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Consider the following floating-point format.

     

    Mantissa is a pure fraction in sign-magnitude form.

The decimal number 0.239 $\times$ 2$^{13}$ has the following hexadecimal representation (without normalization and rounding off):

  1. 0D 24
  2. 0D 4D
  3. 4D 0D
  4. 4D 3D
asked in Digital Logic by Veteran (69k points)
edited by | 2.5k views

In ques B , we just append 3 0's or we calculate three more bit using general method 

0.00111101 *213 = 1.11101000 *210 (  ans is   0 1001010 11101000=4AE8 )  

or 

0.00111101 *213 = 1.11101001 *210 (  ans is   0 1001010 11101001=4AE9 )

@Anil Khatri , Even I am getting 4AE9

Whats the right method?

2 Answers

+24 votes
Best answer

answer = option D in both questions.

0.239 = $(0.00111101)_2$

a)Store exponent = actual + biasing                 

   13  +  64  =  77                              

   (77)10 = (1001101)2       

   ans is  1001101 00111101 =$4D 3D$

b) For normalized representation

    0.00111101 *213

    1.11101 *210

    Store exponent = 10+64=74

    (74)10  =  (1001010)2

    ans  0 1001010 11101000=$4AE8$                

 

answered by Veteran (34.3k points)
edited by
Shouldn't bias be $2^{k-1}-1 = 2^{6}-1 = 63$ ?
......................
I have one doubt in it after converting .239 into binay am getting 11101111.Plz anybody clarify it

My you have converted 239 into binary. 0.239 does not work this way. 5 in binary is 101 but 0.5 in binary is 0.1. Try https://answers.yahoo.com/question/index?qid=20081008152438AACtuGE

yaa i did that mistake only.Thanks allot.
Why bias is 64 ?
Bias is 64..how??
Bias is given. It can be anything till it does the work of converting all negative exponent to positive.
When  x  bits are for exponent then bias  can be   2^(x-1)  . need not to be 2^(x-1)  - 1
When it is mentioned that 64, it is 64....

I think those people who are confused, their confusion is due to relating it with IEEE standard...where for single precision bias is 127 .. & for double precision bias is 1023 ...
what if nothing is mentioned about normalization.Then what to do?
By default we use normalization or not?
By default take Normalized.(b/c computer stored it in normalized form)
the given decimal number is 0.239 × 2^13.

My ques is:

Shouldn't we 1st multiply 0.239 with 8192(=2^13) ie. 0.239* 8192 = 1957.888 (in base 10 form)

then convert it into base 2 form ?
+1 vote
1) (.239)base10 can be written in binary as 00111101 in 8 bit

 exp given 13 so 0001101 in seven bit

sign bit 0 positive no.

hence total 0 0001101 00111101

writing it in decimal we have 0D 4D

ans is B

2) in second

biasing by ( 2^n-1)-1

63+13=76

exp= 1001100

sign bit=0

mantisa since implicit normalization is used here

previously m= 00111101

for implicit  m= 11101000

hence we have 0 1001100 11101000

which is equivalent to 4A E8
answered by (313 points)
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