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Problem

I only need help with part C). What I've done for part B) is square the wave, which makes Vpeak->V^2peak and the waveform in the negative region(under the period axis) become positive AND squared(like the waveform from 3T to 5T). I know Vpk=sqrt(2)Vrms, but that, of course, won't give maximum marks. So I equaled Vrms=sqrt(1/8T(total period) times the areas under the waveform graphs)=V/(sqrt(2))=140(sqrt(2))=198V

C) wants us to to find the pk Voltage to get the same 140V rms value, which is again 198V. My question is: How do we find the area under the graph?. We square the wave as before, obtaining the negative cycle as a squared positive one. I see we have 2 flipped parabolas. I think we can solve this using a definite integral, but that's where I get stuck and can't get the 198V answer.

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  • \$\begingroup\$ "I know Vpk=sqrt(2)Vrms, but that, of course, won't give maximum marks" - Why it won't? \$\endgroup\$ – Eugene Sh. May 26 '16 at 14:19
  • \$\begingroup\$ Tbh, I'm only assuming, but I'd like to know the full derivation just to be safe. \$\endgroup\$ – Kevin Thomson May 26 '16 at 14:20
  • \$\begingroup\$ The relationship of sqrt(2) between vpeak and vrms is only valid for a sinus, not for the square wave of b \$\endgroup\$ – Claudio Avi Chami May 26 '16 at 14:22
  • \$\begingroup\$ @ClaudioAviChami the question is about the sine, no? \$\endgroup\$ – Eugene Sh. May 26 '16 at 14:22
  • \$\begingroup\$ No in b it is not a sine. Read what he did for b \$\endgroup\$ – Claudio Avi Chami May 26 '16 at 14:24
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I know Vpk=sqrt(2)Vrms

That is only valid for certain waveforms. It is well-known to be valid for a sinewave, it is not so well known to be true for the waveform of part B in your question.

BTW you seem to have read the question backwards, it's asking what value of \$V_{peak}\$ is needed to make \$V_{rms} = 140 V\$ NOT what the value of \$V_{rms}\$ is for \$V_{peak}=140V\$

My question is: How do we find the area under the graph?. We square the wave as before, obtaining the negative cycle as a squared positive one.

For simplicity let \$t = \frac{1}{8}T\$ . The equation for the instantanious voltage of the sinewave is then.

$$V=V_{sin}\sin(2\pi t)$$

We calculate the RMS by first squaring, then calculating the mean over one cycle (which since it's a continuous function we do by integration) and then finally square rooting.

$$V_{RMS}=\sqrt{\int_0^1(V_{sin}\sin(2\pi t))^2\mathrm{d}t}$$

Then it's a matter of trig rearrangements and integration to solve that.

But honestly I would expect your examiner to expect you to just know that \$V_{RMS} = \frac{V_{sin}}{\sqrt{2}}\$ and not have to re-derive it in the exam.

I see we have 2 flipped parabolas.

Not really, what you find when you do the trig identies is that squaring a sinewave ends up with a sinewave plus a constant.

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The positive and negative half cycles have the same RMS value, so you can just find the RMS value over the half cycle t=0..4T. Since the waveform goes from 0 to 0 once between 0 and 4T, the equation for the sine is:

\$v(t) = V_{SIN} \sin(\pi t/4T)\$

The RMS value is just

\$V_{RMS} = V_{SIN} \sqrt{\frac{1}{4T}\int_{0}^{4T}{\sin^2(\frac{\pi t}{4T})dt} }\$

which evaluates to \$\frac{V_{SIN} }{\sqrt{2}}\$

You can either look up the integral of sin^2(ax)dx or easily derive it from integrating using the trig identity \$\sin^2(x) \equiv (1/2)( 1- \cos(2x))\$.

You should not call the half-wave sine waveform a parabola- it's not, it's a sine wave and can be treated mathematically in closed-form (exactly) as such. As Peter says the integral looks like a constant plus a sinusoidal waveform (at twice the frequency). Which makes sense if you think about what the power looks like as the voltage changes- it's always positive, hence there must be an offset, and touches zero at the zero crossings.

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The RMS (using piecemeal methods) in part B is found by counting the time-slots containing a non-zero voltage, multiplying by \$V_P^2\$, dividing by 8 (total number of time slots) and then taking the square root. So, there are 4 time-slots containing voltage and this means RMS is: -

\$\sqrt{\dfrac{4V_P^2}8{}}\$ = \$\dfrac{V_P}{\sqrt2}\$

For part C this means \$V_{SIN}\$ = \$V_P\$. It shouldn't strike anyone as being unusual if you consider this picture: -

enter image description here

I've superimposed the square wave (now in red) over the sinewave and you could argue that this is what you would get from an ADC sampling the sinewave at 4 times per cycle after all you wouldn't expect an RMS derived value from those samples to be anything else other than \$\dfrac{V_P}{\sqrt2}\$

If you want the RMS of a sinewave (and I don't intend to do the formal math) you square the sine (below in red) thus: -

enter image description here

Then take the mean value of the blue waveform (sine squared). This is clearly 0.5. Then you take the square root i.e. you get 0.7071. So the peak of the sinewave was 1 and the RMS is \$\dfrac{1}{\sqrt2}\$

This is what RMS is about; you take the root of the mean of the square. If you want proof about squaring a sinewave then this should suffice: -

enter image description here

As you can see, the mean value of \$sin^2(x)\$ is 0.5

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  • \$\begingroup\$ Could you express this using definite integrals (for the sine wave)? \$\endgroup\$ – Kevin Thomson May 26 '16 at 15:04
  • \$\begingroup\$ @KevinThomson, taking the integral of a piecewise constant function is just multiplying the value by the duration for each of the constant segments. \$\endgroup\$ – The Photon May 26 '16 at 15:45

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