RMS definition in Pulsed waveforms

Discussion in 'Math' started by adwina2, Jul 15, 2015.

  1. adwina2

    Thread Starter New Member

    Jul 14, 2015
    18
    2
    I am struggling to completely understand RMS values, and I believe it is a lack of math understanding. The underlying question is what exactly is RMS? All the definitions I find relating to math state RMS is the root of the average squares of a signal. Relating to electronics I find it to be the average value over a period of a signal equivalent to what a constant "DC" signal would produce. For example, I'll reference a PWM type waveform where the duty cycle is 25%, and Vpeak is 12V. A derivation I found for a pulsed DC waveform is:
    [​IMG] -------> [​IMG] --------> [​IMG]
    The above derivation uses voltage, but I believe it would work for current and power as well. T is period and t1 is time hi. Using this equation I get:
    (12V) * sqrt(.25) = 6 volts
    But intuitively I would say the average "DC" value here is 3V(25% of 12V). And if I wanted to find the average over time I would use the same approach as above, but I would not square u(t) and then find the root. This seems like it is related to probabilities where the larger probabilities hold more weight so squaring them skews the average higher. Is there a mathematical reason to use the square of u(t) here besides getting rid of negatives? Can anybody please explain this to me, preferably both mathematically and applied to electronics?
     
  2. adwina2

    Thread Starter New Member

    Jul 14, 2015
    18
    2
    OK, I figured it out. The values from these equations are used to calculate power. So if I take the PWM example where the duty cycle is 25%, intuitively it is:

    P=(V^2/R) * .25 and this equals the derivation of the RMS equation:

    Prms = Vrms^2 / R where Vrms = [V * sqrt(t/T)] ---> Prms = [V * sqrt(t/T)]^2 / R


    And the equations work out nicely if you start with the first equation from the original post and use either P = V^2 / R or P = I^2 * R. Working them out starting with Vavg(t) = ∫(v(t) dt) / (T2-T1) from T2 to T1 and using this in the power equation led to my understanding.
     
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