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+23 votes
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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 (59.6k points)
edited by | 3.2k views
0

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 )

+2
@Anil Khatri , Even I am getting 4AE9

Whats the right method?
0
Whats the meaning of line" mantissa is in sign magnitude form".

Does it mean that in mantissa field 1 bit is reserved for sign.

But here in question already 1 bit was given for sign.
0
I am also not able to understand relevance of this line "mantissa is in sign magnitude form" ,becoz dat would mean 1 bit of mantissa is for sign
0

I am getting 4AE9 for  question 'B'

I mean  : (0.239)10 = (0.00111101001)2

3 Answers

+24 votes
Best answer

Answer is option D in both questions.

$0.239 =$ $(0.00111101)_2$

a) Store exponent = actual + biasing                 

   $13  +  64  =  77$                            

   $(77)10 =$ $(1001101)2$      

   Answer is:  $1001101$  $00111101$ =$4D 3D$

b) For normalized representation

    $0.00111101$ *$2^{13}$

    $1.11101$ *$2^{10}$

    Store exponent $= 10+64=74$

    $(74)10  =$  $(1001010)$$_{2}$

    Answer:  $0$  $1001010$  $11101000$ $=$ $4AE8$                .

answered by Boss (31.7k points)
edited by
+4
Shouldn't bias be $2^{k-1}-1 = 2^{6}-1 = 63$ ?
0
......................
0
I have one doubt in it after converting .239 into binay am getting 11101111.Plz anybody clarify it
+1

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

0
yaa i did that mistake only.Thanks allot.
0
Why bias is 64 ?
0
Bias is 64..how??
0
Bias is given. It can be anything till it does the work of converting all negative exponent to positive.
+2
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 ...
0
what if nothing is mentioned about normalization.Then what to do?
By default we use normalization or not?
0
By default take Normalized.(b/c computer stored it in normalized form)
0
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 (263 points)
0 votes
  • Option d is right.

answered by Boss (23.9k points)
0
Plzz tell me that only in IEEE format the excess value if not mentioned taken as 2^(k-1)-1 otherwise 2^(k-1)


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