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## Karnataka 2nd PUC Computer Science Question Bank Chapter 2 Boolean Algebra

### 2nd PUC Computer Science Boolean Algebra One Mark Questions and Answers

Question 1.

What is another name of Boolean Algebra?

Answer:

Switching Algebra.

Question 2.

What do you understand by the term truth value?

Answer:

The values depicted by logical constants 0 and 1.

Question 3.

What do you understand by the term truth function?

Answer:

Functions which can be determined to be true or false.

Question 4.

What do you meant by binary valued variables?

Answer:

Binary valued variables that can store one at the two values 1 or 0 [ TRUE or FALSE].

Question 5.

What do you understand by logic function?

Answer:

Logic statements or truth functions are combined with the help of Logical operations like AND, OR and NOT to form a compount statement or logicals function,.

Question 6.

Give examples for Logic functions.

Answer:

- X AND Y NOT Z
- A NOT B OR C v

Question 7.

What is meant by tautology & fallacy?

Answer:

- Tautology: If result of any logical statement or expression is always TRUE is V it is called Tautology.
- Fallacy It result at any logical statement or expresion is always FALSE or ‘O’ it is called Fallacy.

Question 8.

Prove the 1 + y is a tautology & O.y is a fallacy.

Answer:

Question 9.

Name the three logical operators.

Answer:

- NOT
- AND
- NOR

Question 10.

What is a truth table? What is its significance?

Answer:

Truth table is a table which represent all the possible values of Logical variables/statements along with all the possible results for the given combinations of values.

Question 11.

What is NOT operator?

Answer:

A logical operator which Inverts/complements the, output.

Question 12.

Write Venn diagram for NOT operator.

Answer:

Question 13.

Write the truth table for NOT operation.

Answer:

table1

Question 14.

What is OR operator?

Answer:

OR operator is Logical Addition.

Question 15.

Write Venn diagram for OR operator.

Answer:

Question 16.

Write the truth table for OR operation.

Answer:

Question 17.

What is AND operator?

Answer:

AND operator is Logical Multiplication.

Question 18.

What is Venn diagram for AND operator.

Answer:

Question 19.

Write the truth table for AND operation.

Answer:

Question 20.

State the Indempotence law.

Answer:

Idempotence law states that “when a variable is combines with itself using OR or AND operator, the output is the same variable”.

Question 21.

Prove Indempotence law using truth table.

Answer:

1. x + x = x

Proof:

2. x . x = x

proof:

Question 22.

Draw logical diagram to represent Indempotence law.

Answer:

1. x + x = x

2. x . x = x

Question 23.

State Involution law.

Answer:

Involution law states that “the complement of a variable is complemented again, we get the same variable”.

Question 24.

Prove the Involution law using truth table.

Answer:

Question 25.

Draw logical diagram to represent Involution law.

Answer:

Question 26.

State complementarity law.

Answer:

Complementarity law states that x + \(\bar{x}\) = 1 and x . \(\bar{x}\) = 0.

Question 27.

Prove complementarity law using truth table.

Answer:

1. x + \(\bar{x}\) = 1.

2. x . \(\bar{x}\) = 0.

Question 28.

Draw logic diagram to represent complementarity law.

Answer:

1. x + \(\bar{x}\) = 1.

2. x . \(\bar{x}\) = 0.

Question 29.

State commutative law.

Answer:

Commutative law states that

- P + Q = Q + P
- P . Q = Q . P

Question 30.

Prove commutative law using truth table.

Answer:

1. P + Q = Q + P

2. P . Q = Q . P

Question 31.

Draw logic diagram to represent commutative law.

Answer:

1. P + Q = Q + P

2. P . Q = Q . P

Question 32.

State Associative law.

Answer:

Association law states that

- x + (y + z) = (x +y) + z → Associative law of Addition.
- x . (y . z) = (x . y) . z → Associative law of multiplication.

Question 33.

Prove Associative law using truth table.

Answer:

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

LHS = RHS Hence proved

Question 34.

Draw logic diagram to represent Associate law.

Answer:

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

Question 35.

State the Distributive law.

Answer:

The Distributive state that

x(y + z) = xy + xz and x + (y . z) = (x + y) (x + z)

Question 36.

Prove the Distributive law using truth table.

Answer:

x(y + z) = xy +xz

LHS = RHS Hence proved

Question 37.

Draw logic diagram to represent Distribution law.

Answer:

x(y + z) = xy + xz

Question 38.

Prove that x + xy = x (Absorption law).

Answer:

LHS = x + xy

= x(1 + y) | | ∴ 1 + y = 1

= x . 1 | | ∴ x . 1 = X

= x

= RHS

Question 39.

Prove that x(x + y) = x (Absorption law).

Answer:

LHS = x(x + y)

= x . x + x . y

= x + x . y

= x(1 + y) | | ∴ 1 + y = l

= x . 1 | | ∴ x . 1 = x

= x

= RHS.

Question 40.

Draw logic diagram to represent Absorption law.

Answer:

1. x + xy = x

2. x(x + y) = x

Question 41.

xy + \(x \bar{y}\) = x

Answer:

LHS = xy + \(x \bar{y}\)

= x(y + \(\bar{y}\))

= x(1) | | ∴ y + \(\bar{y}\) = 1

= x

=RHS

Question 42.

Prove that \((x+y)(x+\bar{y})=x\).

Answer:

Question 43.

Prove that \(x+\bar{x} y=x+y\)

Answer:

Question 44.

What is a minterm?

Answer:

Minterm is a product of ail literals (with or without the bar) within the logic system.

Question 45.

Find the minterm for xy + z.

Answer:

Question 46.

What is a maxterm?

Answer:

A maxterm is a sum of all literals (with or without the bar) within the system.

Question 47.

Find the maxterm for x + \(\overline{\mathbf{y}}\) + z.

Answer:

Question 48.

What is the canonical form of Boolean expression?

Answer:

Boolean expression composed entirely either of minterms or maxterms is referred to as canonical expression.

Question 49.

Given example for a Boolean expression in the sum of minterms form.

Answer:

F(A, B) = \(A B+A \bar{B}+\bar{A} \bar{B}\)

Question 50.

Give an example for a Boolean expression in the sum of maxterms form.

Answer:

F(A, B) = \((A+B)(\bar{A}+B)\)

### 2nd PUC Computer Science Boolean Algebra Two Marks Questions and Answers

Question 1.

Prove algebrically that (x + y) (x + z) = x + yz.

Answer:

Question 2.

Prove algebrically that \(x+\bar{x} y=x+y\).

Answer:

LHS = \(x+\bar{x} y\)

= \((x+\bar{x})(x+y)\)

= (1) (x + y) | | ∴ \(x+\bar{x}=1\)

= x + y

= RHS

Question 3.

Use duality theorem to derive another Boolean relation from \(\mathbf{A}+\overline{\mathrm{A}} \mathbf{B}=\mathbf{A}+\mathbf{B}\).

Answer:

\(\overline{\mathrm{A}} \cdot \mathrm{A}+\overline{\mathrm{B}}=\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}\)

\(\overline{\mathrm{A}} \mathrm{A}+\overline{\mathrm{A}} \overline{\mathrm{B}}=\overline{\mathrm{A}} \overline{\mathrm{B}}\)

\(\mathrm{O}+\overline{\mathrm{A}} \overline{\mathrm{B}}=\overline{\mathrm{A}} \overline{\mathrm{B}}\)

\(=\overline{\mathrm{AB}}=\overline{\mathrm{A}} \overline{\mathrm{B}}\)

Question 4.

What would be the complement of the following,

- \(\overline{\mathbf{A}}(\mathbf{B} \overline{\mathbf{C}}+\overline{\mathbf{B}} \mathbf{C})\)
- \(\mathbf{A} \overline{\mathbf{B}}+\overline{\mathbf{C}} \overline{\mathbf{D}}\)
- \(xy+\bar{y} z+z \bar{z}\)
- \(x+\overline{x y}+\bar{x} \bar{z}\)

Answer:

1. \(\overline{\mathbf{A}}(\mathbf{B} \overline{\mathbf{C}}+\overline{\mathbf{B}} \mathbf{C})\)

2. \(\mathbf{A} \overline{\mathbf{B}}+\overline{\mathbf{C}} \overline{\mathbf{D}}\)

3. \(xy+\bar{y} z+z \bar{z}\)

4. \(x+\overline{x y}+\bar{x} \bar{z}\)

Question 5.

What are the fundamental products for each of the input word ABCD = 0010, ABCD = 110, ABCD = 1110. Write SOP expression.

Answer:

Fundamental products are

0011 = \(\overline{\mathrm{A}} \overline{\mathrm{B}} \mathrm{C} \overline{\mathrm{D}}\)

0110 = \(\overline{\mathrm{A}} \mathrm{B} \mathrm{C} \overline{\mathrm{D}}\)

1110 = \(\mathrm{ABC} \overline{\mathrm{D}}\)

SOP expression is \(\bar{A} \bar{B} C \bar{D}+\bar{A} B C \bar{D}+A B C \bar{D}\)

Question 6.

A truth table has output 1 for each of these inputs. ABCD = 0011, ABCD = 0101, ABCD = 1000, what are the fundamental products and write minterm expression.

Answer:

Fundamental products are

Question 7.

Construct a Boolean function of these variables x,y and z that has an output 1 when exactly two of x,y,z are having values 0, and an output 0 in all other cases.

Answer:

According to the given condition

From truth table Boolean function is

F(x, y, z) = \(\bar{x} \bar{y} z+\bar{x} y \bar{z}+x \bar{y} \bar{z}\)

Question 8.

Construct a truth for three variable A, B and C that will have an output when XYZ = 100, XYZ = 101, XYZ = 110 and XYZ = 111. Write the Boolean expression for logic Network in sop form.

Answer:

From the given condition, truth table is

100 = \(x \bar{y} \bar{z} \quad=A \bar{B} \bar{C}\)

101 = \(x \bar{y} z \quad=A \bar{B} C\)

110 = \(x y \bar{z} \quad=A B \bar{C}\)

111 = xyz = ABC

Boolean expression in SOP form is

F(A, B, C) = \(A \bar{B} \bar{C}+A \bar{B} C^{+} A B \bar{C}+A B C\)

Question 9.

Convert the following expressions to canonical product – of – sums form.

- (A + C)(C + D)
- A(B + C)(C + D)
- (x + y)(y + z)(x + z)

Answer:

1. (A + C)(C + D)

2. A(B + C)(C + D)

3. (x + y)(y + z)(x + z)

Question 10.

Convert the following expression to canonical sum – of – product form,

- \((x+\overline{x} y+\bar{x} \bar{z})\)
- \(y z+\overline{x} y\)
- \(\mathbf{A} \overline{\mathbf{B}}(\overline{\mathbf{B}}+\overline{\mathbf{C}})\)

Answer:

1.

2.

3.

Question 11.

Draw k – maps for the following expressions

- \(\bar{x} \bar{y}+\overline{x}y\)
- \(x y \bar{z}+\bar{x} \ \bar{y}z\)
- \(\bar{x} \bar{y} \bar{z}+\ x \bar{y} \bar{z}+\overline{\mathbf{x}} y \bar{z}\)

Answer:

1. \(\bar{x} \bar{y}+\overline{x}y\)

2. \(x y \bar{z}+\bar{x} \ \bar{y}z\)

3. \(\bar{x} \bar{y} \bar{z}+\ x \bar{y} \bar{z}+\overline{\mathbf{x}} y \bar{z}\)

Question 12.

Draw a general karnaugh – map for four variables A, B, C and D.

Answer:

For 4 – variable k – map

Question 13.

Given the expression in four variables, draw the k – map for the function.

- m
_{2}+ m_{3}+ m_{5}+ m_{7}+ m_{9}+ m_{11}+ m_{13}. - m
_{0}+ m_{2}+ m_{4}+m_{8}+ m_{9}+ m_{10}+ m_{11}+ m_{12}+ m_{13}.

Answer:

1. m_{2} + m_{3} + m_{5} + m_{7} + m_{9} + m_{11} + m_{13}

Let the 4 variables be ABCD. We need 4 variable k – map

2. m_{0} + m_{2} + m_{4} +m_{8} + m_{9} + m_{10} + m_{11} + m_{12} + m_{13}

Let the 4 varaiables be A, B, C & D

k – map

Question 14.

Draw the K – map for the function in three variables given below

- m
_{0}+ m_{2}+ m_{4}+ m_{6}+ m_{7}. - m
_{1}+ m_{2}+ m_{3}+ m_{5}+ m_{7}.

Answer:

1. m_{0} + m_{2} + m_{4} + m_{6} + m_{7}.

Let 3 variables be xyz

2. m_{1} + m_{2} + m_{3} + m_{5} + m_{7}.

Question 15.

Write SOP expression to the function F in the following truth table & draw the logic gate diagram (use OR & AND Gates).

Answer:

SOP expression from the truth table is y = \(\overline{\mathrm{A}} \mathrm{B} \overline{\mathrm{C}}+\overline{\mathrm{A}} \mathrm{BC}+\mathrm{AB} \overline{\mathrm{C}}+\mathrm{ABC}\)

Logic Diagram:

### 2nd PUC Computer Science Boolean Algebra Three Marks Questions and Answers

Question 1.

State and prove any three theorems of Boolean Algebra.

- 0 + x = x
- 0 . x = 0
- 1 . x = x

Answer:

1. 0 + x = x

Statement: Any variable ORed with ‘0’ produces the same output as that at x (variable)

Proof:

If x = 0, LHS = 0 + x

If x = 1, RHS = 0 + x

= 0 + 1

= 1

= x = RHS. Hence proved

2. 0 . x = 0

Statement: Any variable ANDed with ‘0’ provides the output as ‘0’

Proof:

If x = 0, LHS = 0 . x,

= 0 . 0

= 0

= RHD

If x = 1, LHS = 0 . x

= 0 . 1

= 0

= RHS

Hence proved.

3. 1 . x = x

Statement: Any variable ANDed with 1 produce the O/P as that of variable.

Proof:

If x = 0, LHS = 1 . x,

= 1 . 0

= 0

= x

= RHS

If x = 1, LHS = 1 . x

= 1 . 1

= 1

= RHS

Hence proved.

Question 2.

State and prove the Associative law of addition & Multiplication.

Answer:

Associative law of Addition & Multiplication Statement: Associative law states that

- x + (y + z) = (x + y) + z → Associative law of Addition
- x . (y . z) = (x . y).z → Associative law of Multiplication.

Proof:

1. Associative law of Addition.

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

LHS = RHS. Hence Proved

2. Associative law of multiplications.

x . (y . z) = (x . y) . z

LHS = RHS. Hence proved.

Question 3.

State and prove De – morgan’s theorems by the method of perfect induction.

Answer:

De- morgan’s first Theorem:

\(\overline{x+y}=\bar{x} \cdot \bar{y}\)

Statement: De-morgan’s first theorem states that, ” sum of complement of any 2 variables is equal to the product of complements of variables”.

Proof:

LHS = RHS. Hence proved.

De- morgan’s Second Theorem:

\(\overline{x \cdot y}=\bar{x}+\bar{y}\)

Statement: De morgan’s second Theorem states that, ” Product of complement of variables is equal to the sum of complement of variables”.

Proof:

LHS = RHS. Hence proved.

Question 4.

Obtain the minterm expression for the Boolean function F = \(\mathbf{A}+\overline{\mathbf{B}} \mathbf{C}\)

Answer:

Question 5.

Explain with an example how to express a Boolean function in its sum – of – products form.

Answer:

A product term or several product terms logically added, it becomes sum of products (SOP)

Ex: Consider an expression P + Q

1. Convert it into sop form

i.e P + Q = p . 1 + Q . 1

2. In each term if any variable is missing multiply each term by (missing term + complement of missing term).

i.e P + Q = P . 1 + Q . 1

\(=P(Q+\bar{Q})+Q(P+\bar{P})\)

3. Expand the expression

i.e. \(P(Q+\bar{Q})+Q(P+\bar{P})\)

\(=P Q+P \bar{Q}+P Q+Q \bar{P}\)

4. Remove all the duplicates.

i.e. = \(P Q+P \bar{Q}+Q \bar{P}\)

Question 6.

Explain with an example how to express a Boolean function in its product – of – sums form.

Answer:

A sum terms or several sum terms, when logically multiplied, it becomes product of sum (POS)

Ex: Consider an expression (P + Q) (Q + P)

1. Convert it into POS form

i.e (P + Q) (Q + R) = (P + Q + 0) (Q + R + 0)

2. In each term if any variable is missing add (missing term and complement of missing term)

i.e (P + Q + 0) (Q + R + 0) = \((P+Q+R \bar{R})(P \bar{P}+Q+R)\)

3. Expand the expression

\((P+Q+R)(P+Q+\bar{R})(P+Q+R)(\bar{P}+Q+R)\)

Question 7.

Construct the truth table for the minterms & maxterms for three variables and designate the terms.

Answer:

Question 8.

Using the basic gates, construct a logic circuit for the Boolean expression \((\bar{x}+y)\) (x + y) (y + z).

Answer:

Question 9.

Simplify the Boolean expressions and draw logic circuit diagrams of the simplified expressions using only NAND Gates.

Answer:

1.

2.

3.

4.

Question 10.

For a four variable map in w, x, y and z draw the subcubes for

(a) \(\mathbf{W} \mathbf{X} \bar{Y}\)

(b) WX

(c) \(XY \bar{Z}\)

(d) Y

Answer:

Question 11.

Convert the following Product of Sums forms into its corresponding Sum – of – products form using write the truth table. F(x, y, z) = Π(2, 4, 6, 7).

Answer:

F(x, y, z) = Π(2, 4, 6, 7)

From the above table corresponding SOP is

Question 12.

- Reduce the following Boolean expression to the simplest form: A . [B + C. (AB + AC)].
- Given: F(x, y, z) = ∑ (1, 3, 7) then prove that F'(x, y, z) = Π (0, 2, 4, 5, 6)

Answer:

1. Reduce the following Boolean expression to the simplest form: A . [B + C. (AB + AC)].

A[B + ABC + ACC]

2. Given: F(x, y, z) = ∑ (1, 3, 7) then prove that F'(x, y, z) = Π (0, 2, 4, 5, 6)

F(x, y, z) = ∑ (1, 3, 7)

F (x, y, z) = m_{1} + m_{3} + m_{7}

i.e.

m_{1} = \(\bar{x} \bar{y} z\)

m_{2} = \(\bar{x} y z\)

m_{3} = xyz

from first group = \(\bar{x} z\)

from second group = yz

Resultant expression is \(\bar{x} z+y z\) ……………………(1)

Similarly, consider F'(x, y, z) = π(0, 1, 4, 5, 6)

∴ F'(x, y, z) = π(m_{0}, m_{2}, m_{3}, m_{5}, m_{6})

consider k – map is

From first group = z

From second group = \(\bar{x}+y\)

Resultant expression = \(z(\bar{x}+y)=\bar{x} z+y z\) …………….. (2)

Comparing equations (1) and (2)

If, F(x, y, z) = ∑(1, 3, 7) then F’ (x, y, z) = π (0, 2, 4, 5, 6). Hence proved.

### 2nd PUC Computer Science Boolean Algebra Five Marks Questions and Answers

Question 1.

Using maps, simplify the following expressions in four variables W, X, Y and Z.

- m
_{1}+ m_{3}+ m_{5}+ m_{6}+ m_{7}+ m_{9}+ m_{11}+ m_{11}. - m
_{0}+ m_{2}+m_{4}+m_{8}+ m_{9}+ m_{10}+ m_{11}+ m_{12}+m_{13}.

Answer:

1. m_{1} + m_{3} + m_{5} + m_{6} + m_{7} + m_{9} + m_{11} + m_{11}

From first group = \(\overline{\mathrm{W}} \mathrm{Z}\)

From second group = \(\overline{\mathrm{W}} \mathrm X {Y}\)

From third group = \(\mathrm{W} \overline{\mathrm{X}} \mathrm{Z}\)

From fourth group = \(\mathrm{W} \overline{\mathrm{Y}} \mathrm{Z}\)

minimized expression = \(\overline{\mathrm{w}} \mathrm{z}+\overline{\mathrm{w}} \mathrm{X} \mathrm{Y}+\mathrm{w} \overline{\mathrm{X}} \mathrm{Z}+\mathrm{W} \overline{\mathrm{Y}} \mathrm{Z}\)

2. m_{0} + m_{2} +m_{4} +m_{8} + m_{9} + m_{10} + m_{11} + m_{12} +m_{13}.

From first group = \(\overline{\mathrm{x}} \overline{\mathrm{z}}\)

From second group = \(X \overline{Y Z}\)

From third group = \(\mathbf{w} \overline{\mathrm{Y}} \mathbf{z}\)

From fourth group = \(W \bar{X} Z\)

∴ Minimized expression = \(\overline{\mathrm{x}} \overline{\mathrm{z}}+\overline{\mathrm{y} \overline{\mathrm{z}}}+\mathrm{w} \overline{\mathrm{Y}} \mathrm{z}+\mathrm{w} \overline{\mathrm{x}} \mathrm{z}\)

Question 2.

For the Boolean function F and F’ in the truth table, find the following

- List the minterms of the function F and F’
- Express F and F’ in sum of minterms in algebraic form.
- Simplify the functions to an expression with a minimum number of literals.

Answer:

1.

2.

3.

Question 3.

State and prove De – Morgan’s theorem algebrically.

Answer:

To prove De- Morgan’s first theorem, \(\overline{x+y}=\bar{x} \cdot \bar{y}\). we will use complementarity laws.

Let us assume that p = x + y

Where p, x, y are logical variables.

Then, according to complementation law \(p+\bar{p}=1\) 1 and p . p = 0.

i,e, if p, x, y are Boolean variables then this complementarity law must hold for variable p.

In other words, if p i.e, if \(\overline{x+y}=\bar{x} \cdot \bar{y}\) then \((x+y)+\overline{x y}\) must be equal to 1( As x+\bar{x}=1) and \((x+y) \cdot \overline{x y}\) must be equal to 0 (\(x \bar{x}=0\))

Let us prove the 1st part, i.e, \((x+y)+(\overline{x y})+1\)

Now let us prove 2nd part i.e., \((x+y) \cdot(\overline{x y})=0\)

To prove second theorem \((\overline{x y})=\bar{x}+\bar{y}\) we will make use of complementarity law i.e., \(x+\bar{x}=1\) and \(\text { x. } \bar{x}=0\)

If xy’s complement is \(\bar{x}+\bar{y}\) then it must be true that

- \(x y+(\bar{x}+\bar{y})=1\)
- \(x y+(\bar{x}+\bar{y})=0\)

To prove first part i.e, \(x y+(\bar{x}+\bar{y})=1\)

1.

= 1

= RHS

2.

Hence the thoerem proved.

Question 4.

Find the complement of F = X + YZ, then show that F . F’ = 0 and F + F’ = 1.

Answer:

For F = X + YZ

Complement of F, i.e, \(\overline{\mathrm{F}}=\overline{\mathrm{X}} \cdot \overline{\mathrm{Y}}+\overline{\mathrm{Z}}\)

= \(XYZ+X \overline{Y Z}+X \overline{Y} Z+\bar{X} Y Z+\overline{X Y} Z+\overline{X Y Z}+\bar{X} Y \bar{Z}\)

= ∑(0, 1, 2, 3, 4, 5, 6, 7) = 1

Question 5.

- Sate the two Absorption laws of Boolean Algebra, Verify using truth table.
- Simplify using laws of Boolean Algebra. At each step state clarity the law used for simplification.

F = x . y + x . z + x . y . z.

Answer:

1. Absorption law states that

x + xy = x

proof:

1^{st} column = 4^{th} column.

∴ x + xy = x

Hence proved.

2. x(x + y) = x

1^{st} column = 4^{th} column.

∴ x(x + y) = x

Hence proved.

F = xy +xz + xyz | Commutative law

= xy + xyz + xz | Distributive law

= xy (1) + xz | property of 1 i.e. 1 + x = 1

= xy + xz | property of 1 i.e. xy . 1 = xy

= x(y + z) | Distributive law

Question 6.

Given the Boolean function F(X, Y,Z) = ∑(0, 2, 4, 5, 6)

Answer:

F(X, Y,Z) = ∑(0, 2, 4, 5, 6)

F(X, Y, Z) = \(X \bar{Z}+\bar{Z}\)

Question 7.

- State the two complement properties of Boolean algebra. Verify using the truth tables.
- X . (YZ + YZ)

Answer:

1. The two complement properties of Boolean Algebra are

(a) \(X+\bar{X}=1\)

Verification using the table.

(b) \(\mathrm{X} . \overline{\mathrm{X}}=0\)

∴ \(X \cdot(Y \bar{Z}+Y Z)\)

= \(X+\bar{X}=1\)

= X . 1 = X

Question 8.

Given the Boolean function F(A, B, C, D) = ∑(5, 6, 7, 8, 9, 10, 14). Use Karnaugh’s map to reduce the function F using S – O – P from. Write a logic gate diagram for the reduced S – O – P expression.

Answer:

F(A, B, C, D) = ∑(5, 6, 7, 8, 9, 10, 14)

F(A, B, C, D) = \(\mathrm{A} \overline{\mathrm{B}} \overline{\mathrm{C}}+\overline{\mathrm{A}} \mathrm{B} \mathrm{D}+\overline{\mathrm{A}} \mathrm{BC}+\mathrm{AC} \overline{\mathrm{D}}+\mathrm{BC} \overline{\mathrm{D}}\)

Logic Gate Diagram:

Question 9.

Given F (A, B, C, D) = (0, 2, 4, 6, 8, 10, 14). Use Karnaugh map to reduce the function F using P – O – S form write a logic gate diagram for the reduced P – O – S expression.

Answer:

Given F (A, B, C, D) = (0, 2, 4, 6, 8,10, 14) K – map using P – O – S form

P-O-S expression F(A, B, C, D) = \(\mathrm{B}+\mathrm{D} \cdot \overline{\mathrm{C}}+\mathrm{D} \cdot \mathrm{A}+\mathrm{D}\)

Logic Gate Diagram:

Question 10.

Use Karnaugh map to reduce the given functions using SOP form. Draw the logic gate diagrams for the reduced SOP expression. You may use gates with more than two inputs. Assume that the variables and their complements are available as inputs.

Answer:

Consider a function

f(x, y, z) = \(\overline{x} y \overline{z}+x \overline{y z}+x y \bar{z}+x y z\)

This expression has 3 – variables x, y, z. So a 3 – variable K – map should be drawn.

first term \(\overline{x} y \overline{z}\) = 010

second term \(x \overline{y z}\) = 100

third term \(x y \bar{z}\) = 110

fourth term xyz = 111

From first group = \(x \bar{z}\)

From second group = xy

From third group = \(y \bar{z}\)

Minimised SOP expression is \(x \bar{z}+x y+y \bar{z}\)

Logic Circute:

Question 11.

Given the boolean function F(A, B, C, D) = ∑(0, 4, 8, 9, 10, 11, 12, 13, 15). Reduce it by using Karnaugh map.

Answer:

F(A, B, C, D) = I (0, 4, 8, 9, 10, 11, 12, 13, 15).

From first group = \(=\overline{\mathrm{C}} \overline{\mathrm{D}}\)

From second group = AD

From third group = \(A \bar{B}\)

∴ F(A, B, C, D) = \(\overline{\mathrm{C}} \overline{\mathrm{D}}+\mathrm{AD}+\mathrm{A} \overline{\mathrm{B}}\)