# Average & rms for waveforms

1. Sep 21, 2014

### Zondrina

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

1. Find the average and RMS values of the following waveform:

2. Show that the average and RMS values of the following sawtooth are independent of the position of the peak and are given by $0.5$ and $0.577$ of the peak value respectively.

2. Relevant equations

$f_{avg} = \frac{1}{T} \int_0^T f(t) dt$
$f_{rms} = \sqrt{\frac{1}{T} \int_0^T f^2(t) dt}$

3. The attempt at a solution

1. Looking at the function, I see it is a piecewise function with five different pieces in one period. So to compute the average/rms value would require five integrals.

Looking more closely though, I see the function is symmetric about the x-axis over one period. This implies the average value is zero right away, $v_{avg} = 0$. Is there similar logic I can apply to find the RMS value without having to square and integrate five times? For reference I got $v_{rms} = V$.

2. This is the same question as the first really. I'm guessing all the question wants is for me to compute $f_{avg}$ and $f_{rms}$? What do they mean by independent of the position of the peak exactly? Am I over-thinking this one beyond the computations?

2. Sep 21, 2014

### Zondrina

I believe I have some reasonable thoughts about my first question, but you can't really edit your posts anymore, so I apologize for this double.

If I square that $v(t)$ they've given me, the amplitude wont change. Geometrically it would flip the troughs into crests, and all the values will be positive, so that in turn the square root will be properly defined every time.

What I'm looking for is the area over one period of $v^2(t)$, which is two of those positive peaks (or simply twice the area of the first). Then I want to average it over the entire period and take the root. Those operations are harder to interpret geometrically, so I don't think there's a quick shortcut to finding the RMS value. The average value can be simplified through the use of geometry in some circumstances though (like this one).

For the second question, the computations were obvious, but I think I have some intuition now about why the position of the peak doesn't matter. If I moved the peak to point $Y = X + \Delta x$, where $X \leq Y \leq T$, the waveform will still have the same peak, just at a different time. The area under the curve itself wont change because the equations for the lines that define the waveform will change to accommodate it. Hence the average and rms values will be the same.

Does this sound reasonable?

3. Sep 22, 2014

### milesyoung

You only have to consider a half-cycle to determine the RMS value of the waveform - sign doesn't matter. It would be a good idea to find general expressions for the area under the square of those couple of segments that appear in your waveform, i.e. where it's increasing/decreasing linearly or is constant.

No, but you only need those two expressions to determine the RMS value and it'll solve your second assignment as well.

That would be true if your waveform was square instead of trapezoidal.

That argument works for the average value, since the area of a triangle is given by $A = \frac{1}{2}b h$, where $b$ and $h$ is its base and height, respectively, and moving the peak changes neither.

It doesn't hold for the RMS value, however, since, as a counterexample, two waveforms can have the same area under their curves but different RMS values.

4. Sep 24, 2014

### Zondrina

That was the listed answer. I got $0.83 V$ when doing it myself. I decided to compute:

$V_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} v^2(t) dt} = \sqrt{\frac{2}{T} \int_{0}^{\frac{T}{2}} v^2(t) dt}$

since only three integrals would be required instead of all five.

I have $0.816 V$. The RMS value of a trapezoidal waveform is something you can easily look up in a table if you want to check your result.