When designing the circuit with the state table shown below using JK fip flops, then AA $$ $K A=\ldots$. [ (3 Points) \begin{tabular ICCCCCCC) Whline Present State & multicolumn{5}{c} Input } &\multicolumn{2}{c} Next State } & Imulticolumn{1}c} JA } & KA Whline A & $\mathbf {B}$ & $\mathbf{X}$ & $\mathbf {A}$ & $\mathbf {B}$ & & W Whline O&O &&O & & & Whline O&O & 1 & 0 & 1 && W Thline 0 & 1 & 0 & 1 & 0 && Vhline 0 & 1 & 1 & 2 & 1 && W Thline 1 & 0 & 9 & 1 & 0 && W Nhline 1 & 0& 1 & 1 & 1 && W Whline 1 & 1 & 0&1&1 && W Whline 1 &1&1 & 0 & 0 && Whline \end{tabular) \begin{tabular|CC|CCCCC!! Whline $A$ & $B$ & $X$ & $A$ & $B$ && Whline 0 & & & & & & W Whline D & D & 1 & 0&1&& W Whline 0 & 1 & 0&1&0&& Whline 0 & 1 & 1 & 0 & 1 && Whline 1 & & & 1 & 2 && Whline 1 & 0&1&1 & 1 && W Vhline 1 & 1 & 0&1&1 && W Vhline 1 & 1 & 1 & 0& 0 && W Whline \end{tabular) $J_{A}B X, K_{A}A^{\prime] $ $A_{1}=8 x{\prime), K_{1}=B, x$ $J_{1}=B^{\prime} \cdot x*{\prime), k_{1}=A^{\prime}$ $J_{A}B \times K_{A}=A$ SE.SD.002