# Calculating RMS of Simple Waveform

A problem in my textbook gives this task:

A solution is given as:

I tried to work this problem using a different method. I simply calculated the areas of the triangles as if the height of the triangles was A^2. This gives me this result:

I was confused by why these answers should be different. I rearranged the first integral from the given solution into 1/2 * base * height form to see what triangle they are using. It's this:

In other words, if they calculated it with triangles, the base of each triangle is squared as well. Why should the base of the triangle be different? Shouldn't the value of the function simply be squared for each point in time?

• Think of a triangle whee the height coordinate is squared - it isn't just taller, the straight lines become quadratic - parabolas. Commented May 13, 2020 at 5:46
• Hint: what is the result of $\int{x}dx$? Commented May 13, 2020 at 5:55
• Shouldn't the scalers of partial integrals (I mean 1/T) be 1/(2T/3) and 1/(T/3), respectively? Commented May 13, 2020 at 6:48

I simply calculated the areas of the triangles as if the height of the triangles was A^2.

That was an error - the triangles when squared are no longer triangles and the rest of your math is thrown-off because of this. A DC biased triangle wave (blue) when squared (red) looks like this: -

As for the derivation of the RMS of a triangle wave consider this: -

The equation for one section of triangle from time 0 to T can be written:

v(t) = $$\A\cdot (x/T)\$$, if you square the instantaneous value of v you get

$$\v^2(t) = (A^2/T^2)x^2\$$

That is what you need to integrate.

I hope this will be enough demonstration for those who are looking for that, in general all triangle/sawtooth waveforms gives the same value when it comes to the average (vpk/2) and rms (vpk/√ 3)