Gateforum Test Series: Digital Logic - Floating Point Representation

Question

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

Comments

A ?

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Answer is $A$

Answer 1

Mantissa = $1.m$

$m=\underbrace{111..111111}_{m\,1^{'s}}=2^{m}-1$

$0.m = (2^m-1)*2^{-m}=1-2^{-m}$

Mantissa = $1.m=1+(1-2^{-m})=(2^{m+1}-1)\times2^{-m}$

So, answer should be $(A)$

Correct me if i'm wrong

Comments

Thanks, got it

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please clear my doubt

in second step

$0.m=(2^m -1 ) *2^{-m} =1-2^{-m}$

$2^{-m}$ comes from ?

if we shift  ' . '  to left of m then it should be $2^{m}$  ...

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because you can write 0.111 as 111 * $2^{-3}$