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A 32-bit floating-point number is represented by a 7-bit signed exponent, and a 24-bit fractional mantissa. The base of the scale factor is 16,

The range of the exponent is ___________, if the scale factor is represented in excess-64 format.
asked in Digital Logic by Veteran (112k points) | 389 views

3 Answers

+4 votes
Best answer
Minimum number $= 0$ , maximum number $= 127$,

Given excess $64$ so, bias number is $64$,

Range will be $0-64=-64$ to $(127-64=63) 63$.

Here, in question given that base is $16$, so the actual value represented will be $(-1)^s ( 0.M) \times16 ^{E-\text{bias}}.$
answered by Boss (16k points)
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+1
why (0.M) and not (1.M) ?
0
It is not mentioned weather normalized or denormalized.
By default if its not explicitly mentioned we consider normalized?
So it should be (1.M)
0
here it is given that exponent is signed bit representation then what is need of doing  biasing.
0
I am also stuck at this point. Did you understand why biasing is needed in the signed representation?
0 votes
Range = 2^126 to 2^127
answered by Junior (837 points)
0
exponent will be in between 00000001 to 11111110

right?
0 votes

It is asked , range of the exponent in excess 64 format , i.e. double precession format

Single Precession: SEEEEEEE EMMMMMMM MMMMMMMM MMMMMMMM
Double Precession: SEEEEEEE EEEEMMMM MMMMMMMM MMMMMMMM MMMMMMMM MMMMMMMM MMMMMMMM MMMMMMMM

Now for positive or negative normalize real number exponent range must be in between 000.....01  to 11.....10

Now for single precession it will be 0000 0001 to 1111 1110

and in double precession it will be 0000 0000 001 to 1111 1111 110

http://steve.hollasch.net/cgindex/coding/ieeefloat.html

https://gateoverflow.in/194521/dl-ieee-754-single-precision-range

https://gateoverflow.in/82469/coa-floating-point-arithmatic

answered by Veteran (103k points)
edited by

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