Consider the following floating-point format.Mantissa is a pure fraction in sign-magnitude form.The normalized representation for the above format is specified as follows. The mantissa has an implicit 1 preceding the binary (radix) point. Assume that only 0s are padded in while shifting a field.The normalized representation of the above number a)0A 20b)11 34c)49 D0d)4A E8Correct answer is option 'D'. Can you explain this answer? - EduRev Computer Science Engineering (CSE) Question

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Consider the following floating-point format.

Mantissa is a pure fraction in sign-magnitude form.
The normalized representation for the above format is specified as follows. The mantissa has an implicit 1 preceding the binary (radix) point. Assume that only 0’s are padded in while shifting a field.

The normalized representation of the above number

a)
0A 20
b)
11 34
c)
49 D0
d)
4A E8

Correct answer is option 'D'. Can you explain this answer?

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Consider the following floating-point format.Mantissa is a pure fracti...

For finding normalised representation we need to find unnormalised one first..So we have , 0.239 * 2¹³ as the number ..So we find the binary equivalent of 0.239 till 8 digits as capacity of mantissa field is 8 bits..
So following the procedure we have , 0.239 * 2 = 0.478
0.478 * 2 = 0.956
0.956 * 2 = 1.912
0.912 * 2 = 1.824
0.824 * 2 = 1.648
0.648 * 2 = 1.296
0.296 * 2 = 0.512
0.512 * 2 = 1.024
We stop here as we have performed 8 iterations and hence 8 digits of mantissa of unnormalised number is obtained.So we have

Mantissa of given number = 0011 1101
So the number can be written as : 0.00111101 * 213 Now we need to align the mantissa towards left to get normalised number..And in the question it is mentioned that during alignment process 0's will be padded in the right side as a result of mantissa alignment to left..
So to get normalised number , we align to left 3 times , so new mantissa = 11101 000
So exponent will also decrease by 3 hence new exponent = 10
So normalised number = 1.11101000 * 2¹⁰
So actual exponent = 10
Given excess 64 is used..So bias value = 64

So exponent field value = 74
And of course sign bit = 0 being a positive number.
Thus the final representation of number = 0 1001010 11101000 = 0100 1010 1110 1000
= (4AE816)
Hence D) should be the correct answer.

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Consider the following floating-point format.Mantissa is a pure fraction in sign-magnitude form.The normalized representation for the above format is specified as follows. The mantissa has an implicit 1 preceding the binary (radix) point. Assume that only 0’s are padded in while shifting a field.The normalized representation of the above number a)0A 20b)11 34c)49 D0d)4A E8Correct answer is option 'D'. Can you explain this answer?

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