# Fourier Series

 Question 1
The discrete time Fourier series representation of a signal $x[n]$ with period $N$ is written as $x[n]=\sum_{k=0}^{N-1}a_ke^{j(2kn\pi/N)}$. A discrete time periodic signal with period $N=3$, has the non-zero Fourier series coefficients: $a_{-3}=2$ and $a_4=1$. The signal is
 A $2+2e^{-\left ( j\frac{2\pi}{6}n \right )} \cos \left ( \frac{2\pi}{6}n \right )$ B $1+2e^{\left ( j\frac{2\pi}{6}n \right )} \cos \left ( \frac{2\pi}{6}n \right )$ C $1+2e^{\left ( j\frac{2\pi}{3}n \right )} \cos \left ( \frac{2\pi}{6}n \right )$ D $2+2e^{\left ( j\frac{2\pi}{6}n \right )} \cos \left ( \frac{2\pi}{6}n \right )$
GATE EE 2022   Signals and Systems
Question 1 Explanation:
Given: $x(n)=\sum_{k=0}^{N-1}a_ke^{j\frac{2 \pi kn}{N}}$
$\therefore \; x(n)=a_0+a_1e^{\frac{j2\pi n}{3}}+0+... \;\;\;...(1)$
We have,
$a_k=a_{k+N}$
$a_0=a_3$
$a_1=a_4=1$
$a_{-3}=a_0=2$
Put n = 0 in eq. (1)
$x(0)=a_0+a_1=2+1=3$
Put n = 1 in eq. (1)
$x(1)=a_0+a_1e^{j 2 \pi /3}=2+1\left ( \cos \frac{2\pi}{3} +j \sin \frac{2\pi}{3}\right )=2+(-0.5+j0.866)=1.5+j0.866$
These two conditions satisfy by the option (B). Hence, option (B) will be correct.
 Question 2
A periodic function $f(t)$, with a period of $2\pi$, is represented as its Fourier series,
$f(t)=a_0+\sum_{n=1}^{\infty }a_n\cos nt+\sum_{n=1}^{\infty }a_n\sin nt$.
If
$f(t)=\left\{\begin{matrix} A \sin t, & 0\leq t\leq \pi\\ 0,& \pi\lt t\lt 2\pi \end{matrix}\right.$
the Fourier series coefficients $a_1 \; and \; b_1$ of $f(t)$ are
 A $a_1 =\frac{A}{\pi} ; \; b_1=0$ B $a_1 =\frac{A}{2} ; \; b_1=0$ C $a_1 =0 ; \; b_1=\frac{A}{\pi}$ D $a_1 =0 ; \; b_1=\frac{A}{2}$
GATE EE 2019   Signals and Systems
Question 2 Explanation:
\begin{aligned} T_0 &=2\pi \Rightarrow \;\omega _0=\frac{2\pi}{T_0}=1 \\ \text{Now, }b_1 &=\frac{2}{T_0}\int_{0}^{T_0}f(t) \sin \omega _0 t \; dt \\ &= \frac{2}{2 \pi}\int_{0}^{2\pi}f(t) \sin t \; dt\;\;[\because \omega _0=1]\\ &= \frac{1}{\pi}\int_{0}^{\pi} A \sin t \cdot \sin t\; dt \\ &= \frac{A}{2 \pi}\int_{0}^{\pi}2 \sin ^2 t \; dt \\ &= \frac{A}{2 \pi}\int_{0}^{\pi}[1-\cos 2t] \; dt \\ &=\frac{A}{2 \pi} \left [ t-\frac{\sin 2t}{2} \right ]_0^{\pi} \\ &= \frac{A}{2 \pi} [(\pi-0)-(0-0)]=\frac{A}{2}\\ a_1&=\frac{2}{T_0}\int_{0}^{T_0}f(t) \cos \omega _0 t \; dt \\ &= \frac{2}{2 \pi}\int_{0}^{2\pi}f(t) \cos t \; dt \\ &= \frac{1}{\pi}\int_{0}^{\pi} A \sin t \cdot \cos t\; dt \\ &= \frac{A}{2 \pi}\int_{0}^{\pi} \sin 2t \; dt=0 \end{aligned}

 Question 3
Let the signal $x(t)=\sum_{k=-\infty }^{+\infty }(-1)^{k}\delta (t-\frac{k}{2000})$ be passed through an LTI system with frequency response $H(\omega)$ , as given in the figure below The Fourier series representation of the output is given as
 A $4000+4000cos(2000\pi t)$ $+4000cos(4000\pi t)$ B $2000+2000cos(2000\pi t)$ $+2000cos(4000\pi t)$ C $4000cos(2000\pi t)$ D $2000cos(2000\pi t)$
GATE EE 2017-SET-1   Signals and Systems
Question 3 Explanation:
\begin{aligned} x(t)&=\sum_{k=-\infty }^{\infty } (-1)^K \delta \left ( t-\frac{K}{2000} \right )\\ &= \sum_{k=-\infty }^{\infty } (-1)^K \delta (t-KT)\\ \text{where, } &=T=\frac{1}{2000} \end{aligned} Time period:
\begin{aligned} T_0&=2T=\frac{1}{1000}\\ \omega _0&=\frac{2\pi}{T_0}=2000 \pi \; rad \end{aligned}
Since, x(t) is even-halt wave symmetric. So, its expansion will contain only odd harmonics of cos. Therefore, coefficient of fundamental harmonic is
\begin{aligned} a_1&=\frac{2}{T}\int_{-T_0/2}^{T_0/2}x(t) \cos \omega _0 t \; dt\\ &=\frac{4}{T_0}\int_{0}^{T_0/2}\delta (t) \cos \omega _0 t \; dt\\ &=\frac{4}{T_0}\int_{0}^{T_0/2}\delta (t) \cos 0 \; dt\\ &=\frac{4}{T_0}\left [\int_{0}^{T_0/2}\delta (t) \; dt \right ]\\ &=\frac{4}{T_0}=4000 \end{aligned}
Now, frequency components available in expansion are
\begin{aligned} \omega _0,3\omega _0,...\\ 2000\pi, 6000 \pi,... \end{aligned}
As, LTI system given in the question will pass upto $5000\pi$ rad/sec frequency component of input.So, output will have only one component of sfrequency $2000\pi$ rad/sec
Thus, y(t)= expansion of output $=a_1 \cos \omega _0 t=4000 \cos 2000 \pi t$
 Question 4
Consider $g(t)=\left\{\begin{matrix} t-\left \lceil t \right \rceil, & t\geq 0 \\ t-\left \lceil t \right \rceil , & otherwise \end{matrix}\right. , \; \; where\; t \in \mathbb{R}$
Here, $\left \lfloor t \right \rfloor$ represents the largest integer less than or equal to t and $\left \lceil t \right \rceil$ denotes the smallest integer greater than or equal to t. The coefficient of the second harmonic component of the Fourier series representing g(t) is _________.
 A 0 B 1 C 2 D 3
GATE EE 2017-SET-1   Signals and Systems
Question 4 Explanation:
Given that, $g(t)=\left\{\begin{matrix} t-\left \lfloor t \right \rfloor, & t \geq 0\\ t-\left \lceil t \right \rceil & \text{otherwise} \end{matrix}\right.$
where,
$\left \lfloor t \right \rfloor=$ greatest integer less than or equal to 't'.
$\left \lceil t \right \rceil=$ smallest integer greater than or equal to 't'.
Now,  Since, g(t) is nonperiodic. So, there is no Fourier series expansion of this signal and hence no need to calculate harmonic here.
 Question 5
Let f(x) be a real, periodic function satisfying f(-x)=-f(x). The general form of its Fourier series representation would be
 A $f(x)=a_{0}+\sum_{k=1}^{\infty }a_{k}cos(kx)$ B $f(x)=\sum_{k=1}^{\infty }b_{k}sin(kx)$ C $f(x)=a_{0}+\sum_{k=1}^{\infty }a_{2k}cos(kx)$ D $f(x)=\sum_{k=0}^{\infty }a_{2k+1}sin(2k+1)x$
GATE EE 2016-SET-2   Signals and Systems
Question 5 Explanation:
Given that,
$f(-x)=-f(x)$
So, function is an odd function.
So, the fourier series will have sine term only. So,
$f(x)=\sum_{k=1}^{\infty }b_x \sin (kx)$

There are 5 questions to complete.