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Consider an excess - 50 representation for floating point numbers with $4 BCD$ digit mantissa and $2 BCD$ digit exponent in normalised form. The minimum and maximum positive numbers that can be represented are __________ and _____________ respectively.
asked in Digital Logic by Veteran (39.7k points) 253 1302 1929 | 278 views
floating point standard not given.

1 Answer

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Just like in binary we have normalised number of the form  (-1)s * 1.M * 2Base exponent - Bias

where s : sign bit 

         M : Mantissa

Here the digits are in BCD instead , so the positive normalised number representation will be : 9.M * 10Base exponent - Bias

Here bias is given to be excess - 50 meaning that we need to subtract 50 from base exponent field to get the actual exponent..

So maximum mantissa value with 4 BCD digits = 9999

   Maximum base exponent value with 2 BCD digits         =  99

   So maximum actual exponent value possible with 2 BCD digits  =  99 - Bias 

                                                                                                =  99 - 50

                                                                                                =  49

  So magnitude of largest positve number                                    =  9.9999 * 1049

  Similarly , 

  To get minimum positve number  , we have to set mantissa = 0 and  exponent field  = 0

  So doing that we get exponent  =  0 - 50   =  -50

  So magnitude of minimum positive number                               =  9.0000 * 10-50

 

  Therefore , maximum positive number             =          9.9999 * 1049

   Minimum positive number                              =           9.0000 * 10-50

answered by Veteran (87.4k points) 15 57 292
Why took "9.0" and not "1.0" for normalization?
Sir since the field is given to have BCD digits instead of binary digits..Correct me sir if I m wrong..N if so , plz let me know what should be the correction..

AFAIK for normalization any non-zero digit would do.

https://en.wikipedia.org/wiki/Normalized_number

So, in binary only "1" is possible. In decimal "1-9" possible. So, the minimum positive number should change.

I think the following:

2 digits for exponent: Max value in binary 10011001 (i.e 99 in BCD) = 153 in decimal

So, biased exponent range = 0 to 153

 

Now, max mantissa = 1. 1001100110011001 (i.e 9999 in BCD)

 

Hence, max number = 1. 1001100110011001 * 2153-50

and min number = 1. 0000000000000000 * 20-50

 

@Arjun, am I right?

sir, instead of taking it in BCD can't we just say that mantissa is (16 bit) and the exponent is(8 bit) and then specify the maximum by putting all 1's in the field ??

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