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Question
From the logic gate diagram given below, derive the Boolean expression for (1), (2), and Q. Reduce the derived expression.
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Solution
This is the output of a 3-input AND gate with inputs A, B, and C.
(1) = A·B·C
This is the output of a 2-input AND gate.
The top input is A.
The bottom input is the output of an OR gate receiving `barB` and `bar C` (from NOT gates).
(2) = A·(`barB + barC`)
This is the output of the final OR gate.
Q = (1) + (2) = ABC + A(`barB + barC`)
Reduction Steps:
Start with your expression for Q:
Q = ABC + A(`barB + barC`)
Factor out A from the whole expression
Q = A(BC + (`barB + barC`)
Apply De Morgan's Law (`barB + barC = bar(BC)`)
Q = A(BC + `bar(BC)`)
Apply the Complement Law (X + `barX` = 1)
Since BC + `bar(BC)` = 1, the expression becomes
Q = A(1)
Q = A
