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Find the Fourier Series for the Square Wave (Of Period 2π2 \pi

Question 9

Multiple Choice

Find the Fourier series for the square wave (of period 2π2 \pi ) given by f(x) ={1,πx<01,0xπ\mathrm{f}(\mathrm{x}) =\left\{\begin{array}{cc}-1, & -\pi \leq \mathrm{x}<0 \\ 1, & 0 \leq \mathrm{x} \leq \pi\end{array}\right.


A) n=02πsin[(2n+1) x]2n+1=2π[sinx+sin3x3+sin5x5+sin7x7+]\sum_{n=0}^{\infty} \frac{2}{\pi} \cdot \frac{\sin [(2 n+1) x]}{2 n+1}=\frac{2}{\pi}\left[\sin x+\frac{\sin 3 x}{3}+\frac{\sin 5 x}{5}+\frac{\sin 7 x}{7}+\ldots\right]
B) n=04πsin[(2n+1) x]2n+1=4π[sinx+sin3x3+sin5x5+sin7x7+]\sum_{n=0}^{\infty} \frac{4}{\pi} \cdot \frac{\sin [(2 n+1) x]}{2 n+1}=\frac{4}{\pi}\left[\sin x+\frac{\sin 3 x}{3}+\frac{\sin 5 x}{5}+\frac{\sin 7 x}{7}+\ldots\right]
C) n=04πcos[(2n+1) x]2n+1=4π[cosx+cos3x3+cos5x5+cos7x7+]\sum_{n=0}^{\infty} \frac{4}{\pi} \cdot \frac{\cos [(2 n+1) x]}{2 n+1}=\frac{4}{\pi}\left[\cos x+\frac{\cos 3 x}{3}+\frac{\cos 5 x}{5}+\frac{\cos 7 x}{7}+\ldots\right]
D) n=02πcos[(2n+1) x]2n+1=2π[cosx+cos3x3+cos5x5+cos7x7+]\sum_{n=0}^{\infty} \frac{2}{\pi} \cdot \frac{\cos [(2 n+1) x]}{2 n+1}=\frac{2}{\pi}\left[\cos x+\frac{\cos 3 x}{3}+\frac{\cos 5 x}{5}+\frac{\cos 7 x}{7}+\ldots\right]

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