Here is the file I made on RMS. DOes this help make things more clear?

Okay, forget everything you learned about "averages" or "means". The average/mean that you learned are all special cases of the weighted average. THe special case being that all items have equal weightings. Think about how your grade in a class is calculated...

If you have 5 quizzes worth 10% each, and one final worth 50%, then the REGULAR AVERAGE of your quizzes is:

How would you calculate your average mark for the quizzes (not including the final)? Obviously you just go:
(Quiz1 + Quiz2 + Quiz3 + Quiz4 + Quiz5)/5
because you have 5 items of equal weighting. But really, this is just a simplification of the weighted average equation because you have equal weightings. THis simplified expression actually comes from the more general weighted average equation:

(Quiz1*10% + Quiz2*10% + Quiz3*10% + Quiz4*10% + Quiz5*10%)/50%

which simplifies down to

(Quiz1*0.2 + Quiz2*0.2 + Quiz3*0.2 + Quiz4*0.2 + Quiz5*0.2) --->>>
(Quiz1 + Quiz2 + Quiz3 + Quiz4 + Quiz5)*0.2 --->>>
(Quiz1 + Quiz2 + Quiz3 + Quiz4 + Quiz5)/5

But now think of your final grade...it's not simply an average of your quizzes and final exam because your final exam was worth much more than the others (it was weighted more heavily). Since the items now have different weights, no regular average exists. Instead, only the more general weighted average exists:
(Quiz1*10% + Quiz2*10% + Quiz3*10% + Quiz4*10% + Quiz5*10% + Final*50%)/100%

which again simplifies down to

(Quiz1*0.1 + Quiz2*0.1 + Quiz3*0.1 + Quiz4*0.1 + Quiz5*0.1 + Final*0.5)

It seems silly that I wrote 10%, 50%, and 100% and then simplified them to 0.1, 0.5, and 1. BUt this is just because the marks are in percentages which are already a fractional form. But in an example like RMS, these numbers would end up being time segments of the waveform, and 100% would end up being the period of the waveform. So in this case, it obviously isn't so silly, since you are now finding the ratio of
"some time segment" : period
whereas before you were finding the ratio of
"some %" : 100%
which is redundant since % is already a ratio.

Understand? I am a bit curious to what level of schooling you are in right now. Because by the time you get to high school most everyone knows what a weighted average is (if only so they can calculate how well they need to do on the final exam to pass).

THIS IS THE FINAL VERSION OF THIS POST.

Attached Files

RMS.pdf (28.0 KB, 11 views)

 

Oh, you should know that after I edit my posts a LOT after I write them so if you have been reading it, it may be outdated and harder to understand. So if you were reading my previous post and it did not say:
THIS IS THE FINAL VERSION OF THIS POST.
THen you were reading an outdated version that might have been harder to understand and might have had mistakes.

 

thanks dknguyen for the simplification. I learned these stuffs 10 years ago and after that haven't had the chance to practice it and now I'm studying it again. so I'm trying to recall everything. Thanks for your patience

You've mentioned that my answer on my another example is just a weighted average. I think I also performed the square root after.

From: I^2 = (1/T) (T/2) (i/square root of 3)^2 + i^2

To: I = square root of 2/3 * i --> so this one is not correct?



ps. thanks for the PDF File

 

The area for first half + second half of the square function is:

0.25Ti^2 + 0.5Ti^2 = 0.75Ti^2

divide it by the T to normalize, we get:
0.75i^2 = (3*i^2)/4

Take the square root of the whole thing and you get:

i*sqrt(3)/2

There must have been a typo with the brackets. So no, it doesn't seem right. But neither were you. I don't see why your 3 came in so early. THe reason it is there is from the 0.75 = 3/4 after you've added everything up.

SQRT[(1/T){(T/2)(0.25i/square root of 3)^2 + (T/2)i^2}]
I have no idea what is going on in the italics. THe bold, underlined part is totally out the window. It just shouldn't be there.

THe 0.25 is from the fact that the area of the first half of the period is 0.5*(T/2)*i^2. You can do it mathematically using the slope of the line, but I just did it visually. The negative half of the line is the same area as the positive half and since you are squaring it to make it all positive, I just flipped it to above the X axis, and from there you can see it takes one half of a grid square (and one grid square covers T/2 of the period).

 

ok. i also didn't get it here. But if you say that there might be a typo with the brackets I have to check this then with my professor.
btw, i'm trying to solve here the Irms

Thanks again

 

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 ?

Attached Thumbnails

 

 

Yes, that is correct.

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.

 

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

 

The first half function is:
Just changing names, a = î.

f(t) = (1/T)*(4at - Ta)

The second:
g(t) = a

 

hi hayato what do you mean the 2nd here? could you please explain.

thanks

 

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.

 

is this the 2nd part you mean?

please see attached gif.file

Attached Thumbnails

 

 

Yes. That's correct.

 

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