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This is solution manual for Digital Logic Design course. It was helpful for assignment Dr. Archan Singh gave us at Punjab Engineering College. It includes: Decimal, Binary, Octal, Hexadecimal, Numbers, Quadratic, Equation, BCD, Digits
Typology: Exercises
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a) 7562/8 = 945 + 2/8 Ô 2 945/8 = 118 +1/8 Ô 1 118/8 = 14 + 6/8 Ô 6 14/8 = 1 + 6/8 Ô 6 1/8 = 1/8 Ô 1
0.20x8 = 3.2 Ô 3 (7562.45) 10 = (16612.3463) 8
b) (1938.257) 10 = (792.41CA) 16 c) (175.175) 10 = (10101111.001011) 2
Decimal, Binary, Octal and Hexadecimal Numbers from (16) 10 to (31) 10 Dec 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 Bin 1 0000 1 0001 1 0010 1 0011 1 0100 1 0101 1 0110 1 0111 1 1000 1 1001 1 1010 1 1011 1 1100 1 1101 1 1110 1 1111 Oct 20 21 22 23 24 25 26 27 30 31 32 33 34 35 36 37 Hex 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
Decimal Binary Octal Hexadecimal 369.3125 101110001.0101 561.24 171. 189.625 10111101.101 275.5 BD.A 214.625 11010110.101 326.5 D6.A 62407.625 1111001111000111.101 171707.5 F3C7.A
Problem Solutions – Chapter 1
a) (673.6) 8 = (110 111 011.110) (^2) = (1BB.C) (^16) b) (E7C.B) 16 = (1110 0111 1100.1011) (^2) = (7174.54) (^8) c) (310.2) 4 = (11 01 00.10) (^2) = (64.4) 8
a) (BEE) r = (2699) (^10)
By the quadratic equation: r = 15 or r ≈ –16. ANSWER: r = 15 b) (365) r = (194) 10
By the quadratic equation: r = -9 or 7 ANSWER: r = 7
a) (0100 1000 0110 0111)BCD = (4867) 10 = (1001100000011) (^2) b) (0011 0111 1000.0111 0101)BCD = (378.75) 10 = (101111010.11) (^2)
a) (101101101) (^2) b) (0011 0110 0101) (^) BCD c) 0011 0011 0011 0110 0011 0101ASCII
BCD Digits with Odd and Even Parity 0 1 2 3 4 5 6 7 8 9 Odd 1 0000 0 0001 0 0010 1 0011 0 0100 1 0101 1 0110 0 0111 0 1000 1 1001 Even 0 0000 1 0001 1 0010 0 0011 1 0100 0 0101 0 0110 1 0111 1 1000 0 1001
11 × r^2 + 14 × r^1 + 14 × r^0 = 2699 11 × r^2 + 14 × r – 2685 = 0
3 × r^2 + 6 × r^1 + 5 × r^0 = 194 3 × r^2 + 6 × r – 189 = 0