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I'm working through my homework and there's one question that I absolutely cannot figure out.

I'm supposed to implement X = ABC'+AD +A'CD'+A'E'F' With the following constraints:

  • All gates must be 2-input
  • Each level of the circuit must alternate between AND and OR gates
  • The final level must be a single AND gate

For a previous question I had to implement this following the same constraints except that the last gate had to be an OR gate. For whatever reason, that was super easy. I manipulated the equation into X = A(BC'+D)+A'(C'D+E'F'), but I can't figure this out, my final gate is always an OR gate.

How do I approach this?

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    \$\begingroup\$ You're going to need some sort of inverter or NAND or NOR seing as you have ' inputs. \$\endgroup\$
    – Transistor
    Commented Oct 4, 2017 at 18:21

2 Answers 2

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I would not normally answer a homework question, but it sounds like you have done most of it already.

FYI: Your simplification does not match your original equation.

A'CD' <> A(C'D

I'm assuming that is a typo, in the simplification and the original formula is correct.

Assuming you don't need invertors by having the inverse values as inputs, and if you can get to the final gate with an OR, then perhaps just add a final AND as a buffer.

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ I've a hunch that this trivial addition of a gate at the end to meet the requirements is likely to result in the addition of another rule banning the idea. But good catch on the error between the original equation and the so-called "manipulated" one! \$\endgroup\$
    – jonk
    Commented Oct 4, 2017 at 19:41
  • \$\begingroup\$ @jonk :) LOL could be. Or it's a simple... make the student think with a "duh" example exercise. I swear they did that stuff on purpose. \$\endgroup\$
    – Trevor_G
    Commented Oct 4, 2017 at 19:44
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    \$\begingroup\$ I suppose it depends on the level of the course. So could be. (But it is like a joke that's only funny once.) \$\endgroup\$
    – jonk
    Commented Oct 4, 2017 at 19:47
  • \$\begingroup\$ @jonk ya, my prof regularly put little quirks like that in to make you question your own answer. I figured he knew that was part and parcel of being a good engineer. \$\endgroup\$
    – Trevor_G
    Commented Oct 4, 2017 at 19:49
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    \$\begingroup\$ I recall a large project where the assignment included a rather large notebook filled with project details. It was a real situation where client data had been meticulously documented and captured. Our job was to develop our own complete solution. Buried within the volumes of documentation was the phrase, "If you read this, be the first to tell the teacher and you will receive a gold coin." Still have the coin. \$\endgroup\$
    – jonk
    Commented Oct 4, 2017 at 19:57
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OK So you did not like the my cheater answer...

Other method is to convert the current SOP terms into POS terms.

Looking at the SOP terms

\$X = A(BC'+D)+A'(CD'+E'F')\$

\$=> X = A.x + A'.y\$
\$=> X = (A + y).(A' + x)\$ : Equation 1

where :
\$x = B.C' + D\$
\$=>x = (B + D).(C' + D)\$

and:

\$y = (C.D' + E'.F')\$
\$=>y = (C + (E'.F')).(D' + (E'.F'))\$

Substituting x and y back into Equation 1, final POS logic is:

\$=> X = (A + ((C + (E'.F')).(D' + (E'.F')))).(A' + ((B + D).(C' + D)))\$

schematic

simulate this circuit – Schematic created using CircuitLab

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