digital logic Floating point number

Question

A 32-bit floating-point number(follows IEEE STANDARD) is represented by a 8-bit signed exponent, and a 23-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.

Answer 1

the number will be represented in the form of (-1)^s(0.M)*16^(E-64).the range of exponent can be given as (-128 to 127) because exponent is 8 bit signed number(2's complement).am i right?????

Comments

@ BASANT KUMAR  Exponents range from −126 to +127.

Minimum #= -126

Maximum #= 127

In EXCEES-64 =  (-126 - 64)  TO  (127-64) = -190 TO + 63

Is it correct?

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@iarnav , range of exponents will be from -126 to +127 if we take 32-bit single precision floating point number in implicit normalised binary form   because for this form , 

1 $\leq$ E $\leq$ 254 , So, range of exponent will be 1-127 = -126 to 254-127 = +127 because here bias is 127 (i.e. bias is in Excess-127 format) 

But here , excess-64 format is given . so bias = 64

and since single precision floating point number is not mentioned , so we have to take 0 $\leq$ E $\leq$ 255 .so, range of exponent is  from 0 - 64 = -64 to 255-64 = +191.. I am not getting the meaning of "The base of the scale factor is 16" . if you know then please explain.

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@ ankitgupta.1729 "The base of the scale factor is 16" means that exponent is raised to 16.

For example- (-1)^s * 1.M * 16^ (exponent-bias)

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@iarnav , Thank you! can you please tell me that bias should be calculated in 16's complement form or 2's complement form if base is 16 ?

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So  -190 to +63 is correct answer or not.

And also explain the

 The base of the scale factor is 16,