Hybrid Waveform RMS

The RMS integral works the same if we integrate from $0$ to $\pi$ . I also checked the result from $0$ to $2 \pi$ using numerical integration.

$\frac{y_{rms}}{x_{rms}} = 0.9 \\ \frac{y^2_{rms}}{x^2_{rms}} = 0.81 \\ y^2_{rms} = 0.81 \\ \int_0^{\pi} y^2(t) \, dt = 0.81 \, \pi$

Plugging into the integral:

$\int_0^{\pi} y^2(t) \, dt = 0.81 \, \pi \\ = \int_0^{\pi} \Big(\alpha \, sin(t) + 1-\alpha \Big)^2 \, dt = 0.81 \, \pi$

Evaluating the integral yields:

$\Big( \frac{3 \pi}{2} - 4 \Big) \, \alpha^2 - 2(\pi - 2) \, \alpha + \pi - 0.81 \, \pi = 0$

Solving the quadratic yields two values of $\alpha$ , only one of which is in the desired range.

$\boxed{\alpha_1 \approx 0.287} \\ \alpha_2 \approx 2.918$

The idea of an RMS is the equivalent DC value that produces the same average value for the square of the wave. Squaring (\x) gives a constant function as long as we "ignore" the discontinuities at $n\pi$ . (Also, if this function is expanded in Fourier Series, the value at the discontinuity would converge to the average value, not as defined as above, but it makes not practical difference since it'd be a set with measure 0, and contribute nothing to the RMS value.)

Thus, it suffices to show $y_{rms}$ =0.9 for some $\alpha$ .

First, we recognize that the RMS value of $\sin{(2\pi ft)}$ is $\frac{\sqrt{2}}{2}$ , or the average value of $\sin^2{(2\pi ft)}$ is $\frac{1}{2}$ .

Second, we recognize that in a Fourier Series representation of a periodic wave (well behaved enough) the RMS values can be computed in a manner much like a form of Pythagorean Theorem (but for infinite dimensional vectors).

$f(t)=\sum_n{A_n\sin{2\pi f_n t}+B_n\cos{2\pi f_n t}} \rightarrow <f^2>=\sum_{n}\frac{1}{2}(A_n^2+B_n^2)$

Third, the Fourier Series representation of the square wave contains harmonics with $A_n = \frac{4}{(2n+1)\pi}$ .

The function $y(t)$ as given, simply change the harmonic amplitude for the fundamental from $1$ to $\alpha + (1-\alpha)\frac{4}{\pi}$ .

This is where the intuition comes handy. We know $<x^2(t)>$ = 1, and the component of the fundamental is just $\frac{1}{2}A_1^2 = \frac{16}{\pi^2}\frac{1}{2}=\frac{8}{\pi^2}$ . Thus, the mean squared value of the square wave without the fundamental is just $1-\frac{8}{\pi^2}$ .

Now, we put back the fundamental with the new amplitude and obtain an expression for $<y^2(t)>$ :

$<y^2(t)>=\left(\alpha + (1-\alpha)\frac{4}{\pi}\right)^2\frac{1}{2}+(1-\alpha)^2(1-\frac{8}{\pi^2})$ .

Setting the above to $0.9^2 = 0.81$ and solving for $\alpha$ we obtain the value $0.287$ .

$y^2(t)=\alpha^2 \sin^2 t + 2 \alpha (1-\alpha) x(t) \sin t + (1-\alpha)^2 x^2(t)$

$\int_0^{2\pi}{y^2}dt= \alpha^2 \int_0^{2\pi} \sin^2 t dt + 2 \alpha (1-\alpha) (\int_0^{\pi} \sin t dt -\int_{\pi}^{2\pi} \sin t dt ) + (1-\alpha)^2 \int_0^{2\pi}1 dt$

$= \pi \alpha^2 + 2 \alpha (1-\alpha) (2+2) +  2\pi(1-\alpha)^2$

$= (3\pi-8) \alpha^2 + (8- 4\pi) \alpha +  2\pi$

This has to be equal to $(0.9)^2 \int_0^{2\pi}{x^2}dt= 0.81\cdot 2\pi=1.62 \pi$

Solving $\alpha =\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ using $a=3\pi-8, b=8- 4\pi, c=0.38\pi$ and selecting the solution that lies within the given range gives $\alpha =\boxed{0.287}16...$