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Unformatted text preview: the expression (b) is fed as input to the NOT gate, the output of the NOT gate is
A • B -------------- (c)
Now expressions (a) and (c) are fed as input to the second AND gate, whose output will
be (A + B) • (A•B)
Hence, C = (A + B) • (A-B), which is the desired logic equation for the output produced
by the given logic circuit.
Find the Boolean expression for the output of the logic circuit given below.
At point 1, the output of the OR gate is
A + B __________ (a)
At point 2, the output of the NOT gate is
C _________ (b)
At point 3, the output of the NOT gate is
D ________ (c)
The inputs to the AND gate at point 4 are (A + B), C, and D. Hence at point 4, the output
of the AND gate is
(A+B) • C • D ____ (d)
The inputs to the AND gate at point 5 are (A + B), C, and D. Hence at point 5, the output
of the AND gate is
(A + B) •C •D
Finally, the inputs to the OR gate at point 6 are (d) and (e). Hence at point 6, the output of
the OR gate is
(A+B) • C • D+(A + B) • C • D
Hence, E = (A + B) • C • D+(A + B) • C • D,
which is the required Boolean expression for the output of the given logic circuit. Converting Expressions to Logic Circuits
We have just now considered few examples that illustrate the method of deriving
Boolean expression for a given logic circuit. The reverse problem of constructing a logic circuit for a given Boolean expression is also not difficult. The three logic gates - AND,
OR, and NOT are said to be logically complete because any Boolean expression may be
realized using only these three gates. The method of constructing logic circuits for
Boolean expressions using only these three gates is illustrated below with the help of
The Universal NAND Gate
We have seen that AND, OR, and NOT gates are logically complete in the sense that any
Boolean function may be realized using these three gates. However, the NAND gate,
which was introduced in the previous section, is said to be an u...
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This document was uploaded on 04/07/2014.
- Spring '14