Test: Boolean logic - Grade 11 MCQ

Given expression: 

• NOR, 
• NAND,
• {OR and NOT i.e It behaves like NOR Gate } and
• {AND, NOT, It behaves like NAND Gate}

are the universal gates because they may be used to create any logic gate.  
The given option 1, option 2 and option 3 are not sufficient to represent any boolean operation.
Option 1:
{OR , OR} behaves like an OR gate. It can not sufficient for the universal gate.
Option 2:
{XOR} is a simple XOR gate. It can not sufficient for the universal gate.
Option 3:
{AND, AND} behaves like an AND gate. It can not sufficient for the universal gate.
Option 4:  
2×1 MUX or 2X1 multiplexer is functionally complete provided we have external 1 and 0 available.

• For NOT gate, use x  as a select line and use 0 and 1 as inputs.
• For AND gate, use y and 0 as inputs and x as select. With {AND, NOT} any other gate can be made.

Hence the correct answer is 2×1 MUX.

Concept:
Options verification is the better way to solve this type of question.
Option 1:

The given circuit is gives (A+B)I.
Option 2: 

The given circuit is (And-Gate AB) with NAND blocks. It gives AB as the resultant gate.
Option 3:
The given circuit is 

The given circuit is gives (A+B)I.
Option 4:
The given circuit is 

The given circuit is gives (A+B)I.
Hence the correct answer is option 1, option 3 and option 4.

Only Options (2), (3), and (4) are not Equivalent to the given expression.



Option (A):


Option (B):


Option (C):


Option (D):

Idempotent laws:
Boolean logic has idempotence within both AND and OR gates. A logical AND gate with two inputs A will also have an output of A. (1 AND 1 = 1, 0 AND 0 = 0). An OR gate has idempotence because 0 OR 0 = 0, and 1 OR 1 = 1.
Hence, x + x = x and x.x = x
Identity laws:
The first Boolean identity is that the sum of anything and zero is the same as the original “anything.” This identity is no different from its real-number algebraic equivalent: No matter what the value of A, the output will always be the same: when A=1, the output will also be 1; when A=0, the output will also be 0.
Hence, x + 0 = x and x.1 = 0
Dominance laws:
The complement is used in the operations to form these laws. The idea behind these laws is that if the first number is 1 then the negation of 1 is 0. In equation 1, the numbers are opposite therefore in any order: 0 OR 1 is equal to 1.
Hence, x + 1 = 1 and x.0 = 0.
Absorption law:
This law enables a reduction in a complicated expression to a simpler one by absorbing like terms.
A + (A.B) = (A.1) + (A.B) = A(1 + B) = A (OR Absorption Law)
A(A + B) = (A + 0).(A + B) = A + (0.B) = A (AND Absorption Law)
Hence, x.(x+y) = x
Hence the correct answer is A - 3, B - 4, C - 1, D - 2.

All Boolean algebra laws are shown below:

• (x ⊕ y) ⊕ z = x ⊕ (y ⊕ z) 
  
  Therefore exclusive OR is associative and hence (x ⊕ y) ⊕ z = x ⊕ (y ⊕ z) 
• 
  
  (x + y) ⊕ z ≠ x ⊕ (y + z) ∴ is it not a valid identity
• 
  
  x + y = x ⊕ y // if xy = 0
• (xy + x'y')'
  = (x’ + y’).(x+y) // Demorgan’s Law
  = x’y +xy’
  = x ⊕ y

All Boolean algebra laws are shown below

Calculation:

Now using Distributive Law

Given 
This can be written as:

Since  the above expression can be written as:

With C.C = C, we can write:

XOR gate is a gate that gives a true output when the number of true inputs is odd.
Explanation:
Given, x₁ ⊕ x₂ ⊕ x₃ ⊕ x₄ = 0
Where, x₁, x₂, x₃, x₄ are Boolean variables, and ⊕ is the XOR operator
Consider x₁ = 1, x₂ =1, x₃ =1 and x₄= 1
1 ⊕ 1 ⊕ 1 ⊕ 1 = 0
Now, consider all the options one by one.
1) x₁x₂x₃x₄ = 0 [Incorrect]
Here, put the value of x₁,x₂, x₃, x₄ as 1
So, 1.1.1.1 = 1
2) x₁x₃ + x₂ = 0 [Incorrect]
1.1 + 1 =1

So, 0 ⊕ 0 = 0 ⊕ 0,
0 = 0
4) x₁ + x₂ + x₃ + x₄ = 0 [Incorrect]
As, 1+1+1+1 = 1

∴ S1 is FALSE
But

∴ S2 is TRUE
Important Points:

It is a NAND operation and NAND is commutative but not associative.

• The branch of algebra which deals with the values of the variables in the form of the truth values true and false is called Boolean algebra.
• True and false are usually denoted by 1 and 0 respectively.

Explanation
If we take complement, we get negation of the variable but if we again take the complement of complemented variable, we get the same variable.

Laws of Boolean Algebra:

Application:
f(A, B) = ∑ m(0, 1, 2, 3)

From truth table:

F(x, y) = X