Dwight, we haveĬalculating the power directly, wecompute the energy in one period and divide by the period: Note that we squareand multiply by 1 for the DC offset, (i.e., constant offset), a &nu,since this is the average value of 1 2.įor our Fourier series we have only odd coefficientsfor sines:įrom Tables of Integrals and OtherMathematical Data by Herbert B. We calculate the average power in termsof Fourier series coefficients by squaring coefficients and multiplying by 1/2(since this is the average value of a sinusoid squared). )įor k odd, −cos(π k ) = 1, andwe obtain our final answer: (Note that we may even use a value for T that changes with k. Because of shift-flip symmetry, we need only integratefrom 0 to T /2 and double the value. The square wave, having shift-flip symmetry, has only oddnumbered terms.ĭirect calculation gives the coefficientsfor the Fourier series. The square wave, being an odd function,has only sine terms. Sol'n: Consider a square wave, v( t ), of period T that is an odd function (positive with value A from t = 0to t = T /2 and value − A from t = T /2 to t = T ). Ex: Verifythat the power of a square wave calculated directly from equals the power calculated fromFourier coefficients.Īns: Powerin both cases is p = A 2 where A is the amplitude of the square wave.
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