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Boolean Expression to 2-input logic gates only

power13

Oct 26, 2018
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Hello,

I have a Boolean expression and I want to convert it to circuit with only 2-input gates and NOTs.

x'y'z + x'yz' + xy'k' + xyk

Is there any different approach than:

(x'y')z + (x'y)z' + (xy')k' + (xy)k



Thought like this:
and3.png
 

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(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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You could also try using de Morgan's theorem and see if it simplifies anything.

Alternately, just implement a three input gate with two two input gates and invert the inputs as required.

Have you created a truth table for it?
 

power13

Oct 26, 2018
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You could also try using de Morgan's theorem and see if it simplifies anything.

Alternately, just implement a three input gate with two two input gates and invert the inputs as required.

Have you created a truth table for it?

My expression is simplified by Karnaugh map.

Unfortunately I think de Morgan's theorem doesn't helps here.
 

power13

Oct 26, 2018
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The result of K-map is this expression:

x'y'z + x'yz' + xy'k' + xyk

And now I want to draw the circuit with only 2input logic gates and NOTs.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Aaah! I thought that was what you had modeled in the K-map.

Why not start with something like this:

x'y'z + x'yz' + xy'k' + xyk
x'(y'z + yz') + x(y'k' + yk)
 

power13

Oct 26, 2018
10
Joined
Oct 26, 2018
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Aaah! I thought that was what you had modeled in the K-map.

Why not start with something like this:

x'y'z + x'yz' + xy'k' + xyk
x'(y'z + yz') + x(y'k' + yk)

Thank you for your help.

After some reading of Boolean Algebra, I ended up with this expression:

( (NOT X) AND (Y XOR Z) ) OR ( X AND (NOT (Y XOR K)) )

I think it is a good expression with 2 xor, 2 and, 2 not's.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Jan 21, 2010
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Just be aware that a 2 input XOR gate may not be an allowable gate. Check this out, if you can.
 
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