Boolean-Expression
Question 1 |
Which one of the following is NOT a valid identity?
(x + y) ⊕ z = x ⊕ (y + z) | |
(x ⊕ y) ⊕ z = x ⊕ (y ⊕ z) | |
x ⊕ y = x + y, if xy = 0 | |
x ⊕ y = (xy + x'y')' |
Question 1 Explanation:
Let x=1, y=1, z=0.
(x+y) ⊕ z = (1+1)⊕ 0 = 1 ⊕ 0 = 1
x ⊕ (y+z) = 1⊕(1+0) = 1 ⊕ 1 = 0
So,
(x+y) ⊕ z ≠ x ⊕ (y+z)
(x+y) ⊕ z = (1+1)⊕ 0 = 1 ⊕ 0 = 1
x ⊕ (y+z) = 1⊕(1+0) = 1 ⊕ 1 = 0
So,
(x+y) ⊕ z ≠ x ⊕ (y+z)
Question 2 |
The value of a1 + a1.b1 + b1 + a + 0 is :
a1 + b1 | |
b1 + a | |
1 | |
0 |
Question 2 Explanation:
a’ + a’b’ + b’ + a + 0
= a’(a+b’) + b’ + a + 0
= a’ + a + b’ (∵1+x=1)
= 1+b (∵ x’+x=1)
= 1 (∵ 1+x=1)
= a’(a+b’) + b’ + a + 0
= a’ + a + b’ (∵1+x=1)
= 1+b (∵ x’+x=1)
= 1 (∵ 1+x=1)
Question 3 |
According to boolean law: (A')' = ?
0 | |
A | |
(A')'
| |
1 |
Question 3 Explanation:
Complement of any variable even times will give the same variable. And complement of any variable odd times will give the complement of that variable.
Since in the question complement of variable is done 2 times which is even, so will give the same variable.
Since in the question complement of variable is done 2 times which is even, so will give the same variable.
Question 4 |
For following logic diagram which expression is true?
(AB) (AB.)'
| |
((A'B'),(AB.))'
| |
((A'B')+(AB.))' | |
(AB)'.AB |
Question 5 |
Which of the following is equivalent to the boolean expression given below?
A+A'.B+A.B'
A+B | |
A+B' | |
B+A' | |
B'+A'
|
Question 5 Explanation:
A + A’B + AB’
= (A + A’) (A + B) + AB’
= 1 ∙ (A + B) + AB’ (∴ A + A’ = 1)
= (A + B) + AB’
= A + (B + A) (B + B’) (∴ B + B’ = 1)
= A + (B + A) ∙ 1
= A + B + A
= A + B
= (A + A’) (A + B) + AB’
= 1 ∙ (A + B) + AB’ (∴ A + A’ = 1)
= (A + B) + AB’
= A + (B + A) (B + B’) (∴ B + B’ = 1)
= A + (B + A) ∙ 1
= A + B + A
= A + B
Question 6 |
Canonical SOP form of logic expression consists of only ____
Maxterms | |
Laterals
| |
Cell
| |
Minterms |
Question 6 Explanation:
There are 2 steps to derive the Canonical Sum of Products Form from its truth table.
1. A Minterm is a product (AND) term containing all input variables of the function in either true or complemented form. A variable appears in complemented form ~X if it is a 0 in the row of the truth-table, and as a true form X if it appears as a 1 in the row.
Examples
A=0, B=1, C=0 -> ~A*B*~C
A=1. B=0, C=0 -> A*~B*~C
2. The canonical form is obtained by taking the sum (OR) of the minterm of the rows where a 1 appears in the output.
1. A Minterm is a product (AND) term containing all input variables of the function in either true or complemented form. A variable appears in complemented form ~X if it is a 0 in the row of the truth-table, and as a true form X if it appears as a 1 in the row.
Examples
A=0, B=1, C=0 -> ~A*B*~C
A=1. B=0, C=0 -> A*~B*~C
2. The canonical form is obtained by taking the sum (OR) of the minterm of the rows where a 1 appears in the output.
Question 7 |
The simplified SOP(Sum of Products) form of the boolean expression (P+Q’+R’) (P+Q’+R) (P+Q+R’) is
(P’ Q + R) | |
(P + Q’. R’) | |
(P’.Q + R) | |
(P. Q + R) |
Question 7 Explanation:
Question 8 |
A | |
B | |
C | |
A, B and C |
Question 8 Explanation:
There are 8 questions to complete.