Introduction
Boolean functions may be practically implemented by using electronic gates. The following points are important to understand.
- Electronic gates require a power supply.
- Gate INPUTS are driven by voltages having two nominal values, e.g. 0V and 5V representing logic 0 and logic 1 respectively.
- The OUTPUT of a gate provides two nominal values of voltage only, e.g. 0V and 5V representing logic 0 and logic 1 respectively. In general, there is only one output to a logic gate except in some special cases.
- There is always a time delay between an input being applied and the output responding.
Truth Tables
Truth tables are used to help show the function of a logic gate. If you are unsure about truth tables and need guidance on how go about drawing them for logic circuits then check our post for truth table exercises
Logic gates
Digital systems are said to be constructed by using logic gates. These gates are the AND, OR, NOT, NAND, NOR, EXOR and EXNOR gates. The basic operations are described below with the aid of truth tables.
NAND and NOR Gates are called universal Gates, Other Gates can be made by using only NAND or NOR gates. Below given table shows that
Boolean Algebra
The most obvious way to simplify Boolean expressions is to manipulate them in the same way as normal algebraic expressions are manipulated. With regards to logic relations in digital forms, a set of rules for symbolic manipulation is needed in order to solve for the unknowns.
A set of rules formulated by the English mathematician George Boole describe certain proposition whose outcome would be either true or false. With regard to digital logic, these rules are used to describe circuits whose state can be either, 1 (true) or 0 (false).
Boolean Algebra Rules
| x+0=x |
| x+1=1 |
| x.0=0 |
| x.1=x |
| x+x=x |
| x.x=x |
| x+x'=1 |
| x.x'=0 |
| x''=x |
Commutative law
| A+B=B+A |
| A.B=B.A |
Associative law
| (A+B)+C=A+(B+C) |
| (A.B).C=A.(B.C) |
Distributive law
| A+(B+C)=(A+B)+(A+C) |
| A.(B.C)=(A.B)+(A.C) |
Redundance law
| A+A.B=A A+A.B+A.B.C=A |
| A'+A.B'+A.C=A'+B'+C |
Perfect Induction
Perfect induction employs a truth table, which describes the validity of the boolean entity for the possible value combinations of the boolean variables.
Prove the following using Perfect induction
x+xy=x
| X | Y | XY | X+XY |
| 0 | 0 | 0 | 0 |
| 0 | 1 | 0 | 0 |
| 1 | 0 | 0 | 1 |
| 1 | 1 | 1 | 1 |
x(x+y)=x
| X | Y | X+Y | X(X+Y) |
| 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 1 |
x+x'y=x+y
| X | Y | X' | X'Y | X+X'Y | X+Y |
| 0 | 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 | 1 | 1 |
x+(yz)=(x+y).(x+z)
| X | Y | Z | YZ | (X+Y) | (X+Z) | X+(YZ) | (X+Y).(X+Z) |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 |
| 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |
| 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
Prove the above using Boolean Rules.
x+xz=x
LHS=x+xz
=x(1+z)
=x.1
=x
=RHS
x'y+y'z+yz=x'y+z
LHS=x'y+y'z+yz
=x'y+z(y'+y)
=x'y+z.1
=x'y+z
=RHS
AC'D'+AB'D+ACD'=A(B'+D')
LHS=AC'D'+AB'D+ACD'
=A(C'D'+B'D+CD')
=A(B'D+C'D'+CD')
=A(B'D+D'(C'+C))
=A(B'D+D'.1)
=A(B'D+D')
=A(B'+D')
=RHS
A'B'C'+A'C+A'B'C'+A'C'=A'
LHS=A'B'C'+A'C+A'B'C'+A'C'
=A'(B'C'+C+B'C'+C')
=A'(B'C'+B'C'+(C+C'))
=A'(B'C'+B'C'+1)
=A'.1
=A'
=RHS
Using the boolean algebra, simplify the following expression.
A=x'yz'+x'yz+xy'z+xyz
=x'y(z'+z)+xz(y'+y)
=x'y.1+xz.1
=x'y+xz
x=AB+A(B+C)+B(B+C)
=AB+AB+AC+BB+BC
=AB+AC+BB+BC
=AB+AC+B+BC
=AC+B(A+1+C)
=AC+B
A=xy'+x'y+(x+y').x'y
=xy'+x'y+xx'y+x'yy'
=xy'+x'y+(xx')y + x'(yy')
=xy'+x'y+ 0.y+x'.0
=xy'+x'y
=x(y'+y)
=x.1
=x
A=x+y'+x'y+(x+y').x'y
=x+y'+x'y +xx'y+x'yy'
=x+y'+x'y+0.y+x'.o
=x+y'+x'y
=x+y+y'
=x+1
z=ABC(AB+C'(BC+AC))
=ABC(AB+BCC'+ACC')
=ABC(AB+B.0+A.0)
=ABC(AB)
=ABCAB
=AB.AB.C
=AB.C
=ABC
Y=(A+B')(A+C)
=AA+AB'+AC+B'C
=A+AB'+AC+B'C
=A(1+B'+C)+B'C
=A.1 +B'C
=A+B'C
Z=AB+ABC+ABCD+ABCDE
=AB(1+C+CD+CDE)
=AB.1
=AB
S=A'+A.C+B'C
S=A'+C+B'C
S=A'+C
Y=A.C+A.B.C+A.B.C'
Y=A.C + A.B(C+C')
Y=A.C + A.B.(1)
Y=A.C + A.B
De Morgan's Theorem
(x+y)'=x'.y'
(xy)'=x'+y'
Apply De Morgans theorem to the following expressions.
z=(A+B')'
=A'.B''
=A'B
Z=(AB+CD)'
=(AB)'.(CD)'
=(A'+B').(C'+D')
Z=((AB)'.(CD)')'
=(AB)''+(CD)''
=AB+CD
Apply De Morgan's theorems and simplify the given expressions.
z=((A+B').(C'+D))'
=(A+B')'+(C'+D)'
=A'.B''+C''.D'
=A'B+CD'
Z=((XY'.(W'+Y')')'
Z=(XY')'+(W'+Y')''
Z=X'+Y'' +(W'+Y')
Z=X'+Y+W'+Y'
Z=X'+W'+Y+Y'
Z=X'+W'+1
Z=1
Y=(A+(B.C')'+CD)'+BC''
=A'.(B.C')''.(CD)'+BC
=A'.BC'.(CD)'+BC
=A'.BC'.(C'+D')+BC
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