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Consider the following floating-point format.

 

Mantissa is a pure fraction in sign-magnitude form.

The normalized representation for the above format is specified as follows. The mantissa has an implicit $1$ preceding the binary (radix) point. Assume that only $0’s$ are padded in while shifting a field.

The normalized representation of the above number $(0.239 \times 2^{13})$ is:

  1. $0A\;20$
  2. $11\;34$
  3. $49\;D0$
  4. $4A\;E8$
in Digital Logic by Veteran (106k points)
edited by | 2k views
+2

|0|1001010|11101000|

|sign bit|biased exponent|normalized mantissa|

(D) is the correct answer!

0

2 Answers

+29 votes
Best answer

For finding normalised representation we need to find unnormalised one first. So, we have:

$0.239 \times 2^{13}$ as the number. So, we find the binary equivalent of $0.239$ till $8$ digits as capacity of mantissa field is $8$ bits. We follow the following procedure:

  • $0.239 \times 2 = 0.478$
  • $0.478 \times 2 = 0.956$
  • $0.956 \times 2 = 1.912$
  • $0.912 \times 2 = 1.824$
  • $0.824 \times 2 = 1.648$
  • $0.648 \times 2 = 1.296$
  • $0.296 \times 2 = 0.512$
  • $0.512 \times 2 = 1.024$


We stop here as we have performed $8$ iterations and hence, $8$ digits of mantissa of unnormalised number is obtained. Now we have:
Mantissa of given number  $=$  $0011$  $1101$

So, the number can be written as: $0.00111101\times 2^{13}$

Now we need to align the mantissa towards left to get normalised number. And in the question it is mentioned that during alignment process $0's$ will be padded in the right side as a result of mantissa alignment to left.

So, to get normalised number, we align to left $3$ times, to get new mantissa $= 11101 000$

Exponent will also decrease by $3$ hence, new exponent  $=  10$

So, normalised number$ = 1.11101000 \times 2^{10}$

Actual exponent $= 10$

Given excess $64$ is used which means bias value $= 64$

So, exponent field value $=10+64 =  74$

And of course sign $bit = 0$ being a positive number.

Thus the final representation of number 
$\qquad\qquad\qquad=  0\;  1001010\;11101000$
$\qquad\qquad\qquad=  0100\;1010\;1110\;1000$
$\qquad\qquad\qquad=  (4AE8)_{16}$

Hence, (D) should be the correct answer.

by Veteran (102k points)
edited by
0
What is the meaning of line "The mantissa has an implicit 1 preceding the binary (radix) point"??
0
it means that only the fraction part of 1.xxxx... will be stored as mantissa in the format which is . , and that 1 will be automatically appended while retrieving the value.
0

0.296×2=0.592 (this doesn't make any change in answer)

0.592×2=1.084

0
i cannot understand the logic why .296*2 =.512 and not .592..

please explain
0
their is a calculation mistake Habib sir answer, 2nd last step should be 0.296*2=0.592 so last step is 0.592*2=1.084 (this doesn't change the answer)
0
okay,thanks
+1
  • 0.239×2=0.478 = 0
  • 0.478×2=0.956 = 0
  • 0.956×2=1.912 = 1
  • 0.912×2=1.824 = 1
  • 0.824×2=1.648 = 1
  • 0.648×2=1.296 = 1
  • 0.296×2=0.592 = 0
  • 0.592×2=1.184 = 1
  • 0.184×2=0.368 = 0
  • 0.368×2=0.736 = 0
  • 0.736×2=1.472 = 1

So mantissa = 0.00111101001 = (1.1110 1001)x$2^{-3}$.So mantissa part becomes E9 not E8.

0
In IEEE- 754 single precision format where we had 8 bits for exponent we used bias as 2^(8-1)-1= 127 then in this ques why didn't we used exponent bias as 63 instesd of 64.
+1

In IEEE 754 single precision format the bias which is added to exponent is usually referred to as "Excess 127". It is not necessary that bias should always be in the form of 2^n-1. It could be anything. Here they gave the bias as "Excess 64". So it basically means we need to add 64 to the exponent part to get the actual exponent value. @Poonam Arora

0
Thnq so much
0
It should be E8
+4 votes
  • Option d is right

 

by Boss (36.7k points)
0
this is perfect!
Answer:

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