Tuesday, January 25, 2011

Make an exclusive-OR Gate by using NAND's

A NAND B = ¬(AΛB)
A XOR B = ¬AΛB + AΛ¬B

We have to make XOR equation has the form of NAND.
We will use De Morgan's Theorem: ¬AV¬B = ¬(AΛB)
¬AΛ¬B = ¬(AVB)

¬AΛB V AΛ¬B = ¬ (AV¬B) V ¬ (¬AVB)
= ¬¬ ( ¬ (AV¬B) V ¬ (¬AVB) )
= ¬ ( (AV¬B) Λ (¬AVB) )
= ¬ ( (AΛB) V (¬AΛ¬Β) )
= ¬(AΛB) Λ ¬(¬AΛ¬Β)
= ¬(AΛB) Λ (AVΒ)
= (¬(AΛB)ΛΑ) V (¬(AΛB)ΛB)
= ¬¬((¬(AΛB)ΛΑ) V (¬(AΛB)ΛB))
= ¬( ¬ ( ¬(AΛB) Λ Α ) Λ ¬( ¬(AΛB) Λ B ) )



Now we have a NAND "¬(AΛB)" connected to a NAND with A as second
input, connected to a NAND that has one input from the first NAND and B, and both are connected to a NAND.
Note: V = OR, ¬ = NOT, Λ = AND


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