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

Concept:

Laws of Boolean algebra:

All Boolean algebra laws are shown below

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.

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".

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

${\mathrm{P}}_{\mathrm{m}\mathrm{a}\mathrm{x}}=\frac{2^{\mathrm{n}}}{2}=2^{\mathrm{n}-1}$

Example with n = 4:

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

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

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

∴ S1 is FALSE

But

$Q\ne R=\overline{Q}+\overline{R}=\overline{R}+\overline{Q}=R\ne Q$

∴ S2 is TRUE

Important Points 

𝑋 $\ne$ 𝑌 = 𝑋′ + 𝑌′ = (𝑋.𝑌)'  

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

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

Key Points

(AB + C + DC)(AC + BC + D)

= (AB + C[1+ D])(AC + BC + D)

= (AB + C) (AC + BC + D)

= ABC + ABC + ABD + AC + BC + CD

= ABC + ABD + AC + BC + CD

= AC (1 + B) + ABD + BC + CD

= AC + BC + CD + ABD

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

$\overline{X+Y}=\overline{X}.\overline{Y}$

$\overline{XY}=\overline{X}+\overline{Y}$

Let P be the output of 1- NAND gate

$\mathrm{P}=\overline{\mathrm{A}.\mathrm{B}}$

Let Q be the output of 1- NAND gate

$\mathrm{Q}=\overline{\overline{A}+B}$

$\mathrm{Y}=\overline{\mathrm{P}+\mathrm{Q}}=\overline{P}\overline{Q}a$

$\mathrm{Y}=\overline{\stackrel{―}{\mathrm{A}.\mathrm{B}}}.\overline{\stackrel{―}{\overline{A}+B}}$

Y = (A.B). (A̅ + B)

Y = A.B.A̅  + A.B.B

∵ A.A̅  = 0 and B.B = B

Y = AB 

Concept:

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

3) x̅₁ ⊕ x̅₃ = x̅₂ ⊕ x̅₄ [Correct]

Here, x̅₁ = x̅₃ = x̅₂ = x̅₄ = 0,

So, 0 ⊕ 0 = 0 ⊕ 0,

0 = 0

4) x₁ + x₂ + x₃ + x₄ = 0 [Incorrect]

As, 1+1+1+1 = 1