5. (2 pts) What would the number 26.328125 ten be in IEEE 754single precision fl
ID: 3617266 • Letter: 5
Question
5. (2 pts) What would the number 26.328125 ten be in IEEE 754single precision floating point format. You need to follow thefollowing steps:a). Write the above number in binary. (before normalizing it)
b). Write the above number in the normalized format.
c). Compute the biased exponent, and write it in binary.
d). Write its IEEE 754 single precision floating point format inbinary, then in hex.
6. (2 pts) What would the number -47.6875 ten be in IEEE 754 singleprecision floating point format.
You need to follow the following steps:
a). Write the above number in binary. (before normalizing it)
b). Write above number in the normalized format.
c). Compute the biased exponent, and write it in binary.
d). Write its IEEE 754 single precision floating point format inbinary, then in hex.
7. (2 pts) What decimal number would the IEEE 754 single precisionfloating point number 0xC9540000 (this is in hex) be? Write yourfinal answer in scientific notation as m x 10 p where p is aninteger.
Explanation / Answer
please rate -thanks I don't know what you mean by normalized but this answer shouldcover all the above 26.328125 The number is positive so the sign bit or most significant bit is0 26 in binary is 11010 (I’m assuming you know how to dothis) to change .328125 to binary Step 1: Begin with the decimal fraction and multiply by 2. Thewhole number part of the result is the first binary digit to theright of the point. Because .328125*2= 0.65625, the first binary digit to the right ofthe point is a 0. So far, we have .328125=.0??? . . . (base 2) . Step 2: Next we disregard the whole number part of the previousresult (the 0 in this case) and multiply by 2 once again. The wholenumber part of this new result is the second binary digit to theright of the point. We will continue this process until we get azero as our decimal part or until we recognize an infiniterepeating pattern. Because .65625x 2 = 1.3125, the second binary digit to the right ofthe point is a 1. So far, we have ..329125 = .01?? . . . (base 2) . Step 3: Disregarding the whole number part of the previous result(we multiply by 2 once again. The whole number part of the resultis now the next binary digit to the right of the point. Because .3125 x 2 = 0.625, the third binary digit to the right ofthe point is a 0. have .010 One more time and we get that .625x2= 1.25 so we have decimal is.0101 binary continuing until we finish we get 328125 =.010101 so the binary number you are starting with is 11010.010101 Using rules of scientific notation this can be changed to1.1010010101x 24 The exponent, which is in bits 30-23, will be 4+ 127 (we always add127 to the exponent) or 131. 131 in binary is 10000011 thisis 8 bits so no need to add any leading 0’s The fraction or significand, which is in bits 22-0, is gottenby dropping the 1 from the 1.1010010101 above (we drop the 1 sincethe digit before the . will always be one ) and have1010010101 less then the 23 bits needed. Therefore we add trailing0’s. 10100101010000000000000 Therefore 26.328125 is 0 10000011 10100101010000000000000 0100 0001 1101 0010 1010 0000 0000 0000 4 1 D 2 A 0 0 0 --------------------------------------------------- -47.6875 The number is negative so the sign bit or most significant bit is1 47 in binary is 101111 (I’m assuming you know how to dothis) to change .6875 to binary Step 1: Begin with the decimal fraction and multiply by 2. Thewhole number part of the result is the first binary digit to theright of the point. Because .6875*2= 1.375, the first binary digit to the rightof the point is a 1. So far, we have . 6875=.1??? . . . (base 2) . Step 2: Next we disregard the whole number part of the previousresult (the 0 in this case) and multiply by 2 once again. The wholenumber part of this new result is the second binary digit to theright of the point. We will continue this process until we get azero as our decimal part or until we recognize an infiniterepeating pattern. Because .375x 2 = 0.75, the second binary digit to the right of thepoint is a 0. So far, we have .6875= .10?? . . . (base 2) . Step 3: Disregarding the whole number part of the previous result(we multiply by 2 once again. The whole number part of the resultis now the next binary digit to the right of the point. Because .75 x 2 = 1.5, the third binary digit to the right of thepoint is a 0. have .101 One more time and we get that .5x2= 1. so we have decimal is .1011binary so the binary number you are starting with is 101111.1011 Using rules of scientific notation this can be changed to1.011111011x 25 The exponent, which is in bits 30-23, will be 5+ 127 (we always add127 to the exponent) or 132. 132 in binary is 10000100 thisis 8 bits so no need to add any leading 0’s The fraction or significand, which is in bits 22-0, is gottenby dropping the 1 from the 1.011111011 above (we drop the 1 sincethe digit before the . will always be one ) and have011111011 less then the 23 bits needed. Therefore we add trailing0’s. 01111101100000000000000 Therefore -47.6875 is 1 10000100 01111101100000000000000 1100 0010 0011 1110 1100 0000 0000 0000 C 2 3 E C 0 0 0 --------------------------------------------------- 0xC9540000 C 9 5 4 0 0 0 0 1 10010010 10101000000000000000000 sign bit =1 so negative exponent = 146-127=19 significand =1.10101000000000000000000 move decimal point and we have 11010100000000000000.0000 = 86835210but it isnegative so the answer is -868352 so we have
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