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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 by Active (4.7k points)
edited by | 72 views
0
A ?
0
Answer is $A$

1 Answer

+2 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
by Loyal (5.3k points)
0
What you did from 2nd step?
0
calculated the value of 0.m then added 1
0
Why should I consider 0.M, the question only mentions 1.M :(

And how that 0.M is calculated.
+1
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

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