RMS and average current triangle

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Discussion Overview

The discussion focuses on calculating the RMS (Root Mean Square) and average current of a triangular waveform, with specific values provided for the maximum and minimum of the waveform. Participants explore different methods and definitions related to these calculations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant seeks guidance on calculating RMS and average current for a triangular waveform with given maximum and minimum values.
  • Another participant references the definition of RMS and suggests considering how the final answer will be derived.
  • A detailed mathematical approach is provided, including integrals and specific formulas for calculating RMS based on the waveform's characteristics.
  • Further calculations are presented for the average value, leading to a proposed average of 2.15.

Areas of Agreement / Disagreement

The discussion does not show clear consensus on the calculations, as participants present different methods and results without resolving the differences.

Contextual Notes

Participants utilize various mathematical definitions and approaches, but there are no explicit confirmations of the correctness of the calculations or assumptions made during the discussion.

billy fok
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I would like to know how to find RMS and average current of a triangle?
i searched the web. but it focus from 0. which i know.
but my waveform is different from the internet. i have attached the waveform..
Maximum = 2.5 , Minimum = 1.8

i need to find out the RMS and average current
 

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  • Screen Shot 2015-03-27 at 12.26.28 am.png
    Screen Shot 2015-03-27 at 12.26.28 am.png
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Following the definition:

"The RMS over all time of a periodic function is equal to the RMS of one period of the function."

And :

frms=sqrt[integral(f(t)^2/T]

f(t)up=yo+(y1-yo)/(x1-xo)*(x-xo) integrated for x=x1 and x=xo

f(t)down=y1-(y1-yo)/(x1-xo)*(x-x1) for x=2*x1 and x=x1

where T=0.05 xo=0;x1=T/2;yo=1.8;y1=2.5

frms=sqrt{[integral(f(t)up^2+integral(f(t)down^2]/T}

integral(f(t)up^2=aup*x+bup*x^2/2+cup*x^3/3 x=x1 and x=xo

integral(f(t)down^2=adown*x+bdown*x^2/2+cdown*x^3/3 x=2*x1 and x=x1

Let's put k= (y1-yo)/(x1-xo).Then:

aup=yo^2-2*k*yo*xo+k^2*xo^2

bup=2*yo*k-2*xo*k^2

cup=k^2

adown=y1^2+2*y1*k*x1+k^2*x1^2

bdown=-2*y1*k-2*k^2*x1

cdown=k^2

frms=2.16[approx.]
upload_2015-3-30_8-20-57.png
 
billy fok said:
i need to find out the RMS and average current
What have you calculated as its average value?
 
Following:

http://en.wikipedia.org/wiki/Average

AM=1/n*Sum[y(i)]|y(1)=yo to y(n)=y1

Let's take n number of equal distanced points on upright part of the triangle.The number of intervals

will be n-1.Each y[y(i)] value will be y(i)=yo+(i-1)*(yn-yo)/(n-1)

Sum[y(i)]|y(1)=yo to y(n)=y1=n*yo+ (y1-yo)/(n-1)*Sum(1+2+3+..n-1)

Sum(1+2+3+..n-1)=n*(n-1)/2. Then:

Sum[y(i)]|y(1)=yo to y(n)=y1=n*yo+n*(n-1)/2*(y1-yo)/(n-1)=n*yo+n/2*(y1-yo)

AM=1/n*Sum=yo+(y1-yo)/2=(y1+yo)/2

AM=(1.8+2.5)/2=2.15
 

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