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$An alternating current after passing through a rectifier has the form i={[i_(0)sin x",",0 <= x <= pi],[0",",pi <= x <= 2pi]:} where i_(o) is the maximum current and the period is 2pi. Express ' i ' in a Fourier series. [Ans. i=(i_(0))/(2)+(i_(0))/(2)sin x-(2i_(0))/(pi)sum_("even ")^(oo)(cos nx)/(n^(2)-1)$

$3$ . An alternating current after passing through a rectifier has the form $i=\left\{\begin{array}{ll}i_{0} \sin x, & 0 \leq x \leq \pi \\ 0, & \pi \leq x \leq 2 \pi\end{array}\right.$ $\newline$ where $i_{\mathrm{o}}$ is the maximum current and the period is $2 \pi$ . Express ' $i$ ' in a Fourier series. $\newline$ [Ans. $i=\frac{i_{0}}{2}+\frac{i_{0}}{2} \sin x-\frac{2 i_{0}}{\pi} \sum_{\text {even }}^{\infty} \frac{\cos n x}{n^{2}-1}$

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Find derivatives of sine and cosine functions

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3

. An alternating current after passing through a rectifier has the form

i=\left\{\begin{array}{ll}i_{0} \sin x, & 0 \leq x \leq \pi \\ 0, & \pi \leq x \leq 2 \pi\end{array}\right.

\newline

where

i_{\mathrm{o}}

is the maximum current and the period is

2 \pi

. Express '

i

' in a Fourier series.

\newline

[Ans.

i=\frac{i_{0}}{2}+\frac{i_{0}}{2} \sin x-\frac{2 i_{0}}{\pi} \sum_{\text {even }}^{\infty} \frac{\cos n x}{n^{2}-1}

Question Prompt: Question Prompt: Express the given piecewise function for alternating current in a Fourier series.
Step $1$ : Step $1$ : Identify the function to be transformed into a Fourier series. The function is defined piecewise: $\newline$ $i(x) = \begin{cases} i_0 \sin(x) & \text{for } 0 \leq x \leq \pi, \ 0 & \text{for } \pi < x \leq 2\pi. \end{cases}$
Step $2$ : Step $2$ : Write the general form of a Fourier series. A Fourier series is given by: $i(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty}(a_n \cos(nx) + b_n \sin(nx))$ .
Step $3$ : Step $3$ : Calculate the constant term $a_0$ , which is the average value of the function over one period $[0, 2\pi]$ : $\newline$ $a_0 = \frac{1}{\pi} \int_{0}^{2\pi} i(x) \, dx$ $\newline$ $= \frac{1}{\pi} \left(\int_{0}^{\pi} i_0 \sin(x) \, dx + \int_{\pi}^{2\pi} 0 \, dx\right)$ $\newline$ $= \frac{1}{\pi} \left(i_0 \left[ -\cos(x) \right]_{0}^{\pi}\right)$ $\newline$ $= \frac{1}{\pi} \left(i_0 \left[ -\cos(\pi) + \cos(0) \right]\right)$ $\newline$ $= \frac{1}{\pi} \left(i_0 \left[ 1 + 1 \right]\right)$ $\newline$ $= \frac{2i_0}{\pi}.$