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Chapter 2 · 4 hours

Average and RMS Values

IOE past exam questions

Past questions and answers

2 questions set from this chapter. Most asked first.

  • 2081 Baishakh (new course) · 2+4 marks

Define RMS and average value of ac signal. Find the RMS and average value of half wave rectified sinusoidal waveform V = Vm sin ωt.

Answer

Definitions

  • RMS (effective) value: the value of a steady DC current that produces the same heat in a resistor, in the same time, as the AC current. Mathematically, the square root of the mean of the squared values over one period: I_rms = √[(1/T) ∫₀ᵀ i² dt].
  • Average value: the arithmetic mean of all instantaneous values over one period (or half period for a symmetrical wave): V_avg = (1/T) ∫₀ᵀ v dt. It equals the DC component of the wave.

Half-wave rectified sine wave

v = Vm sin ωt   for 0 ≤ ωt ≤ π
v = 0           for π ≤ ωt ≤ 2π
Period = 2π (in terms of θ = ωt)

 Vm |   __           __
    |  /  \         /  \
    | /    \       /    \
  0 +/------\_____/------\____ ωt
    0   π   2π    3π  4π

Average value

V_avg = (1/2π) ∫₀^π Vm sin θ dθ
      = (Vm/2π) [−cos θ]₀^π
      = (Vm/2π) [1 + 1]
      = Vm/π
      ≈ 0.318 Vm

RMS value

V_rms² = (1/2π) ∫₀^π Vm² sin²θ dθ
       = (Vm²/2π) ∫₀^π (1 − cos 2θ)/2 dθ
       = (Vm²/4π) [θ − (sin 2θ)/2]₀^π
       = (Vm²/4π) × π
       = Vm²/4
V_rms  = Vm/2 = 0.5 Vm

Answer: V_avg = Vm/π ≈ 0.318 Vm, V_rms = Vm/2 = 0.5 Vm.

Related factors: form factor = V_rms/V_avg = (Vm/2)/(Vm/π) = π/2 ≈ 1.57; peak factor = Vm/V_rms = 2. For comparison, a full sine wave has V_rms = 0.707 Vm.

  • 2081 Chaitra (new course) · 2+3 marks

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.]

Answer

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.

Questions from Exam papers (ENEX 101) (Two new-course (2080 batch) IOE papers: 2081 Baishakh and 2081 Chaitra). Answers are written for this site; check them against your class notes.

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