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X = 1.0000\ 1101 (269/256)

D = 1.0011\ 0110 (310/256)

Q = 0.1101\ 1110 (222/256)

D[1.3] = 1.001, so we use the ``1.001" column of chart 13.X. This means we select a quotient bit of 2 if the partial remainder is greater than or equal to 3.5, a quotient bit of 1 if the partial is greater or equal to than 1.0, a zero if the partial is greater than or equal to -1.5, -1 if the partial is greater than or equal to -3.75, and a -2 otherwise.

{\small
\begin{center}
\begin{tabular}{cccc}
    Initialization&$D$&$0001.0011\ 0110\ 00$&\\
    &$2D$&$0010.0110\ 1100\ 00$&\\
    &$-D=\overline{D}+1$&$1110.1100\ 1001\ 11$&(+ 1 ulp)\\
    &$-2D=\overline{2D}+1$&$1101.1001\ 0011\ 11$&(+ 1 ulp)\\
    &&&\\
    &$WS_{-1}=X$&$0001.0000\ 1101\ 00$&\\
    &$WC_{-1}$&$0000.0000\ 0000\ 00$&\\
    \hdashline\\
    Step 0: &$WS_{-1}   $&$0001.0000\ 1101\ 00$&\\
            &$WC_{-1}   $&$0000.0000\ 0000\ 0\mathbf{1}$&($W_{msbs}=0001.000\ \text{so}\ q_0=1$)\\
            &$-q_0D     $&$1110.1100\ 1001\ 11$&\\
    \cline{2-3}
            &$sum       $&$1111.1100\ 0100\ 10$&$\ll2$\\
            &$carry     $&$0000.0001\ 0010\ 10$&$\ll2$\\
    \hdashline\\
    Step 1: &$WS_0      $&$1111.0001\ 0010\ 00$&\\
            &$WC_0      $&$0000.0100\ 1010\ 0\mathbf{0}$&($W_{msbs}=1111.010\ \text{so}\ q_1=-1$)\\
            &$-q_1D     $&$0001.0011\ 0110\ 00$&\\
    \cline{2-3}
            &$sum       $&$1110.0110\ 1110\ 00$&$\ll2$\\
            &$carry     $&$0010.0010\ 0100\ 00$&$\ll2$\\
    \hdashline\\
    Step 2: &$WS_1      $&$1001.1011\ 1000\ 00$&\\
            &$WC_1      $&$1000.1001\ 0000\ 0\mathbf{1}$&($W_{msbs}=0010.010\ \text{so}\ q_2=2$)\\
            &$-q_2D     $&$1101.1001\ 0011\ 11$\\
    \cline{2-3}
            &$sum       $&$1100.1011\ 1011\ 10$&$\ll2$\\
            &$carry     $&$0011.0010\ 0000\ 10$&$\ll2$\\
    \hdashline\\
    Step 3: &$WS_2      $&$0010.1110\ 1110\ 00$&\\
            &$WC_2      $&$1100.1000\ 0010\ 0\mathbf{0}$&($W_{msbs}=1111.011\ \text{so}\ q_3=-1$)\\
            &$-q_3D     $&$0001.0011\ 0110\ 00$\\
    \cline{2-3}
            &$sum       $&$1111.0101\ 1010\ 00$&$\ll2$\\
            &$carry     $&$0001.0100\ 0100\ 00$&$\ll2$\\
    \hdashline\\
    Step 4: &$WS_3      $&$1101.0110\ 1000\ 00$&\\
            &$WC_3      $&$0101.0001\ 0000\ 0\mathbf{1}$&($W_{msbs}=0010.011\ \text{so}\ q_4=2$)\\
            &$-q_4D     $&$1101.1001\ 0011\ 11$\\
    \cline{2-3}
            &$sum       $&$0101.1110\ 1011\ 10$&$\ll2$\\
            &$carry     $&$1010.0010\ 0000\ 10$&$\ll2$\\
    \hdashline\\
    Step 5: &$WS_4      $&$0111.1010\ 1110\ 00$&\\
            &$WC_4      $&$1000.1000\ 0010\ 0\mathbf{0}$&($W_{msbs}=0000.001\ \text{so}\ q_5=0$)\\
            &$-q_5D     $&$0000.0000\ 0000\ 00$\\
    \cline{2-3}
            &$sum       $&$1111.0010\ 1100\ 00$&$\ll2$\\
            &$carry     $&$0001.0000\ 0100\ 00$&$\ll2$\\
    \hdashline\\
    Terminate&Quotient&$00.11\ 01\ 11\ 10\ (00\ 1)$
\end{tabular}
\end{center}
}

\end{document}
