2020 A/L Physics Past Paper · Question 18 (English Medium)

Answer: (2) Waveform (2)

What the question asks

We must find: the waveform of the output $X$ of a circuit where $A$, $B$ feed an AND gate, and its output together with $C$ feed an OR gate.

Relevant theory

Only when both A and B switches are closed (NO), the bulb lights. That is only when both A and B inputs are 1, the output F is 1.

Boolean expression for the two-input AND gate is written as, F = A.B

The truth for the two-input OR gate can be given as "When the input A or B is 1, the output is 1.

Boolean expression for the OR gate is, F = A + B

Required equations

$Y = A\cdot B$

$X = Y + C = A\cdot B + C$

Reading the options

All five options are waveforms drawn on the same time axis; let us compute $X$ at each time slot, then compare.

Reasoning and calculation

Starting point: let us divide the time axis into 9 equal slots (0 to 8), matching the changes of $A$. $A = 1$: time slots 1, 3, 5, 7; $B = 1$: time slots 2 through 6; $C = 1$: time slots 1, 7. At time slot 0 everything is $0$, so $X = 0$.

Time slot 1

$X = 1$

$A = 1$, $B = 0$ ⇒ $Y = A\cdot B = 0$; $C = 1$ ⇒ $X = 0 + 1 = 1$.

Time slots 2, 4, 6

$X = 0$

In these slots $A = 0$; so whatever $B$ is, the AND gate's output $Y = 0$. $C$ is also $0$, so both OR-gate inputs are $0$ — the new state is $X = 0$: this is the time slot that separates the pulses.

Time slots 3, 5

$A = 1$, $B = 1$ ⇒ $Y = 1$; so even with $C = 0$, $X = 1$.

Time slot 7

$A = 1$, but $B$ has already returned to $0$ ⇒ $Y = 0$; $C = 1$ ⇒ $X = 1$. Time slot 8: everything is $0$, $X = 0$.

Final state: $X = 1$ in time slots 1, 3, 5, 7 — four pulses, at the same places and the same width as the pulses of $A$.

Matching with the options: the only waveform with four pulses is waveform (2).

Checks

Invariant check: by Boolean algebra, $X = A\cdot B + C$. Here, in every time slot where $C = 1$, $A$ is also $1$; and in every time slot where $B = 0$, $A = 1$ but $C = 1$ too — so for these particular waveforms, $X = A$ — exactly $A$'s four pulses ✓. Second way: drawing $Y$ first gives waveform (5); OR-ing it with $C$ (waveform (4)) gives waveform (2) ✓.

Conclusion

$X = A\cdot B + C$

$Y = A\cdot B$ only in time slots 3, 5; $X = Y + C$ in time slots 1, 3, 5, 7 — four pulses — the answer is (2).

Taking the AND gate's output ($Y$) as the final output.

$Y$ still joins $C$ through the OR gate: $X = Y + C$.