This is my first time posting on any math forum, let alone stackexchange, so I do hope I'm doing everything correct!

Some Background

I'm an engineer, and not a mathematician, although I do enjoy maths which is the reason for this question. In my own time I've been looking at RMS - DC converters within the field of electronics, essentially they take an input of an AC signal, and output a DC voltage. The signal should, ideally, have a mean equal to the RMS of the input signal. For example, a 1Vrms sine wave input should produce a DC voltage of 1V. Due to the nature of the conversion, the output is a signal with a small amount of AC content (ripple) and also a DC offset (due to attenuation, resulting from a low pass filter used to implement the moving average). I've been attempting to derive this error mathematically - which is certainly nothing new. However, it seems i'm struggling with the maths and I hope someone here can help!

The Problem

Determine the error from an RMS-DC converter.

There is a paper available on IEEE Xplore (which I have and can upload here if that is allowed?) that details one method of determining this error. There are two methods of determining RMS, implicit and explicit. The explicit is easier for me to understand so I hope someone can help me with the following:

$V_{RMS}(t) = \sqrt{AVG(V^2_{in}(t))}$

In electronics, the average is done with an operational amplifier, which gives:

$V_{RMS}(t) = \sqrt{-\frac{1}{\tau}\int_{0}^{t}V_{in}(t)^2\delta t}$

The minus sign is due to the practical implementation of the operational amplifier being setup in an inverting operation (input signal is entering the inverting pin (-) on the operational amplifier)

Conventionally, $V_{RMS}()$ is used as the signal, representing voltage. However, the paper uses $e_o(t)$, presumably representing the error, so I will continue with their nomenclature from now on:

$V_{RMS}(t) == e_o(t)$


$V_{in}(t) == e_i(t)$

The paper shows an example circuit:

Example circuit

To which it goes on to say that "it is easy to show that the differential equation for $e_o^2(t)$ can be written as..." (unfortunately, not so easy for me!):

$\tau \frac{\delta e_o^2(t)}{\delta t} + e_o^2(t) = e_i^2(t)$

The paper continues to derive the DC error (through, as I understand, solving the differential equation, and then using some series expansion, and taking the DC term). however, I fail at the first hurdle and can't seem to get the differential equation.

My Workings So Far

From the definition of RMS, I derived what I thought was the differential equation by (I believe it's called) implicit differentiation as follows:

$e_{o}(t) = \sqrt{- \frac{1}{\tau}\int^t_0{e_i^2(t)}\delta t}$

Where $t$ is the time-varying quantity, and $\tau$ is the Resistor-Capacitor time constant, introduced by the R and C in the circuit.

Squaring both sides:

$e^2_{o}(t) = - \frac{1}{\tau}\int^t_0{e_i^2(t)}\delta t$

and then (implicit?) differentiation:

$\frac{\delta}{\delta t} e^2_{o}(t) = - \frac{1}{\tau} \frac{\delta}{\delta t}[\int^t_0{e_i^2(t)}\delta t] $


$\frac{\delta}{\delta t} e^2_{o}(t) = - \frac{1}{\tau} {e_i^2(t)} $

$\tau \frac{\delta e^2_{o}(t)}{\delta t} = - e_i^2(t) $

Which, of course is not the same as the author in the paper. Can anyone see where I have gone wrong?

I understand this is quite a simple question and probably didn't need so much detail, but I thought I would include just in case it's an assumption that's causing the wrong answer.

Thanks in advance!

  • $\begingroup$ Perhaps you could revise your text a bit? You are jumping from $V_{in}$ to $e_i$ and from $V_{RMS}$ to $e_0$. Then there is a minus sign in front of one integral but not in front of the other. Last but not least it's unclear what actually varies when you write $V_{RMS}(t)$, because there are many letters $t$ (and also $\tau$) and it's unclear which one is the variable of integration and which one is the variable of the time-dependent function $V_{RMS}$ (and whether $\tau$ is a constant or not). $\endgroup$ – Christoph Jan 29 at 3:48
  • $\begingroup$ Hi Christoph, thanks for the message. I've revised the question, and I hope to have addressed all your points. Let me know if I've missed anything, coming from an electronics background I forgot to explicitly declare some things (for example, $\tau$ is usually always reserved for time constants). Thank again! $\endgroup$ – JustAnEngineer Jan 29 at 8:24

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