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Average Voltage Problem

Discussion in 'Electronics Homework Help' started by BlackMelon, Jul 14, 2017.

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  1. BlackMelon


    Aug 7, 2012

    I am curious about the equation 2.7(a), and trying to prove it. So, I did my own calculation, and got a different result from the textbook. Did I make a right calculation? Also, I want to know where is the time t = tr? (Since fig 2.2 begins at time t = t(k-1), not t=0)

    Thank You

    PS. The pdf file is cut from the book Advanced Electrical Drives Analysis, Modeling, Control.

    Attached Files:

  2. Harald Kapp

    Harald Kapp Moderator Moderator

    Nov 17, 2011
    The textbook is very probably right, which answers the question about your solution...

    As I see it, your starting point is wrong: You use U(tr) = 1/Ts integral from (tk-1 + tr) to (tk-1 + tr +Ts) Udc/2 dt (from there on your calculation seems valid, but you know, when you run in the wrong direction, it doesn't help to run faster ;)).
    What is required is a calculation of the average voltage between tk-1 and tk-1+Ts. You have a share of -Udc/2 and +Udc/2 within that interval (your equation uses +Udc/2 only.

    Try again, I'm sure you'll get at the solution by yourself once you're headed in the right direction.
  3. Harald Kapp

    Harald Kapp Moderator Moderator

    Nov 17, 2011
    tk-1, tk, tk+1 are used only to indicate the repetitive nature of the signal.
    To simplify calculations you can simply set tk-1 = 0 for the calculations involving the green interval TS or tk=0 for theblue interval. As you noticed in your own calculation these elements cancel when solving the integral and substituting the bounds into the equation. This is explained by the fact that it is not relevant, where in time the signal occurs, only how the different parts of the signal relate to each other.

    To answer your question: t=tr is the point t=tk-1+tr in the diagrams of figure 2.2.
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