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Find the Third-Degree Fourier Polynomial For f(t)=f(t)= {0c\left\{\begin{array}{l}0 \\c\end{array}\right. 2<0<\begin{array}{c}-2 <\\0<\end{array}

Question 29

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Find the third-degree Fourier polynomial for f(t) =f(t) ={0c\left\{\begin{array}{l}0 \\c\end{array}\right.2<0<\begin{array}{c}-2 <\\0<\end{array}x0x2\begin{array}{l}x \leq 0 \\x \leq 2\end{array} , where c is a constant, by writing a new function, g(x) =f(t) g(x) =f(t) , with period 2π2 \pi .


A) c22cπsin(πt2) 2c3πsin(3πt2) \frac{c}{2}-\frac{2 c}{\pi} \sin \left(\frac{\pi t}{2}\right) -\frac{2 c}{3 \pi} \sin \left(\frac{3 \pi t}{2}\right)
B) c2+2cπsin(πt2) +2c3πsin(3πt2) \frac{c}{2}+\frac{2 c}{\pi} \sin \left(\frac{\pi t}{2}\right) +\frac{2 c}{3 \pi} \sin \left(\frac{3 \pi t}{2}\right)
C) c2+2cπsin(t) +2c3πsin(3t) \frac{c}{2}+\frac{2 c}{\pi} \sin (t) +\frac{2 c}{3 \pi} \sin (3 t)
D) c22cπsin(t) 2c3πsin(3t) \frac{c}{2}-\frac{2 c}{\pi} \sin (t) -\frac{2 c}{3 \pi} \sin (3 t)

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