Average value of a pulsed waveform: A signal has a baseline (DC offset) of 3 V, a duty cycle of 20%, and an amplitude of 8 V relative to the baseline during its ON interval. What is the average (mean) output voltage?

Introduction / Context:
Pulse and digital waveforms often ride on a baseline (DC offset). The average value depends on both the offset and how long the waveform spends at its higher level. Understanding how duty cycle and amplitude interact is essential in power electronics, PWM control, and signal conditioning.

Given Data / Assumptions:

• Baseline (offset) voltage V_base = 3 V.
• Amplitude during ON relative to baseline A = 8 V (so ON level is V_base + A).
• Duty cycle D = 20% = 0.2 (fraction of the period at the ON level).
• Assume ideal rectangular transitions; OFF level equals the baseline V_base.

Concept / Approach:
The average of a two-level waveform is the time-weighted mean of its levels. If the OFF level is the baseline and the ON level is baseline + amplitude, the average is V_avg = V_base + D * A. This expression naturally separates the steady offset from the pulsed increment contributed by the duty cycle.

Step-by-Step Solution:

Verification / Alternative check:
Use compact form: V_avg = V_base + D * A = 3 + 0.2 * 8 = 3 + 1.6 = 4.6 V. Both methods agree. This matches intuition: with only 20% ON time, the average lies modestly above the 3 V baseline, not near the 11 V peak.

Why Other Options Are Wrong:

• 4 V: Would correspond to a smaller amplitude or duty cycle; it ignores the specified 8 V amplitude contribution.
• 1.6 V: This is just D * A, omitting the baseline, so it underestimates the mean.
• 11 V: That is the instantaneous ON level, not the average across a period with 20% duty.

Common Pitfalls:

• Forgetting to add the baseline to the duty-weighted amplitude.
• Confusing amplitude with peak-to-peak or with absolute level.

Final Answer:
4.6 V