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Question 1 Explain What is  Denormalized Number

  • Give Example
  • Give Representation in IEEE 754 and excess 64 (if any)

Question 2 How to Convert $(12.625)_10$

$(12.625)_10 \Leftrightarrow (1100.101)_2$   to

  • IEEE 754 Single Precision (With Normalization)
  • IEEE 754 Single Precision (Without Normalization)
  • Excess-64 (With Normalization)
  • Excess-64 (Without Normalization)
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1 Answer

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Question 1
For IEEE754  the value (N) is calculated as follows:

  • For 1 ≤ E ≤ 254, N = (-1)^S × 1.F × 2^(E-127). These numbers are in the so-called normalized form. The sign-bit represents the sign of the number. Fractional part (1.F) are normalized with an implicit leading 1. The exponent is bias (or in excess) of 127, so as to represent both positive and negative exponent. The range of exponent is -126 to +127.
  • For E = 0, N = (-1)^S × 0.F × 2^(-126). These numbers are in the so-called denormalized form. The exponent of 2^-126 evaluates to a very small number. Denormalized form is needed to represent zero (with F=0 and E=0). It can also represents very small positive and negative number close to zero.
  • For E = 255, it represents special values, such as ±INF (positive and negative infinity) and NaN (not a number). This is beyond the scope of this article.

https://www3.ntu.edu.sg/home/ehchua/programming/java/datarepresentation.html


Question 2

IEEE 754 Single Precision (With Normalization)

With Normalization $\Rightarrow (-1)^{S} (1.M)_2 \times 2^{E-\text{bias}} $

$(1.100101)_2 \times 2^{3} $

$E-\text{bias}=\text{True Exponent}$

$E-127=3 \Rightarrow E=130$

$\underbrace{{\color{Red} {\textbf{0}} }}_{\text{sign}} |\underbrace{{\color{Blue} {\textbf{1000 0010}} }}_{\text{Exponent}} |\underbrace{{\color{Green} {\textbf{1001 0100 0000 0000 0000 000}} }}_{\text{Mantissa}}$

IEEE 754 Single Precision (Without Normalization)

Without Normalization $\Rightarrow (-1)^{S} (0.M)_2 \times 2^{E-\text{bias}} $

$(0.1100101)_2 \times 2^{4} $

$E-\text{bias}=\text{True Exponent}$

$E-127=4 \Rightarrow E=131$

$\underbrace{{\color{Red} {\textbf{0}} }}_{\text{sign}} |\underbrace{{\color{Blue} {\textbf{1000 0011}} }}_{\text{Exponent}} |\underbrace{{\color{Green} {\textbf{1100 1010 0000 0000 0000 000}} }}_{\text{Mantissa}}$

  • from hamacher computer organization

Normalized Minimum $\pm N_{min}$

When S= $\pm \Rightarrow 0 or 1$ , E=1 ,M=0

$0 | 0000 0001 | 0000 0000 0000 0000 0000 000|$

$(-1)^{S} \times (1.0)_2 \times 2^{1-127} $

$\pm N_{min}= (-1)^{S} \times (1.0)_2 \times 2^{-126} $

Normalized Maximum $\pm N_{max}$

When S= $\pm \Rightarrow 0 or 1$ , E=254 ,M=$2^{23}-1$

$0 | 1111 1110 | 1111 1111 1111 1111 1111 111$

$(-1)^{S} \times (1.1111 1111 1111 1111 1111 111)_2 \times 2^{254-127} $

$\pm N_{max}= (-1)^{S} \times (1.1.1111 1111 1111 1111 1111 111)_2 \times 2^{127} $

Denormalized Minimum $\pm D_{min}$

When S= $\pm \Rightarrow 0 or 1$ , E=0 ,M=1

$0 | 0000 0000 | 0000 0000 0000 0000 0000 001|$

$\pm D_{min}= (-1)^{S} \times (0.0000 0000 0000 0000 0000 001)_2 \times 2^{-126} $

Denormalized Maximum $\pm D_{max}$

When S= $\pm \Rightarrow 0 or 1$ , E=0 ,M=$2^{23}-1$

$0 | 0000 0000 | 1111 1111 1111 1111 1111 111$

$\pm D_{max}= (-1)^{S} \times (0.1.1111 1111 1111 1111 1111 111)_2 \times 2^{-126} $

https://www3.ntu.edu.sg/home/ehchua/programming/java/datarepresentation.html


Similarly

Excess 64 (With Normalization)

With Normalization $\Rightarrow (-1)^{S} (1.M)_2 \times 2^{E-\text{bias}} $

$(1.100101)_2 \times 2^{3} $

$E-\text{bias}=\text{True Exponent}$

$E-64=3 \Rightarrow E=67$

$\underbrace{{\color{Red} {\textbf{0}} }}_{\text{sign}} |\underbrace{{\color{Blue} {\textbf{1000 011}} }}_{\text{Exponent}} |\underbrace{{\color{Green} {\textbf{1001 0100}} }}_{\text{Mantissa}}$

Excess 64 (Without Normalization)

Without Normalization $\Rightarrow (-1)^{S} (0.M)_2 \times 2^{E-\text{bias}} $

$(0.1100101)_2 \times 2^{4} $

$E-\text{bias}=\text{True Exponent}$

$E-64=4 \Rightarrow E=68$

$\underbrace{{\color{Red} {\textbf{0}} }}_{\text{sign}} |\underbrace{{\color{Blue} {\textbf{1000 100}} }}_{\text{Exponent}} |\underbrace{{\color{Green} {\textbf{1100 1010 }} }}_{\text{Mantissa}}$


Rounding Off

Suppose 3 digits are allowed in mantissa then :-

$\begin{align*} [113. +(-111.)]+7.51 \\ =&[002.]+7.51\\ =&[2.00]+7.51\\ =&9.51\\ \end{align*}$

$\begin{align*} 113. +[(-111.)+7.51] \\ =&113.+[-111.+008.] \\ =&113.+(-103.)\\ =&010.\\ \end{align*}$.

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