Octal-to-binary encoder; A+B'C canonical form

Octal-to-binary encoder

An octal-to-binary (8-to-3) encoder has eight inputs $D_0$–$D_7$, one for each octal digit, and three outputs $Y_2Y_1Y_0$ giving the 3-bit binary code of the active input. Only one input is assumed to be active at a time.

Each output is 1 for the inputs whose code has a 1 in that position:

$$Y_2 = D_4 + D_5 + D_6 + D_7,\quad Y_1 = D_2 + D_3 + D_6 + D_7,\quad Y_0 = D_1 + D_3 + D_5 + D_7$$

 D4 D5 D6 D7 --[OR 4]--> Y2
 D2 D3 D6 D7 --[OR 4]--> Y1
 D1 D3 D5 D7 --[OR 4]--> Y0
 (D0 is not connected: it gives 000)

Limitations: $D_0$ active and "no input active" both give 000, and two active inputs give a wrong code (e.g. $D_3 + D_5$ → 111). A priority encoder solves these.

Canonical form of $A + B'C$

Expand each term with the missing variables using $X + X' = 1$:

$$\begin{aligned} A &= A(B + B')(C + C') = ABC + ABC' + AB'C + AB'C' \\ B'C &= B'C(A + A') = AB'C + A'B'C \end{aligned}$$

Combine and remove the repeated term $AB'C$:

$$A + B'C = A'B'C + AB'C' + AB'C + ABC' + ABC$$ $$= \Sigma m(1, 4, 5, 6, 7)$$

The canonical POS form uses the remaining combinations:

$$A + B'C = \Pi M(0, 2, 3) = (A + B + C)(A + B' + C)(A + B' + C')$$