ok but the answer square root 0f (3)/2 * i for the example i gave, you think the correct one?

 

If you remember some posts back, sqrt(2) / 3 is the answer that I got.

 

I got i*sqrt(2/3).

 

hi Hayato, could you please explain, how did you get it?

thanks

 

hello Hayato I would be happy if you could show me how did you get it to your answer.

thank you very much

 

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.

 

Hi MrAl thank you for the very informative explanation.
I'm not sure which problem here did you take as an example
was it this one :http://www.electro-tech-online.com/a...ngularwave.jpg?

Thanks

 

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

Attached Thumbnails

 

 

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.

 

did you also get the same answer as mine?

 

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):
It should say:
How did you get this Hayato? Every time I've done it I get my answer (obviously).

 

i hope Hayato will also show his solution, to clear things up.
Thanks dknguyen

 

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)

 

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.