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1) The floating point to be used in this problem is an 32-bit IEEE 754, (a) Show the representation of -13.00 and 112 (0.5) 13.00 0.50 (b) Perform Floating point addition of the two numbers show all steps, if you give result only or one step, grade 0) (c) Figure 2 represents the hardware for Floating Point addition. Assume that values of registers in each intermediate step (appeared in bold Italic) have been captured. Provide the binary value of each of these registers in the following table Exponent 1: Fraction 1: Fraction 2: Exponent 2 SmallALU Out: MUX1Out: Ctrl ShiftAmount: BigAlu In1: Big Alu In2:

Explanation / Answer

1(a).

step1: -13.00 convert into-- binary is 1101.0

step2: normalize the binary number ..

                1101.0 -- 1.1010*23 (right shift 3 bit position maximize the number)

step3: the exponent                       3+127=130

                           convert 130 into binary is         1000 0010

step4: in ieee 754 single 32 bit single precision the 32 bits are divided into sin bit,exponent bits,mantissa bits they are : S EEEEEEEE MMMMMMMMMMMMMMMMMMMMMMM

                                                                                 S=SIGN

E=EXPONENT

M=MANTISSA

                     -13 is negative number sign bit is            1

                      8 bits           are used exponent          1000 0010

1           1000 0010       101 0000 0000 0000 0000 0000

Sign     Exponent            Mantissa

step1: 0.5 convert into-- binary is 0.1

step2: normalize the binary number . to move the decimal point one position right to maximize the number

                                0.1 normalized is         1.0*2-1

step3: bias the exponent

                          -1+127=126

                          convert 126 into binary is 0111 1110

step4: in ieee 754 single 32 bit single precision the 32 bits are divided into sin bit,exponent bits,mantissa bits they are : S EEEEEEEE MMMMMMMMMMMMMMMMMMMMMMM

                                                                                 S=SIGN

E=EXPONENT

M=MANTISSA

               0.5 is positive number sign bit is   0 (31 bit )

               eight bits are used exponent          0111 1110

                23 bits are used to mantissa          0000000000000000000000

  

              1             0111 1110          0000 0000 0000 0000 0000 000

            Sign        exponent             mantissa

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