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enter image description here

As a sine wave passes through a circuit, it clips and the waveform in the figure is formed.

How can I find the RMS value of a full wave?

JRE
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Electronx
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    If this is a homework problem, please add the homework tag. – relayman357 Dec 07 '20 at 00:04
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    You could have solved this very easily. – a concerned citizen Dec 07 '20 at 12:59
  • If your teacher is like my teacher it is very difficult to understand. There are people who do not know the meaning of integral, and they are incapable of solving sample questions.The only answer I expect is to find the squares of the areas of the wave (each voltage value generated) with the help of the integral and divide it by the total number of voltages we get (which makes up the value of 2pi). I wouldn't ask if I could figure it out ''very easily''. – Electronx Dec 07 '20 at 14:38
  • If it's indeed a sine wave, and the peak voltage is 15v, the RMS voltage is 15/sqrt(2) = 10.6066 – Hot Licks Dec 07 '20 at 19:26
  • @HotLicks the RMS of the complete sinewave is, the RMS of the sine wave with a bit missing will be lower. – Peter Green Dec 07 '20 at 19:46
  • @PeterGreen - OP said "the RMS value of a full wave". – Hot Licks Dec 07 '20 at 19:58
  • I assumed he meant of a full cycle of the waveform shown in the diagram, not 100% clear phrasing though. – Peter Green Dec 07 '20 at 20:10
  • @HotLicks: I like your way. But the solution is for the sinus part (4 units) the first 3 part are 0. => 15/sqrt(2) / 7 *4 = 6.06. – Bill Dec 07 '20 at 21:39
  • I calculated the RMS value of a full wave in the question with the help of the integral, which is equal to 7.83326 volts. – Electronx Dec 08 '20 at 17:52

3 Answers3

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This might sound a bit strange, but, really:

By calculating the RMS. It's the root mean square! So,

$$\text{RMS}=\sqrt{\frac1{T_{\text{observation}}}\int_{0}^{T_{\text{observation}}} V^2(t)\,\mathrm dt}\,,$$

nothing more, nothing less. It's really that simple!

Marcus Müller
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  • this is the situation I mean. First, I calculate the rms value of pi / 4 wave. then I calculate the rms value of the part in negative alternans. Then how should I do with these two values? – Electronx Dec 06 '20 at 21:55
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    To the OP: The fact that it is clipped doesn't matter. You can't do the RMS calculation of only one part. You need to do the RMS calculation over one whole period. For the portion of the period where the waveform is zero, well, zero squared is also zero. But, when you compute the mean, you need to use the full period. So when summing Vsquare you need to sum over the whole period. And when you divide by Tobservation, you need to divide by one period. – user57037 Dec 06 '20 at 22:12
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    Really good, clear answers here. This example may help as well. – relayman357 Dec 07 '20 at 00:08
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First write an equation for one cycle of your waveform, based on what you can see on the graph (and the assumption this is indeed a sinewave with part clipped out).

$$v = \left\{ \begin{array}{ll} 15\sin \omega t & \text{for }0 <\omega t<a \\ 0 & \text{for }a<\omega t< \pi \\ 15\sin \omega t & \text{for }\pi <\omega t< 2\pi \end{array}\right. $$

Now take the definition of RMS.

$$v_\text{rms}=\sqrt{\frac1{T}\int_{0}^{T} v^2\,\mathrm dt}\,,$$

Where \$T\$ is the time we are calculating the rms over, \$\frac{2\pi}{\omega}\$ in this case.

\$\omega\$ is just an annoying constant that doesn't actually effect the final answer (time-stretching a waveform does not change it's rms), so treat it as 1 to make the algebra simpler.

Then chop-up the time interval and substitute.

$$v_\text{rms}=\sqrt{\frac1{2\pi}\left(\int_{0}^{a} (15\sin t)^2\,\mathrm dt+\int_{a}^{\pi} 0^2\,\mathrm dt+\int_{\pi}^{2\pi} (15\sin t)^2\,\mathrm dt\right)}\,,$$

We can simplify this a bit by removing the zero term and combining the other two terms by taking advantage of the fact that sin is periodic, so the integral over the range \$2\pi\$ to \$2\pi+a\$ is the same as over the interval \$0\$ to \$a\$.

$$v_\text{rms}=\sqrt{\frac1{2\pi}\int_{\pi}^{2\pi+a} 15^2\sin^2t\,\mathrm dt}\,,$$

To solve such an integral you make use of the identity.

$$\sin^2x = \frac{1 - \cos(2x)}{2}$$

When you substitute that in, you should get something you can integrate fairly easily with well-known integration rules.

Peter Green
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Here is how to do it digitally. I think it can be done analytically, also, if desired since it is just a sine wave. First, sample the waveform. Ideally, sample it for exactly one period. Then compute the discrete RMS value.

Let's call the series of samples Y(t). Sample at a reasonable rate. Like 100 samples over the full period. The sampling period is denoted with T. So Y(0) = the first sample, Y(T) is the second sample, Y(2T) is the third, etc.

First compute the series Y^2(t) by squaring each sample individually.

Next compute the mean of Y^(t). The mean is just the average of all the squared samples. Note that the mean is not a series. It is just a single number.

Take the square root of the mean. This is the RMS value of the waveform. RMS is taken to be always positive.

user57037
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