Test: Logic Gates & Boolean Algebra - 2

Concept:
Boolean function:
The variables used in Boolean Algebra only have one of two possible values, a logic “0” and a logic “1” but an expression can have an infinite number of variables all labeled individually to represent inputs to the expression.
For example, variables A, B, C, etc, give us a logical expression of A + B = C, but each variable can ONLY be a 0 or a 1.
The given expression is = x + yz the equivalent expression is,
Option 1: x + z + xy
= x + z + xy
= x(1+y)+ z  (∴ 1 + x = x + 1 = x)
= x + z it is not equivalent x + yz.
Option 2: (x + y)(x + z)
= (x + y)(x + z)
= x + xz + yx + yz
= x(1+ z+ y) + yz  (∴1 + x = x + 1 = x)
= x + yz it is equivalent x + yz.
Option 3: x + xy + yz
= x + xy + yz
= x(1 + y) +yz
= x + yz  it is equivalent x + yz.
Option 4: x + xz + xy
= x + xz + xy
= x(1+z+y) (∴1 + x = x + 1 = x)
= x it is not equivalent x + yz.
Hence the correct answer is option 2 and option 3.

Concept:
A two-input OR gate shows the output as:

A two-input AND gate shows the output as:

Circuit symbol of AND gate

Calculations:
Given: 
Statement I: A two-input OR gate and a two-input AND gate have similar inputs. 
For OR gate output = A + B, AND gate output is = A.B
So, statement 1 is incorrect.
The output of AND gate is shown by its truth table:

According to Statement II: An AND gate has inputs A and B. The input B is always low,
Then, B shows 0 as input then any Value of A can not change the output. So, this statement is incorrect.
Hence, Option (2) is correct that both statements are incorrect.

• AND, OR, NOT gates are the basic gates.
• The logic gates which are derived from the basic gates like AND, OR, NOT gates are known as derived gates. NAND, NOR, XOR, and XNOR are the derived gates.
• A universal gate is a gate which can implement any Boolean function without need to use any other gate type. The NAND and NOR gates are universal gates.

Concept:
K-map:

• K-map (Karnaugh Map) is a pictorial method used to minimize Boolean expression without having to use Boolean Algebra theorems and equation manipulation.
• K-map can be thought of as a special version of a truth table.
• Using K-map, expression with two to four variables are easily minimized.
• K-maps are also referred to as 2D truth tables as each K-map is nothing but a different format of representing the values present in a one-dimensional truth table.
• To simplify a logic expression with two inputs, we require a K-map with 4 cells (= 2²)
• Similarly, a logic expression with four inputs we require a K-map with 16 cells (= 2⁴)
• Each cell within K-map has a definite place value which is obtained by using on encoding technique known as Gray code.
• For n-variable K-map, with 2ⁿ cells, try to group 2ⁿ cells first, then for 2^n-1 cells, next for 2^n-2 cells, and so on until the group contains only 2° cells ie. Isolated bits (if any)
• Also remember, the number of cells in a group must be equal to an integer power to 2 i.e. 1, 2, 4, 8, ...

Calculation:

→ There are no 16 bits group, no 8-bits group, but there are 2-four bits group
→ Eliminate the variables for which the corresponding hit appears within the group as both 0 and 1.

• Group 1 → B̅1 B2
• Group 2 → B1 B̅2

→ Therefore in SOP form (sum of products) output 

(A.A̅) + A
= 0 + A = A
All Boolean algebra laws are shown below:

• AND, OR, NOT gates are the basic gates.
• The logic gates which are derived from the basic gates like AND, OR, NOT gates are known as derived gates. NAND, NOR, XOR, and XNOR are the derived gates.
• A universal gate is a gate which can implement any Boolean function without need to use any other gate type. The NAND and NOR gates are universal gates.

XOR GATE
Symbol:

Truth Table:

Output Equation: 
Key Points: 
1) If B is always High, the output is the inverted value of the other input A, i.e. A̅.
1) The output is low when both the inputs are the same. 
2) The output is high when both the inputs are different.


Y = 1

F = ∑m (0, 2, 3, 5)

To find the complement of Y.

Concept:
De Morgan’s law states that:

Application:

This can be written as:
f = A̅ + B̅ + A̅ B̅ 
f = A̅ (1 + B̅) + B̅
f = A̅ + B̅ 
Again using De-Morgan's property, we get: