Ncp computer appls num sys1 pramod
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The document discusses conversion between binary, octal, hexadecimal, and decimal number systems. It also covers addition, subtraction, multiplication, and division of binary numbers. Finally, it introduces floating-point number representation including sign-magnitude, two's complement, and normalized floating-point formats.
Ncp computer appls num sys1 pramod
- 1. Conversion of Binary, Octal and Hexadecimal Numbers From Binary to Octal Starting at the binary point and working left, separate the bits into groups of three and replace each group with the corresponding octal digit. 100010112 = 010 001 011 = 2138 From Binary to Hexadecimal Starting at the binary point and working left, separate the bits into groups of four and replace each group with the corresponding hexadecimal digit. 100010112 = 1000 1011 = 8B16 From Octal to Binary Replace each octal digit with the corresponding 3-bit binary string. 2138 = 010 001 011 = 100010112 From Hexadecimal to Binary Replace each hexadecimal digit with the corresponding 4-bit binary string. 8B16 = 1000 1011 = 100010112
- 2. Conversion of Decimal Numbers From Decimal to Binary 17 12 34 02 69 12 139 12 8 02 4 02 2 02 1 MSD LSD From Binary to Decimal 100010112 = 1×27 + 0×26 + 0×25 + 0×24 + 1×23 + 0×22 + 1×21 + 1×20 = 128 + 8 + 2 + 1 13910 = 100010112
- 3. Conversion of Fractions Starting at the binary point, group the binary digits that lie to the right into groups of three or four. 0.101112 = 0.101 110 = 0.568 0.101112 = 0.1011 1000 = 0.B816 Problems Convert the following Binary Octal Decimal Hex 10011010 2705 2705 3BC Binary Octal Decimal Hex 10011010 232 154 9A 10111000101 2705 1477 5C5 101010010001 5221 2705 A91 1110111100 1674 956 3BC 5 42 28 338 28 2705 18 10=A 169 916 2705 116
- 4. Add 1 1 1 1 1 0 0 1 1 + 1 0 0 1 + 1 1 1 0 1 1 0 0 0 1 0 0 0 0 1 Subtract 1 1 0 0 0 1 0 0 1 1 - 1 1 1 1 - 1 1 1 1 1 0 0 1 1 0 0 Multiply normally for implementation - add the shifted multiplicands one at a time. 1 1 1 0 = 14 1 1 1 0 * 1 1 0 1 = 13 * 1 1 0 1 1 1 1 0 1 1 1 0 0 0 0 0 + 0 0 0 0 1 1 1 0 0 1 1 1 0 + 1 1 1 0 + 1 1 1 0 1 0 1 1 0 1 1 0 1 0 0 0 1 1 0 + 1 1 1 0 1 0 1 1 0 1 1 0 (8 bits)
- 5. Divide 1 1 0 1 1 1 0 1 1 1 1 ) 1 1 0 0 0 1 0 1 | 1 1 0 1 ) 1 0 1 1 0 0 1 | 1 1 1 1 | 1 1 0 1 | 1 0 0 1 1 0 1 | 1 0 0 1 0 1 | 1 1 1 1 | 1 1 0 1 | 1 0 0 0 1 | 1 0 1 1 | 0 0 0 0 | 0 0 0 0 | 1 0 0 0 1 | 1 0 1 1 1 1 1 1 | 1 0 1 0 0 1 1 1 0 1 ) 1 1 1 1 0 0 1 | 1 1 0 1 | 1 0 0 0 1 | 0 0 0 0 | 1 0 0 0 1 | 0 0 0 0 | 1 0 0 0 1 | 1 1 0 1 | 1 0 0
- 6. Sign-Magnitude 0 = positive 1 = negative n bit range = -(2n-1 -1) to +(2n-1 -1) 4 bits range = -7 to +7 2 possible representation of zero. 2's Complement flip bits and add one. n bit range = -(2n-1 ) to +(2n-1 -1) 4 bits range = -8 to +7 0 0 0 0 = 0 0 0 0 1 = 1 0 0 1 0 = 2 0 0 1 1 0 1 0 0 0 1 0 1 0 1 1 0 0 1 1 1 = 7 1 0 0 0 = -8 1 0 0 1 = -7 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 = -2 1 1 1 1 = -1 Example 1 1 1 0 = 14 0 0 0 1 flip bits 0 0 1 0 add one WRONG this is not -14. Out of range. Need 5 bits 0 1 1 1 0 = 14 1 0 0 0 1 flip bits 1 0 0 1 0 add one. This is -14. Sign Extend add 0 for positive numbers add 1 for negative numbers
- 7. Add 2's Complement 1 1 1 0 = -2 1 1 1 0 = -2 + 1 1 0 1 = -3 + 0 0 1 1 = 3 1 1 0 1 1 ignore carry = -5 1 0 0 0 1 ignore carry = 1 Be careful of overflow errors. An addition overflow occurs whenever the sign of the sum if different from the signs of both operands. Ex. 0 1 0 0 = 4 1 1 0 0 = -4 + 0 1 0 1 = 5 + 1 0 1 1 = -5 1 0 0 1 = -7 WRONG 1 0 1 1 1 ignore carry = 7 WRONG Multiply 2's Complement 1 1 1 0 = -2 1 1 1 0 = -2 * 1 1 0 1 = -3 * 0 0 1 1 = 3 1 1 1 1 1 1 1 0 sign extend to 8 bits 1 1 1 1 1 1 1 0 sign extend to 8 bits + 0 0 0 0 0 0 0 + 1 1 1 1 1 1 0 1 1 1 1 1 1 1 0 1 1 1 1 1 1 0 1 0 ignore carry = -6 + 1 1 1 1 1 0 1 1 1 1 1 0 1 1 0 ignore carry + 0 0 0 1 0 negate -2 for sign bit 1 0 0 0 0 0 1 1 0 ignore carry = 6 1 0 0 1 0 = -14 * 1 0 0 1 1 = -13 1 1 1 1 1 1 0 0 1 0 sign extend to 10 bits + 1 1 1 1 1 0 0 1 0 1 1 1 1 1 0 1 0 1 1 0 ignore carry + 0 0 0 0 0 0 0 0 1 1 1 1 0 1 0 1 1 0 + 0 0 0 0 0 0 0 1 1 1 1 0 1 0 1 1 0 + 0 0 1 1 1 0 negate -14 for sign bit 1 0 0 1 0 1 1 0 1 1 0 ignore carry = 182
- 8. Floating-Point Numbers mantissa x (radix)exponent The floating-point representation always gives us more range and less precision than the fixed-point representation when using the SAME number of digits. 11-bit excess 1023 charactstic Mantissa sign 52-bit normalized fraction Sign exponent Mantissa sign Mantissa magnitude 8-bit excess-127 characteristic Mantissa sign 23-bit normalized fraction General format 32-bit standard 64-bit standard 0 1 12 63 31910 Implied binary point Normalized fraction - the fraction always starts with a nonzero bit. e.g. 0.01 … x 2e would be normalized to 0.1 … x 2e-1 1.01 … x 2e would be normalized to 0.101 … x 2e+1 Since the only nonzero bit is 1, it is usually omitted in all computers today. Thus, the 23-bit normalized fraction in reality has 24 bits. The exponent is represented in a biased form. • If we take an m-bit exponent, there are 2m possible unsigned integer values. • Re-label these numbers: 0 to 2m -1 → -2m-1 to 2m-1 -1 by subtracting a constant value (or bias) of 2m-1 (or sometimes 2m-1 -1). • Ex. using m=3, the bias = 23-1 = 4. Thus the series 0,1,2,3,4,5,6,7 becomes -4,-3,-2,- 1,0,1,2,3. Therefore, the true exponent -4 is represented by 0 in the bias form and -3 by +1, etc. • zero is represented by 0.0 … x 20 . Ex. if n = 1010.1111, we normalize it to 0.10101111 x 24 . The true exponent is +4. Using the 32-bit standard and a bias of 2m-1 -1 = 28-1 -1 = 127, the true exponent (+4) is stored as a biased exponent of 4+127 = 131, or 10000011 in binary. Thus we have 0 | 1 0 0 0 0 0 1 1 | 0 1 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Notice that the first 1 in the normalized fraction is omitted. The biased exponent representation is also called excess n, where n is 2m-1 -1 (or 2m-1 ).