Analyzing Circuits: Fill in F Column Values in Truth Table

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The discussion focuses on evaluating the truth table for the logical expression $F=(xy) \oplus (x \oplus z)$, where $\oplus$ denotes the exclusive OR operation. The initial submission of truth values for F was incorrect, as the user miscalculated the intermediate values. The corrected truth table shows that F outputs 1 only for specific combinations of inputs, specifically when (1, 1, 1) and (0, 0, 1). This highlights the importance of accurately following logical operations in circuit analysis.

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shamieh
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Analyze the following circuit then fill in the F column values in the truth table for the circuit.

View attachment 1418

Just need someone to check my work.

My Answer:
  • X Y Z | F
  • 0 0 0 | 0
  • 0 0 1 | 0
  • 0 1 0 | 0
  • 0 1 1 | 0
  • 1 0 0 | 0
  • 1 0 1 | 0
  • 1 1 0 | 0
  • 1 1 1 | 1

(By the way not sure why my phone is rotating all my pictures, apologies in advance).
 

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Assuming you meant that $F=(xy) \oplus (x \oplus z),$ where the $\oplus$ symbol stands for "exclusive OR", then I get the following (building up the truth table from the beginning):
$$
\begin{array}{c|c|c|c|c|c}
x &y &z &xy &x \oplus z &(xy) \oplus(x \oplus z)\\ \hline
0 &0 &0 &0 &0 &0 \\
0 &0 &1 &0 &1 &1 \\
0 &1 &0 &0 &0 &0 \\
0 &1 &1 &0 &1 &1 \\
1 &0 &0 &0 &1 &1 \\
1 &0 &1 &0 &0 &0 \\
1 &1 &0 &1 &1 &0 \\
1 &1 &1 &1 &0 &1
\end{array}
$$
 
I'm a moron. I was evaluating xy then evaluating y ⊕ z. I see the problem now. Thanks for the help Ach!
 

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