Waveform and Average Values

  • Thread starter Marcin H
  • Start date
  • #1
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Homework Statement


Circle all that are correct

Homework Equations


V(t)=Vocos(wt)

The Attempt at a Solution


I don't really understand what the question wants when they ask for an average value. Average value of what? And how do you find it. There are a lot of questions with graphs like this that ask to find the average value, but I'm not sure how to get that exactly. How do I find the average value of this problem?
New Doc 10_1.jpg
 

Answers and Replies

  • #2
NascentOxygen
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On a graph, the average value is drawn as a horizontal straight line at a level such that there is as much enclosed area above that line and the waveform as there is below that line and the waveform, for one cycle.
 
  • #3
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On a graph, the average value is drawn as a horizontal straight line at a level such that there is as much enclosed area above that line and the waveform as there is below that line and the waveform, for one cycle.
Oh. So do you get the average by just looking at it or is there a formula to get it. Looking at this one it looks like the average would be 0 right? How can I get the average for something like this:
New Doc 10_2.jpg
 
  • #4
NascentOxygen
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The easy ones you can do by inspection or by counting squares, but for the harder ones you need to use integral calculus.

Did you plot a horizontal line at y=0 on that graph, to check that it looks about right?
 
  • #5
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The easy ones you can do by inspection or by counting squares, but for the harder ones you need to use integral calculus.
This is a non-calc based course, so is there an easier way. I've taken up to calc 3 and some diff eq, but I don't think we can use calc for this. (This is EGR 110 - Intro to comp E and EE) Also, the way my teacher does it doesn't make sense... :/

Did you plot a horizontal line at y=0 on that graph, to check that it looks about right?
I looked at one cycle and saw that the triangles would cancel each other out, so it's average would be 0. I think.
 
  • #6
NascentOxygen
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For the triangular waveform, yes, its average is zero.
 
  • #7
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For the triangular waveform, yes, its average is zero.
How about for the other one? Is there a simple way of doing it?
 
  • #8
NascentOxygen
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How about for the other one? Is there a simple way of doing it?
Counting squares. Half will be above the average mark, half below it.
 
  • #9
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Counting squares. Half will be above the average mark, half below it.
Where would the average mark go? Horizontal line at 1Amp?
 
  • #10
NascentOxygen
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Where would the average mark go? Horizontal line at 1Amp?
How many "squares" do you count between that line and the waveform?
 
  • #11
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How many "squares" do you count between that line and the waveform?
Which line? The horizontal line at 1Amp? 7.5 boxes
 
  • #12
NascentOxygen
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Which line? The horizontal line at 1Amp? 7.5 boxes
Why the 1A line?

Below the average consider the squares to have a negative sign, above the average line call them positive. So when you have it correct, the differences cancel.
 

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