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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.
in Digital Logic 1.2k views
0
floating point standard not given.

1 Answer

6 votes

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                                    $=  9.9999$ *$10^{49}$

  Similarly, 

  To get minimum positive 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


edited by
0
Why took "9.0" and not "1.0" for normalization?
0
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..
8

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.

1

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?

0
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 ??
1

@vamp_vaibhav You cannot do this, as in BCD 1111 is invalid. You have to think in terms of BCD representation.

0

@Habibkhan Sir isn't the magnitude of minimum positive number  be 9.0000*10^(-50) in the 3rd last step??.

0

@Habibkhan the min positive number needs to be changed, right? 

To 1.0000×10^(-50)

0

I think what @Sushant Gokhale said is correct. No where it is mentioned that 50 in "excess-50" is BCD. So answer should be as the comment made by @Sushant Gokhale. 

0
Can you please explain more clearly why you have taken 9.M instead of 1.M?
0

In minimum instead of 1 9 is taken because of biasing actually what he did is 


  0            00               0000
sign   Exponent   Mantisa

Now,  converting  $(-1)^0$  * (9.0000)[biasing] * $10^{0-50}$

Hence   9.0000 * $10^{-50}$

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