Deriving RMS value from sinusoidal waveform.

  • Thread starter dE_logics
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  • #1
dE_logics
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I'm having a problem with that integration part.

The average value of i^2 in one cycle = (sum of all i^2 in that period)/(that period).

To derive (sum of all i^2 in that period) we use integration, but that gives the area, how can the area be a substitution for this?...they are different things right?



Ok it might be that I'm reading the wrong source.
 

Answers and Replies

  • #2
Ben Niehoff
Science Advisor
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What is

[tex]\lim_{N \to \infty} \frac{1}{N} \sum_{k=1}^N f(x_k)[/tex]

Hint: try rewriting it as

[tex]\lim_{N \to \infty} \frac{1}{L} \sum_{k=1}^N f(x_k) \frac{L}{N}[/tex]

where L is the total length of the interval in question.
 
  • #3
Bob S
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Remember that the final quantity has to have units of amps, and it has to be independent of the sign of the current. So integrate i(t)^2 dt over the period T, divide the result by T and take the square root. To verify, use i(t) = Izero cos(wt). You should get Izero/sqrt(2).
 
  • #4
dE_logics
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What is

[tex]\lim_{N \to \infty} \frac{1}{N} \sum_{k=1}^N f(x_k)[/tex]

Hint: try rewriting it as

[tex]\lim_{N \to \infty} \frac{1}{L} \sum_{k=1}^N f(x_k) \frac{L}{N}[/tex]

where L is the total length of the interval in question.

:confused: Sorry man...but that went over my head.

The sum should come infinite that way right?
 
  • #5
dE_logics
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Remember that the final quantity has to have units of amps, and it has to be independent of the sign of the current. So integrate i(t)^2 dt over the period T, divide the result by T and take the square root. To verify, use i(t) = Izero cos(wt). You should get Izero/sqrt(2).

Yes that is the standard procedure, but why is integration applied here, the result that integration gives is not desired right?

We need the total length of the curve, and not the area I guess.
 
  • #6
dE_logics
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Can someone help me!?
 
  • #7
Naty1
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RMS values ARE related to areas under curves....but the results are not obvious....

Try reading wikipedia: http://en.wikipedia.org/wiki/Root_mean_square

If that doesn't help, go to the Cartwright website mentioned in the Wikipedia references at the bottom, and look for "...RMS...without Calculus"...that gives a logical approach that shows the steps involved....
 
  • #8
dE_logics
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I did check online resources before posting (including that).


No no...that doesn't help, I'm sorta asking how do you prove that the formula for the RMS values gives such a value that is equivalent to the effective current in DC.
 
  • #9
dE_logics
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Check the attachment...its a PDF, an alternative method to derive the RMS value, but its not correct; though I can see no errors with the methodology.
 
Last edited:
  • #10
jtbell
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A sinusoidal current has a period of T seconds for one complete cycle. During that time, it dissipates the same amount of energy in a resistor as a constant current [itex]I_{rms}[/itex], which dissipates power [itex]P_{rms} = I_{rms}^2 R[/itex]. Therefore

[tex]P_{rms} T = \int_0^T {P(t) dt}[/tex]

[tex](I_{rms}^2 R) T = \int_0^T {I^2(t) R dt}[/tex]

[tex]I_{rms}^2 T = \int_0^T {I_{max}^2 \sin^2 (\omega t) dt}[/tex]

where [itex]\omega = 2 \pi / T[/itex].
 
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  • #11
dE_logics
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That T in the LHS is left over and so proving as a hindrance in the complete solution.
 
  • #12
dE_logics
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And where did that R go? :surprised
 
  • #13
jtbell
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That T in the LHS is left over and so proving as a hindrance in the complete solution.

If you do the integral on the right side correctly, you get a T over there which cancels the T on the left side. Note that [itex]\omega = 2 \pi / T[/itex].

And where did that R go?

Look at both sides of my second step. Hint: R is a constant.
 
  • #14
dE_logics
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Ok...the r problem's gone...thanks.

But the final function after integration is x - sin x cos x...and that x is (2 pi f t)...I replaced t with 1/f....so no T is left over.
 
  • #15
JaWiB
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I may be reading incorrectly but I think there should be a T there since it is the mean value of the current over one period
 
  • #16
jtbell
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But the final function after integration is x - sin x cos x...and that x is (2 pi f t)...I replaced t with 1/f....so no T is left over.

Without seeing the detailed steps that you used to solve the integral, it's hard to say exactly what your problem is. My third equation above should reduce to

[tex] I_{rms}^2 T = \frac{1}{2} I_{max}^2 T [/tex]

in which the T cancels. Maybe you missed a step in the substitution that is needed to solve the integral.
 
  • #17
dE_logics
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[tex]i^2 _{rms} T= i^2 _{max}(\frac{1}{2}(2 \pi f t - cos(2 \pi f t)sin(2 \pi f t))^T _0[/tex]

[tex]i^2 _{rms} T = i^2 _{max}(\pi f t - \frac{cos(2 \pi f t)sin(2 \pi f t)}{2})_0 ^T[/tex]


[tex]i^2 _{rms} T = i^2 _{max}(\pi - \frac{cos(2 \pi)sin(2 \pi)}{2})[/tex]

Sorry, I'm bad at maths.
 
  • #18
jtbell
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It looks like you're trying to find the integral

[tex]\int {\sin^2 (2 \pi f t) dt}[/tex]

by making the substitution [itex]x = 2 \pi f t[/itex] and using the integral

[tex]\int{\sin^2 x dx} = \frac{1}{2}(x - \cos x \sin x)[/tex]

However, you didn't do the substitution completely.

[tex]\int {\sin^2 (2 \pi f t) dt} \ne \int{\sin^2 x dx}[/tex]

because [itex]dt \ne dx[/itex]. You have to apply the substitution to dt also, by using [itex]2 \pi f dt = dx[/itex].

This may not be an actual homework or coursework exercise, but this thread is starting to look like a homework-help type thread, so I'm moving it to one of the "homework help" forums.
 
  • #19
dE_logics
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Oh yes, thanks for notifying.

I'll fix it.
 
  • #20
dE_logics
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Thanks for all the help...problem solved.
 

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