unit 4-part 1 : boolean algebra

Unit 4-Part 1 Boolean Algebra

1

|EE amp EC Department | 3130907 ndash Analog and Digital Electronics

Table of Contents

31 Introduction 3

311 Advantages of Boolean Algebra 3

32 Boolean Algebra Terminology 3

33 Logic Operators 4

34 Axioms or Postulates 4

35 Boolean Algebrarsquos Laws and Theorems 4

351 Reduction of Boolean Expression 10

3511 De-Morganized the following functions 10

3512 Reduce the following functions using Boolean Algebrarsquos Laws and Theorems 10

36 Different forms of Boolean Algebra 12

361 Standard Form 12

3611 Standard Sum of Product (SOP) 12

3612 Standard Product of Sum (POS) 12

362 Canonical Form 12

3621 Sum of Product (SOP) 12

3622 Product of Sum (POS) 13

362 MINTERMS amp MAXTERMS for 3 Variables 14

363 Conversion between Canonical Forms 14

3631 Convert to MINTERMS 14

3632 Convert to MAXTERMS 15

37 Karnaugh Map (K-Map) 17

371 2 Variable K-Map 17

3711 Mapping of SOP Expression 17

3712 Mapping of POS Expression 17

3713 Reduce Sum of Product (SOP) Expression using K-Map 18

3714 Reduce Product of Sum (POS) Expression using K-Map 18




2








3733 Looping of POS Expression 21



3736 Reduce SOP amp POS Expression with Donrsquot Care Combination using K-Map 24





38 Converting Boolean Expression to Logic Circuit and Vice-Versa 26

39 NAND and NOR RealizationImplementation 28

310 Tabulation Quine-McCluskey Method 29

311 GTU Questions 33


3


31 Introduction

Inventor of Boolean algebra was George Boole (1815 - 1864)

Designing of any digital system there are three main objectives 1) Build a system which operates within given specifications 2) Build a reliable system 3) Minimize resources

Boolean algebra is a system of mathematical logic

Any complex logic can be expressed by Boolean function

Boolean algebra is governed by certain rules and laws

Boolean algebra is different from ordinary algebra amp binary number system In ordinary algebra A + A = 2A and AA = A2 here A is numeric value

In Boolean algebra A + A = A and AA = A here A has logical significance but no numeric significance

Table Difference between Binary Ordinary and Boolean system

Binary number system Ordinary no system Boolean algebra

1 + 1 = 1 0 1 + 1 = 2 1 + 1 = 1

- A + A = 2A and AA = A2 A + A = A and AA = A

In Boolean algebra nothing like subtracting or division no negative or fractional numbers

Boolean algebra represent logical operation only Logical multiplication is same as AND operation and logical addition is same as OR operation

Boolean algebra has only two values 0 amp 1

In Boolean algebra If A = 0 then A ne 1 amp If A = 1 then A ne 0

311 Advantages of Boolean Algebra

1 Minimize the no of gates used in circuit 2 Decrease the cost of circuit 3 Minimize the resources 4 Less fabrication area is required to design a circuit 5 Minimize the designerrsquos time 6 Reducing to a simple form Simpler the expression more simple will be hardware 7 Reduce the complexity

32 Boolean Algebra Terminology

1 Variable The symbol which represent an arbitrary elements of a Boolean algebra is known as variable eg F = A + BC here A B and C are variable and it can have value either 1 or 0


4


2 Constant In expression F = A + 1 the first term A is variable and second term 1 is known as constant Constant may be 1 or 0

3 Complement A complement of any variable is represented by a ldquo rdquo (BAR) over any variable

eg Complement of A is 119860 4 Literal Each occurrence of a variable in Boolean function either in a non-

complemented or complemented form is called literal 5 Boolean Function Boolean expressions are constructed by connecting the Boolean

constants and variable with the Boolean operations This Boolean expressions are also known as Boolean Formula

eg F(A B C) = (119860 + 119861) C OR F = (119860 + 119861) C

33 Logic Operators

1 AND Denoted by ∙ (eg A AND B = A ∙ B) 2 OR Denoted by + (eg A OR B = A + B) 3 NOT OR Complement Denoted by ldquo rdquo (BAR) or ( )prime (eg 119860 or (119860)prime)

34 Axioms or Postulates

Axioms 1 0 middot 0 = 0 Axioms 2 0 middot 1 = 0 Axioms 3 1 middot 0 = 0 Axioms 4 1 middot 1 = 1

Axioms 5 0 + 0 = 0 Axioms 6 0 + 1 = 1 Axioms 7 1 + 0 = 1 Axioms 8 1 middot 1 = 1

Axioms 9 1rsquo = 0 Axioms 10 0rsquo = 1

35 Boolean Algebrarsquos Laws and Theorems

1 Complementation Laws

The term complement simply means to invert ie to change 0rsquos to 1rsquos and 1rsquos to 0rsquos Law 1 0rsquo = 1 Law 2 1rsquo = 0 Law 3 If A = 0 then Arsquo = 1

Law 4 If A = 1 then Arsquo = 0 Law 5 Arsquorsquo = A

2 AND Laws

Law 1 A middot 0 = 0 Law 2 A middot 1 = A

Law 3 A middot A = A Law 4 A middot Arsquo = 0

3 OR Laws

Law 1 A + 0 = A Law 2 A + 1 = 1

Law 3 A + A = A Law 4 A + Arsquo = 1


5


4 Commutative Laws

Commutative laws allow change in position of AND or OR variables Law 1 A + B = B + A Proof

This law can be extended to any numbers of variables for eg

A + B + C = B + A + C = C + B + A = C + A + B Law 2 A middot B = B middot A Proof


A middot B middot C = B middot A middot C = C middot B middot A = C middot A middot B 5 Associative Laws

The associative laws allow grouping of variables Law 1 (A + B) + C = A + (B + C) Proof

This law can be extended to any no of variables for eg

A + (B + C + D) = (A + B + C) + D = (A + B) + (C + D)


6


Law 2 (A middot B) middot C = A middot (B middot C) Proof


A middot (B middot C middot D) = (A middot B middot C) middot D = (A middot B) middot (C middot D)

6 Distributive Laws

The distributive laws allow factoring or multiplying out of expressions Law 1 A (B + C) = AB + AC Proof

Law 2 A + BC = (A + B) (A + C) Proof RHS = (A + B) (A + C) = AA + AC + BA + BC = A + AC + BA + BC = A + BC (∵ A(1 + C + B) = A) = LHS

Law 3 A + ArsquoB = A + B Proof LHS = A + ArsquoB = (A + Arsquo) (A + B) = A + B = RHS

7 Idempotence Laws

Idempotence means the same value Law 1 A middot A = A Proof Case 1 If A = 0 A middot A = 0 middot 0 = 0 = A Case 2 If A = 1 A middot A = 1 middot 1 = 1 = A

Law 2 A + A = A Proof Case 1 If A = 0 A + A = 0 + 0 = 0 = A Case 2 If A = 1 A + A = 1 + 1 = 1 = A


7


8 Complementation Law Negation Law Law 1 A middot Arsquo = 0 Proof Case 1 If A = 0 A middot Arsquo = 0 middot 1 = 0 Case 2 If A = 1 A middot Arsquo = 1 middot 0 = 0

Law 2 A + Arsquo = 1 Proof Case 1 If A = 0 A + Arsquo = 0 + 1 = 1 Case 2 If A = 1 A + Arsquo = 1 + 0 = 1

9 Double Negation Involution Law

This law states that double negation of a variables is equal to the variable itself Law Arsquorsquo = A Proof Case 1 If A = 0 Arsquorsquo = 0rsquorsquo = 0 = A Case 2 If A = 1 Arsquorsquo = 1rsquorsquo = 1 = A

Any odd no of inversion is equivalent to single inversion

Any even no of inversion is equivalent to no inversion at all

10 Identity Law Law 1 A middot 1 = A Proof Case 1 If A= 1 A middot 1 = 1 middot 1 = 1 = A Case 2 If A= 0 A middot 0 = 0 middot 0 = 0 = A

Law 2 A + 1 = 1 Proof Case 1 If A= 1 A + 1 = 1 + 1 = 1 = A Case 2 If A= 0 A + 0 = 0 + 0 = 0 = A

11 Null Law Law 1 A middot 0 = 0 Proof Case 1 If A= 1 A middot 0 = 1 middot 0 = 0 = 0 Case 2 If A= 0 A middot 0 = 0 middot 0 = 0 = 0

Law 2 A + 0 = A Proof Case 1 If A= 1 A + 0 = 1 + 0 = 1 = A Case 2 If A= 0 A + 0 = 0 + 0 = 0 = A

12 Absorption Law Law 1 A + AB = A Proof LHS = A + AB = A (1 + B) = A (1) = A = RHS

Law 2 A (A + B) = A Proof LHS = A (A + B) = A middot A + AB = A + AB = A (1 + B) = A = RHS


8


13 Consensus Theorem Theorem 1 A middot B + ArsquoC + BC = AB + ArsquoC Proof LHS = AB + ArsquoC + BC = AB + ArsquoC + BC (A +Arsquo) = AB + ArsquoC + BCA + BCArsquo = AB (1 + C) + ArsquoC (1 + B) = AB + ArsquoC = RHS

This theorem can be extended as AB + ArsquoC + BCD = AB + ArsquoC

Theorem 2 (A + B) (Arsquo + C) (B + C) = (A + B) (Arsquo + C) Proof LHS = (A + B) (Arsquo + C) (B + C) = (AArsquo + AC + ArsquoB + BC) (B + C)

= (0 + AC + ArsquoB + BC) (B + C) = ACB+ACC+ArsquoBB+ArsquoBC+BCB+BCC = ABC + AC + ArsquoB + ArsquoBC + BC + BC

= ABC + AC + ArsquoB + ArsquoBC + BC = AC (1 + B) + ArsquoB (1 + C) + BC = AC + ArsquoB + BC = AC + ArsquoB helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(1)

RHS = (A + B) (Arsquo + C) = AArsquo + AC + BArsquo + BC = 0 + AC + BArsquo + BC = AC + ArsquoB + BC = AC +ArsquoB helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip(2)

Eq (1) = Eq (2) So LHS = RHS

This theorem can be extended to any no of variables (A + B) (Arsquo + C) (B + C + D) = (A + B) (Arsquo + C)

14 Transposition theorem

Theorem AB + ArsquoC = (A + C) (Arsquo +B) Proof RHS = (A + C) (Arsquo +B) = AArsquo + AB + CArsquo + CB = 0 + AB + CArsquo + CB = AB + CArsquo + CB = AB + ArsquoC (∵ AB + ArsquoC + BC = AB + ArsquoC) =LHS

15 De Morganrsquos Theorem Law 1 (A + B)rsquo = Arsquo middot Brsquo OR (A + B + C)rsquo = Arsquo middot Brsquo middot Crsquo Proof


9


Law 2 (Amiddot B)rsquo = Arsquo + Brsquo OR (A middot B middot C)rsquo = Arsquo + Brsquo + Crsquo Proof

LHS

NAND Gate

A

B

AB (AB)rsquo=

RHS

B

A Arsquo

Brsquo

Arsquo + Brsquo

=

Bubbled OR

A

B

(AB)rsquo

A

B

Arsquo + Brsquo

A B AB (AB)rsquo

0 0 0 1

0 1 0 1

1 0 0 1

0 1 1 0

A B Arsquo Brsquo Arsquo+Brsquo

0 0 1 1 1

0 1 1 0 1

1 0 0 1 1

0 1 0 0 0

16 Duality Theorem

Duality theorem arises as a result of presence of two logic system ie positive amp negative logic system

This theorem helps to convert from one logic system to another

From changing one logic system to another following steps are taken 1) 0 becomes 1 1 becomes 0 2) AND becomes OR OR becomes AND 3) lsquo+rsquo becomes lsquomiddotrsquo lsquomiddotrsquo becomes lsquo+rsquo 4) Variables are not complemented in the process

=


10


351 Reduction of Boolean Expression

3511 De-Morganized the following functions

(1) F = [(A + Brsquo) (C + Drsquo)]rsquo (2) F = [(AB)rsquo (CD + ErsquoF) ((AB)rsquo + (CD)rsquo)]rsquo

Sol Sol

F = [(A + Brsquo) (C + Drsquo)]rsquo F = [(AB)rsquo (CD + ErsquoF) ((AB)rsquo + (CD)rsquo)]rsquo

there4 F = (A + Brsquo)rsquo + (C + Drsquo)rsquo there4 F = (AB)rsquorsquo + (CD + ErsquoF)rsquo + ((AB)rsquo + (CD)rsquo)rsquo

there4 F = Arsquo Brsquorsquo + CrsquoDrsquorsquo there4 F = AB + [(CD)rsquo (ErsquoF)rsquo] + [(AB)rsquorsquo (CD)rsquorsquo]

there4 F = ArsquoB + CrsquoD there4 F = AB + (Crsquo + Drsquo) (E + Frsquo) + ABCD

(3) F = [(AB)rsquo + Arsquo + AB]rsquo (4) F = [(P + Qrsquo) (Rrsquo + S)]rsquo

Sol Sol

F = [(AB)rsquo + Arsquo + AB]rsquo F = [(P + Qrsquo) (Rrsquo + S)]rsquo

there4 F = (AB)rsquorsquo middot Arsquorsquo middot (AB)rsquo there4 F = (P + Qrsquo)rsquo + (Rrsquo + S)rsquo

there4 F = ABA (Arsquo + Brsquo) there4 F = PrsquoQrsquorsquo + RrsquorsquoSrsquo

there4 F = AB (Arsquo + Brsquo) there4 F = PrsquoQ + RSrsquo

there4 F = ABArsquo + ABBrsquo

(5) F = [P (Q + R)]rsquo (6) F = [[(A + B)rsquo (C + D)rsquo]rsquo [(E + F)rsquo (G + H)rsquo]rsquo]rsquo

Sol Sol

F = [P (Q + R)]rsquo F = [[(A + B)rsquo (C + D)rsquo]rsquo [(E + F)rsquo (G + H)rsquo]rsquo]rsquo

F = Prsquo + (Q + R)rsquo there4 F = [(A + B)rsquo (C + D)rsquo]rsquorsquo + [(E + F)rsquo (G + H)rsquo]rsquorsquo

F = Prsquo + Qrsquo Rrsquo there4 F = [(A + B)rsquo (C + D)rsquo] + [(E + F)rsquo (G + H)rsquo]

there4 F = ArsquoBrsquoCrsquoDrsquo + ErsquoFrsquoGrsquoHrsquo

3512 Reduce the following functions using Boolean Algebrarsquos Laws and Theorems

(1) F = A + B [ AC + (B + Crsquo)D ] (2) F = A [ B + Crsquo (AB + ACrsquo)rsquo ]

Sol Sol

F = A + B [ AC + (B + Crsquo)D ] F = A [ B + Crsquo (AB + ACrsquo)rsquo ]

there4 F = A + B [ AC + (BD + CrsquoD) ] there4 F = A [ B + Crsquo (AB)rsquo (ACrsquo)rsquo ]

there4 F = A + ABC + BBD + BCrsquoD there4 F = A [ B + Crsquo (Arsquo + Brsquo) (Arsquo + C) ]

there4 F = A + ABC + BD + BCrsquoD there4 F = A [ B + (Arsquo Crsquo + Brsquo Crsquo) (Arsquo + C) ]

there4 F = A (1 + BC) + BD (1 + Crsquo) there4 F = A [ B + (Arsquo CrsquoArsquo + Brsquo CrsquoArsquo) (Arsquo Crsquo C + Brsquo Crsquo C) ]

there4 F = A (1) + BD (1) there4 F = A [ B + (ArsquoCrsquo + Brsquo CrsquoArsquo) (0 + 0) ]

there4 F = A + BD there4 F = A [ B + ArsquoCrsquo ( 1 + Brsquo) ]

there4 F = AB + ArsquoACrsquo

there4 F = AB


11


(3) F = (A + (BC)rsquo)rsquo (ABrsquo + ABC) (4) F = [(A + B) (Arsquo + B)] + [(A + B) (A + Brsquo)]

Sol Sol

F = (A + (BC)rsquo)rsquo (ABrsquo + ABC) F = [(A + B) (Arsquo + B)] + [(A + B) (A + Brsquo)]

there4 F = (Arsquo (BC)rsquorsquo) (ABrsquo + ABC) there4 F = [AArsquo + AB + BArsquo + BB] + [AA + ABrsquo + BA + BBrdquo]

there4 F = (ArsquoBC) (ABrsquo + ABC) there4 F = [0 + AB + ArsquoB + B] + [A + ABrsquo + AB + 0]

there4 F = ArsquoBCABrsquo + ArsquoBCABC there4 F = [B (A + Arsquo + 1)] + [A (1 + Brsquo + B)]

there4 F = 0 + 0 there4 F = B + A

there4 F = 0 there4 F = A + B

(5) F = [(A + Brsquo) (Arsquo + Brsquo)]+ [(Arsquo + Brsquo) (Arsquo + Brsquo)] (6) F = (A + B) (A + Brsquo) (Arsquo + B) (Arsquo + Brsquo)

Sol Sol

F = [(A + Brsquo) (Arsquo + Brsquo)] + [(Arsquo + Brsquo) (Arsquo + Brsquo)] F = (A + B) (A + Brsquo) (Arsquo + B) (Arsquo + Brsquo)

there4 F = [AArsquo + ABrsquo + BrsquoArsquo + BrsquoBrsquo] + [ArsquoArsquo + ArsquoBrsquo

+ BrsquoArsquo + BrsquoBrsquo]

there4 F = (AA + ABrsquo + BA + BBrsquo) (ArsquoArsquo + ArsquoBrsquo + BArsquo +

BBrsquo)

there4 F = [0 + ABrsquo + ArsquoBrsquo + Brsquo] + [Arsquo + ArsquoBrsquo + Brsquo] there4 F = [A (1+ Brsquo + B)] [Arsquo (1 + Brsquo + B)]

there4 F = [Brsquo (A + Arsquo + 1)] + [Arsquo + Brsquo(1)] there4 F = [A(1)] [Arsquo(1)]

there4 F = Brsquo + Arsquo + Brsquo there4 F = AArsquo

there4 F = Arsquo + Brsquo there4 F = 0

(7) F = (B + BC) (B + BrsquoC) (B + D) (8) F = ABrsquoC + B + BDrsquo + ABDrsquo + ArsquoC

Sol Reduce the function to minimum no of literals

F = (B + BC) (B + BrsquoC) (B + D) Sol

there4 F = (BB + BBrsquoC + BBC + BCBrsquoC) (B + D) F = ABrsquoC + B + BDrsquo + ABDrsquo + ArsquoC

there4 F = (B + 0 + BC + 0) (B + D) there4 F = ABrsquoC + B (1 + Drsquo + ADrsquo) + ArsquoC

there4 F = B (B + D) (∵B + BC = B(1 + C) = B) there4 F = ABrsquoC + B + ArsquoC

there4 F = B + BD (∵B (B + D) = BB + BD) there4 F = C (Arsquo + ABrsquo) + B

there4 F = B (∵B + BD = B (1 + D)) there4 F = C (Arsquo + A) (Arsquo + Brsquo) + B

there4 F = C (1) (Arsquo + Brsquo) + B

(9) F = AB + ABrsquoC +BCrsquo there4 F = C (Arsquo + Brsquo) + B

Sol there4 F = ArsquoC + CBrsquo + B

F = AB + ABrsquoC +BCrsquo there4 F = ArsquoC + (C + B) (Brsquo + B)

there4 F = A (B + BrsquoC) + BCrsquo there4 F = ArsquoC + (B + C) (1)

there4 F = A (B + Brsquo) (B + C) + BCrsquo there4 F = ArsquoC + B + C

there4 F = AB + AC + BCrsquo ∵ B + Brsquo = 1 there4 F = C (1 + Arsquo) + B

there4 F = CA + CrsquoB + AB there4 F = B + C

there4 F = CA + CrsquoB ∵ 119888119900119899119904119890119899119904119906119904 119905ℎ119890119900119903119890119898 Here 2 literals are present B amp C


12


36 Different forms of Boolean Algebra

There are two types of Boolean form 1) Standard form 2) Canonical form

361 Standard Form

Definition The terms that form the function may contain one two or any number of literals ie each term need not to contain all literals So standard form is simplified form of canonical form

A Boolean expression function may be expressed in a nonstandard form For example the function

F = (A + C) (ABrsquo + Drsquo)

Above function is neither sum of product nor in product of sums It can be changed to a standard form by using distributive law as below

F = ABrsquo + ADrsquo + ABrsquoC + CDrsquo

There are two types of standard forms (i) Sum of Product (SOP) (ii) Product of Sum (POS) 3611 Standard Sum of Product (SOP)

SOP is a Boolean expression containing AND terms called product terms of one or more literals each The sum denote the ORing of these terms

An example of a function expressed in sum of product is

F = Yrsquo + XY + XrsquoYZrsquo

3612 Standard Product of Sum (POS)

The OPS is a Boolean expression containing OR terms called sum terms Each terms may have any no of literals The product denotes ANDing of these terms

An example of a function expressed in product of sum is

F = X (Yrsquo + Z) (Xrsquo + Y + Zrsquo + W)

362 Canonical Form

Definition The terms that form the function contain all literals ie each term need to contain all literals

There are two types of canonical forms (i) Sum of Product (SOP) (ii) Product of Sum (POS)

3621 Sum of Product (SOP)

A canonical SOP form is one in which a no of product terms each one of which contains all the variables of the function either in complemented or non-complemented form summed together

Each of the product term is called ldquoMINTERMrdquo and denoted as lower case lsquomrsquo or lsquoƩrsquo

For minterms Each non-complemented variable 1 amp Each complemented variable 0

For example 1 XYZ = 111 = m7 2 ArsquoBC = 011 = m3

3 PrsquoQrsquoRrsquo = 000 = m0 4 TrsquoSrsquo = 00 = m0


13


36211 Convert to MINTERM

(1) F = PrsquoQrsquo + PQ (2) F = XrsquoYrsquoZ + XYrsquoZrsquo + XYZ

Sol Sol

F = 00 + 11 F = 001 + 100 + 111

there4 F = m0 + m3 there4 F = m1 + m4 + m7

there4 F = Σm(03) there4 F = Σm(147)

(3) F = XYrsquoZW + XYZrsquoWrsquo + XrsquoYrsquoZrsquoWrsquo

Sol

F = 1011 + 1100 + 0000

there4 F = m11 + m12 + m0

there4 F = Σm(01112)

3622 Product of Sum (POS)

A canonical POS form is one in which a no of sum terms each one of which contains all the variables of the function either in complemented or non-complemented form are multiplied together

Each of the product term is called ldquoMAXTERMrdquo and denoted as upper case lsquoMrsquo or lsquoΠrsquo

For maxterms Each non-complemented variable 0 amp Each complemented variable 1

For example 1 Xrsquo+Yrsquo+Z = 110 = M6 2 Arsquo+B+Crsquo+D = 1010 = M10

36221 Convert to MAXTERM

(1) F = (Prsquo+Q)(P+Qrsquo) (2) F = (Arsquo+B+C)(A+Brsquo+C)(A+B+Crsquo)

Sol Sol

F = (10)(01) F = (100) (010) (001)

there4 F = M2M1 there4 F = M4 M2 M1

there4 F = ΠM(12) there4 F = ΠM(124)

(3) F = (Xrsquo+Yrsquo+Zrsquo+W)(Xrsquo+Y+Z+Wrsquo)(X+Yrsquo+Z+Wrsquo)

Sol

F = (1110)(1001)(0101)

there4 F = M14M9M5

there4 F = ΠM(5914)


14


362 MINTERMS amp MAXTERMS for 3 Variables

Table Representation of Minterms and Maxterms for 3 variables

363 Conversion between Canonical Forms

3631 Convert to MINTERMS

1 F(ABCD) = ΠM(037101415)

Solution

Take complement of the given function

there4 Frsquo(ABCD) = ΠM(1245689111213)

there4 Frsquo(ABCD) = (M1 M2 M4 M5 M6 M8 M9 M11 M12 M13)rsquo

Put value of MAXTERM in form of variables there4 Frsquo(ABCD) = [(A+B+C+Drsquo)(A+B+Crsquo+D)(A+Brsquo+C+D)(A+Brsquo+C+Drsquo)(A+Brsquo+Crsquo+D)

(Arsquo+B+C+D)(Arsquo+B+C+Drsquo)(Arsquo+B+Crsquo+Drsquo)(Arsquo+Brsquo+C+D)(Arsquo+Brsquo+C+Drsquo)]rsquo there4 Frsquo(ABCD) = (ArsquoBrsquoCrsquoD) + (ArsquoBrsquoCDrsquo) + (ArsquoBCrsquoDrsquo) + (ArsquoBCrsquoD) + (ArsquoBCDrsquo)

+ (ABrsquoCrsquoDrsquo) + (ABrsquoCrsquoD) + (ABrsquoCD) + (ABCrsquoDrsquo) + (ABCrsquoD) there4 Frsquo(ABCD) = m1 + m2 + m4 + m5 + m6 + m8 + m9 + m11 + m12 + m13

there4 Frsquo(ABCD) = Σm(1245689111213)

In general Mjrsquo = mj


15


2 F = A + BrsquoC

Solution

A B amp C is missing So multiply with (B + Brsquo) amp (C + Crsquo)

BrsquoC A is missing So multiply with (A + Arsquo) there4 A = A (B + Brsquo) (C + Crsquo)

there4 A = (AB + ABrsquo) (C +Crsquo)

there4 A = ABC + ABrsquoC + ABCrsquo + ABrsquoCrsquo

And BrsquoC = BrsquoC (A + Arsquo)

there4 BrsquoC = ABrsquoC + ArsquoBrsquoC

So F = ABC + ABrsquoC + ABCrsquo + ABrsquoCrsquo + ABrsquoC + ArsquoBrsquoC

there4 F = ABC + ABrsquoC + ABCrsquo + ABrsquoCrsquo + ArsquoBrsquoC

there4 F = 111 + 101 + 110 + 100 + 001

there4 F = m7 + m6 + m5 + m4 + m1


3632 Convert to MAXTERMS

1 F = Σ(14567)

Solution


there4 Frsquo(ABC) = Σ(023)

there4 Frsquo(ABC) = (m0 + m2 + m3)

Put value of MINTERM in form of variables there4 Frsquo(ABC) = (ArsquoBrsquoCrsquo + ArsquoBCrsquo + ArsquoBC)rsquo

there4 Frsquo(ABC) = (A+B+C)(A+Brsquo+C)(A+Brsquo+Crsquo)

there4 Frsquo(ABC) = M0M2M3

there4 Frsquo(ABC) = ΠM(023)

In general mjrsquo = Mj

2 F = A (B + Crsquo)

Solution

A B amp C is missing So add BBrsquo amp CCrsquo

B + Crsquo A is missing So add AArsquo there4 A = A + BBrsquo + CCrsquo


16


there4 A = (A + B + CCrsquo) (A + Brsquo + CCrsquo)

there4 A = (A + B + C) (A + B + Crsquo) (A + Brsquo + C) (A + Brsquo + Crsquo)

And B + Crsquo = B + Crsquo + AArsquo

there4 B + Crsquo = (A + B + Crsquo) (Arsquo + B + Crsquo)

So F = (A + B + C) (A + B + Crsquo) (A + Brsquo + C) (A + Brsquo + Crsquo) (A + B + Crsquo) (Arsquo + B + Crsquo)

there4 F = (A + B + C) (A + B + Crsquo) (A + Brsquo + C) (A + Brsquo + Crsquo) (Arsquo + B + Crsquo)

there4 F = (000) (001) (010) (011) (101)


3 F = XY + XrsquoZ

Solution

there4 F = XY + XrsquoZ

there4 F = (XY + Xrsquo) (XY + Z)

there4 F = (X +Xrsquo) (Y + Xrsquo) (X + Z) (Y + Z)

there4 F = (Xrsquo + Y) (X + Z) (Y + Z)

Xrsquo + Y Z is missing So add ZZrsquo

X + Z Y is missing So add YYrsquo Y + Z X is missing So add XXrsquo

there4 Xrsquo + Y = Xrsquo + Y + ZZrsquo

there4 Xrsquo + Y = (Xrsquo + Y + Z) (Xrsquo + Y + Zrsquo)

And X + Z = Xrsquo + Z + YYrsquo

there4 X + Z = (X + Y + Z) (X + Yrsquo + Z)

And Y + Z = Y + Z + XXrsquo

there4 Y + Z = (X + Y + Z) (Xrsquo + Y + Z)

So F = (Xrsquo + Y + Z) (Xrsquo + Y + Zrsquo) (X + Y + Z) (X + Yrsquo + Z) (X + Y + Z) (Xrsquo + Y + Z)

there4 F = (100) (101) (000) (010)

there4 F = M4 M5 M0 M2



17


37 Karnaugh Map (K-Map)

A Boolean expression may have many different forms

With the use of K-map the complexity of reducing expression becomes easy and Boolean expression obtained is simplified

K-map is a pictorial form of truth table and it is alternative way of simplifying Boolean function

Instead of using Boolean algebra simplification techniques you can transfer logic values from a Boolean statement or a truth table into a Karnaugh map (k-map)

Tool for representing Boolean functions of up to six variables then after it becomes complex

K-maps are tables of rows and columns with entries represent 1rsquos or 0rsquos of SOP and POS representations

K-map cells are arranged such that adjacent cells correspond to truth table rows that differ in only one bit position (logical adjacency)

K-Map are often used to simplify logic problems with up to 6 variables

No of Cells = 2 n where n is a number of variables The Karnaugh map is completed by entering a lsquo1rsquo (or lsquo0rsquo) in each of the appropriate cells

Within the map adjacent cells containing 1s (or 0rsquos) are grouped together in twos fours or eights and so on

371 2 Variable K-Map

For 2 variable k-map there are 22 = 4 cells

If A amp B are two variables then SOP Minterms ArsquoBrsquo (m0 00) ArsquoB (m1 01) ABrsquo (m2 10) AB (m3 11) POS Maxterms A + B (M0 00) A + Brsquo (M1 01) Arsquo + B (M2 10) Arsquo + Brsquo (M3 11)

3711 Mapping of SOP Expression

1 in a cell indicates that the minterm is

included in Boolean expression

For eg if F = summ(023) then 1 is put in cell no 023 as shown below

1

11

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

3712 Mapping of POS Expression

0 in a cell indicates that the maxterm is


For eg if F = ΠM(023) then 0 is put in cell no 023 as shown below

0

00

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1


18


3713 Reduce Sum of Product (SOP) Expression using K-Map

(1) F = m0 + m1 (2) F = ArsquoBrsquo + ABrsquo (3) F = Σ(13) Sol

1

00

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

01

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

0

10

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

F = Arsquo F = Brsquo F = B (4) F = m2 + m3 (5) summ(013) (6) summ(0123) Sol

0

11

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

11

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

11

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

F = A F = Arsquo + AB F = 1

3714 Reduce Product of Sum (POS) Expression using K-Map

(1) F = ΠM(0231) (2) F = (A+B) (Arsquo+B) (A+Brsquo) (3) F = M3M1M2 Sol

0

00

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1

Sol

0

10

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1

Sol

1

00

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1

F = 0 F = A B F = Arsquo Brsquo





0


19



(1) F = ArsquoBrsquoC + ABC + ArsquoBCrsquo (2) F = Σ(167) Sol

Sol

F = ArsquoBrsquoC + ABC + ArsquoBCrsquo F = ArsquoBrsquoC + AB (3) F = ArsquoBrsquoCrsquo + ABCrsquo + ABrsquoCrsquo + ArsquoBC (4) F = Σm(012456) Sol

Sol

F = BrsquoCrsquo + ACrsquo + ArsquoBC F = Brsquo + Crsquo (5) F = m3 + m4 +m6 + m7 (6) F = Σm(37160254) Sol

Sol

F = BC + ACrsquo F = 1


(1) F = (Arsquo+Brsquo+Crsquo) (Arsquo+B+Crsquo) (2) F = M0M3M7 Sol

Sol

F = (Arsquo + Crsquo) F = (A + B + C) (Brsquo + Crsquo)


20


(3) F = ΠM(125) (4) F = ΠM(041573) Sol

Sol

F = (B + Crsquo) (A + B + C) F = (B) (Crsquo) (5) F = (A+B+C)(A+Brsquo+Crsquo)(Arsquo+B+C) (6) ΠM(57032461) Sol

Sol

F = (B + C) (A + Brsquo + Crsquo) F = 0 373 4 Variable K-Map





21


3733 Looping of POS Expression

Looping Groups of Two

Looping Groups of Four


22


Looping Groups of Eight

Examples


23



(1) F = sum (01245689121314) (2) F = ArsquoBrsquoCrsquo + BrsquoCDrsquo + ArsquoBCDrsquo + ABrsquoCrsquo Sol

Sol

F = Crsquo + ArsquoDrsquo + BDrsquo F = BrsquoCrsquo + BrsquoDrsquo + ArsquoCDrsquo (3) F = sum (012357891213) (4) F = sum (01345671315) Sol

Sol

F = ArsquoBrsquo + ACrsquo + ArsquoD F = ArsquoCrsquo + ArsquoD + BD + ArsquoB


24


(5) F = summ (56791011131415) Sol

F = BD + BC + AD + AC


(1) F = ΠM (0125789101415) (2) F = M1 M3 M4 M7 M6 M9 M11 M12 M14 M15 Sol

Sol

F = (B + D) (B + C) (A + Brsquo + Drsquo) (Arsquo + Brsquo + Crsquo) F = (Brsquo + D) (Crsquo + Brsquo) (Drsquo + B)

3736 Reduce SOP amp POS Expression with Donrsquot Care Combination using K-Map

(1) F = summ (156121314) + d (24) (2) F = ΠM (4710111215) middot d (68) Sol

Sol

F = BCrsquo + BDrsquo + ArsquoCrsquoD F = (Brsquo + C + D) (Brsquo + Crsquo + Drsquo) (Arsquo + B + Crsquo)


25






(1) F = summ (023101112131617181920212627) Sol

F = CrsquoD + BrsquoCrsquoErsquo + ABrsquoDrsquo + ArsquoBCDrsquo

(2) F = summ (02469111315172125272931) Sol

F = BE + ADrsquoE + ArsquoBrsquoErsquo


26



(1) F = ΠM (145678914152223242528293031) Sol

F = (Brsquo + C + D) (Crsquo +Drsquo) (A + B + Crsquo) (Arsquo + Brsquo + D) (A + C + D + Ersquo)

38 Converting Boolean Expression to Logic Circuit and Vice-Versa

(1) For the logic circuit shown fig find the Boolean expression and the truth table Identify the gate that given circuit realizes


Sol Sol

Here Output of OR gate will be (A+B) Output of NAND gate will be (AB)rsquo So C will be AND of these two outputs

Here bubble indicates inversion Hence input of top OR gate is Arsquo and Brsquo and hence its output will be Arsquo+Brsquo Output of bottom OR gate will be A+B

there4 C = (A+B) middot (AmiddotB)rsquo there4 Y = (Arsquo+Brsquo)(A+B)

Truth table for the same can be given below Truth table for the same can be given below Input Output

A B A+B AmiddotB (AmiddotB)rsquo (A+B)middot(AmiddotB)rsquo

0 0 0 0 1 0

0 1 1 0 1 1

1 0 1 0 1 1

1 1 1 1 0 0

Input Output

A B Arsquo Brsquo Arsquo+Brsquo A+B (Arsquo+Brsquo)(A+B)

0 0 1 1 1 0 0

0 1 1 0 1 1 1

1 0 0 1 1 1 1

1 1 0 0 0 1 0

From the truth table it is clear that the circuit realizes Ex-OR gate


NOTE NAND = Bubbled OR


27


(3) For the logic circuit shown fig find the Boolean expression

(4) For the given Boolean expression draw the logic circuit F = X + (Yrsquo + Z)

Sol The expression primarily involves three logic gates ie NOT AND and OR To generate Yrsquo a NOT gate is required To generate YrsquoZ an AND gate is required To generate final output OR gate is required

Sol

Here Output of top AND gate will be ABrsquo Output of bottom AND gate will be ArsquoB So C will be OR of these two outputs

there4 C = ABrsquo + ArsquoB

(5) F = sum (01245689121314) Reduce using

k-map method Also realize it with logic circuit or gates with minimum no of gates

(6) F = summ (156121314) + d (24) Reduce using k-map method Also realize it with logic circuit

Sol

Sol

F = Crsquo + ArsquoDrsquo + BDrsquo F = BCrsquo + BDrsquo + ArsquoCrsquoD

Realization of function with logic circuit Realization of function with logic circuit


28


39 NAND and NOR RealizationImplementation

Steps to implement any function using NAND or NOR gate only as below 1 Reduce the given function if necessary 2 For NAND add Bubbles at the outputs of AND gates and at the inputs of OR gates 3 For NOR add Bubbles at the outputs of OR gates and at the inputs of AND gates 4 Add an inverter symbol wherever you created a Bubble 5 Ignore cascading connection of two NOT gates if any are present 6 Replace all gates with NAND gates or NOR gates depending on the type of implementation

(1) F = AB + CD + E Implement given function using (i) NAND gates only (ii) NOR gates only Sol

Function realization using basic logic gates as below

Using NAND gates Using NOR gates Step1 Step1

Step2 Step2

Step3 Step3


29


Step4 Step4

310 Tabulation Quine-McCluskey Method

As we know that the Karnaugh map method is a very useful and convenient tool for simplification of Boolean functions as long as the number of variables does not exceed four

But for case of large number of variables the visualization and selection of patterns of adjacent cells in the Karnaugh map becomes complicated and too much difficult For those cases Quine McCluskey tabulation method takes vital role to simplify the Boolean expression

The Quine McCluskey tabulation method is a specific step-by-step procedure to achieve guaranteed simplified standard form of expression for a function

Steps to solve function using tabulation method are as follow

Step 1 minus Arrange the given min terms in an ascending order and make the groups based on the number of ones present in their binary representations So there will be at most lsquon+1rsquo groups if there are lsquonrsquo Boolean variables in a Boolean function or lsquonrsquo bits in the binary equivalent of min terms

Step 2 minus Compare the min terms present in successive groups If there is a change in only one-bit position then take the pair of those two min terms Place this symbol lsquo_rsquo in the differed bit position and keep the remaining bits as it is

Step 3 minus Repeat step2 with newly formed terms till we get all prime implicants

Step 4 minus Formulate the prime implicant table It consists of set of rows and columns Prime implicants can be placed in row wise and min terms can be placed in column wise Place lsquo1rsquo in the cells corresponding to the min terms that are covered in each prime implicant

Step 5 minus Find the essential prime implicants by observing each column If the min term is covered only by one prime implicant then it is essential prime implicant Those essential prime implicants will be part of the simplified Boolean function

Step 6 minus Reduce the prime implicant table by removing the row of each essential prime implicant and the columns corresponding to the min terms that are covered in that essential prime implicant Repeat step 5 for reduced prime implicant table Stop this process when all min terms of given Boolean function are over


30


(1) Simplify the following expression to sum of product using Tabulation Method

119917(119912 119913 119914 119915) = sum(120782 120783 120784 120785 120786 120788 120789 120783120783 120783120784 120783120787)

Sol

Determination of Prime Implicants


119917(119912 119913 119914 119915) = 119913119914prime119915prime + 119912prime119913prime + 119914119915 + 119912prime119915prime


31


(2) Simplify the following expression to sum of product using Tabulation Method 119917(119912 119913 119914 119915) = 119950(120782 120786 120790 120783120782 120783120784 120783120785 120783120787) + 119941(120783 120784)

Sol


Determination of Essential Prime Implicants

119917(119912 119913 119914 119915) = 119912119913119915 + 119914prime119915prime + 119913prime119915prime


32


(3) Simplify the following expression to sum of product using Tabulation Method 119917(119912 119913 119914 119915) = 120619(120783 120785 120787 120789 120783120785 120783120787)

Sol



119917(119912 119913 119914 119915) = (119912 + 119915prime)(119913prime + 119915prime)


119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol



119917(119912 119913 119914 119915) = (119912prime + 119913prime + 119915prime)(119913 + 119915)(119912prime + 119913prime + 119914)


33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3

A State and explain De Morganrsquos theorems with truth tables

7 7 35_8

A

Apply De Morganrsquos theorem to solve the following (1) [A + (BC)] [AB + ABC] = 0 (2) A [ B + C (AB + AC)] = AB

7 35_8

A Reduce the expression (1) A + B (AC + (B+Crsquo) D) (2) (A + (BC)rsquo )rsquo(ABrsquo + ABC)

4 3512_10

A Demonstrate by means of truth tables the validity of the De Morganrsquos theorems for three variables

3 35_8

A

Simplify the following Boolean functions to a minimum number of literals (i) F(xyz)=xy+xyz+xyzrsquo+xrsquoyz (ii) F(p q r s) = (prsquo+q) (p+q+s)srsquo

4 3512_11

A State and prove De Morganrsquos theorems 7 35_8

A Reduce the expression F = (B+BC) (B+BC)(B+D)

3 3512_11

B Explain minterm and maxterm 3 3 3621_12 3622_13

B Compare SOP and POS 3 3611_12

B

Give examples of standard and nonstandard SOP and POS forms Explain how a NON standard POS expression can be converted in to standard POS expression using example you have given

7 36_12

B Express the Boolean function F=A+AC in a sum of min-terms

7 3631_15

BampC Express ArsquoB + ArsquoC as sum of minterms and also plot K-map

7 3631_15

372_18

C

What are SOP and POS forms of Boolean expressions Minimize the following expression using K-map Y= Σm(457121415) + d( 3810)

7 3611_12 3736_24

C Write short note on K-map OR Explain K-map simplification technique

7 7 37_17


34


3

C Simplify the following Boolean function using K-map F (w x y z) = Σ m(1 3 7 11 15) with donrsquot care d(w x y z) = Σm(025)

3 3736_24

C

Simplify the Boolean function F = ArsquoBrsquoCrsquo+ABrsquoD+ArsquoBrsquoCDrsquo using donrsquot-care conditions d=ABC+ABrsquoDrsquo in (i) sum of products and (ii) product of sums by means of Karnaugh map and implement it with no more than two NOR gates Assume that both the normal and complement inputs are available

7 3736_24

39_28

C Reduce using K-map (i)Σm(56791011131415) (ii)ΠM(156711121315)

4 3734_23 3735_24

C Minimize using K-map f(ABCD) = Σ(134681115) +d(057) also draw MSI circuit for the output

7 3736_24

C

Simplify equation using K-map F(abcd) = Σm(371112131415) Realize the expression with minimum number of gates

4 3734_24

38_26

C

Minimize the following Boolean expression using K- Map and realize it using logic gates F(ABCD)=Σm(0159131415)+d(3471011)

7 3736_24

CampD Compare K-map and tabular method of minimization

3 37_17

310_29

D Simplify following Boolean function using tabulation method F(w x y z) = Σ (0 1 2 8 10 11 14 15)

7 310_29

D Simplify the Boolean function F(x1x2x3x4)=Σm(0578910111415) using tabulation method

7 310_29

E Simplify Y=ArsquoBCDrsquo + BCDrsquo + BCrsquoDrsquo + BCrsquoD and implement using NAND gates only

7 39_28

E Implement following Boolean function using only NAND gates Y=ABCrsquo+ABC+ArsquoBC

4 39_28


2








3733 Looping of POS Expression 21



3736 Reduce SOP amp POS Expression with Donrsquot Care Combination using K-Map 24





38 Converting Boolean Expression to Logic Circuit and Vice-Versa 26

39 NAND and NOR RealizationImplementation 28

310 Tabulation Quine-McCluskey Method 29

311 GTU Questions 33


3


31 Introduction










1 + 1 = 1 0 1 + 1 = 2 1 + 1 = 1











4







eg F(A B C) = (119860 + 119861) C OR F = (119860 + 119861) C

33 Logic Operators










2 AND Laws



3 OR Laws

Law 1 A + 0 = A Law 2 A + 1 = 1



5


4 Commutative Laws








A + (B + C + D) = (A + B + C) + D = (A + B) + (C + D)


6





6 Distributive Laws




7 Idempotence Laws




7















8














9



LHS

NAND Gate

A

B

AB (AB)rsquo=

RHS

B

A Arsquo

Brsquo

Arsquo + Brsquo

=

Bubbled OR

A

B

(AB)rsquo

A

B

Arsquo + Brsquo

A B AB (AB)rsquo

0 0 0 1

0 1 0 1

1 0 0 1

0 1 1 0


0 0 1 1 1

0 1 1 0 1

1 0 0 1 1

0 1 0 0 0

16 Duality Theorem




=


10





Sol Sol






Sol Sol







Sol Sol







Sol Sol









there4 F = AB


11



Sol Sol








Sol Sol





BBrsquo)























12




361 Standard Form














362 Canonical Form










13




Sol Sol

F = 00 + 11 F = 001 + 100 + 111




Sol

F = 1011 + 1100 + 0000

there4 F = m11 + m12 + m0









Sol Sol

F = (10)(01) F = (100) (010) (001)




Sol

F = (1110)(1001)(0101)

there4 F = M14M9M5



14






1 F(ABCD) = ΠM(037101415)

Solution










15


2 F = A + BrsquoC

Solution









there4 F = 111 + 101 + 110 + 100 + 001

there4 F = m7 + m6 + m5 + m4 + m1



1 F = Σ(14567)

Solution










Solution




16








there4 F = (000) (001) (010) (011) (101)


3 F = XY + XrsquoZ

Solution














there4 F = (100) (101) (000) (010)




17




















1

11

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1





0

00

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1


18




1

00

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

01

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

0

10

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1


0

11

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

11

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

11

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1




0

00

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1

Sol

0

10

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1

Sol

1

00

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1






0


19




Sol


Sol


Sol




Sol



20


(3) F = ΠM(125) (4) F = ΠM(041573) Sol

Sol


Sol






21






22



Examples


23




Sol


Sol



24


(5) F = summ (56791011131415) Sol




Sol




Sol



25






(1) F = summ (023101112131617181920212627) Sol


(2) F = summ (02469111315172125272931) Sol



26



(1) F = ΠM (145678914152223242528293031) Sol





Sol Sol






0 0 0 0 1 0

0 1 1 0 1 1

1 0 1 0 1 1

1 1 1 1 0 0

Input Output


0 0 1 1 1 0 0

0 1 1 0 1 1 1

1 0 0 1 1 1 1

1 1 0 0 0 1 0





27





Sol






Sol

Sol




28







Step2 Step2

Step3 Step3


29


Step4 Step4













30



119917(119912 119913 119914 119915) = sum(120782 120783 120784 120785 120786 120788 120789 120783120783 120783120784 120783120787)

Sol





31



Sol





32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


3


31 Introduction










1 + 1 = 1 0 1 + 1 = 2 1 + 1 = 1











4







eg F(A B C) = (119860 + 119861) C OR F = (119860 + 119861) C

33 Logic Operators










2 AND Laws



3 OR Laws

Law 1 A + 0 = A Law 2 A + 1 = 1



5


4 Commutative Laws








A + (B + C + D) = (A + B + C) + D = (A + B) + (C + D)


6





6 Distributive Laws




7 Idempotence Laws




7















8














9



LHS

NAND Gate

A

B

AB (AB)rsquo=

RHS

B

A Arsquo

Brsquo

Arsquo + Brsquo

=

Bubbled OR

A

B

(AB)rsquo

A

B

Arsquo + Brsquo

A B AB (AB)rsquo

0 0 0 1

0 1 0 1

1 0 0 1

0 1 1 0


0 0 1 1 1

0 1 1 0 1

1 0 0 1 1

0 1 0 0 0

16 Duality Theorem




=


10





Sol Sol






Sol Sol







Sol Sol







Sol Sol









there4 F = AB


11



Sol Sol








Sol Sol





BBrsquo)























12




361 Standard Form














362 Canonical Form










13




Sol Sol

F = 00 + 11 F = 001 + 100 + 111




Sol

F = 1011 + 1100 + 0000

there4 F = m11 + m12 + m0









Sol Sol

F = (10)(01) F = (100) (010) (001)




Sol

F = (1110)(1001)(0101)

there4 F = M14M9M5



14






1 F(ABCD) = ΠM(037101415)

Solution










15


2 F = A + BrsquoC

Solution









there4 F = 111 + 101 + 110 + 100 + 001

there4 F = m7 + m6 + m5 + m4 + m1



1 F = Σ(14567)

Solution










Solution




16








there4 F = (000) (001) (010) (011) (101)


3 F = XY + XrsquoZ

Solution














there4 F = (100) (101) (000) (010)




17




















1

11

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1





0

00

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1


18




1

00

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

01

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

0

10

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1


0

11

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

11

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

11

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1




0

00

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1

Sol

0

10

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1

Sol

1

00

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1






0


19




Sol


Sol


Sol




Sol



20


(3) F = ΠM(125) (4) F = ΠM(041573) Sol

Sol


Sol






21






22



Examples


23




Sol


Sol



24


(5) F = summ (56791011131415) Sol




Sol




Sol



25






(1) F = summ (023101112131617181920212627) Sol


(2) F = summ (02469111315172125272931) Sol



26



(1) F = ΠM (145678914152223242528293031) Sol





Sol Sol






0 0 0 0 1 0

0 1 1 0 1 1

1 0 1 0 1 1

1 1 1 1 0 0

Input Output


0 0 1 1 1 0 0

0 1 1 0 1 1 1

1 0 0 1 1 1 1

1 1 0 0 0 1 0





27





Sol






Sol

Sol




28







Step2 Step2

Step3 Step3


29


Step4 Step4













30



119917(119912 119913 119914 119915) = sum(120782 120783 120784 120785 120786 120788 120789 120783120783 120783120784 120783120787)

Sol





31



Sol





32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


4







eg F(A B C) = (119860 + 119861) C OR F = (119860 + 119861) C

33 Logic Operators










2 AND Laws



3 OR Laws

Law 1 A + 0 = A Law 2 A + 1 = 1



5


4 Commutative Laws








A + (B + C + D) = (A + B + C) + D = (A + B) + (C + D)


6





6 Distributive Laws




7 Idempotence Laws




7















8














9



LHS

NAND Gate

A

B

AB (AB)rsquo=

RHS

B

A Arsquo

Brsquo

Arsquo + Brsquo

=

Bubbled OR

A

B

(AB)rsquo

A

B

Arsquo + Brsquo

A B AB (AB)rsquo

0 0 0 1

0 1 0 1

1 0 0 1

0 1 1 0


0 0 1 1 1

0 1 1 0 1

1 0 0 1 1

0 1 0 0 0

16 Duality Theorem




=


10





Sol Sol






Sol Sol







Sol Sol







Sol Sol









there4 F = AB


11



Sol Sol








Sol Sol





BBrsquo)























12




361 Standard Form














362 Canonical Form










13




Sol Sol

F = 00 + 11 F = 001 + 100 + 111




Sol

F = 1011 + 1100 + 0000

there4 F = m11 + m12 + m0









Sol Sol

F = (10)(01) F = (100) (010) (001)




Sol

F = (1110)(1001)(0101)

there4 F = M14M9M5



14






1 F(ABCD) = ΠM(037101415)

Solution










15


2 F = A + BrsquoC

Solution









there4 F = 111 + 101 + 110 + 100 + 001

there4 F = m7 + m6 + m5 + m4 + m1



1 F = Σ(14567)

Solution










Solution




16








there4 F = (000) (001) (010) (011) (101)


3 F = XY + XrsquoZ

Solution














there4 F = (100) (101) (000) (010)




17




















1

11

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1





0

00

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1


18




1

00

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

01

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

0

10

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1


0

11

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

11

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

11

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1




0

00

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1

Sol

0

10

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1

Sol

1

00

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1






0


19




Sol


Sol


Sol




Sol



20


(3) F = ΠM(125) (4) F = ΠM(041573) Sol

Sol


Sol






21






22



Examples


23




Sol


Sol



24


(5) F = summ (56791011131415) Sol




Sol




Sol



25






(1) F = summ (023101112131617181920212627) Sol


(2) F = summ (02469111315172125272931) Sol



26



(1) F = ΠM (145678914152223242528293031) Sol





Sol Sol






0 0 0 0 1 0

0 1 1 0 1 1

1 0 1 0 1 1

1 1 1 1 0 0

Input Output


0 0 1 1 1 0 0

0 1 1 0 1 1 1

1 0 0 1 1 1 1

1 1 0 0 0 1 0





27





Sol






Sol

Sol




28







Step2 Step2

Step3 Step3


29


Step4 Step4













30



119917(119912 119913 119914 119915) = sum(120782 120783 120784 120785 120786 120788 120789 120783120783 120783120784 120783120787)

Sol





31



Sol





32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


5


4 Commutative Laws








A + (B + C + D) = (A + B + C) + D = (A + B) + (C + D)


6





6 Distributive Laws




7 Idempotence Laws




7















8














9



LHS

NAND Gate

A

B

AB (AB)rsquo=

RHS

B

A Arsquo

Brsquo

Arsquo + Brsquo

=

Bubbled OR

A

B

(AB)rsquo

A

B

Arsquo + Brsquo

A B AB (AB)rsquo

0 0 0 1

0 1 0 1

1 0 0 1

0 1 1 0


0 0 1 1 1

0 1 1 0 1

1 0 0 1 1

0 1 0 0 0

16 Duality Theorem




=


10





Sol Sol






Sol Sol







Sol Sol







Sol Sol









there4 F = AB


11



Sol Sol








Sol Sol





BBrsquo)























12




361 Standard Form














362 Canonical Form










13




Sol Sol

F = 00 + 11 F = 001 + 100 + 111




Sol

F = 1011 + 1100 + 0000

there4 F = m11 + m12 + m0









Sol Sol

F = (10)(01) F = (100) (010) (001)




Sol

F = (1110)(1001)(0101)

there4 F = M14M9M5



14






1 F(ABCD) = ΠM(037101415)

Solution










15


2 F = A + BrsquoC

Solution









there4 F = 111 + 101 + 110 + 100 + 001

there4 F = m7 + m6 + m5 + m4 + m1



1 F = Σ(14567)

Solution










Solution




16








there4 F = (000) (001) (010) (011) (101)


3 F = XY + XrsquoZ

Solution














there4 F = (100) (101) (000) (010)




17




















1

11

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1





0

00

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1


18




1

00

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

01

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

0

10

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1


0

11

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

11

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

11

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1




0

00

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1

Sol

0

10

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1

Sol

1

00

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1






0


19




Sol


Sol


Sol




Sol



20


(3) F = ΠM(125) (4) F = ΠM(041573) Sol

Sol


Sol






21






22



Examples


23




Sol


Sol



24


(5) F = summ (56791011131415) Sol




Sol




Sol



25






(1) F = summ (023101112131617181920212627) Sol


(2) F = summ (02469111315172125272931) Sol



26



(1) F = ΠM (145678914152223242528293031) Sol





Sol Sol






0 0 0 0 1 0

0 1 1 0 1 1

1 0 1 0 1 1

1 1 1 1 0 0

Input Output


0 0 1 1 1 0 0

0 1 1 0 1 1 1

1 0 0 1 1 1 1

1 1 0 0 0 1 0





27





Sol






Sol

Sol




28







Step2 Step2

Step3 Step3


29


Step4 Step4













30



119917(119912 119913 119914 119915) = sum(120782 120783 120784 120785 120786 120788 120789 120783120783 120783120784 120783120787)

Sol





31



Sol





32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


6





6 Distributive Laws




7 Idempotence Laws




7















8














9



LHS

NAND Gate

A

B

AB (AB)rsquo=

RHS

B

A Arsquo

Brsquo

Arsquo + Brsquo

=

Bubbled OR

A

B

(AB)rsquo

A

B

Arsquo + Brsquo

A B AB (AB)rsquo

0 0 0 1

0 1 0 1

1 0 0 1

0 1 1 0


0 0 1 1 1

0 1 1 0 1

1 0 0 1 1

0 1 0 0 0

16 Duality Theorem




=


10





Sol Sol






Sol Sol







Sol Sol







Sol Sol









there4 F = AB


11



Sol Sol








Sol Sol





BBrsquo)























12




361 Standard Form














362 Canonical Form










13




Sol Sol

F = 00 + 11 F = 001 + 100 + 111




Sol

F = 1011 + 1100 + 0000

there4 F = m11 + m12 + m0









Sol Sol

F = (10)(01) F = (100) (010) (001)




Sol

F = (1110)(1001)(0101)

there4 F = M14M9M5



14






1 F(ABCD) = ΠM(037101415)

Solution










15


2 F = A + BrsquoC

Solution









there4 F = 111 + 101 + 110 + 100 + 001

there4 F = m7 + m6 + m5 + m4 + m1



1 F = Σ(14567)

Solution










Solution




16








there4 F = (000) (001) (010) (011) (101)


3 F = XY + XrsquoZ

Solution














there4 F = (100) (101) (000) (010)




17




















1

11

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1





0

00

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1


18




1

00

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

01

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

0

10

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1


0

11

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

11

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

11

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1




0

00

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1

Sol

0

10

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1

Sol

1

00

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1






0


19




Sol


Sol


Sol




Sol



20


(3) F = ΠM(125) (4) F = ΠM(041573) Sol

Sol


Sol






21






22



Examples


23




Sol


Sol



24


(5) F = summ (56791011131415) Sol




Sol




Sol



25






(1) F = summ (023101112131617181920212627) Sol


(2) F = summ (02469111315172125272931) Sol



26



(1) F = ΠM (145678914152223242528293031) Sol





Sol Sol






0 0 0 0 1 0

0 1 1 0 1 1

1 0 1 0 1 1

1 1 1 1 0 0

Input Output


0 0 1 1 1 0 0

0 1 1 0 1 1 1

1 0 0 1 1 1 1

1 1 0 0 0 1 0





27





Sol






Sol

Sol




28







Step2 Step2

Step3 Step3


29


Step4 Step4













30



119917(119912 119913 119914 119915) = sum(120782 120783 120784 120785 120786 120788 120789 120783120783 120783120784 120783120787)

Sol





31



Sol





32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


7















8














9



LHS

NAND Gate

A

B

AB (AB)rsquo=

RHS

B

A Arsquo

Brsquo

Arsquo + Brsquo

=

Bubbled OR

A

B

(AB)rsquo

A

B

Arsquo + Brsquo

A B AB (AB)rsquo

0 0 0 1

0 1 0 1

1 0 0 1

0 1 1 0


0 0 1 1 1

0 1 1 0 1

1 0 0 1 1

0 1 0 0 0

16 Duality Theorem




=


10





Sol Sol






Sol Sol







Sol Sol







Sol Sol









there4 F = AB


11



Sol Sol








Sol Sol





BBrsquo)























12




361 Standard Form














362 Canonical Form










13




Sol Sol

F = 00 + 11 F = 001 + 100 + 111




Sol

F = 1011 + 1100 + 0000

there4 F = m11 + m12 + m0









Sol Sol

F = (10)(01) F = (100) (010) (001)




Sol

F = (1110)(1001)(0101)

there4 F = M14M9M5



14






1 F(ABCD) = ΠM(037101415)

Solution










15


2 F = A + BrsquoC

Solution









there4 F = 111 + 101 + 110 + 100 + 001

there4 F = m7 + m6 + m5 + m4 + m1



1 F = Σ(14567)

Solution










Solution




16








there4 F = (000) (001) (010) (011) (101)


3 F = XY + XrsquoZ

Solution














there4 F = (100) (101) (000) (010)




17




















1

11

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1





0

00

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1


18




1

00

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

01

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

0

10

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1


0

11

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

11

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

11

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1




0

00

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1

Sol

0

10

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1

Sol

1

00

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1






0


19




Sol


Sol


Sol




Sol



20


(3) F = ΠM(125) (4) F = ΠM(041573) Sol

Sol


Sol






21






22



Examples


23




Sol


Sol



24


(5) F = summ (56791011131415) Sol




Sol




Sol



25






(1) F = summ (023101112131617181920212627) Sol


(2) F = summ (02469111315172125272931) Sol



26



(1) F = ΠM (145678914152223242528293031) Sol





Sol Sol






0 0 0 0 1 0

0 1 1 0 1 1

1 0 1 0 1 1

1 1 1 1 0 0

Input Output


0 0 1 1 1 0 0

0 1 1 0 1 1 1

1 0 0 1 1 1 1

1 1 0 0 0 1 0





27





Sol






Sol

Sol




28







Step2 Step2

Step3 Step3


29


Step4 Step4













30



119917(119912 119913 119914 119915) = sum(120782 120783 120784 120785 120786 120788 120789 120783120783 120783120784 120783120787)

Sol





31



Sol





32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


8














9



LHS

NAND Gate

A

B

AB (AB)rsquo=

RHS

B

A Arsquo

Brsquo

Arsquo + Brsquo

=

Bubbled OR

A

B

(AB)rsquo

A

B

Arsquo + Brsquo

A B AB (AB)rsquo

0 0 0 1

0 1 0 1

1 0 0 1

0 1 1 0


0 0 1 1 1

0 1 1 0 1

1 0 0 1 1

0 1 0 0 0

16 Duality Theorem




=


10





Sol Sol






Sol Sol







Sol Sol







Sol Sol









there4 F = AB


11



Sol Sol








Sol Sol





BBrsquo)























12




361 Standard Form














362 Canonical Form










13




Sol Sol

F = 00 + 11 F = 001 + 100 + 111




Sol

F = 1011 + 1100 + 0000

there4 F = m11 + m12 + m0









Sol Sol

F = (10)(01) F = (100) (010) (001)




Sol

F = (1110)(1001)(0101)

there4 F = M14M9M5



14






1 F(ABCD) = ΠM(037101415)

Solution










15


2 F = A + BrsquoC

Solution









there4 F = 111 + 101 + 110 + 100 + 001

there4 F = m7 + m6 + m5 + m4 + m1



1 F = Σ(14567)

Solution










Solution




16








there4 F = (000) (001) (010) (011) (101)


3 F = XY + XrsquoZ

Solution














there4 F = (100) (101) (000) (010)




17




















1

11

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1





0

00

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1


18




1

00

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

01

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

0

10

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1


0

11

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

11

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

11

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1




0

00

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1

Sol

0

10

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1

Sol

1

00

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1






0


19




Sol


Sol


Sol




Sol



20


(3) F = ΠM(125) (4) F = ΠM(041573) Sol

Sol


Sol






21






22



Examples


23




Sol


Sol



24


(5) F = summ (56791011131415) Sol




Sol




Sol



25






(1) F = summ (023101112131617181920212627) Sol


(2) F = summ (02469111315172125272931) Sol



26



(1) F = ΠM (145678914152223242528293031) Sol





Sol Sol






0 0 0 0 1 0

0 1 1 0 1 1

1 0 1 0 1 1

1 1 1 1 0 0

Input Output


0 0 1 1 1 0 0

0 1 1 0 1 1 1

1 0 0 1 1 1 1

1 1 0 0 0 1 0





27





Sol






Sol

Sol




28







Step2 Step2

Step3 Step3


29


Step4 Step4













30



119917(119912 119913 119914 119915) = sum(120782 120783 120784 120785 120786 120788 120789 120783120783 120783120784 120783120787)

Sol





31



Sol





32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


9



LHS

NAND Gate

A

B

AB (AB)rsquo=

RHS

B

A Arsquo

Brsquo

Arsquo + Brsquo

=

Bubbled OR

A

B

(AB)rsquo

A

B

Arsquo + Brsquo

A B AB (AB)rsquo

0 0 0 1

0 1 0 1

1 0 0 1

0 1 1 0


0 0 1 1 1

0 1 1 0 1

1 0 0 1 1

0 1 0 0 0

16 Duality Theorem




=


10





Sol Sol






Sol Sol







Sol Sol







Sol Sol









there4 F = AB


11



Sol Sol








Sol Sol





BBrsquo)























12




361 Standard Form














362 Canonical Form










13




Sol Sol

F = 00 + 11 F = 001 + 100 + 111




Sol

F = 1011 + 1100 + 0000

there4 F = m11 + m12 + m0









Sol Sol

F = (10)(01) F = (100) (010) (001)




Sol

F = (1110)(1001)(0101)

there4 F = M14M9M5



14






1 F(ABCD) = ΠM(037101415)

Solution










15


2 F = A + BrsquoC

Solution









there4 F = 111 + 101 + 110 + 100 + 001

there4 F = m7 + m6 + m5 + m4 + m1



1 F = Σ(14567)

Solution










Solution




16








there4 F = (000) (001) (010) (011) (101)


3 F = XY + XrsquoZ

Solution














there4 F = (100) (101) (000) (010)




17




















1

11

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1





0

00

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1


18




1

00

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

01

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

0

10

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1


0

11

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

11

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

11

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1




0

00

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1

Sol

0

10

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1

Sol

1

00

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1






0


19




Sol


Sol


Sol




Sol



20


(3) F = ΠM(125) (4) F = ΠM(041573) Sol

Sol


Sol






21






22



Examples


23




Sol


Sol



24


(5) F = summ (56791011131415) Sol




Sol




Sol



25






(1) F = summ (023101112131617181920212627) Sol


(2) F = summ (02469111315172125272931) Sol



26



(1) F = ΠM (145678914152223242528293031) Sol





Sol Sol






0 0 0 0 1 0

0 1 1 0 1 1

1 0 1 0 1 1

1 1 1 1 0 0

Input Output


0 0 1 1 1 0 0

0 1 1 0 1 1 1

1 0 0 1 1 1 1

1 1 0 0 0 1 0





27





Sol






Sol

Sol




28







Step2 Step2

Step3 Step3


29


Step4 Step4













30



119917(119912 119913 119914 119915) = sum(120782 120783 120784 120785 120786 120788 120789 120783120783 120783120784 120783120787)

Sol





31



Sol





32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


10





Sol Sol






Sol Sol







Sol Sol







Sol Sol









there4 F = AB


11



Sol Sol








Sol Sol





BBrsquo)























12




361 Standard Form














362 Canonical Form










13




Sol Sol

F = 00 + 11 F = 001 + 100 + 111




Sol

F = 1011 + 1100 + 0000

there4 F = m11 + m12 + m0









Sol Sol

F = (10)(01) F = (100) (010) (001)




Sol

F = (1110)(1001)(0101)

there4 F = M14M9M5



14






1 F(ABCD) = ΠM(037101415)

Solution










15


2 F = A + BrsquoC

Solution









there4 F = 111 + 101 + 110 + 100 + 001

there4 F = m7 + m6 + m5 + m4 + m1



1 F = Σ(14567)

Solution










Solution




16








there4 F = (000) (001) (010) (011) (101)


3 F = XY + XrsquoZ

Solution














there4 F = (100) (101) (000) (010)




17




















1

11

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1





0

00

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1


18




1

00

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

01

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

0

10

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1


0

11

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

11

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

11

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1




0

00

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1

Sol

0

10

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1

Sol

1

00

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1






0


19




Sol


Sol


Sol




Sol



20


(3) F = ΠM(125) (4) F = ΠM(041573) Sol

Sol


Sol






21






22



Examples


23




Sol


Sol



24


(5) F = summ (56791011131415) Sol




Sol




Sol



25






(1) F = summ (023101112131617181920212627) Sol


(2) F = summ (02469111315172125272931) Sol



26



(1) F = ΠM (145678914152223242528293031) Sol





Sol Sol






0 0 0 0 1 0

0 1 1 0 1 1

1 0 1 0 1 1

1 1 1 1 0 0

Input Output


0 0 1 1 1 0 0

0 1 1 0 1 1 1

1 0 0 1 1 1 1

1 1 0 0 0 1 0





27





Sol






Sol

Sol




28







Step2 Step2

Step3 Step3


29


Step4 Step4













30



119917(119912 119913 119914 119915) = sum(120782 120783 120784 120785 120786 120788 120789 120783120783 120783120784 120783120787)

Sol





31



Sol





32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


11



Sol Sol








Sol Sol





BBrsquo)























12




361 Standard Form














362 Canonical Form










13




Sol Sol

F = 00 + 11 F = 001 + 100 + 111




Sol

F = 1011 + 1100 + 0000

there4 F = m11 + m12 + m0









Sol Sol

F = (10)(01) F = (100) (010) (001)




Sol

F = (1110)(1001)(0101)

there4 F = M14M9M5



14






1 F(ABCD) = ΠM(037101415)

Solution










15


2 F = A + BrsquoC

Solution









there4 F = 111 + 101 + 110 + 100 + 001

there4 F = m7 + m6 + m5 + m4 + m1



1 F = Σ(14567)

Solution










Solution




16








there4 F = (000) (001) (010) (011) (101)


3 F = XY + XrsquoZ

Solution














there4 F = (100) (101) (000) (010)




17




















1

11

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1





0

00

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1


18




1

00

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

01

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

0

10

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1


0

11

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

11

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

11

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1




0

00

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1

Sol

0

10

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1

Sol

1

00

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1






0


19




Sol


Sol


Sol




Sol



20


(3) F = ΠM(125) (4) F = ΠM(041573) Sol

Sol


Sol






21






22



Examples


23




Sol


Sol



24


(5) F = summ (56791011131415) Sol




Sol




Sol



25






(1) F = summ (023101112131617181920212627) Sol


(2) F = summ (02469111315172125272931) Sol



26



(1) F = ΠM (145678914152223242528293031) Sol





Sol Sol






0 0 0 0 1 0

0 1 1 0 1 1

1 0 1 0 1 1

1 1 1 1 0 0

Input Output


0 0 1 1 1 0 0

0 1 1 0 1 1 1

1 0 0 1 1 1 1

1 1 0 0 0 1 0





27





Sol






Sol

Sol




28







Step2 Step2

Step3 Step3


29


Step4 Step4













30



119917(119912 119913 119914 119915) = sum(120782 120783 120784 120785 120786 120788 120789 120783120783 120783120784 120783120787)

Sol





31



Sol





32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


12




361 Standard Form














362 Canonical Form










13




Sol Sol

F = 00 + 11 F = 001 + 100 + 111




Sol

F = 1011 + 1100 + 0000

there4 F = m11 + m12 + m0









Sol Sol

F = (10)(01) F = (100) (010) (001)




Sol

F = (1110)(1001)(0101)

there4 F = M14M9M5



14






1 F(ABCD) = ΠM(037101415)

Solution










15


2 F = A + BrsquoC

Solution









there4 F = 111 + 101 + 110 + 100 + 001

there4 F = m7 + m6 + m5 + m4 + m1



1 F = Σ(14567)

Solution










Solution




16








there4 F = (000) (001) (010) (011) (101)


3 F = XY + XrsquoZ

Solution














there4 F = (100) (101) (000) (010)




17




















1

11

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1





0

00

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1


18




1

00

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

01

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

0

10

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1


0

11

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

11

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

11

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1




0

00

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1

Sol

0

10

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1

Sol

1

00

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1






0


19




Sol


Sol


Sol




Sol



20


(3) F = ΠM(125) (4) F = ΠM(041573) Sol

Sol


Sol






21






22



Examples


23




Sol


Sol



24


(5) F = summ (56791011131415) Sol




Sol




Sol



25






(1) F = summ (023101112131617181920212627) Sol


(2) F = summ (02469111315172125272931) Sol



26



(1) F = ΠM (145678914152223242528293031) Sol





Sol Sol






0 0 0 0 1 0

0 1 1 0 1 1

1 0 1 0 1 1

1 1 1 1 0 0

Input Output


0 0 1 1 1 0 0

0 1 1 0 1 1 1

1 0 0 1 1 1 1

1 1 0 0 0 1 0





27





Sol






Sol

Sol




28







Step2 Step2

Step3 Step3


29


Step4 Step4













30



119917(119912 119913 119914 119915) = sum(120782 120783 120784 120785 120786 120788 120789 120783120783 120783120784 120783120787)

Sol





31



Sol





32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


13




Sol Sol

F = 00 + 11 F = 001 + 100 + 111




Sol

F = 1011 + 1100 + 0000

there4 F = m11 + m12 + m0









Sol Sol

F = (10)(01) F = (100) (010) (001)




Sol

F = (1110)(1001)(0101)

there4 F = M14M9M5



14






1 F(ABCD) = ΠM(037101415)

Solution










15


2 F = A + BrsquoC

Solution









there4 F = 111 + 101 + 110 + 100 + 001

there4 F = m7 + m6 + m5 + m4 + m1



1 F = Σ(14567)

Solution










Solution




16








there4 F = (000) (001) (010) (011) (101)


3 F = XY + XrsquoZ

Solution














there4 F = (100) (101) (000) (010)




17




















1

11

0

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0

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1A

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0

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18




1

00

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0 1

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1

01

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0

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0

10

1

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1

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1

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0

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1

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1

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0

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0

00

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0

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Sol

0

10

0

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0

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0 1

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A 0

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Sol

1

00

0

B

0

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1A

B

0 1

2 3

A 0

Arsquo 1






0


19




Sol


Sol


Sol




Sol



20


(3) F = ΠM(125) (4) F = ΠM(041573) Sol

Sol


Sol






21






22



Examples


23




Sol


Sol



24


(5) F = summ (56791011131415) Sol




Sol




Sol



25






(1) F = summ (023101112131617181920212627) Sol


(2) F = summ (02469111315172125272931) Sol



26



(1) F = ΠM (145678914152223242528293031) Sol





Sol Sol






0 0 0 0 1 0

0 1 1 0 1 1

1 0 1 0 1 1

1 1 1 1 0 0

Input Output


0 0 1 1 1 0 0

0 1 1 0 1 1 1

1 0 0 1 1 1 1

1 1 0 0 0 1 0





27





Sol






Sol

Sol




28







Step2 Step2

Step3 Step3


29


Step4 Step4













30



119917(119912 119913 119914 119915) = sum(120782 120783 120784 120785 120786 120788 120789 120783120783 120783120784 120783120787)

Sol





31



Sol





32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


14






1 F(ABCD) = ΠM(037101415)

Solution










15


2 F = A + BrsquoC

Solution









there4 F = 111 + 101 + 110 + 100 + 001

there4 F = m7 + m6 + m5 + m4 + m1



1 F = Σ(14567)

Solution










Solution




16








there4 F = (000) (001) (010) (011) (101)


3 F = XY + XrsquoZ

Solution














there4 F = (100) (101) (000) (010)




17




















1

11

0

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0

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1A

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0 1

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0

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18




1

00

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0

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0 1

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Sol

1

01

0

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0

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1A

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0

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1

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0

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0 1

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A 1


0

11

0

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0

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1A

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0 1

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A 1

Sol

1

11

1

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0

B

1A

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0 1

2 3

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A 1

Sol

1

11

1

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0

B

1A

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0 1

2 3

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A 1




0

00

0

B

0

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1A

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0 1

2 3

A 0

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Sol

0

10

0

B

0

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1A

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0 1

2 3

A 0

Arsquo 1

Sol

1

00

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1






0


19




Sol


Sol


Sol




Sol



20


(3) F = ΠM(125) (4) F = ΠM(041573) Sol

Sol


Sol






21






22



Examples


23




Sol


Sol



24


(5) F = summ (56791011131415) Sol




Sol




Sol



25






(1) F = summ (023101112131617181920212627) Sol


(2) F = summ (02469111315172125272931) Sol



26



(1) F = ΠM (145678914152223242528293031) Sol





Sol Sol






0 0 0 0 1 0

0 1 1 0 1 1

1 0 1 0 1 1

1 1 1 1 0 0

Input Output


0 0 1 1 1 0 0

0 1 1 0 1 1 1

1 0 0 1 1 1 1

1 1 0 0 0 1 0





27





Sol






Sol

Sol




28







Step2 Step2

Step3 Step3


29


Step4 Step4













30



119917(119912 119913 119914 119915) = sum(120782 120783 120784 120785 120786 120788 120789 120783120783 120783120784 120783120787)

Sol





31



Sol





32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


15


2 F = A + BrsquoC

Solution









there4 F = 111 + 101 + 110 + 100 + 001

there4 F = m7 + m6 + m5 + m4 + m1



1 F = Σ(14567)

Solution










Solution




16








there4 F = (000) (001) (010) (011) (101)


3 F = XY + XrsquoZ

Solution














there4 F = (100) (101) (000) (010)




17




















1

11

0

Brsquo

0

B

1A

B

0 1

2 3

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A 1





0

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18




1

00

1

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0

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1A

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0 1

2 3

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A 1

Sol

1

01

0

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0

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A 1

Sol

0

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1

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0

B

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0 1

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0

11

0

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0 1

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A 1

Sol

1

11

1

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0

B

1A

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0 1

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A 1

Sol

1

11

1

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0

B

1A

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0 1

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A 1




0

00

0

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0

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1A

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0 1

2 3

A 0

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Sol

0

10

0

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0

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1A

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0 1

2 3

A 0

Arsquo 1

Sol

1

00

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1






0


19




Sol


Sol


Sol




Sol



20


(3) F = ΠM(125) (4) F = ΠM(041573) Sol

Sol


Sol






21






22



Examples


23




Sol


Sol



24


(5) F = summ (56791011131415) Sol




Sol




Sol



25






(1) F = summ (023101112131617181920212627) Sol


(2) F = summ (02469111315172125272931) Sol



26



(1) F = ΠM (145678914152223242528293031) Sol





Sol Sol






0 0 0 0 1 0

0 1 1 0 1 1

1 0 1 0 1 1

1 1 1 1 0 0

Input Output


0 0 1 1 1 0 0

0 1 1 0 1 1 1

1 0 0 1 1 1 1

1 1 0 0 0 1 0





27





Sol






Sol

Sol




28







Step2 Step2

Step3 Step3


29


Step4 Step4













30



119917(119912 119913 119914 119915) = sum(120782 120783 120784 120785 120786 120788 120789 120783120783 120783120784 120783120787)

Sol





31



Sol





32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


16








there4 F = (000) (001) (010) (011) (101)


3 F = XY + XrsquoZ

Solution














there4 F = (100) (101) (000) (010)




17




















1

11

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1





0

00

1

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0

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1A

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0 1

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A 1


18




1

00

1

Brsquo

0

B

1A

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0 1

2 3

Arsquo 0

A 1

Sol

1

01

0

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0

B

1A

B

0 1

2 3

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A 1

Sol

0

10

1

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0

B

1A

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0 1

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A 1


0

11

0

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0

B

1A

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0 1

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A 1

Sol

1

11

1

Brsquo

0

B

1A

B

0 1

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A 1

Sol

1

11

1

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0

B

1A

B

0 1

2 3

Arsquo 0

A 1




0

00

0

B

0

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1A

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0 1

2 3

A 0

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Sol

0

10

0

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0

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1A

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0 1

2 3

A 0

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Sol

1

00

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1






0


19




Sol


Sol


Sol




Sol



20


(3) F = ΠM(125) (4) F = ΠM(041573) Sol

Sol


Sol






21






22



Examples


23




Sol


Sol



24


(5) F = summ (56791011131415) Sol




Sol




Sol



25






(1) F = summ (023101112131617181920212627) Sol


(2) F = summ (02469111315172125272931) Sol



26



(1) F = ΠM (145678914152223242528293031) Sol





Sol Sol






0 0 0 0 1 0

0 1 1 0 1 1

1 0 1 0 1 1

1 1 1 1 0 0

Input Output


0 0 1 1 1 0 0

0 1 1 0 1 1 1

1 0 0 1 1 1 1

1 1 0 0 0 1 0





27





Sol






Sol

Sol




28







Step2 Step2

Step3 Step3


29


Step4 Step4













30



119917(119912 119913 119914 119915) = sum(120782 120783 120784 120785 120786 120788 120789 120783120783 120783120784 120783120787)

Sol





31



Sol





32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


17




















1

11

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1





0

00

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1


18




1

00

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

01

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

0

10

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1


0

11

0

Brsquo

0

B

1A

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0 1

2 3

Arsquo 0

A 1

Sol

1

11

1

Brsquo

0

B

1A

B

0 1

2 3

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A 1

Sol

1

11

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1




0

00

0

B

0

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1A

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0 1

2 3

A 0

Arsquo 1

Sol

0

10

0

B

0

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1A

B

0 1

2 3

A 0

Arsquo 1

Sol

1

00

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1






0


19




Sol


Sol


Sol




Sol



20


(3) F = ΠM(125) (4) F = ΠM(041573) Sol

Sol


Sol






21






22



Examples


23




Sol


Sol



24


(5) F = summ (56791011131415) Sol




Sol




Sol



25






(1) F = summ (023101112131617181920212627) Sol


(2) F = summ (02469111315172125272931) Sol



26



(1) F = ΠM (145678914152223242528293031) Sol





Sol Sol






0 0 0 0 1 0

0 1 1 0 1 1

1 0 1 0 1 1

1 1 1 1 0 0

Input Output


0 0 1 1 1 0 0

0 1 1 0 1 1 1

1 0 0 1 1 1 1

1 1 0 0 0 1 0





27





Sol






Sol

Sol




28







Step2 Step2

Step3 Step3


29


Step4 Step4













30



119917(119912 119913 119914 119915) = sum(120782 120783 120784 120785 120786 120788 120789 120783120783 120783120784 120783120787)

Sol





31



Sol





32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


18




1

00

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

01

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

0

10

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1


0

11

0

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

11

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1

Sol

1

11

1

Brsquo

0

B

1A

B

0 1

2 3

Arsquo 0

A 1




0

00

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1

Sol

0

10

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1

Sol

1

00

0

B

0

Brsquo

1A

B

0 1

2 3

A 0

Arsquo 1






0


19




Sol


Sol


Sol




Sol



20


(3) F = ΠM(125) (4) F = ΠM(041573) Sol

Sol


Sol






21






22



Examples


23




Sol


Sol



24


(5) F = summ (56791011131415) Sol




Sol




Sol



25






(1) F = summ (023101112131617181920212627) Sol


(2) F = summ (02469111315172125272931) Sol



26



(1) F = ΠM (145678914152223242528293031) Sol





Sol Sol






0 0 0 0 1 0

0 1 1 0 1 1

1 0 1 0 1 1

1 1 1 1 0 0

Input Output


0 0 1 1 1 0 0

0 1 1 0 1 1 1

1 0 0 1 1 1 1

1 1 0 0 0 1 0





27





Sol






Sol

Sol




28







Step2 Step2

Step3 Step3


29


Step4 Step4













30



119917(119912 119913 119914 119915) = sum(120782 120783 120784 120785 120786 120788 120789 120783120783 120783120784 120783120787)

Sol





31



Sol





32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


19




Sol


Sol


Sol




Sol



20


(3) F = ΠM(125) (4) F = ΠM(041573) Sol

Sol


Sol






21






22



Examples


23




Sol


Sol



24


(5) F = summ (56791011131415) Sol




Sol




Sol



25






(1) F = summ (023101112131617181920212627) Sol


(2) F = summ (02469111315172125272931) Sol



26



(1) F = ΠM (145678914152223242528293031) Sol





Sol Sol






0 0 0 0 1 0

0 1 1 0 1 1

1 0 1 0 1 1

1 1 1 1 0 0

Input Output


0 0 1 1 1 0 0

0 1 1 0 1 1 1

1 0 0 1 1 1 1

1 1 0 0 0 1 0





27





Sol






Sol

Sol




28







Step2 Step2

Step3 Step3


29


Step4 Step4













30



119917(119912 119913 119914 119915) = sum(120782 120783 120784 120785 120786 120788 120789 120783120783 120783120784 120783120787)

Sol





31



Sol





32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


20


(3) F = ΠM(125) (4) F = ΠM(041573) Sol

Sol


Sol






21






22



Examples


23




Sol


Sol



24


(5) F = summ (56791011131415) Sol




Sol




Sol



25






(1) F = summ (023101112131617181920212627) Sol


(2) F = summ (02469111315172125272931) Sol



26



(1) F = ΠM (145678914152223242528293031) Sol





Sol Sol






0 0 0 0 1 0

0 1 1 0 1 1

1 0 1 0 1 1

1 1 1 1 0 0

Input Output


0 0 1 1 1 0 0

0 1 1 0 1 1 1

1 0 0 1 1 1 1

1 1 0 0 0 1 0





27





Sol






Sol

Sol




28







Step2 Step2

Step3 Step3


29


Step4 Step4













30



119917(119912 119913 119914 119915) = sum(120782 120783 120784 120785 120786 120788 120789 120783120783 120783120784 120783120787)

Sol





31



Sol





32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


21






22



Examples


23




Sol


Sol



24


(5) F = summ (56791011131415) Sol




Sol




Sol



25






(1) F = summ (023101112131617181920212627) Sol


(2) F = summ (02469111315172125272931) Sol



26



(1) F = ΠM (145678914152223242528293031) Sol





Sol Sol






0 0 0 0 1 0

0 1 1 0 1 1

1 0 1 0 1 1

1 1 1 1 0 0

Input Output


0 0 1 1 1 0 0

0 1 1 0 1 1 1

1 0 0 1 1 1 1

1 1 0 0 0 1 0





27





Sol






Sol

Sol




28







Step2 Step2

Step3 Step3


29


Step4 Step4













30



119917(119912 119913 119914 119915) = sum(120782 120783 120784 120785 120786 120788 120789 120783120783 120783120784 120783120787)

Sol





31



Sol





32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


22



Examples


23




Sol


Sol



24


(5) F = summ (56791011131415) Sol




Sol




Sol



25






(1) F = summ (023101112131617181920212627) Sol


(2) F = summ (02469111315172125272931) Sol



26



(1) F = ΠM (145678914152223242528293031) Sol





Sol Sol






0 0 0 0 1 0

0 1 1 0 1 1

1 0 1 0 1 1

1 1 1 1 0 0

Input Output


0 0 1 1 1 0 0

0 1 1 0 1 1 1

1 0 0 1 1 1 1

1 1 0 0 0 1 0





27





Sol






Sol

Sol




28







Step2 Step2

Step3 Step3


29


Step4 Step4













30



119917(119912 119913 119914 119915) = sum(120782 120783 120784 120785 120786 120788 120789 120783120783 120783120784 120783120787)

Sol





31



Sol





32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


23




Sol


Sol



24


(5) F = summ (56791011131415) Sol




Sol




Sol



25






(1) F = summ (023101112131617181920212627) Sol


(2) F = summ (02469111315172125272931) Sol



26



(1) F = ΠM (145678914152223242528293031) Sol





Sol Sol






0 0 0 0 1 0

0 1 1 0 1 1

1 0 1 0 1 1

1 1 1 1 0 0

Input Output


0 0 1 1 1 0 0

0 1 1 0 1 1 1

1 0 0 1 1 1 1

1 1 0 0 0 1 0





27





Sol






Sol

Sol




28







Step2 Step2

Step3 Step3


29


Step4 Step4













30



119917(119912 119913 119914 119915) = sum(120782 120783 120784 120785 120786 120788 120789 120783120783 120783120784 120783120787)

Sol





31



Sol





32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


24


(5) F = summ (56791011131415) Sol




Sol




Sol



25






(1) F = summ (023101112131617181920212627) Sol


(2) F = summ (02469111315172125272931) Sol



26



(1) F = ΠM (145678914152223242528293031) Sol





Sol Sol






0 0 0 0 1 0

0 1 1 0 1 1

1 0 1 0 1 1

1 1 1 1 0 0

Input Output


0 0 1 1 1 0 0

0 1 1 0 1 1 1

1 0 0 1 1 1 1

1 1 0 0 0 1 0





27





Sol






Sol

Sol




28







Step2 Step2

Step3 Step3


29


Step4 Step4













30



119917(119912 119913 119914 119915) = sum(120782 120783 120784 120785 120786 120788 120789 120783120783 120783120784 120783120787)

Sol





31



Sol





32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


25






(1) F = summ (023101112131617181920212627) Sol


(2) F = summ (02469111315172125272931) Sol



26



(1) F = ΠM (145678914152223242528293031) Sol





Sol Sol






0 0 0 0 1 0

0 1 1 0 1 1

1 0 1 0 1 1

1 1 1 1 0 0

Input Output


0 0 1 1 1 0 0

0 1 1 0 1 1 1

1 0 0 1 1 1 1

1 1 0 0 0 1 0





27





Sol






Sol

Sol




28







Step2 Step2

Step3 Step3


29


Step4 Step4













30



119917(119912 119913 119914 119915) = sum(120782 120783 120784 120785 120786 120788 120789 120783120783 120783120784 120783120787)

Sol





31



Sol





32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


26



(1) F = ΠM (145678914152223242528293031) Sol





Sol Sol






0 0 0 0 1 0

0 1 1 0 1 1

1 0 1 0 1 1

1 1 1 1 0 0

Input Output


0 0 1 1 1 0 0

0 1 1 0 1 1 1

1 0 0 1 1 1 1

1 1 0 0 0 1 0





27





Sol






Sol

Sol




28







Step2 Step2

Step3 Step3


29


Step4 Step4













30



119917(119912 119913 119914 119915) = sum(120782 120783 120784 120785 120786 120788 120789 120783120783 120783120784 120783120787)

Sol





31



Sol





32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


27





Sol






Sol

Sol




28







Step2 Step2

Step3 Step3


29


Step4 Step4













30



119917(119912 119913 119914 119915) = sum(120782 120783 120784 120785 120786 120788 120789 120783120783 120783120784 120783120787)

Sol





31



Sol





32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


28







Step2 Step2

Step3 Step3


29


Step4 Step4













30



119917(119912 119913 119914 119915) = sum(120782 120783 120784 120785 120786 120788 120789 120783120783 120783120784 120783120787)

Sol





31



Sol





32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


29


Step4 Step4













30



119917(119912 119913 119914 119915) = sum(120782 120783 120784 120785 120786 120788 120789 120783120783 120783120784 120783120787)

Sol





31



Sol





32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


30



119917(119912 119913 119914 119915) = sum(120782 120783 120784 120785 120786 120788 120789 120783120783 120783120784 120783120787)

Sol





31



Sol





32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


31



Sol





32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


32



Sol





119917(119912 119913 119914 119915) = 119924(120782 120790 120783120782 120783120784 120783120785 120783120787) ∙ 119941(120783 120784 120785) Sol





33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


33


311 GTU Questions

Un

it

Gro

up

Questions

Sum

mer

-15

Win

ter-

15

Sum

mer

-16

Win

ter-

16

Sum

mer

-17

Win

ter-

17

Sum

mer

-18

Win

ter-

18

An

swer

Top

ic N

o_

Pg

No

3


7 7 35_8

A


7 35_8


4 3512_10


3 35_8

A


4 3512_11



3 3512_11



B


7 36_12


7 3631_15


7 3631_15

372_18

C


7 3611_12 3736_24


7 7 37_17


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28


34


3


3 3736_24

C


7 3736_24

39_28


4 3734_23 3735_24


7 3736_24

C


4 3734_24

38_26

C


7 3736_24


3 37_17

310_29


7 310_29


7 310_29


7 39_28


4 39_28

unit 4-part 1 : boolean algebra

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