High School What is the RMS value in this context?

[Frigus]

In this text they are saying that rms value is square root of average value of i² but i am not able to relate this thing to equations they have given.
It feels to me that the equations tell something different from the text.
From text I interpret that if we add each and every value of i² from 0 to T and then divide it by no. of values and then take it's square root then value we will get will be rms value but from equations i interpret that "if we divide the total area of graph of i² and t from 0 to T with T then the take square root of that value then that will be the rms value".
I can't relate how these two things are equivalent.
Please tell me if my question is not clear.

 

[A.T.]

From text I interpret that if we add each and every value of i² from 0 to T and then divide it by no. of values and then take it's square root then value we will get will be rms value but from equations i interpret that "if we divide the total area of graph of i² and t from 0 to T with T then the take square root of that value then that will be the rms value".
I can't relate how these two things are equivalent.

 

[anorlunda]

From text I interpret that if we add each and every value of i² from 0 to T

the total area of graph of i² and t from 0 to T

Those two things are the same. That's why it is equivalent.

To find the total area under a graph from point A to point B using finite steps, simply divide the graph into rectangular bars. The area of each bar is width times height. The sum of those areas for all bars is the area under the curve.

Do you have a different way to calculate the area of graph from 0 to T?

 

[etotheipi]

Basically, if you want to find the average of some function $f(t)$ in the interval $[t_a, t_b]$, then you can split the interval into $n$ regions of width $\Delta t_k = \frac{t_b - t_a}{n} = \frac{T}{n}$, and find an expression for the mean using the discrete formula$$\bar{f} = \frac{\sum_{k=1}^{n} f({t_k}^*)}{n}$$where ${t_k}^* \in [t_{k-1}, t_k]$, and $t_a = t_0$, $t_b = t_n$. Substituting for $n = \frac{t_b - t_a}{\Delta t_k}$,$$\bar{f} = \frac{\sum_{k=1}^{n} f({t_k}^*) \Delta t_k}{t_b - t_a} = \frac{1}{T} \sum_{k=1}^{n} f({t_k}^*) \Delta t_k$$Take limit $n \rightarrow \infty$, and the Riemann sum approaches a Riemann integral,$$\bar{f} = \frac{1}{T} \int_{t_a}^{t_b} f(t) dt$$That's how to obtain the formula for the mean value of a function on a certain interval. The RMS value of a function, like $i(t)$, is defined as "the square root of [the mean of [the function squared]]", i.e. the square root of the mean of $i^2(t)$, in the interval $[t_a, t_b]$. The $\sqrt{\langle i^2 \rangle}$ often a more useful quantity than the mean of the function when there is positive/negative symmetry, since $\langle i \rangle$ is just going to be zero over a whole time period for a sinusoidal current!

 

[DaveE]

I think this is the first time in my life I've seen someone express T as $\int_0^T \,dt$

edit: OK, I'm sure I saw this in my first calculus class. The first and last time I did that integral.

 

[hutchphd]

The ⟨i2⟩ often a more useful quantity than the mean of the function

Well in this case it is a little stronger than this usual rationale for distributions. Explicitly the power is the time average of $i^2$ and that is exactly what you want.

 

[etotheipi]

Well in this case it is a little stronger than this usual rationale for distributions. Explicitly the power is the time average of $i^2$ and that is exactly what you want.

I didn't see a mention of power above, but yes you do have e.g. for a resistor $P = Ri^2 \implies \langle P \rangle = R\langle i^2 \rangle$

 

[Frigus]

Those two things are the same. That's why it is equivalent.

To find the total area under a graph from point A to point B using finite steps, simply divide the graph into rectangular bars. The area of each bar is width times height. The sum of those areas for all bars is the area under the curve.

Do you have a different way to calculate the area of graph from 0 to T?

I tried to do this with some concrete values but i am getting different results.
If we suppose that speed was
2m/s from t=0s to 2s
3m/s from t=2s to 6s
And take rms by squaring each term and dividing by number of terms then value i got is $\frac {13}{2}$ but if i find area of speed² time graph and then divide it by time and then take it's square root then value I get is $\frac {44}{6}$.

 

[Ibix]

I tried to do this with some concrete values but i am getting different results.
If we suppose that speed was
2m/s from t=0s to 2s
3m/s from t=2s to 6s
And take rms by squaring each term and dividing by number of terms then value i got is $\frac {13}{2}$ but if i find area of speed² time graph and then divide it by time and then take it's square root then value I get is $\frac {44}{6}$.

I take it you get 13/2 m/s by squaring the speed in the first two seconds, squaring the speed in the last four seconds, and taking the mean? (Side note: this is the mean square velocity, not the root mean square). If you use this method to calculate the mean velocity (just the mean, not RMS) you get (2+3)/2=2.5 m/s. Is this a correct approach? If not, what would you do differently and how could you apply the same reasoning to the RMS calculation?

 

[Frigus]

I take it you get 13/2 m/s by squaring the speed in the first two seconds, squaring the speed in the last four seconds, and taking the mean? (Side note: this is the mean square velocity, not the root mean square). If you use this method to calculate the mean velocity (just the mean, not RMS) you get (2+3)/2=2.5 m/s. Is this a correct approach? If not, what would you do differently and how could you apply the same reasoning to the RMS calculation?

If mean value by definition means $\frac{sum-of-terms}{no.-Of -terms}$ then by no doubt i will follow it But if i try to find average value by this formula :- $\frac {distance- traveled}{time- taken}$ then i get different value(which is $\frac {8}{3}$),
Due to this i think that dividing area by time is different from dividing sum of terms by no. Of terms.

 

[Ibix]

If mean value by definition means sum−of−termsno.−Of−terms then by no doubt i will follow it But if i try to find average value by this formula :- distance−traveledtime−taken then i get different value(which is 83),
Due to this i think that dividing area by time is different from dividing sum of terms by no. Of terms.

The problem is that the bars in the chart you are summing aren't the same width. By just adding the heights (or heights squared) and dividing by the number, you are implicitly assuming that they are the same width. Read @anorlunda's post #3 again - he tells you how to deal with it.

 

[Frigus]

Thanks to all,
I got it why these are equal.