Average of a rectangular (pulse) waveform with DC baseline: A rectangular wave has a baseline (low level) of 20 V, a peak-to-peak value of 50 V, and a duty cycle of 20%. What is its average voltage over one full period?

Introduction:
Pulse and rectangular waveforms often ride on a DC baseline. Computing the average (DC) component is essential for biasing, measurement, and filter design. When duty cycle and peak-to-peak amplitude are given, you can reconstruct the high and low levels and take a time-weighted average.

Given Data / Assumptions:

• Baseline (low level) Vlow = 20 V.
• Peak-to-peak Vpp = 50 V → Vhigh = Vlow + Vpp = 70 V.
• Duty cycle D = 20% = 0.2 (fraction of period at Vhigh).
• Ideal rectangular shape; negligible rise/fall time.

Concept / Approach:

The average over one period is the time-weighted sum of the levels: Vavg = D * Vhigh + (1 − D) * Vlow. This directly incorporates both the DC baseline and the fraction of time spent at the upper level.

Step-by-Step Solution:

Verification / Alternative check:

As a check, note that Vavg must lie between 20 V and 70 V and be closer to 20 V because the signal stays low 80% of the time; 30 V satisfies both conditions.

Why Other Options Are Wrong:

• 26 V / 40 V / 45 V: Do not match the weighted average using the given duty cycle.
• 20 V: Would imply D = 0% or no pulses above baseline, which is untrue here.

Common Pitfalls:

• Confusing Vpp with amplitude about zero rather than about the given baseline.
• Using arithmetic mean of 20 V and 70 V (45 V) without weighting by duty cycle.

Final Answer:

30 V