I need help with this circuit Boolean algebra question. The question and the image for figure 1 is below.
Write the input/output truth table for the circuit in figure 1. $A$ and $B$ are inputs and $X$ is output.
| $A$ | $B$ | $A∨B$ | $¬B$ | $(A∨B)∧¬B$ | $X (~(A∨B)∧¬B)$ |
|---|---|---|---|---|---|
| 0 | 0 | 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 1 | 1 | 0 |
| 1 | 1 | 1 | 0 | 0 | 1 |
I am asked to build a truth table based on the figure (figure shown) that has 1 OR gate, 2 NOT gates, and 1 AND gate. Based on the figure and information given, I have set up my truth table. My question: Is the first row of my truth table setup correctly? $A, B, A∨B , ¬B, (A∨B)∧¬B, X (~(A∨B)∧¬B)$. I am unsure if I did it correctly since some of my other classmates have received an $X$ output of $1,1,1,1$
Your truth table is correct. However, I am assuming that $$X((A \lor B)\land \not B $$ You are taking its negotation right? As there is a "NOT" before the X.
However, checking the A and B inputs with the end result X, it yields 1 1 0 1, which is correct.