Recent content by i_madini

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    3 input XOR gate using NAND gate only ( Logic )

    Thank you, Sir. I really appreciate it .. And that little secret is so helpful to switch between gates .. But what I'm interesting in is to find the output at every single NAND gate .. Starting from X1, X2, X3, X4, X5, X6, X7, and Y. These X's is shown in attachment. I'm...
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    How to simplify this boolean algebra (logic)

    Let's assume we have : F = A'+AB = A'(B+B') + AB # B+B'= 1, will not affect the equation. = A'B + (A'B' + AB) # Cancel each other. = A'B By the same way :)
  3. I

    3 input XOR gate using NAND gate only ( Logic )

    We can use NAND gate only to get ( XOR gate ) of 2 input ( A and B ) : By using (4) NAND gate : The output of 1'st NAND : (AB)' The output of 2'nd NAND : ((AB)'.A)' = (A'B) The output of 3'rd NAND : ((AB)'.B)' = (AB') The output of the hole circuit will be: ((A'B).(AB'))' = AB'...
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    How to simplify this boolean algebra (logic)

    h = (abc)'(abd)'(acd)'(bcd)' = (a'+b'+c')(a'+b'+d')(a'+c'+d')(b'+c'+d') = (a'+b'+c'+d)(a'+b'+c'+d')(a'+b'+c+d')(a'+b'+c'+d')(a'+b+c'+d')(a'+b'+c'+d')(a+b'+c'+d') (a'+b'+c'+d') =(a'+b'+c'+d)(a'+b'+c+d')(a'+b+c'+d')(a+b'+c'+d')(a'+b'+c'+d') I hope it is correct..