Animated Solution for Physics - Semiconductors: Draw the output signal $Y$ in the given combination of gates

When you are presented with a logic circuit and continuous input waveforms, the problem might look intimidating at first glance. However, by breaking it down into two distinct phases—simplifying the logic and extracting the digital states—the solution becomes incredibly straightforward.

Let's first decode the logic circuit given to us. We have two inputs, $A$ and $B$. Notice how each input first passes through a NOT gate, which is represented by the triangle with a bubble. These gates invert our signals, giving us $\overline{A}$ and $\overline{B}$.

Then, these inverted signals are fed into an OR gate. Therefore, the Boolean expression for the output $Y$ is:

Does this look familiar? By De Morgan's Law, we know that the OR operation of two inverted inputs is exactly equivalent to the NAND operation of the original inputs:

This is a massive simplification! It means our entire circuit is simply a NAND gate. Instead of inverting signals and then adding them, we can just multiply $A$ and $B$ and invert the final result.

Now, let's extract the digital states of our input signals $A$ and $B$ from the given waveforms for each one-second interval. When the waveform is raised, the state is HIGH ($1$). When it is on the axis, the state is LOW ($0$).

Interval $(0-1)\text{s}$: $A=1$, $B=0$ Interval $(1-2)\text{s}$: $A=1$, $B=1$ Interval $(2-3)\text{s}$: $A=0$, $B=0$ Interval $(3-4)\text{s}$: $A=1$, $B=0$ Interval $(4-5)\text{s}$:* $A=1$, $B=1$

Let's compute the output $Y$ using our NAND logic ($Y=\overline{A\cdot B}$) for each interval:

For $(0-1)\text{s}$: $Y=\overline{1\cdot 0}=\overline{0}=1$ For $(1-2)\text{s}$: $Y=\overline{1\cdot 1}=\overline{1}=0$ For $(2-3)\text{s}$: $Y=\overline{0\cdot 0}=\overline{0}=1$ For $(3-4)\text{s}$: $Y=\overline{1\cdot 0}=\overline{0}=1$ For $(4-5)\text{s}$:* $Y=\overline{1\cdot 1}=\overline{1}=0$

We have successfully traced the entire output waveform! The signal is HIGH ($1$) for the first second, LOW ($0$) for the next, HIGH ($1$) for two seconds, and LOW ($0$) again.

Comparing our generated waveform for $Y$ with the given options, we can clearly see that it perfectly matches option (d). Always look for ways to simplify the logic circuit using Boolean algebra before diving into the waveforms!