An ideal comparator is fed with a sine wave of 4Vpp (peak-to-peak) with zero DC component as shown in Figure. The reference voltage of the comparator is 1 V. What is the duty cycle of the output Vout?

Explanation:

Step-by-Step Solution:

Step 1: Analyze the Sine Wave Input

The sine wave is given by the equation:

V(t) = A × sin(ωt)

Where:

• A: Amplitude of the sine wave = 2V
• ω: Angular frequency of the sine wave

The sine wave oscillates between -2V and +2V, crossing the reference voltage (1V) twice per cycle.

Step 2: Determine When the Sine Wave Exceeds the Reference Voltage

To find the time intervals during which the sine wave is greater than the reference voltage (1V), we solve the equation:

A × sin(ωt) > V_ref

Substituting the values:

2 × sin(ωt) > 1

Dividing through by 2:

sin(ωt) > 0.5

From trigonometric principles, sin(ωt) > 0.5 occurs between two angles:

ωt = π/6 and ωt = 5π/6

Therefore, the sine wave exceeds the reference voltage for the duration:

Δt = (5π/6 - π/6)/ω = (4π/6)/ω = (2π/3)/ω

Step 3: Calculate the Duty Cycle

The total time period of the sine wave is:

T = 2π/ω

The duty cycle is the ratio of the time the output is high (Δt) to the total time period (T):

Duty Cycle = Δt / T

Substituting the values:

Duty Cycle = [(2π/3)/ω] / [2π/ω]

Simplifying:

Duty Cycle = (2π/3) / (2π) = 1/3

Expressing this as a decimal:

Duty Cycle = 0.3333

Final Answer: The duty cycle of the output signal is 0.3333.

Hence the correct answer is 33.33 %