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Average and rms values of a given signal

  1. Nov 18, 2015 #1
    1. The problem statement, all variables and given/known data
    I am in an Intro to EE class and we were given a homework assignment including the following question:

    Find the average and rms values of the signal:
    x(t)=3cos(7ωt)+4
    I have formulas for average power and rms but they include V & I, however the signal was given as x(t). I understand that this is arbitrary.. or something but basically I'm extremely lost and need help.

    Thanks

    2. Relevant equations

    Pav = 1/T ∫_0^T p(t)dt = 1/T ∫_0^T ½VI cos(θ)dt + 1/T ∫_0^T ½VI cos(2ωt-θ)dt
    Pav = ½VI cos(θ)

    3. The attempt at a solution

    I don't even know where to start.
     
  2. jcsd
  3. Nov 18, 2015 #2

    gneill

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    Staff: Mentor

    The problem wants the average and rms values for the given signal. Power doesn't enter into the problem. Just apply the definitions of average and rms to the given function.

    Hint: Both definitions expect you to integrate over one period of a repeating function. You can go to the trouble of working out the time interval that represents one period and integrate over that, dividing by that time intervaland so on, or, since your function is based on a trig function you can integrate over one angular period instead of time. That is, the period is ##2 \pi## and you integrate over the angle from ##0## to ##2 \pi##. No need to muss around with ##\omega t##.
     
  4. Nov 18, 2015 #3
    so I did
    ∫_0^2π 3cos(7x)+4 dx = 3/7 sin(7x) + 4x = 3/7sin(7*0) + 4(0) - 3/7 sin(7*2π) - 4(2π)
    --> Pav = -8π

    I feel as if that is wrong
     
  5. Nov 18, 2015 #4

    gneill

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    The argument of the cosine should just be ##\theta##. Not 7x. And don't forget to divide by the period. And it's not a power that you're finding, its the average of signal x(t), so maybe call it ##x_{av}##.

    $$x_{av} = \frac{1}{T} \int_0^T \{ 3 cos(7 \omega t)+4 \} ~dt ~~→~~ \frac{1}{2 \pi} \int_0^{2 \pi} \{ 3 cos(\theta)+4 \} ~d \theta$$
     
  6. Nov 18, 2015 #5
    okay so after using the formula you gave me I got xav = -4

    that feels weird too.. is there supposed to be a constant after I integrate?
     
  7. Nov 18, 2015 #6

    berkeman

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    Can you show your work? Where did the negative sign come from?
     
  8. Nov 18, 2015 #7
  9. Nov 18, 2015 #8

    berkeman

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    Why does the last term have a minus sign?
     
  10. Nov 18, 2015 #9
    I may have finished the integral incorrectly... I haven't done integrals intensely since calc 2 (2 years ago) ...

    I did ...
    say y(t) is the integral I determined... the 3/2π sin ... etc
    to solve a definite integral... y(0) - y(2π)

    is that incorrect? I'm going to look it up right now
     
  11. Nov 18, 2015 #10
    yeah its the other way around, y(2π) - y(0)

    so the answer I get now is xav = 4

    is that correct?
     
  12. Nov 18, 2015 #11

    berkeman

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    Most of it's right except for how you handled the constant at the end. You should end up with 4*2π/2π - 0*2π/2π...
     
  13. Nov 18, 2015 #12

    berkeman

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    Yes. :smile:
     
  14. Nov 18, 2015 #13
    awesome! Thanks

    now for the second part, how do I determine the rms?
     
  15. Nov 18, 2015 #14

    berkeman

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    That's for you to figure out! :smile:

    What is the equation definition of the RMS value of a function? Your textbook should show it, or use wikipedia to look it up.
     
  16. Nov 18, 2015 #15
    I found a formula that says
    xrms = x/√2
    so... xrms = ## \frac{3}{\sqrt 2} cos{(7 \omega t)} + \frac{4}{\sqrt 2} ##

    yes?
     
  17. Nov 18, 2015 #16

    berkeman

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    It might be right, but I don't think so. If it were just the cos() term, then yes, the RMS value is the peak value divided by √2. But having the constant term there means you need to use the full equation for calculating the RMS of the addition of two signals. Just look for the equation that has the square root of the sum of the squares in it... :smile:
     
  18. Nov 18, 2015 #17
    I can only find Vrms and Irms equations

    I also found ## \sqrt{\frac{1}{T} \int_0^T{A^2 sin^2(\omega t) dt}} ##

    If I use 2π for T like I did in the first part, does this formula work?

    Also for A do I use 3 since it's in front of the cos() in the given signal equation? What do I do about the 4 at the end of that?

    Thank you
     
  19. Nov 18, 2015 #18

    berkeman

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    To calculate the RMS value of a function, you need to square the function, average it over a period, and then take the square root. So more like this:

    ## \sqrt{\frac{1}{T} \int_0^T{(3cos(7ωt)+4)^2 dt}} ##
     
  20. Nov 18, 2015 #19
    ImageUploadedByPhysics Forums1447882540.489669.jpg

    This feels wrong
     
  21. Nov 18, 2015 #20

    berkeman

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