Fig. 1 is a logic circuit, draw its truth table and derive its logic expression.

Huu hapa ni ufumbuzi wa swali la mzunguko wa mantiki (logic circuit):

a) Fig. 1 ni mzunguko wa mantiki.

• i) Chora jedwali lake la ukweli (truth table).

  Jedwali la ukweli linaonyesha matokeo ya mzunguko kwa kila mchanganyiko unaowezekana wa pembejeo.
  Kwanza, tunapata matokeo ya kila lango la mantiki:
  Lango la NOT (inverter) hutoa $\bar{A}$.
  Lango la AND (lango la 3) hutoa $Y_3 = \bar{A} \cdot B$.
  Lango la AND (lango la 2) hutoa $Y_2 = A \cdot B$.
  Lango la OR (lango la 4) hutoa $X = Y_3 + Y_2$.

  Jedwali la ukweli ni:

  $$\begin{array}{|c|c|c|c|c|c|} \hline \text{A} & \text{B} & \bar{A} & Y_3 = \bar{A} \cdot B & Y_2 = A \cdot B & X = Y_3 + Y_2 \\ \hline 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 1 & 1 \\ \hline \end{array}$$

• ii) Pata usemi wake wa mantiki (logic expression).

  Kutokana na uchambuzi wa mzunguko:
  Pembejeo A hupitia lango la NOT, ikitoa $\bar{A}$.
  Lango la 3 (AND) hupokea $\bar{A}$ na B, hivyo matokeo yake ni $Y_3 = \bar{A} \cdot B$.
  Lango la 2 (AND) hupokea A na B, hivyo matokeo yake ni $Y_2 = A \cdot B$.
  Lango la 4 (OR) hupokea matokeo ya lango la 3 na lango la 2, hivyo matokeo yake ya mwisho X ni:

  $$X = Y_3 + Y_2$$

  Badilisha $Y_3$ na $Y_2$ kwenye usemi:

  $$X = (\bar{A} \cdot B) + (A \cdot B)$$

  Tumia sheria ya usambazaji (distributive law) ya aljebra ya Boolean:

  $$X = B \cdot (\bar{A} + A)$$

  Tunajua kuwa $\bar{A} + A = 1$:

  $$X = B \cdot 1$$ $$X = \boxed{B}$$