Gateforum Test Series: Digital Logic - Floating Point Representation

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b?
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in normalized form, there is a explicit 1 before mantissa, which is not the case for un-normalised number. so find out binary representation and shift the decimal before 1st 1, adjust the exponent accordingly. try it as i said, you will get b.

2.25  = 10.01 * 2^0 = 0.1001 * 2^2

In un-normalized version , number is represented as  (-1)^s 0.M x 2^(E-63)

here M = 1001

Now E-63 = 2 => E = 65.

And, s=1 , since number is Negative.

So the representation would be ,

1 100100000000000000000000 1000001

where s=1

M=100100000000000000000000

and E-63 which is 65 is 1000001,

from extreme right ,  make groups of 4.

After making groups , one can easily see it is , C8000041
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Perfect brother (Y)
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Nice thank you. It was an easy stuff but always confused me.

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What is the largest mantissa we can store in floating-point format if the size of the mantissa field is m-bit and exponent field is e-bit? The mantissa is normalized and has an implied $1$ in the left of the point. Normalized form of mantissa is 1.M