數位系統作業2
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B073040047 楊志璿
## Chapter 5
### 5.8
$T_A = A + B$ and $T_B = A' + B$
$A(t+1) = (A+B)'A + (A+B)A'$
$B(t+1) = (A'+B)'B + (A'+B)B'$

Hence we can drow the table and the state diagram:

### 5.12

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### 5.18
$A(t+1) = J_Ax'+K_A'x$
$B(t+1) = J_Bx'+K_B'x$

Hence the result is:
$J_A = K_A = E(B \oplus F)'$
$J_B = K_B = E$

## Chapter 6
### 6.17


Hence
$J_0 = K_0 = E$
$J_1 = K_1 = Q_1'E$
$J_2 = K_2 = Q_2'Q_1'E$
$J_3 = K_3 = Q_3'Q_2'Q_1'E$

### 6.24
Hence
$T_0 = Q_0'Q_2'+Q_0Q_2$
$T_1 = Q_0 \oplus Q_1$
$T_2 = Q_1 \oplus Q_2$