Semi-circular Sine Wave Substitute

Calculus Level 3

It is common knowledge that for a sine wave, the ratio of the peak value to the root mean square (RMS) value is 2 \sqrt{2} . Suppose we construct a periodic signal resembling a sine wave, composed of half circles placed end to end, with oscillating polarity.

For this signal, the ratio of the peak value to the RMS value is:

Peak RMS = A B \frac{\text{Peak}}{\text{RMS}} = \sqrt{\frac{A}{B}}

If A A and B B are positive co-prime integers, determine A + B A + B .


The answer is 5.

Steven Chase by Steven Chase , 37, USA

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1 solution

Tapas Mazumdar
May 19, 2018

The radius of the circular waveform is one-fourth the time period of the wave, hence T = 4 R T = 4R . The circular waveform in terms of cartesian coordinates x x and y y can be given by

y 2 + ( x − R ) 2 = R 2 , 0 ≤ x ≤ 2 R y 2 + ( x − 3 R ) 2 = R 2 , 2 R ≤ x ≤ 4 R y^2 + (x-R)^2 = R^2 \ \ , \ \ 0 \le x \le 2R \\ y^2 + (x-3R)^2 = R^2 \ \ , \ \ 2R \le x \le 4R

Thus the RMS value for the wave can be given as

RMS value = 1 4 R ∫ 0 4 R y 2 d x = 1 4 R [ ∫ 0 2 R ( R 2 − ( x − R ) 2 ) d x + ∫ 2 R 4 R ( R 2 − ( x − 3 R ) 2 ) d x ] = 2 3 R \begin{aligned} \text{RMS value} &= \sqrt{\dfrac{1}{4R} \int_0^{4R} y^2 \,dx} \\ &= \sqrt{\dfrac{1}{4R} \left[ \int_0^{2R} (R^2 - (x-R)^2) \,dx + \int_{2R}^{4R} (R^2 - (x-3R)^2) \,dx \right] } \\ &= \sqrt{\dfrac23} R \end{aligned}

while the peak value is simply R R . Thus required ratio is 3 2 \sqrt{\dfrac32} giving the answer as 5 \boxed{5} .

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