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enter image description hereI've been trying to solve this circuit problem(and understand it frankly), and I wanted to double check my thought process with the community helpfully.

After running the circuit out, I have $A+ \bar A B$ for the top half, with the bottom half as $\bar A C$, the issue I have is the last gate is an exclusive-or, so combined(my attempt) was the result $A + \bar A B+ \bar A C$... is this right?

Given this, how do I begin to convert that to a truth table with the exclusive. If I start with for ex, A= 1, B=0, C=0, I end up with 1 as the output?

Following that example, I came up with the outputs as all 1's except for the values A=0, B=0, C=0...

MJD
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tim belk
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1 Answers1

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Your suggestion of $$A + \bar A B+ \bar A C$$ would be right if the final gate were a regular inclusive or ($+$), but it isn't; it is an exclusive or ($\oplus$). These are different. I've written the truth tables for both of them below:

$$\begin{array}{cc|cc} A & B & A+B & A\oplus B \\ \hline 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & \color{maroon}{0} \\ \end{array}$$

Because you were treating that final gate as $+$ instead of as $\oplus$, you didn't get the right truth table.

Note that $A+\bar AB$ can be simplified. Do you know the Karnaugh map technique? It is particularly simple in this case.

MJD
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  • I do not, I understand the output you posted on the exc/or being false when equal(1,1)=0, but does that relate to the boolean terms? – tim belk Aug 28 '14 at 17:56
  • If you don't know that Karnaugh map technique, just make a truth table for $A+\bar AB$ and see if you recognize it as being the same as the truth table of a simpler expression. I don't understand what you are asking when you say “does that relate to the boolean terms”. – MJD Aug 28 '14 at 17:58
  • So A+ A'B simplified becomes A+B right? Which gives the final as A+B + A'C ? – tim belk Aug 28 '14 at 18:26
  • The final gate is $\oplus$, not $+$. $+$ means or, but the final gate is exclusive or, which is different. – MJD Aug 28 '14 at 19:32
  • so A+B (exc or) A'C, then that makes the output different where if it's equal(0,0,0) or (1,1,1) its the reverse right? – tim belk Aug 28 '14 at 21:28