give detailed solutions of this question,for 12 th grade physics student---**Question Stem:**
The output (Y) of the given logic implementation is similar to the output of an/a ________ gate.
**Chart/Diagram Description:**
Type: Digital Logic Circuit Diagram
Main Elements:
- Inputs: Two inputs labeled A and B. Input A is connected to two different points in the circuit. Input B is connected to one point.
- Logic Gates:
- A 2-input NOR gate: Inputs are A and B. Output is connected to the left input of the final gate. The NOR gate symbol is a crescent shape with a circle (inversion bubble) at the output.
- A 2-input NAND gate: Inputs are A and B. Specifically, the top input of this NAND gate is connected to input B, and the bottom input is connected to the other connection point of input A. Output is connected to the upper input of the final gate. The NAND gate symbol is a D-shape with a circle (inversion bubble) at the output.
- A 2-input AND gate: Inputs are connected to the outputs of the NOR gate and the NAND gate. The output of the AND gate is the final output Y. The AND gate symbol is a D-shape.
- Output: Labeled Y. Connected to the output of the final AND gate.
- Connections: Lines connect inputs A and B to the inputs of the NOR and NAND gates. The outputs of the NOR and NAND gates are connected to the inputs of the AND gate. The output of the AND gate is the final output Y.
**Options:**
(1) AND
(2) NAND
(3) OR
(4) NOR
视频信息
答案文本
视频字幕
We have a digital logic circuit with inputs A and B. The circuit contains a NOR gate, a NAND gate, and a final AND gate. We need to determine which standard logic gate the output Y is equivalent to.
Let's analyze the Boolean expressions. The NOR gate output is X1 equals A plus B with a bar over it. The NAND gate output is X2 equals A dot B with a bar over it. The final AND gate combines these: Y equals X1 dot X2, which becomes the expression shown.
Now we apply De Morgan's Laws to simplify the expression. First law: A plus B with a bar equals A bar dot B bar. Second law: A dot B with a bar equals A bar plus B bar. Substituting these into our Y expression and using the distributive property, we get A bar dot B bar plus A bar dot B bar. This simplifies to just A bar dot B bar, which by De Morgan's Law in reverse equals A plus B with a bar over it. This is exactly the expression for a NOR gate!
Let's verify our result with a truth table. For a NOR gate, the output is 1 only when both inputs are 0, and 0 for all other combinations. When A equals 0 and B equals 0, A plus B equals 0, so Y equals 1. For all other input combinations, A plus B equals 1, so Y equals 0. This confirms that our circuit output Y behaves exactly like a NOR gate. Therefore, the answer is option 4: NOR.
To summarize what we've learned: We analyzed a complex digital logic circuit and used Boolean algebra to determine its equivalent gate. By applying De Morgan's Laws and simplifying the expressions, we found that the output Y equals A plus B with a bar over it, which is exactly the expression for a NOR gate. This demonstrates how complex circuits can often be simplified to basic logic gates.