Average value derivation (1 Viewer)

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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! :smile:
 
I don't think anything is attached.
 

HallsofIvy

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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 [itex]\left[0, \pi\right][/itex] is
[tex]\frac{\int_0^\pi sin(x) dx}{\pi}= \frac{-cos(\pi)+ cos(0)}{\pi}= \frac{2}{\pi}[/tex].

The average value of sin(x), on the interval [itex]\left[\pi, 2\pi\right][/itex] is [itex]-2/\pi[/itex] and, of course, the average value of sin(x) over a full perion, [itex]\left[0, 2\pi\right][/itex] is 0.
 
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
[tex]\int_a^b f(x)dx[/tex]
How many sample points are there?
[tex]\int_a^b dx[/tex]
Thus the average of a function [tex]f(x)[/tex] over the interval [a, b] is then
[tex]
\frac{\int_a^bf(x)dx}{\int_a^bdx} = \frac{\int_a^bf(x)dx}{b - a}
[/tex]
 
HallsofIvy said:
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.........
 

Attachments

The first thing, although it is not a mistake, is the integration. It is correct, but here is another way to do it.
[itex]
\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*}
[/itex]

The second thing is finding the average power. This is found by dividing the total energy delivered by the time period of operation.
[itex]
\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*}
[/itex]
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
[itex]
\left(\frac{1}{2}Ri_{max}^2\right)\cdot(1 sec)
[/itex]
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.
 
derek e said:
The energy delivered in one second is
[itex]

\left(\frac{1}{2}Ri_{max}^2\right)\cdot(1 sec)

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

So for one second it should be [itex]\frac{Ri _{max} ^2}{T}[/itex] right? i.e a function of T, rest of it will get eliminated (just as done in the PDF).
 
The energy delivered from 0 seconds to T/2 seconds is the integral of the power from 0 to T/2.
 
:rofl:

Yes I know.
 
Your derivation gives the same thing as in my derivation in that PDF.
 
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.
 
You are mixing time units, then.
 
Last edited:
:confused:

Can you please tell me where am I going wrong?

Appreciate the help. :smile:
 
"... so the average energy delivered per unit time will be - [blah]"
This is correct. This is also the average power. Solve for [itex]i_{avg}[/itex].
 
Ok suppose I take [itex]

\left(\frac{1}{2}Ri_{max}^2\right)\cdot(1 sec)

[/itex] as correct, then it will return [itex]i_{max} \sqrt \frac{1}{2}[/itex]...that is wrong.
 
Ok suppose I take [itex]

\left(\frac{1}{2}Ri_{max}^2\right)\cdot(1 sec)

[/itex] as correct, then it will return [itex]i_{max} \sqrt \frac{1}{2}[/itex]...that is wrong.
It would help if you stated the problem exactly. Some useful relations are
[itex]
\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*}
[/itex]
I assume you are looking for [itex]i_{rms}[/itex] since it looks like the last thing written in the pdf. If so, then
[itex]i_{rms} = \frac{i_{max}}{\sqrt{2}}[/itex]
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?
 
[itex]P_{avg} &= i_{rms}^2R[/itex]
:confused::surprised:cry:


[itex]
P_{avg} = v_{avg} i_{avg}
[/itex]
[itex]
P_{rms} = v_{rms} i_{rms}
[/itex]

If

[itex]
P_{avg} &= i_{rms}^2R
[/itex]

Then

[itex]v_{avg} i_{avg} = v_{rms} i_{rms}[/itex]

Is this true???
 
why use a half period?
Average value for the whole time period will be 0.
 
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.
 
Last edited:
Oh.

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


I'll trying without squaring it.

i.e using v*i formula.
 
Last edited:

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