Boolean Algebra MCQ Quiz in मल्याळम - Objective Question with Answer for Boolean Algebra - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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

F(A, B, C) = C + AB’C’

F(A, B, C) = (C + C’)(C + AB’)

1. (x ⊕ y) ⊕ z = x ⊕ (y ⊕ z) 

Therefore exclusive OR is associative and hence (x ⊕ y) ⊕ z = x ⊕ (y ⊕ z) 

2.

(x + y) ⊕ z ≠ x ⊕ (y + z) ∴ is it not a valid identity

3.

x + y = x ⊕ y / if xy = 0

4.

(xy + x'y')'

= (x’ + y’).(x+y) / Demorgan’s Law

= x’y +xy’

The given circuit gives the output:

Z(a, b, c) = \(a + \;\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\leftharpoonup$}} \over b} \;c\) 

Expanding it into canonical form to obtain the minterms

Z(a, b, c) = \(a\left( {b + \;\bar b} \right)\left( {c + \bar c} \right) + \left( {a + \bar a} \right)\bar b\;c\) 

= \(abc + ab\;\bar c + a\bar b\;c + a\bar b\bar c + \bar a\bar b\;c\)

After rearranging the canonical terms, this corresponds to min-terms: ∑ (1,4, 5, 6, 7)

Alternate solution:

The output of the circuit is \(a + \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\leftharpoonup$}} \over b} \;c\) 

K Map for this Boolean expression

The above K Map corresponds to min-terms: ∑ (1,4, 5, 6, 7)

From truth table:

F(X, Y) = XY̅ + XY

= X[Y̅ + Y]

\(f\left( {X,Y,Z,W} \right)\; = X\bar YZ\bar W + \bar X\bar Y\bar Z\bar W + X\bar Y\bar Z\bar W + \bar X\bar YZ\bar W + \;\bar XY\bar ZW + \bar XYZW + \;XY\bar ZW + XYZW\)

\(f\left( {X,Y,Z,W} \right)\; = \bar X\bar Y\bar Z + \bar X\bar YZ\bar W + \;\bar XY\bar ZW + \bar XYZW + X\bar Y\bar Z\bar W + X\bar YZ\bar W + \;XY\bar ZW + XYZW\)

\(f\left( {X,Y,Z,W} \right)\; = \bar X\bar Y\bar W\;\left( {Z + \bar Z} \right) + \;\bar XYW\left( {Z + \bar Z} \right) + X\bar Y\bar W\left( {Z + \bar Z} \right) + \;XYW\left( {Z + \bar Z} \right)\)

\(f\left( {X,Y,Z,W} \right)\; = \bar X\bar Y\bar W\; + X\bar Y\bar W + \;\bar XYW + \;XYW\)

\(f\left( {X,Y,Z,W} \right)\; = \left( {\bar X + X} \right)\left( {\bar Y\bar W\;} \right) + \;\left( {\bar X + X} \right)\left( {YW} \right)\)

\(f\left( {X,Y,Z,W} \right)\; = \bar Y\bar W + \;YW\;\;\)

\(f\left( {X,Y,Z,W} \right)\; = Y \odot W\)

XNOR gate with NAND gates:

Hence 5 NAND gates is needed

Tips and Tricks:

The above Boolean function can also be solved using K-map:

X = ABC + A̅B + ABC̅

Rearranging:

X = ABC + ABC̅ + A̅B

X = AB(C + C̅) + A̅B

X = AB + A̅B   [C + C̅ = 1]

X = (A + A̅)B  

X = 1.B   [A + A̅ = 1]

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

Calculation:

Let F(A, B) = A + A̅ .B + A.B̅

F(A, B) = A (1 + B̅) + A̅.B

F(A, B) = A + A̅.B

F(A, B) = (A + A̅).(A + B)

F(A, B) = A + B 

A $ B = A̅ + B (1)

W = A + B̅ (2)

Y = W̅ $ A̅

Y = W + A̅ (From 1)

Y = A + B̅ + A̅ (From 2)

The correct answer is "option 1".

EXPLANATION:

Option 1: TRUE

The truth table X is:

F is equal to X.

Option 2: FALSE

The truth table X+Y is:

F is not equal to X+Y.

Option 3: FALSE

The truth table X’Y + XY' is:

F is not equal to X’.Y +X.Y'.

Option 4: FALSE

The truth table Y is :

 F is not equal to Y.

Hence, the correct answer is "option 1".

From truth table, the sum of minterms

F(X, Y) = X.Y̅ + X.Y = X(Y̅ + Y) = X