Minimization of Boolean Expression MCQ Quiz in मल्याळम - Objective Question with Answer for Minimization of Boolean Expression - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

The correct answer is "option 4".

EXPLANATION:

Option 1: FALSE

De Morgan's law equations are : 

(P.Q)'  →  P' + Q'

(P+Q)' →  P' . Q'

Option 2: FALSE

Associative law equations are : 

P + (Q + R)  → (P + Q) + R

P . (Q . R)    → (P . Q) . R

Option 3: FALSE

Commutative law equations are : 

P + Q  → Q + P

P . Q   →  Q . P

Option 4: TRUE

Distributive law equation is : 

P + Q . R → (P+Q) . (P+R)

Hence, the correct answer is "option 4".

Concept:

Laws of Boolean algebra:

All Boolean algebra laws are shown below

In a n variable Boolean function, the maximum number of prime implicant is given by:

\({{\rm{P}}_{{\rm{max}}}} = \frac{{{2^{\rm{n}}}}}{2} = {2^{{\rm{n}} - 1}} \)

Example with n = 4:

Maximum number of prime applicants = 2n-1 = 24 -1 = 8

The correct answer is A - II, B - IV, C - I, D - III

Key Points

Concept:-

Dual of Boolean function:- It is the expression one obtains by interchanging addition and multiplication and interchanging 0's and 1's. The dual of the function F is denoted as Fd.

Explanation:-

To get a dual of any Boolean Expression, replace-

OR with AND i.e. + with .

AND with OR i.e. .with +

Consider the function: F(x, y, z) = x + yz

The dual of this function is-

Fd (x,y,z) = x(y + z).

Additional InformationSelf-Dual Function:- When a function is equal to its dual, it is called a Self-Dual Function.

Given Y = A (A̅ + C) (A̅B + C) (A̅BC + C̅)

This can be written as:

Y = (A.A̅ + AC) (A̅B + C) (A̅BC + C̅)

Since A.A̅ = 0, the above expression can be written as:

Y = (AC) (A̅B + C) (A̅BC + C̅)

Y = (AC.A̅B + AC.C) (A̅BC + C̅)

With C.C = C, we can write:

Y = (AC) (A̅BC + C̅)

Y = AC.A̅BC + AC.C̅ 

Y = 0 + 0

Y = 0

De Morgan’s’ Law

\(\overline {X + Y} = \bar X.\bar Y\)

\(\overline {XY} = \bar X + \bar Y\)

Let P be the output of 1- NAND gate

\({\rm{P}} = \overline {{\rm{A}}.{\rm{B}}}\)

Let Q be the output of 1- NAND gate

\({\rm{Q}} = \overline {\bar A + B} \)

\({\rm{Y}} = \overline {{\rm{P}} + {\rm{\;Q}}} = \bar P\bar Qa\)

\({\rm{Y}} = {\rm{\;}}\overline {\overline {{\rm{A}}.{\rm{B}}} \;} .\;\overline {\overline {\bar A + B} } \;\;\)

Y = (A.B). (A̅ + B)

Y = A.B.A̅  + A.B.B

∵ A.A̅  = 0 and B.B = B

Y = AB 

\(\left( {P \ne Q} \right) \ne R = \overline {\left( {\bar P + \bar Q} \right)} + \bar R = PQ + \bar R\)

\(\\ P \ne \left( {Q \ne R} \right) = \bar P + \overline {\left( {\bar Q + \bar R} \right)} = \bar P + Q\bar R \)

∴ S1 is FALSE

But

\(Q \ne R = \bar Q + \bar R = \bar R + \bar Q = R \ne Q\)

∴ S2 is TRUE

Important Points 

𝑋 \(\ne\) 𝑌 = 𝑋′ + 𝑌′ = (𝑋.𝑌)'  

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

\(\begin{array}{l} \left( {\bar A + \bar B} \right)\left[ {\overline {A\left( {B + C} \right)} } \right] + A\left( {\bar B + \bar C} \right)\\ = \left( {\bar A + \bar B} \right)\left[ {\bar A + \left( {\overline {B + C} } \right)} \right] + A\left( {\bar B + \bar C} \right) \end{array}\)

\(\begin{array}{l} = \left( {\bar A + \bar B} \right)\left[ {\bar A + \bar B\bar C} \right] + A\bar B + A\bar C\\ = \bar A + \bar A\bar B\bar C + \bar A\bar B + \bar B\bar C + A\bar B + A\bar C \end{array}\)

\(\begin{array}{l} = \bar A\left( {1 + \bar B\bar C + \bar A\bar B} \right) + \bar B\bar C + A\bar B + A\bar C\\ = \bar A + \bar B\bar C + A\bar B + A\bar C \end{array}\)

= A̅ + A.B̅ + B̅.C̅ + A.C̅

= A̅ + B̅   + B̅.C̅ + A.C̅

= A̅ + B̅ (1 + C̅) + A.C̅

= A̅ + B̅ + A.C̅

=A̅ + C̅ + B̅ 

\(\begin{array}{l} = \bar A + \bar B + \bar C\\ = \overline {ABC} \end{array}\)

Step 1:

Step 2:

Step 3:

It can be represented by:

• one 3-input NAND gate
• one 2-input NAND gate and one 2-input AND gate
• three 2-input NAND gate

Concept-

• 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.

\(F\left( A \right) = \overline {\bar A} = A\)