 # Logic Functions Logical functions can be expressed in several ways: ▫ Truth table ▫ Logical expressions ▫ Graphical form prepared by:eng.Rula Amjed.

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Logic Functions Logical functions can be expressed in several ways: ▫ Truth table ▫ Logical expressions ▫ Graphical form prepared by:eng.Rula Amjed

3-input majority function ABCF 0000 0010 0100 0111 1000 1011 1101 1111 Logical expression form F = A B + B C + A C prepared by:eng.Rula Amjed

Logical Equivalence All three circuits implement F = A B function prepared by:eng.Rula Amjed

 One of the primary requirements when dealing with digital circuits is to find ways to make them as simple as possible. This requires that complex logical expressions be reduced to simpler expressions that produce the same results under all possible conditions. The simpler expression can then be implemented with a smaller, simpler circuit, which in turn saves the price of the unnecessary gates, reduces the number of gates needed, and reduces the power and the amount of space required by those gates.  One tool to reduce logical expressions is the mathematics of logical expressions, introduced by George Boole in 1854 and known today as Boolean Algebra. The rules of Boolean Algebra are simple and straight-forward, and can be applied to any logical expression. The resulting reduced expression can then be readily tested with a Truth Table Boolean Algebra prepared by:eng.Rula Amjed

Boolean Algebra

Boolean Algebra (cont.) prepared by:eng.Rula Amjed

To prove a distributive law prepared by:eng.Rula Amjed

Example :draw the flowing function: F1 = X + Y‘Z Solution : prepared by:eng.Rula Amjed

Simplify the following Boolean functions 1)x(x' + y) 2)x + x'y - 3)(x + y).(x + y‘) Solution : prepared by:eng.Rula Amjed

Complement of a Function (A + B + C + D )' = A'B'C'D' (ABCD)' = A' + B' +C' + D' Example1: find the complement of the following function F1 = x'yz' + x'y'z Solution: F1 = (x'yz' + x'y'z)' = (x'yz')'. (x'y'z)' = (x+y'+z). (x+y+z') prepared by:eng.Rula Amjed

Example2: find the complement of the following functions F1 = (x+y'+z').(x'+y+z).(x'+y'+z') Solution: F1 =((x+y'+z').(x'+y+z).(x'+y'+z'))' =(x+y'+z')' + (x'+y+z)' + (x'+y'+z')' = (x'yz) + (xy'z') + (xyz) prepared by:eng.Rula Amjed

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