Cuestion 19 (Aritmetica, IEEE 754)

Image 2005Sept1AC19
Solucion:

Convertimos el numero hexadecimal en binario.


\begin{displaymath}414A 0000_{16}=\underbrace{0100}_{4}\underbrace{0001}_{1}\und...
...\underbrace{0000}_{0}\underbrace{0000}_{0}\underbrace{0000}_{0}\end{displaymath}



$signo$ E m
1bit 8bits 23bits
0 10000010 10010100000000000000000



\begin{displaymath}2^{E-127}\cdot 1.m\end{displaymath}




\begin{displaymath}E=10000010_{2}=130\end{displaymath}


\begin{displaymath}m=10010100\end{displaymath}




\begin{displaymath}2^{130-127}\cdot 1.10010100=2^{3}\cdot 1.10010100=1100.10100_{2}=12.625\end{displaymath}



Vamos a comprobarlo con el Mathematica:

IntegerLength[16^^414A0000, 2]

BaseForm[16^^414A0000, 2]

E1 := N[2^^10000010]
E2 := E1 - 127
E1
E2
2^E2 2^^1.10010100







2011-07-31