Overview of Last Lecture!
• Boolean Functions!
• Examples!
• Boolean Functions Practical examples!
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Boolean Function Forms-Boolean Algebra and Digital Logic Design-Lecture Slides, Slides of Digital Logic Design and Programming
This course includes logic operators, gates, combinational and sequential circuits are studied along with their constituent elements comprising adders, decoders, encoders, multiplexers, as well as latches, flip-flops, counters and registers. This lecture includes: Boolean, Function, Forms, Gate, Implementation, Minimization, Algebraic, Manipulations, Complements, Transformed
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2011/2012
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Download Boolean Function Forms-Boolean Algebra and Digital Logic Design-Lecture Slides and more Slides Digital Logic Design and Programming in PDF only on Docsity!
Overview of Last Lecture!
- Boolean Functions!
- Examples!
- Boolean Functions Practical examples!
Today’s Lecture Outline!
- Gate Implementation!
- Minimization of function!
- Algebraic Manipulations!
- Complement of a Boolean function!
Gate Implementation (Examples)! A B C F 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 1
Gate Implementation (Examples)!
A B F 0 0 1 0 1 0 1 0 0 1 1 0
The following is an example of a non-minimized function:! F 2 = x’y’z + x’yz + xy’! Minimization of the F 2 ! !
Minimization of the F
!
The function can be minimized as follows: x’y’z + x’yz + xy’ = = x’z · (y’ + y) + xy’ by postulate: 4(a) = x’z · 1 + xy’ 5(a) = x’z + xy’ 2(b)
Algebraic Manipulation!
- By reducing the number of terms, the number of literals (single variable) or both in a Boolean function, it is possible to obtain a simpler circuit, as each term requires a gate and each variable within the term designates an input to the gate! ! F 1 = x’y’z + x’yz + xy’ contains 3 terms and 8 literals! F 2 = x’z + xy’ contains 2 terms and 4 literals.!
Example Manipulations!
- The following are some example
manipulations:!
- x(x’ + y) = ?! " x(x’ + y) = xx’ + xy = 0 + xy = xy! !
- x + x’y = ?! " x + x’y = (x + x’)(x + y) = 1(x + y) = x + y!
- (x + y)(x + y’) = ?! (x + y)(x + y’) = x + xy + xy’ + yy’ = x(1 + y + y’) = x!
- xy + x’z + yz = ?! " " " "= xy + x’z + yz(x + x’)! " " " "= xy + x’z + xyz + x’yz! " " " "= xy(1 + z) + x’z(1 + y)! " " " "= xy + x’z!
- (x + y)(x’ + z)(y + z) = ?! " " " " "= (x + y)(x’ + z)!
Complement of a Function (Example)!
If F 1 = A+B+C Then F 1 ’ =(A+B+C)' = (A+X)’ let B+C = X = A'X' by DeMorgan's = A'(B+C)' = A'(B'C') by DeMorgan's = A'B'C' associative
Complement of a Function (More Examples)!
(x'yz' + x'y'z)' = (x'yz')' (x‘y'z)' = (x+y'+z) (x+y+z') [x(y'z'+yz)]' = x' + ( y'z'+yz)' = x' + (y'z')' (yz)' = x' + (y+z) (y'+z') A simpler procedure take the dual of the function (interchanging AND and OR operators and 1’s and 0’s) and complement each literal. {DeMorgan’s Theorem}! x'yz' + x'y'z! "The dual of function is (x'+y+z') (x'+y'+z)! "Complement of each literal: (x+y'+z)(x+y+z')!