Question Details

For the given circuit, the input digital signals are applied at the terminals A, B and C. What would be the output at the terminal y ?

Options

Show Answer

Correct Answer :

Option C

Option 3

Solution :

The correct option is Option 3.

Logic Circuit Analysis:
The given digital logic circuit consists of two primary logic gates:

An AND gate with inputs connected to terminals A and B. Let its intermediate output be $X$ .
An OR gate with one input connected to the intermediate output $X$ and the other connected to terminal C. The output of this OR gate is terminal y.

Based on this configuration, we can write the Boolean expression for the intermediate output $X$ as:
$X = A \cdot B$
And the final output y at the terminal is:
$y = X + C = (A \cdot B) + C$

Step-by-Step Interval Evaluation:
From the given timing waveforms of the input digital signals A, B, and C, we identify the logic level (where 5V represents logic 1 and 0V represents logic 0) for each time interval, and calculate the output y:

Interval $0 < t < t_{1}$ : Inputs are A = 0, B = 0, C = 0.
AND gate: $X = 0 \cdot 0 = 0$
OR gate: $y = 0 + 0 = 0$ (Output is 0V)
Interval $t_{1} < t < t_{2}$ : Inputs are A = 1, B = 0, C = 0.
AND gate: $X = 1 \cdot 0 = 0$
OR gate: $y = 0 + 0 = 0$ (Output is 0V)
Interval $t_{2} < t < t_{3}$ : Inputs are A = 1, B = 1, C = 0.
AND gate: $X = 1 \cdot 1 = 1$
OR gate: $y = 1 + 0 = 1$ (Output is 5V)
Interval $t_{3} < t < t_{4}$ : Inputs are A = 0, B = 1, C = 1.
AND gate: $X = 0 \cdot 1 = 0$
OR gate: $y = 0 + 1 = 1$ (Output is 5V)
Interval $t_{4} < t < t_{5}$ : Inputs are A = 0, B = 0, C = 1.
AND gate: $X = 0 \cdot 0 = 0$
OR gate: $y = 0 + 1 = 1$ (Output is 5V)
Interval $t_{5} < t < t_{6}$ : Inputs are A = 1, B = 0, C = 1.
AND gate: $X = 1 \cdot 0 = 0$
OR gate: $y = 0 + 1 = 1$ (Output is 5V)
Interval $t_{6} < t < t_{7}$ : Inputs are A = 0, B = 1, C = 0.
AND gate: $X = 0 \cdot 1 = 0$
OR gate: $y = 0 + 0 = 0$ (Output is 0V)

Conclusion:
The output y is low (0V) from time $0$ to $t_{2}$ , switches to high (5V) at $t_{2}$ and remains high until $t_{6}$ , and then returns to low (0V) from $t_{6}$ to $t_{7}$ . This matches the waveform shown in Option 3.