I have this problem. I am trying to convert my measurement resuls that I got from my analyzer from RMS units to pk-pk units.The signal is modulated with square signal. I already have a factor to multiply my result by so I get the right answer. But the problem is, that I do not understand how this factor was calculated.

I know this: $$U_{true}=U_{rms} k$$

where $k= \frac{\pi\sqrt2}2$.

Everywhere I look it says that pk-pk value of square signal is equal to $U_{rms}$. But it doesn't work in this case. Could anybody explain me how to derive this coefficient?

So the whole thing looks like this:

I have an a system consisting of analyzer, two controllers and a detector which is mounted on the controllers. Detector is being moved over the XY area and is scanning the cross-section of my lightray which is coming from a source. A signal from my detector goes straight to my analyzer which saves the RMS value of the signal into an array for later use. Now, after I finished all my scanning, I want to work with my data, and first thing I was told to do was to convert all my RMS units to peak values and I was given a value of k mentioned above. And I am trying to understand how was this value calculated, this is the story...

  • $\begingroup$ Hi! Welcome. We can't answer your question, because, well, you forgot to actually ask a question! (please help me is not a question) $\endgroup$ – Marcus Müller Jan 30 '17 at 18:20
  • $\begingroup$ Is $U_{true}$ your pk-to-pk voltage? (I have some issues with calling it "truer" than others, but that really doesn't matter here). It would make it a bit easier for us to answer if you explained what "the signal" is, and what you mean with "modulated with square signal". I'd really love to see some "my signal is $x(t)=\ldots$" in your question! $\endgroup$ – Marcus Müller Jan 30 '17 at 18:39
  • $\begingroup$ (you need to put the $ around your formula, not just in front :) ) Ah, cool, could you edit your question and say that (not everyone wants to read the comments)? And also, add a formula for your signal! $\endgroup$ – Marcus Müller Jan 30 '17 at 18:43
  • $\begingroup$ "A signal from my detector goes straight to my analyzer" <-- of that signal, what is the formula? Without knowing that, there can't be an RMS-to-amplitude conversion, possibly. $\endgroup$ – Marcus Müller Jan 30 '17 at 19:05
  • $\begingroup$ "The signal is modulated with square signal". Do you mean that your signal is a square wave, or is modulated by a square wave. If the former, suggest you say so explicitly. If the later, what type of modulation, and what is the original underlying signal? $\endgroup$ – Daniel K Jan 30 '17 at 23:13

A square wave is a waveform in which the amplitude alternates between $A$ and $-A$ (being $A>0$). The RMS value of a waveform is calculated as

$$f_{rms}=\sqrt{\frac{1}{T}\int\limits_0^T [f(t)]^2 \ \mathrm{d}t}$$

As in the case of a square waveform, either $f(t)=A$ or $f(t)=-A$, then $[f(t)]^2=A^2$ for any value of $t$.

So this leads to

$$f_{rms}=\sqrt{\frac{1}{T}\int\limits_0^T A^2 \ \mathrm{d}t} = \sqrt{\frac{A^2}{T}\int\limits_0^T \ \mathrm{d}t} = \sqrt{\frac{A^2}{T}T} = A$$

And that's why the RMS value of a square wave is the same as the peak value. The coefficient you wrote does not correspond to a square wave.


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