=0, Output = \( A \cdot 0 = 0 \) (LOW). The output is fixed, but it's LOW, not HIGH.} \\ \bullet & \text{OR: Output = \( A+B \). If B=0, Output = \( A+0 = A \). The output depends on A, so it's not inhibited.} \\ \bullet & \text{NAND: Output = \( (A \cdot B)' \). If B=0, Output = \( (A \cdot 0)' = (0)' = 1 \) (HIGH). The output is forced to a fixed state (HIGH) whenever any input is LOW.} \\ \end{array}\] This perfectly matches the description of a NAND gate.