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The expression I came up with this circuit is A'B + A'CD + C, would the output change to AB' + AC'D' + C' since it is inverted? I'm assuming the D input compliments and cancels out? What would the logical expression actually be? I am confused about the inverter on the output.

NOT output

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inverting A'B + A'CD + C won't result in AB' + AC'D' + C'. Because \$\overline{(A+B)} = \overline{A}\ \overline{B} \ne \overline{A} + \overline{B}\$. See De Morgan's Laws for details.

Now coming to your circuit, the inputs for 3rd AND (inst5) gate are \$D\$, \$\overline{D}\$ and B hence the output of 3rd AND gate is \$D\overline{D}B = 0\$.

So the output of OR gate (inst6) will be \$\overline{A}B + \overline{A}CD\$. reducing which you will get \$\overline{A}(B+CD)\$.

The final output will be \$\overline{(\overline{A}(B+CD))} = A+\overline{(B+CD)} = A + \overline{B}(\overline{C}+\overline{D}) = A + \overline{B}\ \overline{C} + \overline{B}\ \overline{D} \$

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Once you have an expression, output at OR3 for example, you can follow De Morgan's Laws to determine what the inverted expression becomes.

Looks like I did the wrong expression in the original post:

A'B + A'CD + C = (A'B + A'CD + C)' = (A'B)'(A'CD)'(C)' = (A+B')((A+(CD)')C'

= (A+B')(A+C'+D')C'

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  • \$\begingroup\$ I see the output you have for inst5 is C? Wouldn't the output actually be B? \$\endgroup\$ – user46422 Jun 28 '14 at 13:45
  • \$\begingroup\$ @user46422 the output for inst5 will be 0. \$\endgroup\$ – nidhin Jun 28 '14 at 14:14
  • \$\begingroup\$ @user46422, I took your expression you evaluated and assumed it was the output of inst6. I then evaluated the inversion. Before the edit I mistakenly evaluated the inversion of your second expression. \$\endgroup\$ – sherrellbc Jun 28 '14 at 14:14

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