# Fourier Transform

 Question 1
The discrete-time Fourier transform of a signal $x[n]$ is $X(\Omega)=1(1+\cos \Omega) e^{-j \Omega}$. Consider that $x_{p}[n]$ is a periodic signal of period $N=5$ such that

\begin{aligned} x_{p}[n] & =x[n], \text { for } n=0,1,2 \\ & =0, \text { for } n=3,4 \end{aligned}

Note that $x_{p}[n]=\sum_{k=0}^{N-1} a_{k} e^{j \frac{2 \pi}{N} k m}$. The magnitude of the Fourier series coefficient $a_{3}$ is ____ (Round off to 3 decimal places).
 A 0.038 B 0.025 C 0.068 D 0.012
GATE EE 2023   Signals and Systems
Question 1 Explanation:
Given : $x_{p}(n)$ is a period signal of period $N=5$.
$x_{p}(n)=\left\{\begin{array}{cc} x(n), & \text { for } n=0,1,2 \\ 0, & \text { for } n=3,4 \end{array}\right.$

and $\mathrm{x}(\mathrm{n}) \stackrel{\text { DTFT }}{\longrightarrow} \mathrm{X}(\Omega)=(1+\cos \Omega) \mathrm{e}^{-\mathrm{j} \Omega}$

We have,
\begin{aligned} a_{K} & =\frac{X\left(\frac{2 \pi}{N} K\right)}{N} \\ & =\frac{\left[1+\cos \left(\frac{2 \pi}{N} K\right)\right] \mathrm{e}^{-\mathrm{j} \frac{2 \pi}{N} R}}{N} \end{aligned}

For, $\mathrm{N}=5$ and $\mathrm{K}=3$
\begin{aligned} a_{3} & =\frac{\left[1+\cos \left(\frac{2 \pi \times 3}{5}\right)\right] \mathrm{e}^{-\mathrm{j} \frac{2 \pi}{5} \times 3}}{5} \\ \left|a_{3}\right| & =\frac{1}{5}\left[1+\cos \frac{6 \pi}{5}\right] \\ & =0.0382 \end{aligned}
 Question 2
The Fourier transform $X(\omega)$ of the signal $x(t)$ is given by

\begin{aligned} X(\omega) & =1, \text { for }|\omega| \lt W_{0} \\ & =0, \text { for }|\omega| \gt W_{0} \end{aligned}

Which one of the following statements is true?
 A $x(t)$ tends to be an impulse as $\mathrm{W}_{0} \rightarrow \infty$ B $x(0)$ decreases as $W_{0}$ increases C At $\mathrm{t}=\frac{\pi}{2 \mathrm{~W}_{0}}, \mathrm{x}(\mathrm{t})=-\frac{1}{\pi}$ D At $\mathrm{t}=\frac{\pi}{2 \mathrm{~W}_{0}}, \mathrm{x}(\mathrm{t})=\frac{1}{\pi}$
GATE EE 2023   Signals and Systems
Question 2 Explanation:
Given,
$\quad X(\omega)=\left\{\begin{array}{l}1, \text { for }|\omega|W_{0}\end{array}\right.$

$X(\omega)=\operatorname{rect}\left(\frac{\omega}{2 W_{0}}\right)$

We know,
$\operatorname{Arect}\left(\frac{\mathrm{t}}{\tau}\right)=\operatorname{A\tau S}\left(\frac{\omega \tau}{2}\right)$

By duality,
$\operatorname{A} \tau \mathrm{Sa}\left(\frac{\mathrm{t} \tau}{2}\right) \leftrightarrow 2 \pi \operatorname{Arect}\left(\frac{\omega}{\tau}\right)$

Given,
$2 \pi \mathrm{A}=1$
$\Rightarrow \quad A=\frac{1}{2 \pi}$
$\therefore \quad \mathrm{x}(\mathrm{t})=\frac{\tau}{2 \pi} \mathrm{Sa}\left(\frac{\mathrm{t} \tau}{2}\right)$, where $\tau=2 \mathrm{~W}_{0}$
Thus,
\begin{aligned} x(t)&=\frac{W_0}{\pi} \mathrm{Sa}(W_0 t)\\ &=\frac{1}{\pi}\frac{\sin W_0 t}{t}\\ \therefore \quad \text{At } t&=\frac{\pi}{2W_0}, x(t)=\frac{1}{\pi t} \end{aligned}
From rectangular function, At $W \Rightarrow \infty ,X(w)=1$
Taking inverse fourier transform $x(t)=\delta (t)$

Option (A) will be correct.

 Question 3
Let an input $x(t)=2 \sin (10 \pi t)+5 \cos (15 \pi t)+7 \sin (42 \pi t)+4 \cos (45 \pi t)$ is passed through an LTI system having an impulse response,
$h(t)=2\left ( \frac{\sin (10 \pi t)}{\pi t} \right ) \cos (40\pi t)$
The output of the system is
 A $2 \sin (10 \pi t)+5\cos (15 \pi t)$ B $7 \sin (42 \pi t)+5\cos (15 \pi t)$ C $7 \sin (42 \pi t)+4\cos (45 \pi t)$ D $2 \sin (10 \pi t)+4\cos (45 \pi t)$
GATE EE 2022   Signals and Systems
Question 3 Explanation:
Given: $h(t)=2\left ( \frac{\sin (10 \pi t)}{\pi t} \right ) \cos 40 \pi t$
Fourier transform of signal $\frac{\sin (10 \pi t)}{\pi t}$ is given by

Now, impulse response
$h(t)=2\left ( \frac{\sin (10 \pi t)}{\pi t} \right )\cdot \left ( \frac{e^{+j40\pi t}+e^{-j40\pi t}}{2} \right )$
Using property, $e^{-j\omega _0t}x(t)\rightleftharpoons X(\omega +\omega _0)$
Therefore, Fourier transform of impulse response

Cut-off frequencies,
$\omega_L=30 \pi rad/sec$
$\omega_H=50 \pi rad/sec$
Thus, output of the system $=7 \sin 42 \pi t+4 \cos 45 \pi t$
 Question 4
Consider a continuous-time signal $x(t)$ defined by $x(t)=0$ for $\left | t \right |> 1$, and $x\left ( t \right )=1-\left | t \right | for \left | t \right |\leq 1$. Let the Fourier transform of $x(t)$ be defined as $X\left ( \omega \right )=\int\limits_{-\infty }^{\infty }\:x\left ( t \right )e^{-j\omega t}\:dt$. The maximum magnitude of $X\left ( \omega \right )$ is ___________.
 A 1 B 2 C 3 D 4
GATE EE 2021   Signals and Systems
Question 4 Explanation:
Fourier transform, $F(\omega)=A \tau S a^{2}\left(\frac{\omega \tau}{2}\right)$
\begin{aligned} \text { As } A=1, \tau=1 \\ H \omega) &=S a^{2}\left(\frac{\omega}{2}\right) \\ \left.F(\omega)\right|_{\text {peak }} &=F(0)=S a^{2}(0)=1 \end{aligned}
$\because$ Peak value of sampling function occurs at $\omega=0$
Peak value $=1$

 Question 5
Let $f(t)$ be an even function, i.e. $f(-t)=f(t)$ for all t. Let the Fourier transform of $f(t)$ be defined as $F\left ( \omega \right )=\int\limits_{-\infty }^{\infty }\:f\left ( t \right )e^{-j\omega t}dt$. Suppose $\dfrac{dF\left ( \omega \right )}{d\omega }=-\omega F\left ( \omega \right )$ for all $\omega$, and $F(0)=1$. Then
 A $f\left ( 0 \right )\lt 1$ B $f\left ( 0 \right ) \gt 1$ C $f\left ( 0 \right )= 1$ D $f\left ( 0 \right )= 0$
GATE EE 2021   Signals and Systems
Question 5 Explanation:
$f(t) \rightleftharpoons F(\omega)=\int_{-\infty}^{\infty} f(t) e^{-j \omega t} d t$
The following informations are given about $f(t) \rightleftharpoons F(\omega).$
$\begin{array}{l} (i) f(t)=f(-t)\\ (ii) \left.F(\omega)\right|_{\omega=0}=1\\ (iii) \frac{d F(\omega)}{d \omega}=-\omega F(\omega)\\ \text{From }(iii), \frac{d F(\omega)}{d \omega}+\omega F(\omega)=0 \end{array}$
By solving the above linear differential equations, (by mathematics)
\begin{aligned} \ln F(\omega) &=-\frac{\omega^{2}}{2} \\ \Rightarrow \qquad\qquad F(\omega) &=K \cdot e^{-\omega^{2} / 2} \\ \text { Put } \omega=0, \qquad\qquad F(0) &=K \\ \Rightarrow \qquad\qquad 1 &=K \text { (from info. } \\ \text { From (iv), } \qquad\qquad F(\omega) &=e^{-\omega^{2} / 2} \\ \text{As we know}, e^{-a t^{2}}, a \gt 0 &\rightleftharpoons \sqrt{\frac{\pi}{a}} e^{-\omega^{2} / 4 a}\\ \text{At }a=\frac{1}{2}, \qquad\qquad\quad e^{-t^{2} / 2} &\rightleftharpoons \sqrt{\frac{\pi}{1 / 2}} \cdot e^{-\omega^{2} / 2}\\ f(t)&=\frac{1}{\sqrt{2 \pi}} e^{-t^{2} / 2} \rightleftharpoons e^{-\omega^{2} / 2}=F(\omega)\\ \text { Thus, }\qquad\qquad f(t)&=\frac{1}{\sqrt{2 \pi}} e^{-t^{2} / 2} \\ \text { At } t=0, \qquad\qquad f(0)&=\frac{1}{\sqrt{2 \pi}} \lt 1 \end{aligned}

There are 5 questions to complete.