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how to determine the rms value

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No, tHe answer I got was I*sqrt(3)/2

THis is how I got it:

The area of the first little triangles is:
One Triangle = 0.5*base*height = 0.5*(T/4)*i^2
But there are two of them, so the total area of the first half is double that:
Both Triangles = 2*0.5*(T/4)*i^2 = (T/4)*i^2
Area of the second half (the square) is simply:
Square = base * height = (T/2)*i^2
Add up all the area and you get:
Total Area = (T/4)*i^2 + (T/2)*i^2 = (3/4)Ti^2
Divide by the total of the period to find the average area over time and you get:
Average Area = (1/T)(3/4)Ti^2 = (3/4)i^2

FInally square root the whole thing to get:
RMS = i*sqrt(3)/2

This here is a typo. (from the first post where I calculated it out a bunch of posts back):
If you remember some posts back, sqrt(2) / 3 is the answer that I got.
It should say:
If you remember some posts back, I*sqrt(3)/2 is the answer that I got.

I got i*sqrt(2/3).
How did you get this Hayato? Every time I've done it I get my answer (obviously).
 
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Sorry for the delay. Now I'm away from home, so I do not use the computer often.

Here is how I got to the solution:

Your wave period is T.
The amplitude is "i".

The wave is composed by two parts.
One is from 0 to T/2
And other is from T/2 to T.

The first part expression, "f(t)", is (1/T)*(4it - Ti).
The second part, "g(t)", is just "i", it is a constant.

To find the RMS value, you have to square the expressions and then integrate:

[f(t)]² = 16.i².t²/T² - 8.i².t/T + i² -> from 0 to T/2
[g(t)]² = i² -> from T/2 to T

When you integrate ∫[f(t)]²dt from 0 to T/2, you get:
2.i².T/3 - i²T + i².T/2

When you integrate ∫[g(t)]²dt from T/2 to T, you get:
+ i²T - i².T/2

Then you sum the both:
2.i².T/3

Then you divide per T and take the sqrt:
RMS = i*sqrt(2/3)

-> Remember that RMS = sqrt( [1/T] * ∫[f(t)]² dt)
 
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Hello again,


Just to note, the correct answer is Irms=I*sqrt(2/3) as noted by some of
the other posts. This has been verified using four different techniques,
two of which will be shown here.


Here are two methods to calculate the rms value of that wave.
The first method uses the definition of an rms value, and the
second method (far below) uses a shortcut knowing something
about the properties of rms values for different wave shapes.

Just to note, the wave being worked on here is as follows:

From t=0 to T/2 the wave goes from -I to +I (slanted), and
from t=T/2 to T the wave is a constant equal to +I.



METHOD #1

For the slanted part of the wave, we can use the slope intercept
form of a line to get the required equation of that line from
0 to T/2:

i=mx+b

The slope m is:

m=(i2-i1)/(t2-t1)=(I-(-I))/(T/2-0)=2*2*I/T=4*I/T

(note: the way we use this is similar to the "Two Point Form" of a line)

The intercept constant b is simply:

b=-I

and x is of course equal to t:

x=t

Substituting all these into the slope intercept form of a line above:

i=mx+b

and calling the amplitude I, we get:

i=(4*I/T)*t+(-I)=4*t*I/T-I

So now we have the equation for the slanted line segment:

is=4*t*I/T-I

The equation for the horizontal section is simply:

ih=I

Now squaring both we get:

is^2=I^2*(T^2-8*t*T+16*t^2)/T^2

and

ih^2=I^2

Now integrating is^2 from 0 to T/2 we get:

iis=(I^2*T)/6

and integrating ih from T/2 to T we get:

iih=(I^2*T)/2

The rms value is now:

Irms=sqrt(1/T*(iis+iih))=sqrt(1/T*I^2*T*(1/6+1/2))=sqrt(I^2*(2/3))=I*sqrt(2/3)


METHOD #2:

The way they approached this problem in that linked equation picture was
different however. They used the rms property of triangle waves and of
simple horizontal (pulse) waves, and the fact that rms values add as
the square root of the sum of squares of the individual rms components
times their individual durations.

The component of the triangle wave is:
c1=I/sqrt(3)
(note that you have to know this property of triangular shaped waves
beforehand in order to start the solution the way they did in that pic)

and the component of the pulse wave is:
c2=I
and after squaring we get
c1=I^2/3
and
c2=I^2

and now taking into account that each component only exists for 1/2 the
total time period we multiply by 1/2:

c1=(1/2)*I^2/3
and
c2=(1/2)*I^2

and now we add them:

sum=(1/2)*(I^2/3+I^2)=(1/2)*I^2*(4/3)=I^2*(2/3)

and now taking the square root we get:

Irms=sqrt(I^2*(2/3))=I*sqrt(2/3)


In the original document, another way of taking into account that the waves
are only present for 1/2 the total time period is to multiply by duration T/2 and
then we have to also multiply by 1/T.
 
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I wonder what was not right about the way I did it.


Hi there,


I like the way you approached this problem in looking at the areas.
I think however that you just need to modify that technique a little,
where instead of looking at the area under f(x) you need to look
at the area under (f(x))^2.
In other words, instead of finding the area of those little triangles
shown in the pic of the wave (which have three straight line sides)
you would need to first square them for all time and this would yield little
triangles with one curved side and two straight sides.

I'd like to see you do this again making that one little modification
and am betting your idea will work after that and you'll get the very
same answer. I really do hope you try this again i'd like to see the
result too.
 
I did do that. I squared the Y-axis values (or height) before multiplying them by 1/2*base. I just stopped saying the "square of the function" because I've said it like about 20 times in my previous posts and it got very confusing to follow the post with such a frequent wordy phrase.

The area of the first little triangles is:
One Triangle = 0.5*base*height = 0.5*(T/4)*i^2
But there are two of them, so the total area of the first half is double that:
Both Triangles = 2*0.5*(T/4)*i^2 = (T/4)*i^2
Area of the second half (the square) is simply:
Square = base * height = (T/2)*i^2
Add up all the area and you get:
Total Area = (T/4)*i^2 + (T/2)*i^2 = (3/4)Ti^2
Divide by the total of the period to find the average area over time and you get:
Average Area = (1/T)(3/4)Ti^2 = (3/4)i^2

FInally square root the whole thing to get:
RMS = i*sqrt(3)/2
 
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Hi again,

Well, what you have done is made bigger triangles by squaring the height of the
triangle alone, when you really have to square the entire triangle over time t.
When you use a formula like 0.5*base*height, that means you are sill using
a triangle that has three *straight line* sides (ie a normal looking triangle)
when the triangles that really represent the area have only two sides that are
actually straight and the part of the triangle that is normally slanted (and straight)
becomes a curve instead of a straight line.
You'll have to figure out some way to calculate the area of that new pseudo
triangle (it isnt really a triangle anymore).

Here is an illustration showing the difference this really makes:
 

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Ahhh, i see now. A sloped line that is squared...obviously becomes a curve! I see now. It does not scale linearily.
 
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Hi again,


Yes :) and i posted a pic to illustrate this but i think you replied while
i was drawing it.

You might still find a way to work with that new type of triangle
though.
 

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Hi thank you for the solution, and also for the explanation here:
here is another approach of solving this problem.

The Delta signal between 0 & T/2 has already the RMS of i /sqrt (3). The constant value between T/2 and T has the RMS of i
Both value will be squared, with the time T/2 multiplied, and add
The sum must be at the root, therefore the result is: i * sqrt (2/3)

thanks again!
 
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