What Are the Average and RMS Values of Waveforms?

In summary: It's ##\frac{\sqrt{a^2 + b^2 + ab}}{\sqrt{3}}## for a waveform with amplitude ##a## and base length ##b##.It doesn't hold for the RMS value, however, since, as a counterexample, two waveforms can have the same area under their curves but different RMS values.I will think about this more.Do you have any specific questions about it? Do you understand why the argument doesn't hold for the RMS value?In summary, the conversation discusses two questions related to finding the average and RMS values of given waveforms. The first question involves finding a shortcut to compute the RMS value without having to square and integrate five times. The second question asks
  • #1
STEMucator
Homework Helper
2,076
140

Homework Statement



I had two small questions about the following problems.

1. Find the average and RMS values of the following waveform:

Screen Shot 2014-09-21 at 9.37.52 AM.png


2. Show that the average and RMS values of the following sawtooth are independent of the position of the peak and are given by ##0.5## and ##0.577## of the peak value respectively.

Screen Shot 2014-09-21 at 9.38.08 AM.png


Homework Equations



##f_{avg} = \frac{1}{T} \int_0^T f(t) dt##
##f_{rms} = \sqrt{\frac{1}{T} \int_0^T f^2(t) dt}##

The Attempt at a Solution



1. Looking at the function, I see it is a piecewise function with five different pieces in one period. So to compute the average/rms value would require five integrals.

Looking more closely though, I see the function is symmetric about the x-axis over one period. This implies the average value is zero right away, ##v_{avg} = 0##. Is there similar logic I can apply to find the RMS value without having to square and integrate five times? For reference I got ##v_{rms} = V##.

2. This is the same question as the first really. I'm guessing all the question wants is for me to compute ##f_{avg}## and ##f_{rms}##? What do they mean by independent of the position of the peak exactly? Am I over-thinking this one beyond the computations?
 
Physics news on Phys.org
  • #2
I believe I have some reasonable thoughts about my first question, but you can't really edit your posts anymore, so I apologize for this double.

If I square that ##v(t)## they've given me, the amplitude won't change. Geometrically it would flip the troughs into crests, and all the values will be positive, so that in turn the square root will be properly defined every time.

What I'm looking for is the area over one period of ##v^2(t)##, which is two of those positive peaks (or simply twice the area of the first). Then I want to average it over the entire period and take the root. Those operations are harder to interpret geometrically, so I don't think there's a quick shortcut to finding the RMS value. The average value can be simplified through the use of geometry in some circumstances though (like this one).

For the second question, the computations were obvious, but I think I have some intuition now about why the position of the peak doesn't matter. If I moved the peak to point ##Y = X + \Delta x##, where ##X \leq Y \leq T##, the waveform will still have the same peak, just at a different time. The area under the curve itself won't change because the equations for the lines that define the waveform will change to accommodate it. Hence the average and rms values will be the same.

Does this sound reasonable?
 
  • #3
Zondrina said:
1. Looking at the function, I see it is a piecewise function with five different pieces in one period. So to compute the average/rms value would require five integrals.
You only have to consider a half-cycle to determine the RMS value of the waveform - sign doesn't matter. It would be a good idea to find general expressions for the area under the square of those couple of segments that appear in your waveform, i.e. where it's increasing/decreasing linearly or is constant.

Zondrina said:
Is there similar logic I can apply to find the RMS value without having to square and integrate five times?
No, but you only need those two expressions to determine the RMS value and it'll solve your second assignment as well.

Zondrina said:
For reference I got vrms=Vv_{rms} = V.
That would be true if your waveform was square instead of trapezoidal.

Zondrina said:
The area under the curve itself won't change because the equations for the lines that define the waveform will change to accommodate it. Hence the average and rms values will be the same.
That argument works for the average value, since the area of a triangle is given by ##A = \frac{1}{2}b h##, where ##b## and ##h## is its base and height, respectively, and moving the peak changes neither.

It doesn't hold for the RMS value, however, since, as a counterexample, two waveforms can have the same area under their curves but different RMS values.
 
  • #4
milesyoung said:
You only have to consider a half-cycle to determine the RMS value of the waveform - sign doesn't matter. It would be a good idea to find general expressions for the area under the square of those couple of segments that appear in your waveform, i.e. where it's increasing/decreasing linearly or is constant.No, but you only need those two expressions to determine the RMS value and it'll solve your second assignment as well.That would be true if your waveform was square instead of trapezoidal.That argument works for the average value, since the area of a triangle is given by ##A = \frac{1}{2}b h##, where ##b## and ##h## is its base and height, respectively, and moving the peak changes neither.

It doesn't hold for the RMS value, however, since, as a counterexample, two waveforms can have the same area under their curves but different RMS values.

That would be true if your waveform was square instead of trapezoidal.

That was the listed answer. I got ##0.83 V## when doing it myself. I decided to compute:

##V_{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} v^2(t) dt} = \sqrt{\frac{2}{T} \int_{0}^{\frac{T}{2}} v^2(t) dt}##

since only three integrals would be required instead of all five.

It doesn't hold for the RMS value, however, since, as a counterexample, two waveforms can have the same area under their curves but different RMS values.

I will think about this more.
 
  • #5
Zondrina said:
That was the listed answer.
That's odd. Must be an error.

Zondrina said:
I got ##0.83 V## when doing it myself.
I have ##0.816 V##. The RMS value of a trapezoidal waveform is something you can easily look up in a table if you want to check your result.
 

Related to What Are the Average and RMS Values of Waveforms?

1. What is the difference between average and rms for waveforms?

The average value of a waveform is the sum of all the data points divided by the total number of data points. It gives a measure of the central tendency of the waveform. On the other hand, the rms (root mean square) value is the square root of the average of the squared values of the waveform. It represents the effective or DC value of the waveform, taking into account both positive and negative values.

2. How do you calculate the average and rms for a waveform?

To calculate the average value of a waveform, you need to sum up all the data points and divide by the total number of data points. For example, if you have a sine wave with 10 data points, you would add all the values and divide by 10. To calculate the rms value, you need to square each data point, find the average of the squared values, and then take the square root of that average.

3. What is the significance of average and rms in waveforms?

The average value of a waveform is useful in determining the central tendency of the data and can be used to compare different waveforms. The rms value, on the other hand, is important in calculating power and energy in electrical circuits. It is also used to determine the effective voltage or current in AC circuits.

4. Can the average and rms values of a waveform be the same?

No, the average and rms values of a waveform are generally not the same. The only instance where they can be equal is if the waveform has a DC component, in which case the rms value will be equal to the average value. Otherwise, the rms value will always be greater than or equal to the average value.

5. How do average and rms values change with different types of waveforms?

The average and rms values of a waveform will vary depending on the type of waveform. For example, a sine wave will have an average value of 0, while the rms value will be equal to the peak value divided by the square root of 2. In contrast, a square wave will have an average value equal to its peak value, and the rms value will be equal to the peak value itself. These values will change for other types of waveforms such as triangular, sawtooth, or pulse waves.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
4
Views
6K
  • Engineering and Comp Sci Homework Help
Replies
8
Views
4K
Replies
2
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
6
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
26
Views
5K
  • Engineering and Comp Sci Homework Help
Replies
20
Views
4K
  • Engineering and Comp Sci Homework Help
Replies
11
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
9
Views
1K
  • Classical Physics
Replies
11
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
29
Views
3K
Back
Top