+25 votes
3.5k views

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
edited | 3.5k 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

+27 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 (31.9k points)
edited
+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.
+3
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 (24.2k 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:

+15 votes
2 answers
1
+14 votes
2 answers
2
+14 votes
4 answers
3
+16 votes
1 answer
5
+31 votes
7 answers
6
+12 votes
4 answers
7