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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

in Digital Logic
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A ?
0
Answer is $A$

1 Answer

3 votes
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
0
What you did from 2nd step?
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calculated the value of 0.m then added 1
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Why should I consider 0.M, the question only mentions 1.M :(

And how that 0.M is calculated.
2
1.m = 1 + 0.m

1111 = $(2^4-1)$, 111.1 = $\frac{(2^4-1)}{2}$

So, 0.1111 = $\frac{(2^4-1)}{2^4}$
0
Thanks, got it
0
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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