Animated Solution for Physics - Semiconductors: Find the truth table for the function $Y$ of $A$ and $B$ represented in the following figure.

Look closely at the logic circuit provided. Our goal is to determine the truth table for the entire system. To do this systematically, we first need to identify the individual logic gates present.

In this circuit, we have three distinct gates: an AND gate, a NOT gate, and an OR gate.

Now, let's trace the inputs $A$ and $B$ as they flow through the circuit. Input $A$ goes directly into the top terminal of the AND gate. Input $B$ splits into two paths. One branch goes directly into the bottom terminal of the AND gate. Therefore, the output of this AND gate is simply the logical product of its inputs, which is $A\cdot B$.

Let's follow the other branch of input $B$. It passes through the NOT gate. The NOT gate's sole job is to invert the signal, so its output becomes $\bar{B}$.

Finally, these two intermediate signals—$A\cdot B$ from the AND gate and $\bar{B}$ from the NOT gate—are fed into the OR gate. The OR gate performs logical addition on its inputs. Thus, our final output equation for $Y$ is:

With our master equation ready, we can construct the truth table by substituting all four possible combinations of $A$ and $B$.

Case 1: When $A=0$ and $B=0$. Substituting these into our equation gives $Y=0\cdot 0+\bar{0}$. This simplifies to $0+1$, which equals $1$.

Case 2: When $A=0$ and $B=1$. Substituting these gives $Y=0\cdot 1+\bar{1}$. This simplifies to $0+0$, which equals $0$.

Case 3: When $A=1$ and $B=0$. Substituting these gives $Y=1\cdot 0+\bar{0}$. This simplifies to $0+1$, which equals $1$.

Case 4: When $A=1$ and $B=1$. Substituting these gives $Y=1\cdot 1+\bar{1}$. This simplifies to $1+0$, which equals $1$.

Compiling these results, our final outputs for the sequence $(0,0),(0,1),(1,0),(1,1)$ are $1,0,1,1$. This matches perfectly with option (b).

Boolean Magic

Could we have solved this even more elegantly? Absolutely, by harnessing the power of Boolean algebra!

We can apply the Distributive Law, which states that $X+Y\cdot Z=(X+Y)\cdot (X+Z)$. Applying this to our equation $Y=\bar{B}+A\cdot B$, we get:

Since a variable ORed with its complement is always $1$ (i.e., $\bar{B}+B=1$), the equation simplifies beautifully to:

You can directly compute the truth table from this much simpler expression, saving precious time during an exam!