A periodic function f(t), with a period of 2 \pi, is represented as its Fourier series, f(t)=a_{0}+\sum_{n-1}^{\infty} a_{n} \cos n t+\sum_{n-1}^{\infty} b_{n} \sin n t If {f}( {t})=\\begin{array}{c}{A\end{array} \sin {t}\text{,}, 0 \leq 1 \leq \pi ;0\text{,},\pi t 2\pi the Fourier series coefficients a_{1} and b_{1} of f(t) are

{a}_{1}=0 ; {b}_{1}=\frac{ {A}}{2}

{a}_{1}=\frac{ {A}}{\pi} ; {b}_{1}=0

GATE Electrical Engineering (EE) 2019 Solved Paper

Question : 41 of 65

Marks: +1, -0

A periodic function

f(t)

, with a period of

2 π

, is represented as its Fourier series,

f(t)=a_{0}+∑↙{n-1}↖{∞} a_{n} \cos n t+∑↙{n-1}↖{∞} b_{n} \sin n t

{f}( {t})=\{\table {A} \sin {t}\text ",", 0 ≤ 1 ≤ π ;0\text ",",π\;t\;2π

the Fourier series coefficients

a_{1}

and

b_{1}

f(t)

are

Solution:

Half wave rectifier;

{EFS}: {X}_{1}=-\;{ {j} {A}}/{4}=\;{1}/{2}( {a}_{1}- {j} {b}_{1})

{a}_{1}=0 ;-\;{ {j} {A}}/{4}=\;{- {jb}_{1}}/{2} ⇒ {b}_{1}=\;{ {A}}/{2}

Go to Question:

Prev Question

Next Question