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Well, I've to find the RMS (root mean square) over all time of the following function:

enter image description here

In the picture I plotted a case where I used values for components in the full bridge rectifier with capacitor filter but I want to solve the general case.

From $0\le t\le\frac{1}{4f}$ the function that describes the waveform is:

$$\left|u\sin\left(2\pi ft\right)\right|\tag1$$

When the sine wave is at its top, the function turn into a exponential function (because the capacitor discharges):

$$u\exp\left(-\frac{t}{\tau}\right)\tag2$$

The point where the sine wave returns is:

$$\left|u\sin\left(2\pi ft\right)\right|=u\exp\left(-\frac{t}{\tau}\right)\space\Longleftrightarrow\space t:=T=\dots\tag3$$

Where $\frac{1}{2f}<T<\frac{3}{4f}$

And then the proces starts over again: the exponential function comes back when the sine wave is at its top.

Now, I need to find the RMS over all time so I wrote:

$$\lim_{\text{n}\to\infty}\sqrt{\frac{1}{\text{n}}\int_0^\text{n}\left(\text{y}\left(t\right)\right)^2\space\text{d}t}\tag4$$

And in order to find $\int_0^\text{n}\left(\text{y}\left(t\right)\right)^2\space\text{d}t$ I wrote:

$$\int_0^\text{n}\left(\text{y}\left(t\right)\right)^2\space\text{d}t\stackrel{?}{=}\int_0^\frac{1}{4f}\left(\left|u\sin\left(2\pi ft\right)\right|\right)^2\space\text{d}t+\text{n}\cdot\int_0^T\left(u\exp\left(-\frac{t}{\tau}\right)\right)^2\space\text{d}t+\text{n}\cdot\int_T^\frac{3}{4f}\left(\left|u\sin\left(2\pi ft\right)\right|\right)^2\space\text{d}t\tag5$$

The question is, am I right and how do I proceed?

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  • $\begingroup$ what is the RMS? $\endgroup$ – uniquesolution Feb 15 '18 at 14:33
  • $\begingroup$ @uniquesolution Root Mean Square, see my edit. $\endgroup$ – Jan Feb 15 '18 at 14:34
  • $\begingroup$ It should suffice to calculate the RMS over a single period (the transient phenomenon at the beginning does not matter), which should be write-able as a sum of two integrals: one where the integrand is the square of the sine, one where the integrand is the square of the exponential. $\endgroup$ – Wouter Feb 15 '18 at 14:36
  • $\begingroup$ what is the function $y$? $\endgroup$ – uniquesolution Feb 15 '18 at 14:36
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    $\begingroup$ @jan Yes you can. If you truly take the $n\rightarrow\infty$ limit, the contribution of the transient phenomenon is zero. $\endgroup$ – Wouter Feb 15 '18 at 14:39
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You are calculating an integral $$\lim_{n\rightarrow\infty}\sqrt{\frac{1}{n}\int_0^n y(t)^2 dt}$$ But your $y(t)$ is periodic aside from a transient phenomenon $$y(t)=\left\{\begin{matrix}y_{\text{transient}}(t) & t <T_0 \\ y_{\text{periodic}}(t-T_0) & T_0<t<T_0+P \\ y_{\text{periodic}}(t-(T_0+P)) & T_0+P<t<T_0+2P \\ \vdots\end{matrix}\right.$$ So the original integral becomes $$\lim_{n\rightarrow\infty}\sqrt{\frac{1}{n}\int_0^n y(t)^2 dt}=\lim_{n\rightarrow\infty}\sqrt{\underbrace{\frac{1}{n}\int_0^{T_0} y_{\text{transient}}(t)^2 dt}_{\rightarrow 0\text{ as }n\rightarrow\infty} + \frac{1}{n}\int_{T_0}^{n} y(t)^2 dt}$$ what remains is periodic.

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  • $\begingroup$ And $T_0$ is my equation $\left(3\right)$? $\endgroup$ – Jan Feb 15 '18 at 15:22
  • $\begingroup$ Yes. (but note that your eq. (2) probably needs a temporal translation, if you want it to equal the sine at its peak) $\endgroup$ – Wouter Feb 15 '18 at 15:28
  • $\begingroup$ I do not understand what you mean by that?! $\endgroup$ – Jan Feb 15 '18 at 15:35
  • $\begingroup$ You don't want $u\exp(-t/\tau)$ but $u\exp(-(t-t_{peak})/\tau)$ where $t_{peak}$ is the time where your sine peaks. otherwise your function is discontinuous. $\endgroup$ – Wouter Feb 15 '18 at 15:45
  • $\begingroup$ I see, but how can the last existing integral be rewritten using this? Or must I ask is my equation $\left(5\right)$ right? $\endgroup$ – Jan Feb 15 '18 at 15:59

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