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Which one of the following options is the correct Fourier series of the periodic function f(x) described below: f(x) = { 0 if −2 < x < −1 { 2k if −1 < x < 1; period = 4 { 0 if 1 < x < 2

1. f(x) = k/2 + (2k/π) (cos(πx/2) − (1/3) cos(3πx/2) + (1/5) cos(5πx/2) + …)
2. f(x) = k/2 + (2k/π) (sin(πx/2) − (1/3) sin(3πx/2) + (1/5) sin(5πx/2) + …)
3. f(x) = k + (4k/π) (cos(πx/2) − (1/3) cos(3πx/2) + (1/5) cos(5πx/2) + …)
4. f(x) = k + (4k/π) (sin(πx/2) − (1/3) sin(3πx/2) + (1/5) sin(5πx/2) + …)

Correct answer: f(x) = k + (4k/π) (cos(πx/2) − (1/3) cos(3πx/2) + (1/5) cos(5πx/2) + …)

Solution

The pulse has height 2k on -1<x<1 with period 4, so the DC term is (1/4)*2k*2 = k, and a_n = (4k/(n*pi))*sin(n*pi/2) giving 4k/pi for n=1. Since f is even the series uses cosines: f = k + (4k/pi)(cos(pi x/2) - (1/3)cos(3pi x/2) + ...). That is option C, not the stored choice.

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