Digital Electronics Practice Book (Sem-III 2024) PDF

L J Institute of Engineering and Technology, Ahmedabad. Digital Electronics (DE) Practice Book (Sem-III 2024) Note: This Practice Book is only for reference purpose. LJU Test question p...

L J Institute of Engineering and Technology, Ahmedabad. Digital Electronics (DE) Practice Book (Sem-III 2024) Note: This Practice Book is only for reference purpose. LJU Test question paper may not be completely set from Practice Book. unit_num Sr. No. question_text answer_text marks option A option B option C option D ber 1 The given hexadecimal number (1E.53)16 is equivalent to (35.684)8 (36.246)8 (34.340)8 (35.599)8 1 b 1 ____________. 2 1 The octal number (651.124)8 is equivalent to ______. a 1 (1A9.2A)16 (1B0.10)16 (1A8.A3)16 (1B0.B0)16 3 1 The octal equivalent of the decimal number (417)10 is _____. a 1 (641)8 (619)8 (640)8 (598)8 4 1 (170)10 is equivalent to ____________. c 1 (FD)16 (DF)16 (AA)16 (AF)16 5 1 Convert (214)8 into decimal. a 1 (140)10 (141)10 (142)10 (130)10 6 1 Convert (0.345)10 into an octal number. b 1 (0.16050)8 (0.26050)8 (0.19450)8 (0.24040)8 7 1 Convert the binary number (01011.1011)2 into decimal. a 1 (11.6875)10 (11.5874)10 (10.9876)10 (10.7893)10 8 1 Convert binary to octal: (110110001010)2 =? b 1 (5512)8 (6612)8 (4532)8 (6745)8 1 Which of the following is not a positional number system? Roman Number Octal Number Binary Number System Hexadecimal 9 a 1 System System Number System 10 1 The value of radix in binary number system is _____________. d 1 4 1 10 2 11 1 The binary equivalent of the decimal number 10 is __________. c 1 10 100 1010 101 12 1 The octal equivalent of 1100101.001010 is ______. b 1 624.12 145.12 154.12 145.21 13 1 The hexadecimal number for (95.5)10 is _______. (5F.8) 16 1 1 The hexadecimal number ‘A0’ has the decimal value equivalent to 14 160 1 ______. 15 1 The decimal equivalent of hex number 1A53 is _______. 6739 1 16 1 _________be the decimal equivalent of 111011.10. 1 17 1 Given that (16)10 = (100)x, find the value of x. d 2 X=10 X=1 X=8 X=4 18 1 (4433)5 = ( )10 = ( )2 2 1 1) (673.124)8 = ( )2 2) (4522.25)10 = ( )2 19 4 3) (FACE)16 = ( )10 4) (10101010)2 = ( )8 = ( )16 1 Convert following Octal Number to Hexadecimal and Binary 414, 574, (10C,17C,1D5.54)16 , 725.25. (100001100, 20 101111100, 3 111010101.010101)2 1 Convert the number (435)7into equivalent radix-10 and radix-4 number 21 3 system. 1 Do as directed: 22 3 (10110100)2= ( ?)gray = ( ? )XS-3 = ( ? )BCD 1 (i) Convert (75)10 = (____________)2 23 (ii) Convert (101011)2 = (_____)10 3 (iii)Convert (10101101)2 = (_____)16 = (______)8 1 Convert the decimal number 225.225 to binary, octal and hexadecimal. 24 3 1 Convert following numbers. (a)(4021.2)5 = ( )10 25 (b) (B65F)16= ( )10 2 (c) (630.4)8 = ( )10 (d) (41)10 = ( )8  1 Convert the following Hexadecimal numbers to Octal. 26 (a) 4F7.A8 (b) BC70.0E (c) 42FD 3 1 Convert following Hexadecimal Number to Decimal 27 (2856,4095,3880)10 2 B28, FFF, F28 28 1 (1011011101101110)2 = ( )16 (B76E)16 1 29 1 Convert the decimal number 187 to 8-bit binary. a 1 (10111011)2 (11011101)2 (10111101)2 (10111100)2 1 Convert the decimal number 250.5 to base 3, base 4, base 7 and base 8. 30 4 1 Convert (4BAC)16= ( )8 = ( )4 = ( )2 = ( )10. Show all steps of 31 2 conversion. 32 1 Express decimal number 60.875 into binary form. 1 1 Convert the following numbers form given base to the base indicates. 1. (AEF2.B6)16 = (_____)2 2. (674.12)8 = (______)10 33 2 3. (110110.1011)2 = (______)16 34 1 Express decimal number 10.875 into binary form. 2 35 1 Convert the Decimal Number 412.5 to base 3, 4 and 7 3 36 1 Nibble means________ b 1 2 bit 4 bit 8 bit 16 bit 37 1 Convert the binary number (1001.0010)2 to decimal. b 2 90.125 9.125 125 12.5 1 i. Convert (FFFF)16=(________)10. 38 2 ii. Convert (125.625)10=(________)2. 39 1 (52)10 = ( )2 (110100)2 1 40 1 (436)8 = ( )16 1 41 1 (5C7)16 = ( )10 (1479)10 1 42 1 (11011.101)2 = ( )10 (27.625)10 1 43 1 Convert following 1. (4E7.2)16 = (?)8 2. (521.3)8 = (?)2 2 44 1 Convert (10101101)2 = (_____)16 = (______)8 2 45 1 (734)8 = ( )16 d 1 C1D DC1 1CD 1DC 46 1 (1111.11)2 = ( ? )8 = ( ? )10 2 47 1 (AEF2.B6)16 = (_______)2 1 48 1 (674.12)8 = (________)10 1 49 1 (110110.1011)2 = (________)16 1 50 1 (101001011)2 = ( )10 1 1 (11101110)2 =( )8 51 1 52 1 (68)10 =( )16 1 1 Consider the equation (123)5 = (x8)y with x and y as unknown. The 53 3 1 number of possible solutions is _________. 54 1 (1217)8 is equivalent to______ c 1 (1217)16 (2297)10 (028F)16 (0B17)16 1 (73)x (in base x number system) is equal to (54)y (in base y number 8,16 8,10 10,12 8,11 55 d 2 system), the possible values of x and y are 1 Which of the following number is not allowed in radix – 7 (base 7) 666 739 461 124 56 b 1 system? 57 1 If (2.3)4 + (1.2)4 = y4, then value of y in base 4 system. a 1 10.1 10.01 10.2 1.02 58 1 If (2.3)4 + (1.2)4 = Y10, then value of Y in base 10 system? a 2 4.25 10.1 10.01 3.5 59 1 The Octal equivalent of hexadecimal number AB.CD is ______ a 1 253.632 253.314 526.314 526.632 60 1 Which number system has a base 16? d 1 Octal Binary Decimal Hexadecimal  1 What is Digital Electronics? Engineering of Field of Engineering of All of the devices that electronics devices that mentioned 61 d 1 produce digital involving the accepts digital signal study of digital signals signal 1 Convert the following numbers as directed: (130)10= ( )2 (10000011)2 (10000010)2 (10000111)2 (111001)2 62 b 1 63 1 Convert the following numbers as directed: (1011011)2= ( )10 d 1 (91)8 (52)10 (41)10 (91)10 64 1 Convert (B65F)16 = ( )10. c 1 (44526)10 (78864)10 (46687)10 (48756)10 65 1 Convert (41)10 = ( )2. d 1 (110011)2 (111000)2 (101010)2 (101001)2 66 1 Convert decimal number (43)10 to binary. c 1 (110110)2 (101010)2 (101011)2 (111001)2 67 1 Convert octal number (234)8 to hexadecimal. a 1 (9C)16 (8C)16 (9C)8 (9C)2 68 1 Convert decimal number (0.252)10 to binary. b 1 (0.00001)2 (0.010000)2 (1.010000)2 (10.00100)2 69 1 Convert the Hexadecimal numbers to Octal: 4F7.A8 2 70 1 Convert the Hexadecimal numbers to Octal: BC70.0E 2 71 1 A number (1217)8 can be represented as ______ d 1 (656)10 (01010001111)2 (28F)16 Both B and C 72 1 The number (100000)2 would appear just immediately after – d 1 (37)8 (1F)16 (11111)2 All A,B and C 1 (F23.23)16 = ( ? )8 = ( ? )2 7442.106, 7442.106, 7443.106, 7443.106, 73 c 1 111100100011.00100 111110100011.0010 111100100011.001000 111100000001.0010 011 0011 11 0011 74 1 If (154)b / (14)b = (8)10 then what is the value of radix ‘b’ ? b 1 10 7 12 8 1 How many number of ‘1’ would be appeared in binary representation of - 7 5 6 9 75 d 1 3×512+5×64+7×8+3 ? 1 Convert (DEAD)16 into equivalent radix-10 and radix-8 number system. 76 2 1 Consider the equation (129)5 = (x9)y with x and y as unknown. The not possible 5 3 4 77 a 1 number of possible solutions is _________. 78 1 convert (FFFF)16 to decimal & octal numbers respectively. d 1 a) 65235 &177787 b)62325 & 177878 c)65335 & 177878 d)65535 &177777 79 1 The binary equivalent of (11.6275)10 is equals to ______. d 1 a)101.11011 b)1011.1001 c)101.0011 d)1011.1010 1 Convert the hexadecimal number (9999.9999) to octal number. a) 1414631.463144 b) 463144.114631 c) 141361.114631 d) 114631.463144 80 d 1 1 Which of the following is equivalent to decimal number of (BFA0) a) 90456 b) 40956 c) 49056 d) 56049 81 c 1 hexadecimal number ? 1 Consider the following statements a) i,ii,iii b) i, ii only c) i, iii only d) ii, iii only i) octal number is 1/4th the length of corresponding binary number ii) Hexadecimal is 1/3rd length of corresponding binary number 82 iii) In 10 bit binary number, 512 is the weight of 2nd digit from a 1 MSB.Which of statement/s is/are not correct from above ? 1 Convert the following numbers as per required: (i) (965.198)10 = ( ? )16 83 3 (ii) (3B.4)16 = ( ? )8 (iii) (1110011100.110001)2 = ( ? )10 = ( ? )16 84 2 The 2’s complement of the number 1101101 is _________. 10011 1 2 2's complement of any binary number can be calculated by adding 1's 85 False 1 complement twice. 86 2 The 2’s complement of the number 1101110 is ______. 0010010 1  2 𝑆𝑢𝑏𝑠𝑡𝑟𝑎𝑐𝑡 (45)8 from (66)8 using 2's complement method. 87 2 2 𝑆𝑢𝑏𝑠𝑡𝑟𝑎𝑐𝑡 (45)10 from (93)10using 1’s Complement Method. 88 (48)10 2 2 Subtract using 2’s complement method. 89 2 (10010 − 10011)2 2 Perform subtraction of (78 − 58)10using 2’s complement method. 90 2 2 Using 10’s complement, subtract : (72532-3250)10 91 2 92 2 Using 2’s complement, subtract : (1010100 − 1000100)2 2 2 Perform the subtraction with the following numbers using 1’s 93 4 complement and 2’s complements: (a) 11010-1101, (b) 10010-10011 2 Perform the operation of subtractions with the following binary number 94 using 2’s complement (i) 10010 – 10011 (ii) 100 – 110000 (iii) 11010 – 6 10000 2 Find the 10’s complement of the following: (1) (6106)10 (2) 95 (935)10 4 2 Perform following subtraction using 2’s complement method. (11010)2 – 96 2 (10000)2 2 97 Subtract (32)10 from (58)10 using 8-bit 2's complement arithmetic. 2 98 2 Substract 14 from 46 using 8-bit 2's complement arithmetic. 2 2 99 Substract 27.50 from 68.75 using 12-bit 1's complement arithmetic. 2 2 Add the following binary numbers: 100 1. 11011 + 1101 2 2. 1010.11 + 1101.0 + 1001.11+1111.11 2 Substract the following binary numbers. 101 2 1. 1100.10-111.01 2. 10110-1011 2 102 The addition of these binary numbers 101001+ 010011 would generate: C 1 101110 000111 111100 010100 2 Each 1 to a 0 Each 0 to 1 Each 1 to 0 and each 0 None of Above 103 The 1's complement of a binary number is obtained by changing ____ C 1 to 1 2 Substract the following hexadecimal numbers using the 1's complement arithmetic. 104 4 (i) 4816 − 2616 (ii) 4516 − 7416 2 Substract the following hexadecimal numbers using the 2's complement 105 arithmetic. 4 (i) 6916 − 4316 (ii) 2716 − 7316 2 Substract the following decimal numbers using the 9's and 10's complement arithmetic. 106 4 (i) 27410 − 8610 (ii) 376.310 − 765.610  2 Substract the following decimal numbers using 2's complement 107 arithmetic: 4 1. 46 - 19 2. 36.75 - 89.5 2 Substract the following numbers using 9's complement: 745.81 - 436.62 108 2 2 Substract the following numbers using 10's complement: 416.73 - 109 2928.54 2 2 Express -45 and -73.75 in 2's complement form. 110 2 2 Add 47.25 to 55.75 using binary arithmetic. 111 2 2 Add -75 to +26 using 8 bit 2's complement arithmetic. 112 2 2 Substract the following decimal numbers using 1's complement 113 arithmetic: 4 1. 46-84 2. 63.75-17.5 2 Find out Y, if B = 1 and A = square wave 114 1 115 2 Which gate is equivalent to bubbled OR gate? D 1 AND XOR NOT NAND 116 2 A NOT gate has… A 1 1 input and 1 output 2 input 1 output 1 input 2 output none of above 2 If a 3-input NOR gate has eight input possibilities, how many of those 2 1 7 8 117 B 1 possibilities will result in a HIGH output 2 AND NAND NOR OR If a signal passing through a gate is inhibited by sending a LOW into one 118 B 1 of the inputs, and the output is HIGH, the gate is a(n): 2 119 Implement Boolean expression for Ex-OR gate using NAND gates only. 3 120 2 The output of a ____ gate is only 1 when all of its inputs are 1. C 1 OR NOT AND EXOR 2 The inputs of a NAND gate are connected together. The resulting circuit OR AND NOT NONE OF ABOVE 121 C 1 is ………… 2 AND NAND NOT NONE OF ABOVE 122 The NOR gate is OR gate followed by ……………… C 1 2 NOT OR NOR NONE OF ABOVE 123 The NAND gate is AND gate followed by ………………… A 1 2 Exclusive-OR (XOR) logic gates can be constructed from ………..logic OR GATES ONLY AND gates and NOT OR gates and NOT 124 C 1 gates. gates AND, OR, NOT Gates gates 2 The basic logic gate whose output is the complement of the input is OR AND INVERTER NONE OF ABOVE 125 C 1 …………. 2 EXOR NAND NOR BOTH (B) AND (C) 126 The universal gate is ……………… D 1 127 2 The NAND gate output will be low if the two inputs are _____. 1&1 1 2 The output of a logic gate is 1 when all its inputs are at logic 0. the gate (a NOR or an EX- 128 1 is either ____________. NOR) 129 2 _______ AND gates are required to realize Y = CD+EF+G B 1 1 2 3 4  2 The following waveform pattern is for _____ 130 EXOR GATE 1 2 The following waveform pattern is for _______ 131 OR GATE 1 2 The 8-input XOR circuit shown has an output of Y = 1. Correct input combination below (ordered A – H) is _________. 132 1 2 Find the logic required at R input. 133 R=1 1  2 For a given logic circuit, if A=B=1, and C=D=0. Find output Y 134 0 1 2 Show that 𝐴 ⊕ 𝐵 = 𝐴𝐵′ + 𝐴′ 𝐵 and construct the corresponding logic 135 3 diagrams using truth table 2 Show that 𝐴 ⊙ 𝐵 = 𝐴𝐵 + 𝐴′ 𝐵' = (A⊕B)' = (AB'+A'B)' and 136 3 construct the corresponding logic diagrams using truth table 2 𝐴 ⊕ using 𝐵 137 Show that (A+B)(AB)' is equivalent to truth table 3 2 Derive that logic expression that equals to 1 only when the two bit binary 138 numbers A and B have the same value. Draw the logic diagram and 3 construct the truth table to verify the logic. 2 𝐴⊙𝐵 139 Show that AB + (A+B)' is equivalent to using truth table 3 2 Find the logical equivalent of the following expression: 140 5 (a) 𝐴 ⊕ 0 (b) 𝐴 ⊕ 1 (c ) 𝐴 ⊙ 0 (d) 𝐴 ⊙ 1 (e ) 0 ⊕ 𝐴′ 2 Draw the logic diagram and construct the truth table for following 141 expression: X= A+B+ (CD)' 3 2 Draw the logic diagram and construct the truth table for following 142 expression: Y= (AB) (A+B)' + (EF)' 3 2 Draw the logic diagram and construct the truth table for following 143 expression: Z = (A'B+CD'+ABC)' 3 2 11000011 00110100 00100000 00110011 A calculator performs the binary addition where the 2's complement 144 binary number 1101100 is obtained with no carry , convert the same into C 1 BCD code which is used to display on LCD screen of a calculator. 2 Excess-3 is also known as _____ Positive weighted Negative weighted Cyclic code Self Complementing 145 D 1 code code Code 2 146 Solve: (11000111)𝑋𝑆−3 = ()𝐺𝑅𝐴𝑌 2 2 01010 to 10101 01111 to 11111 10000 to 100000 01011 to 10111 The specific non-weighted code is used in shaft encoder for the process based instrumentation and data acquistion system that initiates encoder 147 A 1 sequence with 01111 to the second sequence as 11111 then evaluate would be the equivalent binary number as per sequence? 2 The transmitter transmits the 8-bit of information in form of data packet 1 0 01 don’tcare as D0 (LSB) to D7 (MSB) where D0 defines the odd parity bit, D1 as 148 start bit ='1', D2 to D5 as data bits = (8)2421 number and D6 and D7 as B 1 stop bits='01' respectively in a transreceiver digital communication system. Evaluate D0  2 Which of the following is true? ONLY 1 ONLY 2 ONLY 3 ALL OF THE 1. (A+B).(AB)' = A XOR B ABOVE 149 D 1 2. AB+(A+B)' = A XNOR B 3. A XOR B' = A XNOR B 150 2 (000100100011)𝐵𝐶𝐷 = ()2 1111011 1 2 Both 1 and 2 both 2 and 3 all are true all are false Which of the following statements are true? 1. The codes 8421,2421,5211,3321,4311 are some of the positively weighted codes. 151 C 1 2. The codes 642-3, 631-1, 84-2-1, 74-2-1 are some of the negatively weighted codes. 3. Non-weighted codes do not obey the position-weighting principle. 2 10101 to 10111 00000 to 11111 11001 to 11010 11010 to 11011 The specific non-weighted code is used in shaft encoder for the process based instrumentation and data acquistion system that initiates encoder 152 C 1 sequence with 10101 to the second sequence as 10111 then evaluate would be the equivalent binary number as per sequence? 2 Both 1 and 2 both 2 and 3 all are true all are false Which of the following statements are true? 1. An XOR gate produces an output 1 only when the inputs are not equal. 153 2. An XNOR gate produces an output 1 only when the inputs are equal. C 1 3. An XOR is also called anti-coincidence gate. 4. An XNOR is also called coincidence gate. 154 2 The excess-3 code of decimal 7 is represented by______. B 1 0111 1010 0011 0100 155 2 Covert the Gray code 1101 to binary 1001 1 2 Do as directed: 1. Convert the gray code 11011 into decimal. 156 2. Convert the decimal 2493 into XS-3 code. 4 3. Convert the decimal 1525 into gray code. 4. (10101101)2 = ()𝐵𝐶𝐷 = ()𝐺𝑅𝐴𝑌 2 Convert decimal 86 into BCD, excess-3 and Gray code. 157 3 2 Convert (96)10 to its equivalent Gray code and EX-3 code 158 4 2 Convert (10000110)𝐵𝐶𝐷 to decimal, binary & octal. 159 3 2 (396)10 = ()𝐵𝐶𝐷 = ()𝐺𝑅𝐴𝑌 = ()𝑋𝑆−3 160 3 2 Show truth table for the following circuit. 161 3  2 B2, 0100 0011 2B, 0100 0011 2B, 0011 0100 B2, 0100 0100 Decimal 43 in Hexadecimal and BCD number system is respectively 162 B 1 ______________ 2 Implement truth table for the following circuit. 163 5 2 Implement truth table for the following circuit. 164 4 2 Implement truth table for the following circuit. 165 3 2 Solve: (101011)2 + (001111)2 = ( )10 = ( )16 166 2 2 The 2's complement of the binary number 1001001 in XS-3 is ____ 167 1 2 Given A = P XOR Q, B= P XOR Q' and C= PQ, find output Y = A OR B 168 2 OR C as per data. 169 2 Solve: (10100101)𝑋𝑆−3 = ()𝐺𝑅𝐴𝑌 2  2 A processor started the operation with reflected code as one bit of change at a time when code is moving from one state to another. The 170 1 obtained code in one state as 111011 and went to 111001. Convert corresponding code into binary equivalent. 2 The transmitter transmits the 8-bit of information in form of data packet 1 0 01 don’tcare as D0 (LSB) to D7 (MSB) where D0 defines the odd parity bit, D1 as 171 start bit ='1', D2 to D5 as data bits = (F)16 number and D6 and D7 as A 1 stop bits='01' respectively in a transreceiver digital communication system. Evaluate D0 2 Subtract 745.81 from 436.62 given in decimal using 9’s complement 172 2 method. 173 2 Solve: (101011)2 - (001111)2 = (? )10 = (? )16 2 174 2 The quantity (337)10 can be represented in Excess-3 code as: D 1 100000011001 11000110111 11000111010 11001101010 2 all 1, 4, 5 only 1 & 4 all 1, 2 ,5 only 2 & 3 Which of the following statements is/are correct for Excess-3 code? 1. It is a self-complementing code. 2. The binary sum of a code and its 1’s complement is equal to 1111. 175 C 1 3. It is a weighted code. 4. The binary sum of a code and its 2’s complement is equal to 1111. 5. Complement can be generated by inverting each bit pattern. 2 The given logic circuit represents - 4-bit binary to 4-bit gray to binary 4-bit binary to gray 4-bit XS-3 to decimal decimal converter converter converter converter 176 C 1 2 Unit distance code is the other name of - Sequential code Cyclic Code Self complementing 2421 BCD code 177 B 1 code 178 2 Which of the following number/s is/are in invalid format? D 1 (CAFE)16 (5208)8 (110000101101)XS-3 Both B and C 179 2 Which of the following is not an invalid BCD code? C 1 1011 1010 1001 1100 180 2 An Excess-3 code for the number (FD)16 is given by _________ C 1 0010 0101 0011 0011 0010 1000 0101 1000 0110 0101 1001 0110 2 Subtract (3250)10 from (72532)10 using 10’s complement method. 181 2 2 If a 3-input NAND gate has eight input possibilities, how many of those 7 1 3 4 182 A 1 possibilities will result in a HIGH output 2 The 2421 and Excess-3 code of decimal 6 is represented by______. 1100 and 1001 1100 and 1100 0110 and 1001 1100 and 0110 183 A 1 2 The transmitter transmits the 8-bit of information in form of data packet 8 2 0 1 as D0 (LSB) to D7 (MSB) where D0 defines the even parity bit, D1 as 184 start bit ='1', D2 to D5 as data bits = (5)10 in 84-2-1 code and D6 and D7 D 1 as stop bits='10' respectively in a transreceiver digital communication system. Evaluate D0  2 Which Logic operation should be performed between A and B for a XNOR XOR OR NAND given output X? 185 A 1 2 Substract 75.125 from 84.625 using 12-bit 1's complement arithmetic. 186 2 2 BCD equivalent code of octal number (1431)8 is a)100101110011 b) 011100000010 c) 100100111001 d) 011110010011 187 d 1 2 Convert XS-3 number (11000111) into gray code a) 1110011 b)0110110 c)1100111 d)1110001 188 d 1 2 Subtract (741.41)16 from (436.6)16 by using decimal (r-1) complement 189 3 method. 2 Perform substraction of A – B using binary diminished radix 190 complement , where A is octal number (75) & B is octal number (53). 3 2 Do as directed: a) Convert the gray code 11011 into decimal. b) 191 2 Convert the decimal 2493 into XS-3 code 2 Do as directed : a)Convert the decimal 1525 into gray code. b) convert 192 2 (10101101)2 to 8421 code & gray code 2 A processor started the operation with reflected code as one bit of 100001 to 110101 101110 to 110101 111111 to 110101 0001 to 111111 change at a time when code is moving from one state to another. The 193 a 1 obtained code in one state as 110001 and went to 101111. Convert corresponding code into binary equivalent? 2 Convert (19)10 into XS-3 code and gray code. a) (01010000) xs-3 & b) (01001100) xs-3 c) (01001101) xs-3 & d) (01001110) xs-3 194 b 1 11010 & 11010 11000 & 10010 2 The decimal equivalent of the excess-3 number 110010100011.01110101 1253.75 861.75 970.42 1132.87 195 c 1 is ______. 196 2 The 10’s complement of (715)8 is _____ d 1 359 953 540 539 2 (98.75)10 is equivalent to _____ (142.6)8 (62.C)16 (10011000.01110101)B All of above 197 d 1 CD 198 2 Assign the proper odd parity bit to the code 111001. b 1 1111011 1111001 111111 0111011 2 Which of the following is false? Only 1 Only 2 Only 3 All are false I. A 1-bit gray code has two code words 0 and 1 representing decimal numbers 0 and 1. respectively. 199 II. 2's complement of a 2's complement of a number is the c 1 number itself. III. Excess-3 code for 369 is 011110011100. 2 The 2’s complement of the decimal number (-541)10 in hexadecimal is DEE DDD DE3 DED 200 c 1 ___. 201 2 (1111011)GRAY = ( )XS-3 a 1 10110101 11111111 10111100 10100101 2 Which gates are called universal gates? Design exclusive-OR gate 202 3 having minimum gates using one of the universal gates. 3 What is the use of boolean identities? Minimizing the Maximizing the To evaluate logical Searching of logical 203 A 1 boolean expression boolean expression identity expression 204 3 __________ is used to implement boolean function. C 1 Logical notation Arithmatic logic Logic gates expressions 205 3 The boolean function A + BC is a reduced form of ________ B 1 AB + BC (A + B)(A + C) A’B + AB’C (A + C)B 206 3 Simplify Y = AB’ + (A’ + B)C. A 1 AB’ + C AB + AC A’B + AC’ AB + A 207 3 Complement of the expression A’B + CD’ is _________ B 1 (A’ + B)(C’ + D) (A + B’)(C’ + D) (A’ + B)(C’ + D) (A + B’)(C + D’) 208 3 (A + B)(A’ B’) = ? A 1 0 1 AB AB' 209 3 DeMorgan’s theorem states that _________ A 1 (AB)’ = A’ + B’ (A + B)’ = A’ * B A’ + B’ = A’B’ (AB)’ = A’ + B 3 In boolean algebra, the OR operation is performed by which properties? Associative Commutative 210 D 1 Distributive properties All of the Mentioned properties properties 211 3 The expression for Absorption law is given by _________ A 1 A + AB = A A + AB = B AB + AA’ = A A+B=B+A 212 3 According to boolean law: A + 1 = ? A 1 1 A 0 A' 213 3 The involution of A is equal to _________ A 1 A A' 1 0 214 3 A(A + B) = ? D 1 AB 1 0 A 215 3 Find the simplified expression A’BC’+AC’. C 1 B A+C (A+B)C’ AB 216 3 (X + Z)(X + XZ’) + XY + Y = ?. D 1 XY+Z’ Y+XZ’+Y’Z X’Z+Y X+Y 217 3 A’(A + BC) + (AC + B’C) = ?. D 1 (AB’C+BC’) (A’B+C’) (A+ BC) C 218 3 Simplify the expression XZ’ + (Y + Y’Z) + XY. C 1 (1+XY’) YZ + XY’ + Z’ (X + Y +Z) XY’+ Z’ 219 3 Y’ (X’ + Y’) (X + X’Y) = ? A 1 XY’ X’Y X+Y X’Y’ 3 If an expression is given that x+x’y’z=x+y’z, find the minimal 220 C 1 y’ + z xz + y’ x+z x’ + y expression of the function F(x,y,z) = x+x’y’z+yz? 221 3 Simplify XY’ + X’ + Y’X’. C 1 X’ + Y XY’ (XY)’ Y’ + X 3 Minimize the Boolean expression using Boolean identities: 222 A 1 B(AC)’ + AC’ AC’ + B’ ABC + B’ + C BC’ + A’B A′B+ABC′+BC’+AB′C′. 3 Minimize the following Boolean expression using Boolean 223 D 1 A + B + C’ AC’ + B B + AC A(B’ + C) identities.F(A,B,C) = (A+BC’)(AB’+C) 224 3 Minimization of function F(A,B,C) = AB(B+C) is _________ D 1 AC B+C B` AB 225 3 Algebra of logic is termed as ______________ B 1 Numerical logic Boolean algebra Arithmetic logic Boolean number 3 Boolean algebra can be used ____________ For designing of the In building logic Building algebraic 226 A 1 Circuit theory digital computers symbols functions 3 A ________ value is represented by a Boolean expression. 227 D 1 Positive Recursive Negative Boolean 228 3 A + AB + ABC + ABCD + ABCDE + …. = ____________ B 1 1 A A+AB AB 3 The minimum number of literal obtained on simplifying the expression 229 ABC + A’C + AB’C + A’BC are _____. C 1 A'C+BC A(A+B') C A'B+AC' 3 Reduce the expression: A+B (AC+(B+C')D) 230 2 3 Reduce the expression: (A+(BC)')’(AB'+ABC) 231 3 3 Minimize the Boolean expressions: X = ( (A’B’C’)’ + (A’B)’ )’. 232 2 3 Find the complement of the following boolean function and reduce to a 233 minimum numbers of literals: B'D + A'BC' +ACD + A'BC 3 3 Draw the logic diagram of the following function. Use one OR gate and 234 3 one AND gate only. Y = (A+B)(A+C)  3 Given boolean function F = XY + X'Y'+Y'Z. Implement it with only OR 235 and NOT gates. 4 3 Prove that a dual of Ex-OR is also its complement. 236 4 3 The logic expression (A'+B) (A+B') can be implemented by giving the 237 D 1 NOR Gate NAND Gate XOR Gate XNOR Gate inputs A and B to a two input _____ 238 3 Which of the given boolean expression is incorrect? B 1 A + A'B = A + B A + AB = B (A+B)(A+C)=A+BC (A+B')(A+B)=A 239 3 The simplified form of boolean expression (X+Y+XY)(X+Z) is C 1 X+Y+Z XY+YZ X+YZ XZ+Y 240 3 AB + A'C + BC = AB + A'C represents which theorem? A 1 Consensus Transposition De Morgan's None 3 The simplified form of boolean expression (A'BC+D)(A'D+B'C') can be 241 A 1 A'D+B'C'D AD+BC'D (A'+D)(B'C+D') AD'+BCD' written as 3 Determine the output X of a logic circuit shown in figure. Simplify the output expression using boolean laws and theorems. Redraw the logic circuit with the simplified expression. Simplified Expression 242 3 is: AB' 3 Consider the following boolean expressions: I: x.y + x'y' II: XOR (x', y') 243 III: XOR (x', y) C 1 I and II I, II and III I and III II and III Which of the above expressions represents exclusive NOR operation? 3 Consider the boolean function: f = (a+bc)(pq+r). Complement (f)' of a 244 B 1 (a'+b'c')(p'q'+r') a'(b'+c')+(p'+q')r' (a'+b'c')+(p'q'+r') a'b'c'+p'q'r' function f is - 3 Derive the logic expression for the given logic diagram 245 A 1 C(A + B)DE [C(A + B)D + E'] [[C(A + B)D]E'] ABCDE 246 3 Most Boolean reductions result in an equation in only one form. A 1 TRUE FALSE 3 According to property of Commutative law, the order of combining initial result of final result of mid-term result of 247 B 1 None terms does not affect _________ combination combination combination 3 According to boolean algebra which of the following relation is not 248 D 1 X(YZ) = (XY)Z X(Y+Z) = XY + XZ X+XZ = X X(X+Y)=1 valid? 249 3 Simplify the following expression: F = AB'C + AB'C'+ABC A 1 F = AC + AB' F = A'C + AB F = AC' + AB F = AC' + AB'  3 Which of the following options correctly represents the consensus law AB + A'C + BC = A'B + A'C + BC = AB + A'C + BC = AB A'B + A'C + BC = 250 A 1 of Digital Circuits? AB + A'C AB + A'C + (AC)' AB + (AC)' 3 Verify the equation by using boolean algebra: AB + AC + BC' = AC 251 4 + BC' 3 Verify the equation by using boolean algebra: (AB + BC + CA)' = A'B' 252 4 + B'C' + C'A' 3 Find out the minimized form of a logical expression (A'B'C' + A'BC' + 253 A'C' + BC' + A'B 3 A'BC + ABC'). 254 3 The boolean expression (X+Y)(X+Y')+(X'Y'+X')' simplifies to A 1 X Y XY X+Y 3 Simplify the following boolean expression to four literals: F = A'C + 255 AB+C+D 3 C'D + B'C + AB 256 3 The reduced form of complement of equation f = [(ab)'a][(ab)'b] B 1 0 1 a'b+ab' a'b' 257 3 The reduced form of dual of equation f = [(ab)'a][(ab)'b] B 1 0 1 a'b+ab' a'b' 258 3 Which of the given boolean expression is incorrect? D 1 (A'+B')' = AB (A+ (B')')' = A' B' (A'B')' = A+B (A'B'')' = A'+B 259 3 The boolean equation x = [(A+B')(B+C)]B can be simplified to - C 1 X = A'B X = AB' X = AB X = A'B' 3 Find the output boolean function for the given logic circuit. 260 A 1 X = A'BC X = AB'C X = ABC X = A'B'C 3 The output Y of the logic circuit is - 261 A 1 1 0 X X' 3 Simplify the following boolean expression and realize it using NOR 262 gates only: 3 𝑌 = 𝐴𝐵ത + 𝐴𝐵𝐶ҧ + 𝐴𝐵𝐶𝐷 + 𝐴𝐵𝐶 𝐷 ഥ 3 Simplify the following boolean expression and realize it using NAND 263 gates only: 3 𝐴𝐵𝐶[𝐴𝐵 + 𝐶ҧ 𝐵𝐶 + 𝐴𝐶 ] 3 Match the appropriate pairs: 1-B, 2-A, 3-D, 4-C 1-D, 2-A, 3-C, 4-B 1-D, 2-A, 3-B, 4-C 1-C, 2-A, 3-D, 4-B (1) x+(y+z)=(x+y)+z (A) Commutative Law 264 (2) x+y=y+x (B) Absorption Law C 1 (3) x+xy=x (C) Complementation law (4) (x')'=x (D) Associative Law 265 3 Simplify: ҧ + 𝐴𝐵𝐶ത B+AC 2 f = 𝐴𝐵 + 𝐴𝐵𝐶 + 𝐴𝐵 3 Minimize the following function using boolean algebra: 266 ҧ f = 𝐴𝐵𝐶𝐷 + 𝐴𝐵𝐶ҧ 𝐷 ഥ +𝐴𝐵𝐶𝐷 ҧ + 𝐴𝐵𝐶𝐷 + 𝐴𝐵𝐶 𝐷 ҧ ഥ + 𝐴𝐵ത 𝐶𝐷 2 ത + 𝐴𝐵𝐶𝐷 + 𝐴𝐵𝐶ത 𝐷 ഥ 267 3 Simplify: ത + (𝐴 + 𝐵𝐶) f = 𝐴 + 𝐵𝐶 ത =1 3  3 Simplify the following boolean expression which represents the output of a logical decision circuit 268 x+y 2 f w, x, y, z = x + xyz + 𝑥𝑦𝑧 ҧ + 𝑤𝑥 + 𝑤𝑥 ഥ + 𝑥𝑦 ҧ 3 Simplify the following boolean expression which represents the output 269 of a logical decision circuit ABC'+D 3 f A, B, C, D, E = (AB + C + D)(𝐶ҧ + 𝐷)(𝐶ҧ + 𝐷 + 𝐸) 3 Prove the following identity: 270 𝐴𝐵 + 𝐴ҧ + 𝐴𝐵 = 0 2 3 Simplify the following function by using boolean algebra: 271 3 Y = AB'C'D + A'B'D + BCD' + A'B + BC' 3 Simplify the following function by using boolean algebra: 272 AB+A'B'C 2 Y = (AB + A'C + BC)(A+B'+AB') 3 Construct a logic circuit to give an output without any reduction in number of gates. Give the logic gates output step by step. 273 3 ҧ 𝑋 = (𝐴𝐵 + 𝐴𝐶)(𝐴𝐷 + 𝐶) 274 3 Find out the complement of boolean expression : AB (B'C+AC) 2 3 The boolean function Y = AB + CD is to be realized using only 2-input 275 NAND gates. The minimum number of NAND gates required is - B 2 2 3 4 5 3 Realize the given logic circuit into appropriate boolean expression and also minimize it. 276 3 3 Make a suitable logic expression from the given truth table and also minimize it up to a single literal remaining. 277 2  3 _______ NAND gates required to implement the function : 278 3 𝐹 = (𝑋ത + 𝑌)(𝑍 ത + 𝑊) 3 The boolean expression for the given circuit is _____ A {F + (B + C) (D + A + F + [(B + C) (D + 279 A 1 A [F + (B + C) (DE)] A [F + (BC) (DE)] E)} E)] 3 DeMorgan's first theorem shows the equivalence of ______ OR gate and NOR gate and NOR gate and NAND NAND gate and 280 B 1 Exclusive OR gate Bubbled AND gate gate NOT gate 281 3 ((AB)'+(AC)')' = ________ B 1 A+B+C ABC A'BC (A+B+C)' 3 Consider four distinct input boolean variables, in which first two and last two variables are ANDed. Then after output of both operation is Ored 282 2 together. Implement the circuit based on the given information and also find its boolean expression. 3 What is the boolean expression for Y in above circuit ? 283 2 3 Find out the simplified boolean equation for the given logic circuit. (A + B) (C + D) (E + 284 A 1 A+B+C+D+E+F ABCDEF ABC + DEF F) 3 Four inputs A, B, C, D are fed to a NOR gate. The output of NOR gate is 285 fed to the two successive inverter. The final output is given by - B 1 A+B+C+D (A+B+C+D)' ABCD (ABCD)' ____________  3 For the logic circuit of the given figure, the minimized expression is - 286 A 3 Y = (AB'C)' Y = A+B+C Y=A+B Y = ABC 3 Choose the correct statement for 'literal' - (I) Literal is a primed variable only. (II) The minimization of the number of literals and the number of terms results in a circuit with less equipment 287 (III) We can not reduce the number of literals in any given Boolean B 1 Only I and III Only II Only II and IV All are correct function. (IV) In the Boolean equation y=a^' b+b^' c+c'd, the terms a’b , b’c and c’d are known as literals. 3 Simplify the boolean expression to a minimum number of literals: 288 2 y(wz'+wz)+xy 3 Reduce the following boolean function upto minimum five literals (including primed and unprimed variables): 289 𝐴𝐵𝐶 + 𝐴′ 𝐵′ 𝐶 + 𝐴′ 𝐵𝐶 + 𝐴𝐵𝐶 ′ + 𝐴′ 𝐵′ 𝐶′ 2 3 Reduce the following boolean function upto minimum four literals (including primed and unprimed variables): 290 2 ҧ 𝐶ҧ + 𝐴𝐶𝐷 + 𝐴𝐵𝐶 ത + 𝐴𝐵 𝑋 = 𝐵𝐷 ҧ 3 Find the complement of F = x +yz, then show that F∙F' = 0 and F+F' = 1 291 2 3 Find the complement of the following Boolean functions and reduce 292 them to a minimum number of literals: 3 (𝐴′ + 𝐶)(𝐴′ + 𝐶 ′ )(𝐴 + 𝐵 + 𝐶 ′ 𝐷)′ 3 Write the Boolean expression for output X for the logic circuit shown in figure. Also show the ouput of each stage of logic gates: 293 3  3 Write the Boolean expression for output X for the logic circuit shown in figure. Also show the ouput of each stage of logic gates: 294 3 3 Construct a logic circuit using INVERTER, AND and OR gates for the Boolean expression 295 3 3 A network of cascaded inverters shown in fig. If a logic 1 is applied to A, Determine the logic levels from point B through F. 296 2 3 Using boolean laws and rules, simplify the expression: Z= 297 3 A'BC + AB'C' + A'B'C' + AB'C + ABC 3 Using boolean laws and rules, simplify the expression: Z= 298 3 (AB+AC)' + A'B'C and prove that Z = A' + B'C' 3 Demorganize the expression: 299 2 𝐴 + 𝐵 𝐶ഥ 𝐷 ഥ + (𝐸 + 𝐹) ത 3 Simplify the boolean expression: 300 3 T[x,y,z] = (x+y) (x'(y'+z'))'+x'y'+x'z'. 3 The circuit shown in Fig. is used to implement the function Z = f (A, B) = A + B. What values should be selected for I and J ? 301 3  3 Determine the logic function realized by the circuit shown in Fig. Also minimize the output function f using laws boolean algebra. 302 3 3 Determine by means of truth table, the validity of De Morgan’s theorem 303 3 for three variables: (ABC)' = A' + B' + C' 3 A switch board contains four electrical switches which are connected in a sequence in a room: 1st switch as main power supply switch of the room, 2nd switch belongs to a fan, 3rd switch belongs to a TV and 4th switch belongs to 0 W bulb. The main power supply switch is also 304 directly connected to the 0W bulb. Design the boolean expression of the 3 switch board of the room such that user can operate switch board as per his requirement. Also show how the boolean expression can be minimized. 3 Boolean arithmetic is a : way to express logic terrible fraud very difficult fast way to solve statements in a perpetrated by calculation used in problems around the traditional philosophers to astronomy house 305 A 1 mathematical disprove things they equation format don't agree with 3 Determine and minimize the Boolean expression for the output, X of a logic circuit shown in figure. Also give your comment on the minimized output. Which logic gate is represented by simplified output? 306 AND gate 3 3 Simplify the following boolean expression using laws of Boolean algebra. 307 3 f(w,x,y,z)=x+x'yz+x'yz+wx+w'x+x'y Y=AB'+ABC'+ABCD+ABCD' 3 As per _______ theorem, PQ+P’R+QR = _______ Transposition, 308 C 1 Absorption, PQ+P’R Consensus, PQ+P’R Consensus, PQ+PR’ PQ+QR 309 3 Which of the given boolean expression is incorrect? A 1 (A'B'')' = A'+B (A'B')' = A+B (A+ (B')')' = A' B' (AB’)’ = A’+B  3 Which of the following statements are true? 1. An XOR gate produces an output 1 only when the inputs are not equal. 2. An XNOR gate produces an output 1 only when the inputs are equal. 3. An XOR is also called anti-coincidence gate. 310 4. An XNOR is also called coincidence gate. D 1 Both 1 and 2 Both 3 and 4 Both 2 and 3 All are true 3 Evaluate which one is correct? 311 C 1 Only 1 Only 2 Only 3 Only 4 3 If a 3-input NAND gate has eight input possibilities, how many of those 312 A 1 1 7 5 6 possibilities will result in a LOW output? 3 The given logic diagram describes the process of an industrial operation that depends on distinct values of A, B and C. Find the boolean expressions of each level (Level 1 to level 7) also define following logic diagram belongs to which equivalent logic gate? 313 4 3 Consider the following boolean expressions and Check whether the Only II and IV are Only III is correct Only I, II and III are All are correct given statement/s is/are Correct or not. correct correct 314 I: MN’ + M’ + M’N’ = (MN)’ C 1 II: Q’ (P’ + Q’) (P + P’Q) = PQ’ III: AB (A + B’) + A (B + B’) = A IV: A’ + A’B + A’BC + A’BCD + A’BCDE + …. = A’B 3 The circuit shown in Fig. is used to implement the function Z = f (A, B) I= A' AND J= B' I= 1 AND J=B' I=A AND J= B I= B' AND J= A' = (AB)'. What values should be selected for I and J ? 315 Both A & B 1  3 Given W = [(A XOR B) XNOR 1], X= [(A XOR B') XOR 0], Y= [(A XNOR B) XOR 1] and Z= [(A’ XOR B) XNOR 0]. Find Output 316 Function = W OR X OR Y OR Z as per data and simplify it using basic 2 theorems and construct simplified function with basic logic gates. 3 Determine the output F of a logic circuit shown in figure with truth table and logical expression. Simplify the output expression using boolean laws and theorems. Redraw the logic circuit of simplified expression with two gates only. 317 3 3 If a signal passing through a gate is inhibited by sending a LOW into one a) AND b) NAND c) OR d) NOR 318 of the inputs, and the output is HIGH, the gate is a(n): b 1 3 Choose correct statement/s from following a) Only I & II b) Only I & III c) Only II & III d)All of the above i) an inverter performs complementation operation ii) process performed by inverter is known as inversion 319 iii) an inverter contains two or more input terminal and one output a 1 terminal. 3 if the boolean expression P’Q + QR + PR is minimized , the expression a) P’Q + QR b) P’Q + PR c) QR + PR d) P’Q+ QR + PR 320 becomes b 1 321 3 A + AB = A ; A(A+B) = A represents which law ? b 1 a) commutative b) Absorption c) Consensus d) Transposition 3 Exclusive-OR (XOR) logic gates can be constructed from ………..logic a)AND, OR, NOT b)OR Gate & NOT c)NOR & NOT Gate d) a & c both 322 d 1 gates gate 3 The circuit shown below generates the function of F= __________ a) x⊕y b) 0 c) xȳ + yx + ȳx d) x + y 323 a 1 3 Transmitter transmit 8 bit of information in form of data packet D0 a) 1 b) 11 c) 10 d) 0 (LSB) to D7 (MSB) where D0 Defines EVEN PARITY BIT, D1 as start 324 bit ‘1’ , D2 to D5 as data bits (E)16 number & D6 & D7 as stop bit ‘10’ a 1 respectively in transreceiver digital communication system. evalue the D0 = ________  3 In the figure shown, the output Y is required to be Y = A.B + C'.D'.The a) AND & OR b) OR & NAND c) NAND & OR d) NOR & OR gates G1 and G2 must be respectively 325 d 1 3 Draw the logic diagram of boolean expression (A+B)(CD)'(E+F)' G' & 326 its complement seperately using only two input basic logic gates ( do not 3 use NAND , NOR , X-OR , X-NOR logic gates ) 3 Which of the following is true? Both 1 and 2 Both 1 and 3 Both 2 and 3 All are true 1. (A+B). (AB)' = A XOR B 327 d 1 2. AB+(A+B)' = A XNOR B 3. A XOR B' = A XNOR B 3 Simplify the Boolean expression. 328 2 AB+(AC)’+AB’C(AB+C) 3 The given logic diagram depends on values of A & B. Find the Boolean expressions of each level till output Y and also define following logic diagram belongs to which equivalent gate? 329 4 3 Match the following a) 1-P, 2-Q, 3-R, 4-S b) 1-Q, 2-P, 3-R, 4-S c) 1-S, 2-R, 3-P, 4-Q d) 1-Q, 2-P, 3-S, 4-R 1) A+A’ P. 0 330 2) A⊕ A Q. 1 b 1 3) 0 ⊕ A R. A 4) A’A’ S. A’ 331 3 MN(M + N’) + M(N + N’) = _____ a 1 a) M b) N c) M' d) N'  3 Find out the Boolean Expression for Logic Diagram given below and simplify the output in the minimal expression, also implement the simplified expression using the AOI logic. 332 4 3 Match the following with correct logic expression: 1-x, 2-z, 3-w, 4-y 1-x, 2-w, 3-z, 4-y 1-w, 2-z, 3-x, 4-y 1-y, 2-w, 3-z, 4-x Column A Column B 1. A’ ʘ A ⊕ A’ w. 1 333 2. A’ ⊕ 0 ⊕ A x. A’ b 1 3. A ⊕ A’ ʘ A y. 0 4. A’ ʘ 1 ʘ A z. A 334 4 The Sum of all Minterm is ______ B 1 0 1 2 4 335 4 The Product of all Maxterm is ________ A 1 0 1 2 4 336 4 The minterm related to F(w,x,y,z)= m6 is A 1 w'xyz' wxyz wx'y'z wx'yz 337 4 The Maxterm related to F(w,x,y,z)= M7 is A 1 w+x'+y'+z' w'+x+y+z w+x+y+z w'+x+y'+z 338 4 Convert SOP into POS : F(A,B,C)= ∑m( 1,3,7) D 1 π M(0,2,4,6) π M(2,4,5,6) π M(0,2,4) π M(0,2,4,5,6) 4 Convert POS into SOP: F(w,x,y,z) = πM( 0,1,2,6,10,12,14,15) ∑m( ∑m( ∑m( 3,4,5,7,8,9,11,13) ∑m( 3,4,5,8,9,11,13) 339 C 1 3,4,5,7,8,9,11,13,16) 0,3,4,5,7,8,9,11,13) 340 4 Express the boolean function F= A + B'C in sum of minterm 2 4 Express the Boolean function F = AB + A’C in a product of maxterm. 341 2 342 4 Convert into Product-of-Maxterms: A(A’+B)(C’) 2 343 4 Convert into Sum-of-Minterms: A’ + B + CA 2 4 Express Function in Product of Maxterms F(x,y,z) = ( xy + z ) ( y + xz ) 344 2 4 Express the Boolean function in sum of minterms F 345 2 (A,B,C)=(A'+B)(B'+C) 4 Express the Boolean function in sum of minterms F (A,B,C,D)= 346 2 D(A'+B)+B'D 4 Express the Boolean function in sum of minterms and product of max terms. 347 3 F (A,B,C,D)= (A+B'+C)(A+B')(A+C'+D')(A'+B+C+D')(B+C'+D') 4 Express the Boolean function in sum of minterms and product of max 348 terms. 3 F (X,Y,Z) = (XY+Z)(Y+XZ)  4 Express the Boolean function in sum of minterms and product of max 349 terms. 3 F (X,Y,Z) = 1 4 Express the Boolean function in sum of minterms and product of max 350 terms. 3 F (W,X,Y,Z) = Y'Z+WXY'+WXZ'+ W'X'Z 351 4 Expand A'+ B' to minterm and maxterm 2 352 4 Expand A+BC'+ABD'+ABCD to minterm and maxterm 2 4 Convert the following Boolean function in product of maxterms and sum 353 3 of minterms: F(m,n,o,p) = n'o'p+nop+mop'+m'n'o+m'no'p 354 4 Expand A(A'+B)(A'+B+C') to minterm and maxterm 2 355 4 Convert the following to other canonical form F( x,y,z) = ∑( 1,3,7) 2 4 Convert the following to other canonical form F( 356 2 A,B,C,D) = ∑( 0,2,6,11,13,14) 4 Convert the following to other canonical form F( X,Y,Z) = π (0,3,6,7) 357 2 4 Convert the following to other canonical form F( A,B,C,D) = π 358 2 (0,1,2,3,4,6,12) 4 A Karnaugh map is an abstract form of _______ diagram organized as a Venn Cycle Point Block 359 A 1 matrix of squares 4 Which of the following code is employed by K-Map for simplification of XS-3 Gray BCD Parity 360 B 1 Boolean Expression? 4 The 3-variable Karnaugh Map (K-Map) has _______ cells for min or 3 6 8 2 361 C 1 max terms 4 Which of the following is NOT considered for forming groups in K- Diagonal Rolling Vertical Horizontal 362 A 1 map? 4 There are 3 Variable in the boolean function and the value of the 3 8 1 0 363 function is 1. Find the number of cells in the K-Map which will contain a B 1 1 when SOP expression is used 4 There are 3 Variable in the boolean function and the value of the 3 8 1 0 364 function is 1. Find the number of cells in the K-Map which will contain a D 1 0 when SOP expression is used _____ 4 Given F(A, B, C, D) = ∑ m(0, 1, 2, 6, 8, 9, 10, 11) + ∑ d(3, 7, 14, 15) is Independent of Depedent on 2 Depedent on 3 variable Depedent on 4 365 a Boolean function, where m represents min-terms and d represents don’t B 1 variables variable variable cares. The minimized logical fuction F is _____ 4 In simplification of a Boolean function of n variables, a group of 2^m m – 1 literals less m + 1 literals less n + m literals n – m literals adjacent 1s leads to a term with than the total number than the total number 366 D 1 of variables of variables 4 Looping on a K-map always results in the elimination of _________ Variables within the Variables that Variables within the Variables within the loop that appear only remain unchanged loop that appear in loop that appear only 367 C 1 in their within the loop both complemented in their complemented form and uncomplemented uncomplemented form form 4 Digital input signals A, B, C with A as the MSB and C as the LSB are 1 (A + C) (A' +C') A'C' + AC A'C + AC' used to realize the Boolean function F = m0 + m2 + m3 + m5 + m7, 368 where mi denotes the ith minterm. In addition, F has a don’t care for m1 A 1 + m4 + m6. The simplified expression for F is given by:  4 Simplify the Boolean function using K-Map: F(w,x,y,z) = Σ 369 3 (0,1,2,4,5,6,8,9,12,13,14) 4 Simplify the Boolean function using K-Map: F(w,x,y) = Σ (0,1,3,4,5,7) 370 2 4 Obtain the simplified expression using K-Map in sum of product for the 371 Boolean functions 3 F= Σ(0,1,4,5,10,11,12,14). 4 Simplify the Boolean function F(x,y,z)= ∑m(0,1,2,3,4,5,6) using K- map. 372 3 Explain groups 4 Simplify the following Boolean function using K-map 373 3 F( w,x,y,z) = Σ( 1, 3, 7 , 11 , 15 ) 4 Simplify the following Boolean function using K-Map. 374 4 F=A′B′C′+B′CD′+A′BCD′+AB′C. 4 Obtain the simplified expressions in sum of products using K-map: x’z + 375 4 w’xy’ + w(x’y + xy’) 4 Simplify following boolean function using K-Map: Y = Σm(1, 5, 7, 9, 11, 376 3 13, 15) 4 377 Simplify following boolean function using K-Map: Y = Σm(0, 2, 3, 5) 2 4 Simplify following boolean function using K-Map: Y = Σm(1, 3, 5, 9, 378 3 11,13) 4 Simplify following boolean function using K-Map: Y = Σm(0, 2, 5, 6, 7, 379 3 8, 10, 13, 15) 4 Simplify following boolean function using K-Map: Y = Σm(1, 5, 6, 7, 11, 380 3 12, 13, 15) 4 Simplify following boolean function using K-Map: Y = Σm(1, 3, 4, 5, 7, 381 3 9, 11, 13, 15) 4 Simplify Boolean function using K-Map: F( w,x,y,z) = ∑( 382 3 1,3,5,8,9,11,15) 4 Simplify the Boolean Function with Karnaugh map: F(w,x,y,z) = 383 3 Σ(0,1,2,4,5,6,8,9,12,13,14,15) 4 Simplify the Boolean Function with Karnaugh map: F = 384 4 A’B’C’+B’CD’+A’BCD’+AB’C’ 4 Simplify the Boolean function F(x,y,z)= Σ (0,2,4,5,6) using K- 385 2 map.Explain groups. 4 Simplify following boolean function using K-Map: Y = AB'C'D' + 386 4 AB'C'D + AB'CD + AB'CD' 387 4 Simplify using K-map F= AB'C + AB'C'D + ABC'D + ABC 4 388 4 Simplify using K-map F= A'B'C + A'BC + ABC + ABC' 3 4 Minimize following Boolean function using K-map: X(A,B,C,D) = Σ 389 3 m(0, 1, 2, 3, 5, 7, 8, 9, 11, 15) 390 4 Simplify the Boolean function, F=Σm(0,1,2,5,8,9,10) 3 4 Minimize following Boolean function using K-map F = Σ m(1, 2, 4, 6, 7, 391 4 11, 15) + Σ d(0, 3) 4 Minimize the following function using K-Map F = 392 4 Σm(0,2,6,10,11,12,13) + d(3,4,5,14,15) 4 Simply the Boolean Function using K-map : F(W,X,Y,Z) = ∑(1, 3, 7, 393 4 11, 15) with don’t care conditions d(W,X,Y,Z) = ∑(0, 2, 5) 4 Reduce the given function using K-map F(A,B,C,D ) = Σm 394 4 (0,1,3,7,11,15) + Σd ( 2,4)  4 Minimize the following logic function using K-maps F(A,B,C,D) 395 4 =∑m(1,3,5,8,9,11,15) + d(2,13) 4 Minimize the following function using K-map 396 4 F (w,x,y,z) = Σm (0,1,2,3,6,7,13,14) + Σd (8,9,10,12) 4 Minimize the following functions using K Map F 397 3 =ΠM(0,4,9,10,11,14,15) 4 Minimize the logic function F (A,B,C,D) = π M (1, 2, 3, 8, 9, 10, 11, 14) 398. d (7, 15) Use Karnaugh map. Draw the logic circuit for simplified 4 function using NOR gates only. 4 Simplify Boolean function F (w,x,y,z ) = Σ (0,1,2,4,5,6,8,9,12,13,14 ) 399 4 using K-map and Implement it using NAND gates only 4 Solve the following Boolean functions by using K-Map. Implement the 400 simplified function by using logic gates F =(w,x,y,z) 4 =∑(0,1,4,5,6,8,9,10,12,13,14) 4 Reduce the expression F = ∑m(0,2,3,4,5,6) using K-map and implement 401 4 using NAND gates only. 402 4 Implement the function F=∑(0,6) with NAND gates only. 3 403 4 Implement the function F=∑(0,6) with NOR gates only 3 4 Obtain the simplified expressions in SOP for the following Boolean function using K-Map method. And implement it using NAND gate. 404 4 F(A,B,C,D) = ABC+AB’C+BCD’+A’CD 4 Reduce using mapping the expression ∑m(0, 1, 2, 3, 5, 7, 8, 9, 10, 12. 405 4 13) and implement it with any universal logic. 4 Reduce using mapping the expression πM(2, 8, 9, 10, 1, 12, 14) and 406 4 implement it with any universal logic 4 Reduce using mapping the expression ∑m(1, 5, 6, 12,13, 14)+ d(2,4) and 407 4 implement it in universal logic. 4 Reduce using mapping the expression ∑m(0,1, 3, 5, 6, 12,13, 14)+ 408 4 d(2,7,8,15) and implement it in universal logic. 4 Reduce using mapping the expression ∑m(0, 1, 4, 7, 13, 14) + d(5, 8, 15) 409 4 and implement it in universal logic. 4 Reduce using mapping the expression πM(0,1 ,3 4, 5, 7, 10, 13, 14, 410 4 15)and implement it in universal logic. 4 Reduce using mapping the expression πM(2, 4, 6, 8, 10, 12, 15) and 411 4 implement it in universal logic. 4 Reduce using mapping the expression πM(0, 1, 4, 6, 8, 9, 11) and d(2, 7, 412 4 13) and implement it in universal logic. 4 Reduce using mapping the expression πM(2, ,4 ,5 7, 9, 12) and d(0, 1, 6) 413 4 and implement it in universal logic. 4 The inputs to a computer circuit are the 4 bits of the binary number A3A2A1A0 The circuit is required to produce a 1 if any of the following conditions hold. 1. The MSB is 1 and all other bits are 0. 414 2. A2 is a 1 and the all other bits are 0. 5 3. Any one of the 4 bits are 0. 4. A1 and A0 is 1 and other bits are 0 Obtain a minimal expression using K-map and implement it with NAND gate  4 A8A4A2A1 is an Binary input to a logic circuit whose output is a 1 when A8=0, A4=0 and A2=1,or when A8=0 and A4=1.Design the 415 5 simplest possible logic circuit with NAND gate only using K-map in SOP form. 4 The ________ is a very useful and convenient tool for simplification of Quine McCluskey K-map SOP K-map POS k-map 416 A 1 Boolean functions for large numbers of variables. tabulation method 4 The implicants which will definitely occur in the final expression are Prime Non-prime Essential Prime implic