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Fourier Series Example

Module by: Don Johnson. E-mail the author

Summary: This module provides an example of the Fourier Series representation of a half-wave rectified sinusoid.

Example 1

Let's find the Fourier series representation for the half-wave rectified sinusoid.

st={sin2πtT  if  0t<T20  if  T2t<T s t 2 t T 0 t T 2 0 T 2 t T
(1)
Begin with the sine terms in the series; to find b k b k we must calculate the integral
b k =2T0T2sin2πtTsin2πktTdt b k 2 T t 0 T 2 2 t T 2 k t T
(2)
The key to evaluating such integrals is the classic trigonometric identities.
sinαsinβ=12(cosαβcosα+β) α β 1 2 α β α β
(3)
cosαcosβ=12(cosα+β+cosαβ) α β 1 2 α β α β
(4)
sinαcosβ=12(sinα+β+sinαβ) α β 1 2 α β α β
(5)
Using these identities turns our integral of a product of sinusoids into a sum of integrals of individual sinusoids, which are much easier to evaluate.
2T0T2sin2πtTsin2πktTdt=1T0T2cos2π(k1)tTcos2π(k+1)tTdt={12  if  k=10  if  k (k2)kN 2 T t 0 T 2 2 t T 2 k t T 1 T t 0 T 2 2 k 1 t T 2 k 1 t T 1 2 k 1 0 k k 2 k
(6)
Thus,
b 1 =12 b 1 1 2
(7)
b 2 = b 3 ==0 b 2 b 3 0
(8)

On to the cosine terms. The average value, which corresponds to a 0 a 0 , equals 1π 1 The remainder of the cosine coefficients are easy to find, but yield the complicated result

a k ={(2π)1k21  if  k240  if  k is odd a k 2 1 k 2 1 k 2 4 0 k is odd
(9)

Thus, the Fourier series for the half-wave rectified sinusoid has non-zero terms for the average, the fundamental, and the even harmonics. Plotting the Fourier coefficients reveals at what component frequencies the half-wave rectified sinusoid has energy ( Figure 1 ). Furthermore, this figure shows what the Fourier series sum looks like with these coefficients as we add more and more terms. Presumably, you now believe more in the Fourier series.

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