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43 lines
2.0 KiB
TeX
43 lines
2.0 KiB
TeX
\subsubsection{Properties}
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The advantage of having denormalized values is that 0 can be represented as the bit-field with all $0$s. Further, this enforces equidistant points for values close to $0$, whereas normalized values increase in distance as they move further from $0$.
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\content{Example} $8$b Floating Point table to visualize the different cases.
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$$
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8\text{b precision Floating Point:}\quad \underbrace{0}_s \underbrace{0000}_c \underbrace{000}_m
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$$
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\renewcommand{\arraystretch}{1.2}
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\begin{center}
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\begin{tabular}{llllll}
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\hline
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Case & $s$ & $e$ & $m$ & $E$ & Value \\
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\hline
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\multirow{6}{*}{Denormalized}
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& 0 & 0000 & 000 & $-6$ & $0$ \\
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& 0 & 0000 & 001 & $-6$ & $\frac{1}{8}\cdot\frac{1}{64}=\frac{1}{512}$ \\
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& 0 & 0000 & 010 & $-6$ & $\frac{2}{8}\cdot\frac{1}{64}=\frac{2}{512}$ \\
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& & & $\vdots$ & & $\vdots$ \\
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& 0 & 0000 & 110 & $-6$ & $\frac{6}{8}\cdot\frac{1}{64}=\frac{6}{512}$ \\
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& 0 & 0000 & 111 & $-6$ & $\frac{7}{8}\cdot\frac{1}{64}=\frac{7}{512}$ \\
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\hline
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\multirow{9}{*}{Normalized}
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& 0 & 0001 & 000 & $-6$ & $\frac{8}{8}\cdot\frac{1}{64}=\frac{8}{512}$ \\
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& 0 & 0001 & 001 & $-6$ & $\frac{9}{8}\cdot\frac{1}{64}=\frac{9}{512}$ \\
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& & & $\vdots$ & & $\vdots$ \\
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& 0 & 0110 & 110 & $-1$ & $\frac{14}{8}\cdot\frac{1}{2}=\frac{14}{16}$ \\
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& 0 & 0110 & 111 & $-1$ & $\frac{15}{8}\cdot\frac{1}{2}=\frac{15}{16}$ \\
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& 0 & 0111 & 000 & $0$ & $\frac{8}{8}\cdot 1 = 1$ \\
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& 0 & 0111 & 001 & $0$ & $\frac{9}{8}\cdot 1 = \frac{9}{8}$ \\
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& 0 & 0111 & 010 & $0$ & $\frac{10}{8}\cdot 1 = \frac{10}{8}$ \\
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& & & $\vdots$ & & $\vdots$ \\
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& 0 & 1110 & 110 & $7$ & $\frac{14}{8}\cdot 128 = 224$ \\
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& 0 & 1110 & 111 & $7$ & $\frac{15}{8}\cdot 128 = 240$ \\
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\hline
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Special
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& 0 & 1111 & 000 & n/a & $\infty$ \\
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\hline
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\end{tabular}
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\end{center}
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\renewcommand{\arraystretch}{1.0}
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