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Average power of sinusoidal signal

  1. Sep 4, 2009 #1
    1. The problem statement, all variables and given/known data
    Question: Consider the sinusoidal signal:
    A*cos([tex]\omega[/tex]t + [tex]\phi[/tex])

    Determine the average power



    2. Relevant equations
    This is my first real attempt in signals and I am really confused with the question...
    I guess my question would be am I suppose to take the P = lim as t-> [tex]\infty[/tex] [tex]\frac{1}{t}[/tex] [tex]\int[/tex] A*cos([tex]\omega[/tex]t + [tex]\phi[/tex]) ?

    3. The attempt at a solution
    I have no attempted solution yet. I am really just trying to see how to start this problem and then go from there!!

    Please help. Thanks
     
  2. jcsd
  3. Sep 4, 2009 #2
    The average value of a function is \frac{1}{t}\int f(t) \Delta t. For sinusoidal functions, we typically average over a period.
    Also, power is proportional to amplitude squared.
     
  4. Sep 5, 2009 #3

    Redbelly98

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    More information is required here.

    Is "the signal" a voltage or a current? How much resistance is the signal connected to?
     
  5. Sep 16, 2009 #4
    The average power of the signal x(t) is given by:

    P = lim T ---> infinity * (1/T) * [ integration of x2(t) over (0 to T) with respect to t]

    please ditinguish between "T" and "t". "T" is the period of the signal and "t" is the time.

    If you integrate [A cos (ωt + ф)]2 over (0 to T) and take the limit as T approaches infinity you will have the answer. The result should be [A2/2]

    Hint: T in the denominator will be simplified with the T in the numerator after the integration and you have no need to substitue any value instead of T.
     
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