Solutions - Homework 1 (Due date: September 25 th ) Presentation and clarity are very important! Show your procedure!
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1 c 10 =0 c 9 =0 c 8 =0 c 7 =0 c 6 =0 c 5 =0 c 10 =1 c 9 =1 c 8 =1 c 7 =0 c 6 =1 c 5 =1 c 4 =1 c 8 =1 c 7 =1 c 6 =0 c 5 =0 c 8 =0 c 7 =0 c 6 =0 c 5 =0 c 8 =1 c 7 =1 c 6 =1 c 5 =0 c 4 =1 b 7 =0 b 6 =0 b 5 =0 b 4 =0 b 3 =1 b 2 =1 b 1 =1 b 0 =0 c 7 =1 c 6 =1 c 5 =1 c 7 =0 c 6 =0 c 5 =1 ELECTRICAL AND COMPUTER ENGINEERING DEPARTMENT, OAKLAND UNIVERSITY Solutions - Homework 1 (Due date: September 25 th ) Presentation and clarity are very important! Show your procedure! PROBLEM 1 (10 PTS) Calculate the result of the additions and subtractions for the following fixed-point numbers UNSIGNED SIGNED UNSIGNED: SIGNED: PROBLEM 2 (10 PTS) Multiply the following signed fixed-point numbers: x x x x x x Instructor: Daniel Llamocca
2 x x x x x x PROBLEM 3 (15 PTS) Get the division result (with x = 4 fractional bits ) for the following signed fixed-point numbers: : To unsigned and then alignment, a = 6: Append x = 4 zeros: Integer Division: = = = Q = , R = Qf = (x = 4) Final result (2C): = : To unsigned and then alignment, a = 6: = = Append x = 4 zeros: Integer Division: Q = 10, R = 1110 Qf = 10.(x = 4) Final result (2C): = : To unsigned (denominator) and then alignment, a = 3: = Append x = 4 zeros: Integer Division: Q = , R = 10 Qf = (x = 4) Final result (2C): = 2C( ) = Instructor: Daniel Llamocca
3 : To unsigned (denominator) and then alignment, a = 6: Append x = 4 zeros: Integer Division: Q = 100, R = Qf = (x = 4) Final result (2C): = 0.01 = PROBLEM 4 (5 PTS) We want to represent numbers between and 179. What is the fixed point format that requires the fewest number of bits for a resolution better or equal than ? 2C representation for integers: 2 n 1 to 2 n 1 1. For 2 n , we have that n 9, so we pick n = 9. For the fractional part, we select the number of fractional bits p that make the resolution better or equal than : Then the Fixed Point format required in [20 11]. 2 p p p = 11 PROBLEM 5 (10 PTS) Complete the table for the following floating point formats (which resemble the IEEE-754 standard) with 16, 24, 48 bits. Only consider ordinary numbers. Exponent Significant bits (E) bits (p) Min Max Range of e Range of significand [ 30,31] [1, ] [ 62,63] [1, ] [ 510,511] [1, ] PROBLEM 6 (20) Calculate the decimal values of the following floating point numbers represented as hexadecimals. Show your procedure. Single (32 bits) Double (64 bits) F DECADE FDEAD378 3DE ABBAA FACEB0E8 7FF32B5A ACCEDE78 FA09D3784D089B7D 80DEADBEE D3787F FA0BEBE80BEEF0A FDB FA09D37809ABC0DE FF ABC0DE DECAFC0FFEE00800 F : e + bias = = 240 e = = 113 Mantissa ([24 23]) = = X = = DECADE: e + bias = = 1 e = = 126 Mantissa = = X = = FDEAD378: e + bias = 1111 = 251 e = = 124 Mantissa = = X = = DE38866: e + bias = 0111 = 123 e = = 4 Mantissa = = X = = Instructor: Daniel Llamocca
4 800ABBAA: e + bias = = 0 Denormal number e = 126 Mantissa = = X = = FACEB0E8: e + bias = = 245 e = = 118 Mantissa = = X = = FF32B5A: e + bias = = 255, f 0 X = NaN ACCEDE78: e + bias = 0001 = 89 e = = 38 Mantissa = = X = = FA09D3784D089B7D: e + bias = = 1952 e = = 929 Mantissa ([53 52]) = = X = = DEADBEE : e + bias = = 13 e = = 1010 Mantissa = = X = = D3787F888800: e + bias = = 0 Denormal number e = 1022 Mantissa = = X = = FA0BEBE80BEEF0A0: e + bias = = 1952 e = = 929 Mantissa = = X = = FDB: e + bias = = 1175 e = = 152 Mantissa = = X = = FA09D37809ABC0DE: e + bias = = 1952 e = = 929 Mantissa = = X = = FF ABC0DE: e + bias = = 2040 e = = 1017 Mantissa = = X = = DECAFC0FFEE00800: e + bias = 100 = 1516 e = = 493 Mantissa = = X = = Instructor: Daniel Llamocca
5 PROBLEM 7 (30) Calculate the result of the following operations with 32-bit floating point numbers. Truncate the results when required. When doing fixed-point division, use 8 fractional bits. Show your procedure. FA FF800FAD CA09D F800FEA + 09ABC0DE 5A09D FDB FC09D F DE32B5A FF DE D959 FA09D378-09ABC0DE FA09D300 4D FE FA09DF00 7F A09D300 0BEEF ABC0DE FF FE0000 FA09D X = FA FF800FAD: FF800FAD: e + bias = = 255, f 0 FF800FAD = NaN X = # + NaN = NaN X = 7F800FEA + 09ABC0DE: 7F800FEA: e + bias = = 255, f 0 7F800FEA = NaN X = NaN + # = NaN X = FC09D F800000: 7F800000: e + bias = = 255, f = 0 7F = + X = # + = + X = 3DE D959: 3DE38866: e + bias = 0111 = 123 e = = 4 Mantissa = DE38866 = D959: e + bias = 01 = 102 e = = 25 Mantissa = D959 = X = X = X = X = = = X = = 3DE3886A X = CA09D : CA98D378: e + bias = = 148 e = = 21 Mantissa = DE38866 = : e + bias = = 0 Denormal number e = = 0 X = = X = = CA09D378 X = 5A09D FDB: 5A09D378: e + bias = 0100 = 180 e = = 53 Mantissa = A09D378 = Instructor: Daniel Llamocca
6 40490FDB: e + bias = = e = = 1 Mantissa = FDB = X = X = The division by 2 52 requires more than p + 1 = 24 bits for proper representation. Thus, we approximate the second operand by 0. X = = X = = 5A09D378 X = 7DE32B5A FF800000: FF800000: e + bias = = 255, f = 0 FF = X = # ( ) = + X = 7F X = FA09D378 09ABC0DE: FA09D378: e + bias = = 244 e = = 117 Mantissa = FA09D378 = ABC0DE: e + bias = = 19 e = = 108 Mantissa = ABC0DE = X = X = The division by requires more than p + 1 = 24 bits for proper representation. Thus, we approximate the second operand by 0. X = = X = = FA09D378 X = FA09D300 4D080000: FA09D300: e + bias = = 244 e = = 117 Mantissa = FA09D300 = D080000: e + bias = 10 = 154 e = = 27 Mantissa = D = X = X = e + bias = = 271 > 254 In this case, there is an overflow. The value X is assigned to. X = = FF X = FE: : e + bias = = 0 Denormal number e = = 0 X = 0 # = 0 X = Instructor: Daniel Llamocca
7 X = FA09DF00 7F800000: 7F800000: e + bias = = 255, f = 0 7F = + X = ( #) = X = = FF X = 7A09D300 0BEEF000: 7A09D300: e + bias = = 244 e = = 117 Mantissa = A09D300 = BEEF000: e + bias = 0001 = 23 e = = 104 Mantissa = BEEF000 = X = X = = = e + bias = = 141 = X = = 4680A35F (four bits were truncated) X = : : e + bias = = 146 e = = 19 Mantissa = = : e + bias = = 128 e = = 1 Mantissa = BEEF000 = X = Alignment: = = Append x = 8 zeros: Integer division Q = 110, R = Qf = Thus: X = = = = e + bias = = 145 = X = = 489B0000 X = ABC0DE: : e + bias = = 0 Denormal number e = = 0 X = 0 # = 0 X = Instructor: Daniel Llamocca
8 X = FF FE0000: FF800000: e + bias = = 255, f = 0 FF = X = # = X = FF X = FA09D FA09D300: e + bias = = 244 e = = 117 Mantissa = FA09D300 = : e + bias = = 144 e = = 17 Mantissa = = X = Alignment: = = Append x = 8 zeros: Integer division Q = , R = Qf = Thus: X = = = = = e + bias = = 226 = X = = F Instructor: Daniel Llamocca
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