RMS and average current triangle

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To find the RMS and average current of a triangular waveform with a maximum of 2.5 and a minimum of 1.8, the RMS value is calculated using the integral of the squared function over one period, resulting in an approximate RMS of 2.16. The average current is determined by taking the mean of the maximum and minimum values, yielding an average of 2.15. The calculations involve defining the waveform segments and applying integration formulas. The key formulas for RMS and average current are referenced from Wikipedia. Understanding these calculations is essential for analyzing non-standard waveforms effectively.
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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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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