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Showing posts with label Boolean Algebra. Show all posts
Showing posts with label Boolean Algebra. Show all posts

## Simplifies Boolean Algebra through truth table or Vice-Versa using POS

1)Derived truth table through function(Boolean Expression) using POS(Product of Sum):-

Simple Rules:- In POS 1=(')like A' and 0 like A,In POS we grasp 0 instead of 1.

Q.1)Z(ABC)=(A+B+C').(A+B'+C).(A'+B+C')

Sol.):-

ABCZ
0001
0010
0100
0111
1001
1010
1100
1110

Q.2)Z(ABC)=(A+B+C).(A+B'+C).(A'+B+C').(A'+B'+C).(A'+B'+C')

Sol.):-

ABCZ
0000
0011
0100
0111
1001
1010
1100
1110

2)Derived function through truth table using POS:-

Q.1):-

ABCZ
0001
0010
0101
0111
1000
1011
1101
1110

Sol.) (A+B+C').(A'+B+C).(A'+B'+C')

Q.2):-

ABCZ
0001
0010
0100
0111
1001
1010
1100
1110

Sol.) (A+B+C').(A+B'+C).(A'+B+C')

## Simplifies Boolean Algebra through truth table or Vice-Versa using SOP

1)Derived truth table through function using SOP(Sum Of Product):-

Simple Rules:- In SOP 0=(')like A' and 1 like A,In SOP we grasp 1 instead of o.

Q.1)Z(ABC)=ฮฃ(3,5,6,7)

Sol.)Z(ABC)=A'BC+AB'C+ABC'+ABC

A B C Z
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 1
1 1 0 1
1 1 1 1

Q.2)Z(ABC)=ฮฃ(0,1,5,7)

Sol.)Z(ABC)=A'B'C'+A'B'C+AB'C+ABC

ABCZ
0001
0011
0100
0110
1000
1011
1100
1111

2)Derived function through truth table using SOP:-

Q.1):

ABCZ
0001
0011
0100
0110
1000
1011
1100
1111

Sol.)Z(ABC)=A'B'C'+A'B'C+AB'C+ABC

Q.2):-

ABCZ
0000
0010
0100
0111
1000
1011
1101
1111

Sol.) Z(ABC)=A'BC+AB'C+ABC'+ABC

## Simplification of Boolean Expressions

Q)Simplify the below Boolean expressions

1) x+(x'y)
2)xy+xy'

Ans 1) x+(x'y)
=(x+x')(x+y)
=1.(x+y)
=(x+y)

Ans 2)xy+xy'
=x(y+y')
=x.1
=x

This above examples are quite simple,so now we discuss some good/quite tough questions.

Q)Simplify the below Boolean expressions

1)x'y'z'+x'yz'+xy'z'+xyz'
2)a[b+c(ab+ac)']
3)(A+B)(A+B')(A'+B)

Ans 1)x'y'z'+x'yz'+xy'z'+xyz'
=x'z'(y'+y)+xz'(y'+y)
=x'z'1+x.z'.1
=z'.1
=z'

Ans 2)a[b+c(ab+ac)'
=a[b+c(ab+ac)']
=a[b+c(b+c(ab)'.(ac)']
=a[b+c(a'a'+a'c'+a'b'+b'c')]
=a[b+c(a'(1+c')+a'b'+b'c')]
=a[b+c(a'.1+a'b'+b'c')]
=a[b+c(a'(1+b')+b'c')]
=a[b+a'c+a'b'c+0]
=ab+aa'c+aa'b'c
=ab+0+0
=ab

Ans 3)(A+B)(A+B')(A'+B)
=(A.A+AB'+AB+B.B')(A'+B)
=(A+AB'(B'+B)+0)(A'+B)
=(A+A+B)(A'+B)
=AA'+AB+AA'+AB
=AB+AB
=AB

## Laws of Boolean Algebra

Now I discuss the different-2 laws of Boolean Algebra.

Law 1:-AND(Multiple law)

a)x.0=0
b)x.1=x
c)x.x=x
d)x.x'=0

a)x+0=0
b)x+x=x
c)x+1=1
d)x+x'=1

Law 3:-NOT(Negative law)

a)A'+A'=1
b)A.A'=0
c)A''=A

Law 4:-Commutative Law

a)A+B=B+A
b)A.B=B.A

Law 5:-Associative Law

a)(A+B)+C=A+(B+C)
b)(A.B).C=A.(BC)

Law 6:-Distributive Law

a)A.(B+C)=A.B+A.C
b)A+(B.C)=(A+B).(A+C)

Law 7:-Absorption Law:-A+AB=A

Law 8:-Auxiliary Law

a)A+A'B=A+B
b)A(A'+B)=AB

Law 9:-Dem organ's Law

a)(A+B)'=A'.B'
b)(A'.B')=(A+B)'