number system examples.pdf

1, convert the following binary numbers to decimal equivalents: A, 001100 B, 000011 C, 011100 D, 111100 E, 111111 Answer For the binary representation of y = {...b2 b1b0.b−1b−2 b−3...} , the value of Y is i y= bi × 2 i A, 001100 = 0 × 2 5 + 0 × 2 4 + 1 × 2 3 + 1...

1, convert the following binary numbers to decimal equivalents: A, 001100 B, 000011 C, 011100 D, 111100 E, 111111 Answer For the binary representation of y = {...b2 b1b0.b−1b−2 b−3...} , the value of Y is i y= bi × 2 i A, 001100 = 0 × 2 5 + 0 × 2 4 + 1 × 2 3 + 1 × 2 2 + 0 × 21 + 0 × 2 0 = 8 + 4 = 12 5 4 3 2 1 0 B, 000011 = 0 × 2 + 0 × 2 + 0 × 2 + 0 × 2 + 1 × 2 + 1 × 2 = 2 + 1 = 3 5 4 3 2 1 0 C, 011100 = 0 × 2 + 1 × 2 + 1 × 2 + 1 × 2 + 0 × 2 + 0 × 2 = 16 + 8 + 4 = 28 5 4 3 2 1 0 D, 111100 = 1 × 2 + 1 × 2 + 1 × 2 + 1 × 2 + 0 × 2 + 0 × 2 = 32 + 16 + 8 = 60 5 4 3 2 1 0 E, 111111 = 1 × 2 + 1 × 2 + 1 × 2 + 1 × 2 + 1 × 2 + 1 × 2 = 32 + 16 + 8 + 4 + 2 + 1 = 63 2, Convert the following binary numbers to their decimal equivalents: A, 11100.001 B, 110011.10011 C, 101010101010.1 Answer For the binary representation of y = {...b2 b1b0.b−1b−2 b−3...} , the value of Y is i y= bi × 2 i A, 11100.001= 1 × 2 4 + 1 × 2 3 + 1 × 2 2 + 0 × 21 + 0 × 2 0 + 0 × 2 −1 + 0 × 2 −2 + 1 × 2 −3 = 28 + 0.125 = 28.125 B, 110011.10011= 1 × 2 5 + 1 × 2 4 + 0 × 2 3 + 0 × 2 2 + 1 × 21 + 1 × 2 0 + 1 × 2 −1 + 0 × 2 −2 + 0 × 2 −3 + 1 × 2 −4 + 1 × 2 −5 = 51 + 0.5 + 0.0625 + 0.03125 = 51.59375 C, 101010101010.1= 1 × 211 + 1 × 2 9 + 1 × 2 7 + 1 × 2 5 + 1 × 2 3 + 1 × 21 + 1 × 2 −1 = 2048 + 512 + 128 + 32 + 8 + 2 + 0.5 = 2730.5 3, Convert the following decimal numbers to their binary equivalents A, 64 B, 128 C, 256 D, 100 E, 111 F, 145 G, 255 Answer A, Quotient Remainder 64/2 32 0 32/2 16 0 16/2 8 0 8/2 4 0 4/2 2 0 2/2 1 0 1/2 0 1 6410 = 1000000 2 B, 12810 = 10000000 2 C, 25610 = 100000000 2 D, Quotient Remainder 100/2 50 0 50/2 25 0 25/2 12 1 12/2 6 0 6/2 3 0 3/2 1 1 1/2 0 1 10010 = 1100100 2 E, Quotient Remainder 111/2 55 1 55/2 27 1 27/2 13 1 13/2 6 1 6/2 3 0 3/2 1 1 1/2 0 1 11110 = 11011112 F, 4510 = 100100012 G, 255 Quotient Remainder 255/2 127 1 127/2 63 1 63/2 31 1 31/2 15 1 15/2 7 1 7/2 3 1 3/2 1 1 1/2 0 1 25510 = 111111112 4, Convert the following decimal numbers to their binary equivalents A, 34.75 B, 25.25 C, 27.1875 A, 34.75 the integer part (34) convert to binary format Quotient Remainder 34/2 17 0 17/2 8 1 8/2 4 0 4/2 2 0 2/2 1 0 1/2 0 1 3410 = 100010 2 the fraction part (0.75) covert to binary format product integer part 0.75x2 1.5 1 0.5x2 1.0 1 0.7510 = 0.112 34.7510 = 100010.112 B, 25.25 the integer part is 25, convert to binary format Quotient Remainder 25/2 12 1 12/2 6 0 6/2 3 0 3/2 1 1 1/2 0 1 2510 = 110012 the fraction part is 0.25, convert to binary format product integer part 0.25x2 0.5 0 0.5x2 1.0 1 => 0.01 0.2510 = 0.012 25.2510 = 11001.012 C, 27.1875 the integer part is 27, convert to binary format Quotient Remainder 27/2 13 1 13/2 6 1 6/2 3 0 3/2 1 1 1/2 0 1 2710 = 110112 the fraction part is 0.1875, convert to binary format product integer part 0.1875x2 0.375 0 0.375x2 0.75 0 0.75x2 1.5 1 0.5x2 1.0 1 0.187510 = 0.00112 27.187510 = 11011.00112 5, Convert the following hexadecimal number to their decimal equivalents a, C b, 9F c, B52 d, F117 e, ABCD f, 1111.1 g, 888.8 h, EBA.C Answer For the hexadecimal representation of y = {...x 2 x1 x0.x −1 x − 2 x −3...} , the value of Y is i y= xi × 16 i a, C = 12 × 16 0 = 12 1 0 b, 9F= 9 × 16 + 15 × 16 = 159 2 1 0 c, B52 = 11 × 16 + 5 × 16 + 2 × 16 = 2898 3 2 1 0 d, F117 = 15 × 16 + 1 × 16 + 1 × 16 + 7 × 16 = 61719 3 2 1 0 e, ABCD = 10 × 16 + 11 × 16 + 12 × 16 + 13 × 16 = 43981 3 2 1 0 −1 f, 1111.1 = 1 × 16 + 1 × 16 + 1 × 16 + 1 × 16 + 1 × 16 = 4369.0625 2 1 0 −1 g, 888.8 = 8 × 16 + 8 × 16 + 8 × 16 + 8 × 16 = 2184.5 2 1 0 −1 h, EBA.C 14 × 16 + 11 × 16 + 10 × 16 + 12 × 16 = 3770.75 6, Convert the following decimal numbers to their hexadecimal equivalents a, 80 Quotient Remainder 80/16 5 0 5/16 0 5 8010 = 5016 b, 2560 Quotient Remainder 2560/16 160 0 160/16 10 0 10/16 0 10 256010 = A0016 c, 65536 Quotient Remainder 65536/16 4096 0 4096/16 256 0 256/16 16 0 16/16 1 0 1/16 0 1 6553610 = 1000016 d, 204.125 the integer part 204, convert to hexadecimal format Quotient Remainder 204/16 12 12 12/16 0 12 20410 = CC16 the fraction part 0.125, convert to hexadecimal format product integer part 0.125x16 2.0 2 0.12510 = 0.216 204.12510 = CC.216 e, 631.25 the integer part 631, convert to hexadecimal format Quotient Remainder 631/16 39 7 39/16 2 7 2/16 0 2 63110 = 27716 the fraction part 0.25, convert to hexadecimal format product integer part 0.25x16 4.0 4 0.2510 = 0.416 531.2510 = 277.416 f, 100000.00390625 the integer part 100000, convert to hexadecimal format Quotient Remainder 100000/16 6250 0 6250/16 390 10 390/16 24 6 24/16 1 8 1/16 0 1 10000010 = 186 A016 the fraction part is 0.00390625, convert to hexadecimal format product integer part 0.00390625x16 0.0625 0 0.0625x16 1.0 1 0.0039062510 = 0.0116 100000.0039062510 = 186 A0.0116