# Homework Help: Exam Revision: Stuck on RMS derivation from voltage waveform diagram

1. Jun 15, 2012

### jars121

1. The problem statement, all variables and given/known data

I've spent the last hour trying to get this, and it's starting to drive me crazy! I'm not sure how to approach the waveform, as there are two forms? I'm guessing that the period in this case is 4 seconds, and they've included a second wave to reiterate that?

2. Relevant equations

I started out using the general RMS equation:

V2RMS = √((1/T)*∫V2dt)

3. The attempt at a solution
I split the voltage into the 3 components ie. 0<t<1, 1<t<3 and 3<t<4.
I integrated each of these sections, and then calculated the RMS by adding each component, squared, within a square root.

ie. VRMS, total = √(VRMS, A)2 + VRMS, B)2 +VRMS, C)2)

I ended up getting ~12, and I know that the answer is 3.3325V.

Can anyone point me in the right direction, as well as give some general advice on approaching these type questions?

Cheers

Last edited: Jun 15, 2012
2. Jun 15, 2012

### cepheid

Staff Emeritus
Welcome to PF! I can't see the image in your post, but if I right click it and select "Open in new window," then it works.

By definition, a waveform is a periodically repeating signal. If you just include one of those bumps, you don't have a wave, you just have a pulse! In a wave, those pulses repeat at regular intervals...forever. That's what periodic means.

No, the period is not 4 seconds. The period is the time required for one full cycle of the oscillation. In other words, it's the time required for you to get back to the exact same feature on the waveform that you started with.

For example: At t = 0, the voltage is 0, and is just beginning to rise up to its maximum value. At what time, t, does this exact same situation occur again?

Alternatively, at t = 2, the voltage is halfway through its plateau of +5 V.How many seconds later does this exact situation occur again?

At t = 4, the voltage is +5 V and is just beginning to fall back down to its minimum value. How many seconds later does this exact same thing occur again?

The answer to these three questions is exactly the same, because in all three cases, I'm asking you to tell me the time interval between two repeated features of the signal, which by definition is the period.

Once you know the period, you can integrate the waveform over this period. You're right that you'll have to do it piecewise and break up the integration interval into smaller sub-intervals. The reason is that the functional form of v(t) changes with time. Sometimes it's constant and high, other times it's constant and low. Still at other times, it's a linear function with a positive slope, and at other times it's a linear function with a negative slope.

Last edited: Jun 15, 2012
3. Jun 15, 2012

### jars121

Thanks for that mate! I've done a couple of other similar problems, and realised my mistake. The period is obviously 6, and I was doing something strange with the integration additions as well. Cheers!

4. Jun 16, 2012

### NewtonianAlch

I guess you're studying for ELEC2134 then, haha. Join the UNSW electrical group, this problem was posted yesterday.

5. Jun 16, 2012

### jars121

I may be

I'll head on over there now (if I can find it) and see what's going on. Cheers!

6. Jun 16, 2012

### jars121

Ok, I'm struggling. Can you post a link to the group? Cheers

7. Jun 16, 2012