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In circuit analysis it is stated that the RMS (Root Mean Square) value of a waveform which consists of a sum of sinusoids of different frequencies, is equal to the square root of the sum of the squares of the RMS values of each sinusoid. Therefore, if

$$i(t)=I_0+I_1\cos(w_1t+\theta_1)+I_2\cos(w_2t+\theta_2)+....+I_n\cos(w_nt+\theta_n)$$

then the RMS value of i(t) is equal to

$$I_{RMS}=\sqrt{I_{0\;RMS}^2+I_{1\;RMS}^2+....+I_{n\;RMS}^2}$$

However, a proof is not given and the RMS formula is:

$$I_{RMS}=\sqrt{\frac{1}{T}\int_{-T/2}^{T/2}i^2(t)dt}$$

According to Period of sum of sinusoids if frequency ratios are irrational the period may become infinite. Here is what I have done considering this fact (just for two terms):

$$I_{RMS}^2=\lim_{T \to \infty}\frac{1}{T}\int_{-T/2}^{T/2}(I_1\cos(w_1t+\theta_1)+I_2\cos(w_2t+\theta_2))^2dt$$

$$I_{RMS}^2=\lim_{T \to \infty}\left( \frac{I_1^2}{2}\cos(2\theta_1)\frac{\sin(w_1T)}{w_1T}+\frac{I_1^2}{2}+\frac{I_2^2}{2}\cos(2\theta_2)\frac{\sin(w_2T)}{w_2T}+\frac{I_2^2}{2} \right)=\frac{I_1^2}{2}+\frac{I_2^2}{2}$$

Since $I_{1\;RMS}=I_1/\sqrt2$, $I_{2\;RMS}=I_2/\sqrt2$.... the result is as expected:

$$I_{RMS}^2=I_{1\;RMS}^2+I_{2\;RMS}^2$$

However, my solution is based on the assumption that the total period is infinite. How can we show above statement is true when all the frequencies are integer multiples of each other? In this case total period should be finite.

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1 Answers1

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In your setup, assume that there is a period $T_0$ such as $$ \omega_i = \frac {2\pi}{p_i T_0} $$for distincts integers $p_1, \dots , p_n$.

The square of the signal is then $$ I_0^2 + I_1^2\cos^2(\omega_1t + \theta_1) + \dots + I_n^2\cos^2(\omega_nt + \theta_n) + \text{ cross products} $$

In this case a period of the signal is $T = p_1\dots p_nT_0$.

  1. Let us take care of the cross products, which do not contribute to the total: $$ \int_0^T \cos(\omega_it + \theta_i) \cos(\omega_jt + \theta_j) dt \\= \frac12 \int_0^T \left[ \cos(\omega_it + \theta_i + \omega_jt + \theta_j) + \cos(\omega_it + \theta_i - \omega_jt - \theta_j) \right] dt \\= \frac12 \left[ \frac{\sin(\omega_it + \theta_i + \omega_jt + \theta_j)} {\omega_i + \omega_j} + \frac{\sin(\omega_it + \theta_i - \omega_jt - \theta_j)} {\omega_i - \omega_j} \right]_0^T = 0 $$ here the important thing is that $\omega_i\neq \omega_i$ when $i\neq j$.

  2. Now for the squares, the computation is the same: $$\int_0^T I_i^2\cos^2(\omega_it + \theta_i) dt =\frac12 I_i^2 \int_0^T \left[ \cos(2\omega_it + 2\theta_i ) + \cos( 0) \right] dt = \frac12 I_i^2 T $$ as $\cos(0) = 1$.

Now divide by $T$ and take the square root: you find that $$I_{RMS}^2= I_0^2 + \sum_{i=1}^n \frac 12 I_i^2$$

Chris Chiasson
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mookid
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  • Thank you for your answer but I can't get this equation $$\int_0^T \left[ \cos(2\omega_it + 2\theta_i ) + \cos( 0) \right] dt = \frac12 I_i^2 T$$ Could you please explain it explicitly? –  Oct 20 '14 at 07:06
  • Sorry I get it now.

    $$\int_0^T \left[ \cos(2\omega_it + 2\theta_i ) + \cos( 0) \right] dt = \frac{\sin(2w_iT)\cos(2\theta_i)+\sin(2\theta_i)\cos(2w_iT)}{2w_i}+T-\frac{\sin(2\theta_i)}{2w_i}$$

    Since $2w_iT=\frac{4\pi}{p_iT_0}T$ and $\frac{T}{p_iT_0}$ is an integer $k_i$ the result becomes:

    $$=\frac{\sin(2\theta_i)}{2w_i}+T-\frac{\sin(2\theta_i)}{2w_i}=T$$

    Thanks a lot.

     –  Oct 20 '14 at 07:45