Answer the following Algebraic equations
Question
Answer the following Algebraic equations
1. Prove the identity of the following Boolean equations, using basic rules of Boolean algebra. So i.e. AB +A B+AB=A+B
2. Simplify the following Boolean expression:
X Y + XY+XYZ
3. Reduce the following Boolean expression to a minimum number of literals: X Y(Z W+W) + ZW+ A9 4. And For the Boolean function E and F, as given in the following truth table:
X Y Z E F 0 0 0 1 0 0 0 1 0 1 0 1 0 1 1 0 1 1 0 0 1 0 0 1 0 1 0 1 0 1 1 1 0 1 0 1 1 1 0 1
4. Express E and F in sum-of item an algebraic form.
5. So Simplify the following Boolean expression using K-map
F(a,C) = AC + AB + ABC
6. And Simplify the following Boolean function by using K map
F(A,B,C,D) = /m(0,2,3,5,7,8,1 0,1 1,14,15)
Summary
We have given 6 different questions with algebraic equations. And we have to answer them.
Explanation
In the algebraic expression true is always represented by 1. And false is represented by 0.
So X will represent 1. And X’ will represent 0
So X and X’ are complements of each other.
1.
A B’ + A’ B’ +AB
= (A B’ + A B’) + A’ B’ + AB ….Using A+A=A
= (A B’ + AB) + (A B’ + A’ B’)
= A(B’ + B) + (A + A’)B …..Using X = X’=1
= A+ B’
2.
X’Y’ + X’Y + XYZ
=(X’Y’ + X’Y)+(X’Y + XYZ)
= X'(Y’Y)+ Y(X’+XZ)
= X’1 + Y (X’+X)(X’+Z)
= X’ + Y.1 (X’+Z)
= X’ + Y X’ + YZ
= X'(1+Y)+YZ
3.
X’Y(Z’W +W’) + Y(X’ ZW + X)
= X’Y (W’+Z’)(W’+W)+Y(X+X’)(X+ZW)
= X’Y(W’ +Z’).1 + Y(X+ ZW)
= Y [ (W’X’ + Z X’) + (X + ZW)]
= Y [ (W’X’ + X) + ZX’ +ZW ]
= Y [ (X + X’) (X +W’) + Z(X’ +W) ]
= Y [ X + W’ + ZX’ + ZW]
= Y [ (X + ZX’) + (W’ + ZW) ]
= Y (X+Z +W+Z)
= Y (X+W+Z)
= XY +WY +ZY
4.
E = Σ(M0, M2, M4, M6)
= ΣM(0,2,4,6)
= X’Y’Z’ + X’ Y Z’ + XYZ’
F= ΣM (1,2,5,7)
= X’Y’Z + X’YZ’ + XY’Z + XYZ
5.
F(A,B,C) = AC + AB + ABC
= AC(B+B’) + AB(C_C’) + ABC
= ABC + AB’C +ABC + ABC’ + ABC
= AB’C + ABC’ + ABC
= ΣM (5,6,7)
| B’C’ | B’C | BC | BC’ | |
| A’ | 0 | 0 | 0 | 0 |
| A | 0 | 1 | 1 | 1 |
F= AC +AB
6.
F(A,B,C,D) = Σ M (0,2,3,5,7,8,10,11,14,15)
| C’D’ | C’D | CD | CD’ | |
| A’B’ | 1 | 0 | 1 | 1 |
| A’B | 0 | 1 | 1 | 0 |
| AB | 0 | 0 | 1 | 1 |
| AB’ | 1 | 0 | 1 | 1 |
F= CD+ AC + B’D’ + A’BD
Also read, Randomly generate 50 integers within the range of 0 – 999.
