How is the average value for a sinewave derived?

[dE_logics]

There's an attachment in this post, that show my attempts.


So how is it actually derived for a sinewave? And if possible please debug the PDF attached.

I'm sort getting the reasons for the derivation given in my book.


Thanks!

 

[derek e]

I don't think anything is attached.

 

[HallsofIvy]

I don't see why this is in the physics section rather than the mathematics section- its purely a mathematics question.

If f(x) is any integrable function defined on the interval [a, b], then its "average value" is its integral from a to b divided by the length of the interval, b- a.

In particular, the average value of sin(x), on the interval $\left[0, \pi\right]$ is
$$\frac{\int_0^\pi sin(x) dx}{\pi}= \frac{-cos(\pi)+ cos(0)}{\pi}= \frac{2}{\pi}$$.

The average value of sin(x), on the interval $\left[\pi, 2\pi\right]$ is $-2/\pi$ and, of course, the average value of sin(x) over a full perion, $\left[0, 2\pi\right]$ is 0.

 

[derek e]

You find the average of an integrable function over an interval the same you find other averages, you sum and then divide. The function f tells us the weight of a sample point, x. This gives a total value of
$$\int_a^b f(x)dx$$
How many sample points are there?
$$\int_a^b dx$$
Thus the average of a function $$f(x)$$ over the interval [a, b] is then
$$\frac{\int_a^bf(x)dx}{\int_a^bdx} = \frac{\int_a^bf(x)dx}{b - a}$$

 

[dE_logics]

I don't see why this is in the physics section rather than the mathematics section- its purely a mathematics question.

It would have been a mathematics question if it had an ambiguity towards calculus; here calculus is used (without any problems in its application) and the real problem is with the derivation.



Why have you used the electrical angle?...I mean electrical angle looses out on information about time and the average value should be a function of time, ok not necessarily for v and i, but for P, it should return the wrong values, since P is itself a function of t. What will the mechanical angle give is average energy delivered per cycle, which will be independent of time.

I think I've done a very bad job conveying the above message.



Sorry for the missing attachment, I'll attach it now...

 

[derek e]

The first thing, although it is not a mistake, is the integration. It is correct, but here is another way to do it.
$\begin{equation*} \begin{split} \int_0^{\frac{T}{2}}P(t)dt &= Ri_{max}^2\int_0^{\frac{T}{2}}\sin^2(\omega t)dt \\ &= Ri_{max}^2\left[\frac{T}{2}\left[\frac{1}{T}\int_0^{T}\sin^2(\omega t)dt\right]\right] \\ &= Ri_{max}^2\left[\frac{T}{2}\left[\frac{1}{2}\right]\right] \\ &= \frac{1}{4}Ri_{max}^2T \\ \end{split} \end{equation*}$

The second thing is finding the average power. This is found by dividing the total energy delivered by the time period of operation.
$\begin{equation*}\begin{split} P_{avg} &= \left(\frac{1}{4}Ri_{max}^2T\right) / \left(\frac{T}{2}\right) \\ &= \frac{1}{2}Ri_{max}^2 \end{split}\end{equation*}$
Notice that the units are correct. When in doubt, the fist thing to check is your units. The interpretation of this equation is the average energy delivered per unit time. The energy delivered in one second is
$\left(\frac{1}{2}Ri_{max}^2\right)\cdot(1 sec)$
It's a little hard for me to understand exactly where you are trying to go with these things, but that might be enough to send you in the right direction.

 

[dE_logics]

The energy delivered in one second is
$\left(\frac{1}{2}Ri_{max}^2\right)\cdot(1 sec)$

But $(\frac{1}{2}Ri_{max}^2)$ is the energy delivered in T/2 seconds right?

So for one second it should be $\frac{Ri _{max} ^2}{T}$ right? i.e a function of T, rest of it will get eliminated (just as done in the PDF).

 

[derek e]

The energy delivered from 0 seconds to T/2 seconds is the integral of the power from 0 to T/2.

 

[dE_logics]

:rofl:

Yes I know.

 

[dE_logics]

Your derivation gives the same thing as in my derivation in that PDF.

 

[dE_logics]

I think I've realized the issue here, but under the operation of division of the integrated function, we will again get time but not in terms of SI units, it will be energy delivery per T/2 seconds; so it will mean the same thing even after realization of this point...I think.

 

[derek e]

You are mixing time units, then.

 

[dE_logics]

Can you please tell me where am I going wrong?

Appreciate the help.

 

[derek e]

"... so the average energy delivered per unit time will be - [blah]"
This is correct. This is also the average power. Solve for $i_{avg}$.

 

[dE_logics]

Ok suppose I take $\left(\frac{1}{2}Ri_{max}^2\right)\cdot(1 sec)$ as correct, then it will return $i_{max} \sqrt \frac{1}{2}$...that is wrong.

 

[derek e]

Ok suppose I take $\left(\frac{1}{2}Ri_{max}^2\right)\cdot(1 sec)$ as correct, then it will return $i_{max} \sqrt \frac{1}{2}$...that is wrong.

It would help if you stated the problem exactly. Some useful relations are
$\begin{equation*}\begin{split} P_{avg} &= i_{rms}^2R \\ i_{rms} &= \sqrt{\frac{1}{T}\int_0^Ti^2(t)dt} \\ \end{split}\end{equation*}$
I assume you are looking for $i_{rms}$ since it looks like the last thing written in the pdf. If so, then
$i_{rms} = \frac{i_{max}}{\sqrt{2}}$
for a sinusoid. If not, then, again, it is hard for me to see exactly where you are trying to go with these things. For example, why use a half period?

 

[dE_logics]

$P_{avg} &= i_{rms}^2R$

$P_{avg} = v_{avg} i_{avg}$
$P_{rms} = v_{rms} i_{rms}$

If

$P_{avg} &= i_{rms}^2R$

Then

$v_{avg} i_{avg} = v_{rms} i_{rms}$

Is this true?

 

[dE_logics]

why use a half period?

Average value for the whole time period will be 0.

 

[derek e]

The convention for the average power is to take rms values in circuit applications.
Also, are you talking about the integral of sin^2 being 0 over the period? :tongue:.
sin^2 is always positive, right? All the integration is doing is adding up positive numbers. Things can't cancel to give you zero. You must be thinking of integrating sin.

 

[dE_logics]

Oh.

So actually taking sin^2 means taking the RMS value.I'll trying without squaring it.

i.e using v*i formula.

 

[dE_logics]

Still more chaos :uhh:

Attached the PDF.

 

[derek e]

Oh.

So actually taking sin^2 means taking the RMS value.


I'll trying without squaring it.

i.e using v*i formula.

I don't think you will get very far since V=iR. Taking the rms (root-mean-square) of a value means you perform the operations in the name in reverse order. So, you take a value, square it, find the mean of the squared value and then take the square root of the result. The final value is the rms.

 

[dE_logics]

Yes, I realized that :tongue2: thought the results are a bit different...good to see I got that figure atleast.

So what should I do?

In fact by this it has been proved that the average value = RMS value.


One more problem!

 

[dE_logics]

Ok...so I have no other option but to leave this derivation.No problem...doesn't cost that much...7-8 mks at most.

 

[derek e]

I think you need to slow down, reread what we have written, reread the problem statement and carefully think through it.