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Average and RMS of sawtooth waveform

Short answer

The waveform rises linearly from 10 to 20 over period T. Its average value is (10+20)/2 = 15. The RMS value is √(1/T ∫₀ᵀ (10+10t/T)² dt) = 15.28 (independent of T).

For the periodic signal given below, find the average and rms value. Assume T = 10 ms. [Figure: sawtooth waveform y(t): it rises linearly from y = 10 at t = 0 (point A) to y = 20 at t = T (point B), then drops vertically back to y = 10 (point C) at t = T, and repeats with period T (rising again to 20 at 2T, etc.). The dashed line through A, E, C marks the level y = 10; D is a point on the ramp at time t with height y above the time axis.]

Equation of the waveform

In one period (0 ≤ t < T) the wave rises linearly from 10 to 20:

y(t) = 10 + (20 − 10) t/T = 10 + 10t/T
T = 10 ms

Average value

Y_avg = (1/T) ∫₀ᵀ (10 + 10t/T) dt
      = (1/T) [10t + 5t²/T]₀ᵀ
      = (1/T) (10T + 5T)
      = 15

This is also clear from the shape: a straight ramp from 10 to 20 has mean (10 + 20)/2 = 15.

RMS value

Put x = t/T (x goes from 0 to 1, dt = T dx):

Y_rms² = (1/T) ∫₀ᵀ (10 + 10t/T)² dt
       = 100 ∫₀¹ (1 + x)² dx
       = 100 [(1 + x)³/3]₀¹
       = 100 × (8 − 1)/3
       = 233.33
Y_rms  = √233.33 = 15.275

Answer: Average value = 15, RMS value ≈ 15.28 (in the units of y, e.g. volts).

The value T = 10 ms does not change the result: the average and RMS of a periodic wave depend only on its shape over one period, not on how long the period is. RMS (15.28) is slightly larger than the average (15) because squaring gives more weight to the larger values near 20.

✓ Answer checked by Claude Opus 5.5 (high effort)

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