How Do You Calculate RMS and Average Voltages for Different Waveforms?

[pjrobertson]

Homework Statement

I need to find the average and RMS voltages for the waves (See attached pictures) but I'm not sure exactly how to do it as I don't know how to get an equation for V(t)

Homework Equations

I Know:

V_RMS = [tex]\sqrt{\frac{1}{T}\int V(t)^2 dt}[/tex] between 0 and T

And for Avg it's just V_AVG = [tex] \frac{1}{T}\int V(t) dt[/tex] between 0 and T

The Attempt at a Solution

I thought I could find the averages by finding the total area for 1 period, then dividing by 1 period. This works (I think, as I don't have the answers)

Giving for the Triangular wave: [tex]\frac{T}{3} \times \frac{10}{T} = 3.3V[/tex]

and for the second sawtooth type wave: [tex] \frac{\frac{1}{2}\frac{T}{3} \times 1}{T} = \frac{1}{6}V[/tex]


Thanks

 

[Redbelly98]

Looks good for V_AVG.

For V_RMS, you'll need to write equations for the waveforms. To do that, you may choose any point to represent t=0.

 

[pjrobertson]

That's the problem, I don't know what to write the equations as. The triangular wave has 3 functions. One for 0 - [tex]\frac{T}{3}[/tex] (upwards slanting triangle) one for 0 - [tex]\frac{2T}{3} [/tex] (downwards slanting triangle) and one for the last part of the waveform, which is just V(t) = 0.

What are the waveforms for the first 2 sections? And can they be summed like so?:

V_RMS = [tex]\sqrt{\frac{1}{T}\int V_A(t)^2 dt} + \sqrt{\frac{1}{T}\int V_B(t)^2 dt}[/tex] the first integral being from 0 to T/3 and the second from T/3 to 2T/3

 

[Redbelly98]

That's pretty much the idea, except that it would be

[tex]
\sqrt{\frac{1}{T}\int V_A(t)^2 dt \ + \ \frac{1}{T}\int V_B(t)^2 dt}
[/tex]

with, as you said, "the first integral being from 0 to T/3 and the second from T/3 to 2T/3"

 

[pjrobertson]

I still can't seem to figure out the RMS values. The problem I have is determining the functions V(t) for each section of the wave.

I have an exam on this tomorrow so any help would be appreciated!

 

[Redbelly98]

Note that the function is composed of straight line segments. Do you remember, from high school algebra,

y = mx + b​

where m is the slope and b is the y-intercept?

For example . . .

. . . for the triangular wave, and 0 ≤ t ≤ T/3:

v(t) = m t + b​

What is the slope m in this interval (0 ≤ t ≤ T/3)?
What is the intercept b in this interval?

Find m and b in this interval, and you'll have the function.

Do the same for the interval T/3 ≤ t ≤ 2T/3, and you'll have v(t) in that interval as well.

 

[pjrobertson]

Hmm...
using v(t) = m t + b
I get:
[tex]V_A(t) = \frac{30t}{T}[/tex]
[tex]V_B(t) = 20 - \frac{30t}{T}[/tex]

So I tried integrating all of this, then taking the root (long and tiring process!) and I got V_RMS = 4.71V

Does this sound about right?

Here's the integration I did (without the square root)

[tex]\sqrt{\frac{30^2}{T^3}\int t^2 dt + \frac{1}{T}\int 400t dt - \frac{1}{T^2}\int 600t^2 dt + \frac{1}{T^3}\int 300t^3 dt}[/tex]

With the first being between 0 and T/3 and the last 3 parts being between T/3 and 2T/3
(there's 3 parts because of squaring [tex]20 - \frac{30t}{T}[/tex])

I hope this is right!

 

[Redbelly98]

You're on the right track, but I do see a couple of errors.

You seem to have expanded (20 - 30t/T)² incorrectly, in particular the "middle" term's coefficient of -600 is wrong.

Also, there is an extra factor of t in 3 of your integrals.