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Using the complex form, find the Fourier series of the function. (30%)

f(x) = 1, 2k -. 25 <= x <= 2k + ,25, k E Z

User Hemaolle
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Answer:

The Fourier series of a periodic function f(x) with period 2L can be expressed as:

f(x) = a0/2 + Σ[n=1 to ∞] (ancos(nπx/L) + bnsin(nπx/L))

where

a0 = (1/L) ∫[-L,L] f(x) dx

an = (1/L) ∫[-L,L] f(x)*cos(nπx/L) dx

bn = (1/L) ∫[-L,L] f(x)*sin(nπx/L) dx

In this case, we have f(x) = 1 for 2k - 0.25 <= x <= 2k + 0.25, and f(x) = 0 otherwise. The period is 0.5, so L = 0.25.

First, we can find the value of a0:

a0 = (1/0.5) ∫[-0.25,0.25] 1 dx = 1

Next, we can find the values of an and bn:

an = (1/0.5) ∫[-0.25,0.25] 1*cos(nπx/0.25) dx = 0

bn = (1/0.5) ∫[-0.25,0.25] 1*sin(nπx/0.25) dx

Since the integrand is odd, we have:

bn = (2/0.5) ∫[0,0.25] 1*sin(nπx/0.25) dx

Using the substitution u = nπx/0.25, du/dx = nπ/0.25, dx = 0.25du/(nπ), we get:

bn = (4/nπ) ∫[0,nπ/4] sin(u) du = (4/nπ) (1 - cos(nπ/4))

Therefore, the Fourier series of f(x) can be written as:

f(x) = 1/2 + Σ[n=1 to ∞] [(4/nπ) (1 - cos(nπ/4))] * sin(nπx/0.25)

for 2k - 0.25 <= x <= 2k + 0.25, and f(x) = 0 otherwise.

User Mehrdadep
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