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+25 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
in Digital Logic by Veteran (52.1k points)
edited by | 4.6k 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?
+2
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.
+1
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

0
The sign is already there at the 15th bit. So no need to reserve 1 bit from mantissa's field.

3 Answers

+30 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:  $0$  $1001010$  $11101000$ $=$ $4AE8$                .

by Boss (31.1k points)
edited by
+7
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
+3

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??
+1
Bias is given. It can be anything till it does the work of converting all negative exponent to positive.
+4
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 ?
0
For normalised operation, I am gettng 0.239 = 0.00111101001. By using this answer comes 4A39. Please correct me if I am wrong
0

@srestha mam,

I understand the 1st part.

but in 2nd part, why are we padding 0's???  Instead of that we can do 3 more iteration of binary fraction to decimal conversion.

I mean 1st we get, 0.001111010 * 2^13.

In 2nd part we can do like this 1.11101010 * 2^10.  here instead of padding 0's I'm doing 3 more iteration of binary to decimal conversion.

what's wrong in that.  please clarify

0

@MRINMOY_HALDER

How r u doing binary operation in a right shift??

0

0.239 in binary has endless string right???

then In unnormalized from, we can write 0.001111010 * 2^13. This is clear.

Now this 0.001111010 has no end, I mean I can keep converting binary to decimal conversion.

0.00111101010011........

so, I can write 1.11101010 * 2^10 instead of 1.11101000 * 2^10 .

basically I want to say that instead of padding 0's after shifting decimal point, I can take more bits which will be generated by conversion of 0.239 to binary. beacuse conversion of 0.239 is not fixed or not having any last point.

Hope you got my doubt. 

0

basically I want to say that instead of padding 0's after shifting decimal point  I can take more bits which will be generated by conversion of 0.239 to binary.

Can right shift  work like this?

0
No, but we can make 8 bits in mantissa by padding 0's or taking next consecutive bits ???

from 0.001111010, after shifting if we're assuming this - 1.11101, then at the same time we're also assuming that 0.239 = 0.001111010. which is wrong becuase by 0.001111010 we can't represent 0.239, right??
0
Mantissa will be only 8 bits. We cannot take more than 8 consecutive bits.  After that we do shifting and shifting with padding 0.

We cannot violate this
+2 votes
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
by (269 points)
0 votes
  • Option d is right.

by Boss (34.9k points)
+1
Plzz tell me that only in IEEE format the excess value if not mentioned taken as 2^(k-1)-1 otherwise 2^(k-1)
0
EXCESS==>bias=2^(k-1)

other ===>bias=2^(k-1)-1
Answer:

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