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

Solution :

The correct output waveform is given by:

Step-by-step Explanation:

1. Understand the Logic Circuit Function:
From the given circuit diagram, we have three inputs $A$ , $B$ , and $C$ connected through logic gates as follows:
- The top gate is an AND gate receiving inputs $A$ and $B$ . Its output is $A \cdot B$ .
- The bottom gate is a NAND gate receiving inputs $B$ and $C$ . Its output is $\overset{&#AF;}{B \cdot C}$ .
- The final gate is an OR gate combining both outputs. Therefore, the total output $y$ is given by the Boolean expression:

$y = (A \cdot B) + \overset{&#AF;}{B \cdot C}$

2. Analyze the Input Signals over Time Intervals:
Let a voltage of $5 V$ represent logic state $1$ (HIGH) and $0 V$ represent logic state $0$ (LOW).

By observing the timing diagram from the given image across different time intervals, we can determine the state of each input and calculate $y$ :

- For $t < t_{1}$ :
$A = 0$ , $B = 0$ , $C = 1$
$A \cdot B = 0 \cdot 0 = 0$
$\overset{&#AF;}{B \cdot C} = \overset{&#AF;}{0 \cdot 1} = \overset{&#AF;}{0} = 1$
$y = 0 + 1 = 1$ ( $5 V$ )

- For interval $t_{1}$ to $t_{2}$ :
$A = 1$ , $B = 0$ , $C = 1$
$A \cdot B = 1 \cdot 0 = 0$
$\overset{&#AF;}{B \cdot C} = \overset{&#AF;}{0 \cdot 1} = \overset{&#AF;}{0} = 1$
$y = 0 + 1 = 1$ ( $5 V$ )

- For interval $t_{2}$ to $t_{3}$ :
$A = 0$ , $B = 1$ , $C = 0$
$A \cdot B = 0 \cdot 1 = 0$
$\overset{&#AF;}{B \cdot C} = \overset{&#AF;}{1 \cdot 0} = \overset{&#AF;}{0} = 1$
$y = 0 + 1 = 1$ ( $5 V$ )

- For interval $t_{3}$ to $t_{4}$ :
$A = 1$ , $B = 1$ , $C = 0$
$A \cdot B = 1 \cdot 1 = 1$
$\overset{&#AF;}{B \cdot C} = \overset{&#AF;}{1 \cdot 0} = \overset{&#AF;}{0} = 1$
$y = 1 + 1 = 1$ ( $5 V$ )

- For interval $t_{4}$ to $t_{5}$ :
$A = 0$ , $B = 0$ , $C = 1$
$A \cdot B = 0 \cdot 0 = 0$
$\overset{&#AF;}{B \cdot C} = \overset{&#AF;}{0 \cdot 1} = \overset{&#AF;}{0} = 1$
$y = 0 + 1 = 1$ ( $5 V$ )

- For interval $t_{5}$ to $t_{6}$ :
$A = 1$ , $B = 0$ , $C = 1$
$A \cdot B = 1 \cdot 0 = 0$
$\overset{&#AF;}{B \cdot C} = \overset{&#AF;}{0 \cdot 1} = \overset{&#AF;}{0} = 1$
$y = 0 + 1 = 1$ ( $5 V$ )

- For $t > t_{6}$ :
$A = 0$ , $B = 0$ , $C = 1$
$y = 1$ ( $5 V$ )

Conclusion:
Since the output terminal $y$ remains at logic state $1$ ( $5 V$ ) continuously for all time intervals, the output waveform is a constant straight line at $5 V$ .