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Find average and rms values of the waveform

  1. Sep 5, 2013 #1
    1. The problem statement, all variables and given/known data
    The problem is located here http://www.chegg.com/homework-help/...ter-11-problem-26e-solution-9780073529578-exc


    2. Relevant equations
    Rms=sqrt(1/T∫ from 0 to T of i(t)^2 DT )
    Average= couldn't find an equation in book??

    3. The attempt at a solution
    Ok so I did it on my own and I'm trying to figure out why in the solution that the person posted. Is i(t) only written for 3 time intervals 0<t<2, 2<t<3, and 3<t<5. Why is there not a 4th 5<t<6 where i(t) would equal -9 A. BTW this is the first part if it wasn't clear. Furthermore how is T=5 I would think it would be 2 as the square wave gets its maximum at -9 and it lasts for 2 and 3? Also what is the equation for average value of current? And more of the same on the second part of the problem why did they only do parts of the waveform?
     
  2. jcsd
  3. Sep 5, 2013 #2
    Or is the solution incorrect
     
  4. Sep 5, 2013 #3
    here is my work
     

    Attached Files:

  5. Sep 5, 2013 #4

    gneill

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    Staff: Mentor

    It's a repeating waveform. The average (or rms) over one cycle (period) is the same as over all periods. So you need to identify one period and work with that. Looks to me like the time interval from 0 to 3 would cover one period of the repeating signal (the interval from 3 to 6 is an exact repetition).

    The mean is calculated in the same way that the mean is calculated for the rms, only the function isn't squared. That is,

    $$mean = \frac{1}{T}\int_{t_o}^{t_f}i(t)dt$$
     
  6. Sep 5, 2013 #5
    ah so T=3 then because it covers the time from its minimum to its max.
     
  7. Sep 5, 2013 #6

    gneill

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    Staff: Mentor

    T = 3 because it covers one period of the waveform. A period is the repeating unit of the waveform.
     
  8. Sep 5, 2013 #7
    ok thanks
     
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