Electrical – Peak and RMS voltage for a sine wave

rmssinesquarevoltagewaveform

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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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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Electronic – Sine wave to square wave – Schmitt trigger

I would like to suggest that the comparator circuit shown by @hwengmgr could use some small improvements over what was posted.

Improvements include:

Bigger capacitor value to reduce filter effect of the capacitor at frequency.
Larger value resistors in dividers to reduce current load on source supply.
Added hysteresis at the comparator to make the circuit less sensitive to noise that may be on the input waveform.
Changed pullup resistor on output of comparator to a more sane value for 1mA load instead of 33mA.

enter image description here

This was simulated with LT-Spice. I used a model for a comparator that was built into the package library. You can substitute a comparator of choice as long as it is specified for operation at a Vdd of 3.3V.

The addition of the hysteresis feature necessitated the swapping of the '+' and '-' inputs of the comparator which causes an inversion of the output waveform. If this is a problem it can be easily flipped with a logic element. If you select a dual comparator the other half of the package could be used to invert the signal.

Electrical – How to ‘average’ a sine wave of RMS values

First of all, you should rephrase your question like: How to get an average value from measurings. The term RMS makes just confusion as it makes no difference if you want to get an average of peak, rms,...whtever values. Simplest method is a moving avearage filter, you need a circular buffer FIFO where you fill the measurings. At each new input, you discard the oldest and compute the new average recursively. https://en.wikipedia.org/wiki/Moving_average

Average = Average + NewestValue/n - OldestValue/n; purge OldestValue from FIFO and insert NewestValue in.

NewValue: xyz;
Average: xyz or float or double;
FIFO: array[0..63] of xyz; //curcular buffer of 64 elements
Pointer: int;

//at init:
FIFO[from 0 to 63]=NewValue;
Average = NewValue;
Pointer = 0;

//at each measuring
Average = Average + NewValue/64 - FIFO[Pointer]/64;
FIFO[Pointer] = NewValue; // purge last and overwrite with new
Pointer= (Pointer + 1) AND 63; //increment pointer with roll over, circular buffer

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Electronic – Sine wave to square wave – Schmitt trigger

Electrical – How to ‘average’ a sine wave of RMS values

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