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how to determine the rms value

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oops by the way you said the area of the first half are you referring to this
please see attach

thanks again

ps. i forgot, so you're answer is i * square root of (3/2) or is it only i * square root of (3) / 2 ?
 

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  • triangularwave.jpg
    triangularwave.jpg
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oops by the way you said the area of the first half are you referring to this
please see attach

Yes, that is correct.

ps. i forgot, so you're answer is i * square root of (3/2) or is it only i * square root of (3) / 2 ?

THe two is not in the square root. It comes from the fact that the 4 was in the square root and sqrt(4) = 2. It has been simplified that way.
 
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hi thank you once again. I have to check this again with my professor.

I wish you a very Merry Christmas and a happy New Year. I'll let you know what he would the final answer be.

regards
 
Your wave is divided in 2 parts.

The first part is a function of time, the wave grows as the times goes by, isn't it?

The second part is a constant. Even if times goes by, the wave remains the same, with the same value.
 
thanks. i'm still trying to understand why is the first have 0.25T i^2 as what dknguyen mentioned. If the area of triangle is = 1/2 bh. What would be the base here?
does that mean: 1/2 b * T/2? for the first triangle?
 
Yeah, I ran into this problem when doing the problem for you as well and it took me a while to realize what I did it two different geometric ways and they were giving me two different answers. ANd here it is:

"THe area of a function is the area between the function and the ****X-AXIS****" Not when the "big" triangle looks like it has bottomed out because if you look closely it never really does. Your brain just draws a line there to form a full triangle when there really isn't one.

So there is actually the positive and negative triangle to calculate the area for (and remember the triangles are squared before the area calculation so they will be positive and won't cancel to zero when you add them up.)

Of course, if you just do everything based off the equation of the line, you don't need to worry about this since it all takes it into account. But if you do it geometrically, you do
 
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RMS stands for "the Root of the Mean of the Square".
This means we must first square the function, then
calculate the mean, and finally calculate the square
root of that.


This is calculated like this:

Vrms=sqrt(1/T*integrate[0 to T]((f(t))^2) dt)

where

f(t) is the function and may be a simple value as in this problem
and that first gets squared as
Square=(f(t))^2
and then we calculate the mean with
Mean=1/T*integrate[0 to T](Square)dt
and then take the square root
Vrms=sqrt(Mean)


Because the specified wave has two parts, we have to divide the sections
up into two parts. The first part goes from 0 to 4 seconds, and the
second part goes from 4 to 8 seconds.
The first part has amplitude of 3, and the second part has amplitude of -1.
Since we have two parts, we use two integrals instead of one and make sure
we get the times correct:

Vrms=sqrt(1/8*(integrate[0 to 4](3^2)dt+integrate[4 to 8]((-1)^2)dt))

and first to simplify the constants a little we end up with:

Vrms=sqrt(1/8*(integrate[0 to 4](9)dt+integrate[4 to 8](1)dt))

Since the integral of a constant K is K*t+C, we replace the constants with
those values and run the dependent variable t between the two limits of
integration for each part:

Vrms=sqrt(1/8*((9*t+C1)from[0 to 4] + (1*t+C2)from[4 to 8]))

and now all that is left to do is the calculations as shown:

Vrms=sqrt(1/8*(((9*4+C1)-(9*0+C1)) + (1*8+C2)-(1*4+C2)))

and since the constants C1 and C2 cancel out we are left with:

Vrms=sqrt(1/8*(((9*4)-(9*0)) + (1*8)-(1*4)))

now doing some simple math and noting

(1*8)-(1*4)=(1*4)

we get:

Vrms=sqrt(1/8*((9*4)+(1*4)))

and note this is now in the same form as the linked picture has shown,

and the final answer is the same:

Vrms=sqrt(1/8*(40))=sqrt(40/8)=sqrt(5)=2.2360 approximately.

Some problems require splitting the integral up into even more parts than two.
It all depends on how well you can describe the function or functions for
the wave. If you cant do it all in one shot, split the integral.
 
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Hi again,


No sorry, it was for the original problem with the squareish wave.
Did you want to do that wave too? We would basically start the
same way only define the wave sections slightly differently.
You want to do that wave too then?
 
Hi again,


No sorry, it was for the original problem with the squareish wave.
Did you want to do that wave too? We would basically start the
same way only define the wave sections slightly differently.
You want to do that wave too then?

Hi MrAl that would be nice. I already got the answer which is sqrt (2/3) * i but I don't know how to arrive to this answer.
dknguyen had another answer as well.
I tried the method mentioned here and I'm getting another answer:
A solution from our class was this: please see attached

I don't understand why it appears to be like this: 1/T and T/2.

Thanks for your time and help
 

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  • DSC02185.JPG
    DSC02185.JPG
    137.6 KB · Views: 178
THe 1/T is the period and you are dividing by the period to find the average over the period (the weighted average). THe T/2 is the weighting of each value (the period is divided into two halves).

What I don't understand is where the root 3 comes from. WHen I did it, it didn't show up until the very end from simplification. It's not obvious to me how it shows up right away. Also...my answer is different too.
 
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