GATE EE · Signals And Systems

Fourier Transform Practice Questions

Generate GATE-level questions on Fourier Transform. Focus on: 1. Continuous-time Fourier Transform (CTFT) and its properties. 2. Magnitude and Phase spectra. 3. Applications in LTI systems and filter analysis.

21 questions · 20 PYQs · 0 AI practice · GATE EE 2027

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Q1GATE 2025NAT

An ideal low pass filter has frequency response given by

H(j \omega)= \begin{cases}1, & |\omega| \leq 200 \pi \\ 0, & \text { otherwise }\end{cases}

Let $h(t)$ be its time domain representation. Then $h(0)=$ ___________ (round off to the nearest integer).

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📖 Explanation

Step 1: Frequency response

H(j\omega) = \begin{cases} 1, & |\omega| \leq 200\pi \\ 0, & \text{otherwise} \end{cases}

Step 2: Find impulse response $h(t)$ by inverse Fourier transform $h(t) = \frac{1}{2\pi} \int_{-\infty }^{\infty } H(j\omega) e^{j\omega t} d\omega$ But since $H(j\omega) = 1$ for $|\omega| \leq 200\pi$ , and 0 otherwise: $h(t) = \frac{1}{2\pi} \int_{-200\pi}^{200\pi} e^{j\omega t} d\omega$ Step 3: Evaluate the integral This is a standard Fourier inverse transform result: $h(t) = \frac{1}{2\pi} \cdot \frac{2 \sin(200\pi t)}{t} = \frac{200 \sin(200\pi t)}{\pi t}$ This is the sinc function scaled: $h(t) = 200 \cdot \text{sinc}(200 t)$ where $\text{sinc}(x) = \frac{\sin(\pi x)}{\pi x}$ Step 4: Find $h(0)$ Use the standard limit: $\lim_{t \to 0} \frac{\sin(200\pi t)}{\pi t} = 200$ So: $h(0) = \frac{200 \sin(0)}{\pi \cdot 0} = 200$ Final Answer: 200

Q3GATE 2023MCQ

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?

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📖 Explanation

Given, [latex]\quad X(\omega)=

\left\{\begin{array}{l}1, \text { for }|\omega| W_{0}\end{array}\right.

[/latex] $X(\omega)=\operatorname{rect}\left(\frac{\omega}{2 W_{0}}\right)$ 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.

Q7GATE 2021MCQ

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

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📖 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}

Q9GATE 2016MCQ

Suppose $x_{1}(t) \; and \; x_{2}(t)$ have the Fourier transforms as shown below. Which one of the following statements is TRUE?

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📖 Explanation

By observing $X_1(j \omega )$ and $X_2(j \omega )$ , we can say that they are not conjugate symmetric. Since, the fourier transform is not conjugate symmetric the signal will not be real. So, $x_1(t), x_2(t)$ are not real. Now the fourier transform of $x_1(t) \cdot x_2(t)$ will be $\frac{1}{2\pi}X_1(j \omega )*X_2(j \omega )$ and by looking at $X_1(j \omega )$ and $X_2(j \omega )$ , we can say that $X_1(j \omega ) *X_2(j \omega )$ will be conjugate symmetric and thus $x_1(t) \cdot x_2(t)$ will be real. By observing $X_1(j \omega )$ and $X_2(j \omega )$ , we can say, $X_2( \omega )=X_1(- \omega )\overset{IFT}{\rightarrow}x_2(t)=x_1(-t)\;\;...(i)$ Now, $X_1(j \omega )$ is real. Therefore, $x_1(t)$ will be conjugate symmetric.

\begin{aligned} \text{i.e., } x_1(t)&=x_1^*(-t) \\ \downarrow t&= -t\\ x_1(-t)&= x_1^*(t)\\ \text{Now, }x_1(t)\cdot x_2(t) &= x_1(t)\cdot x_1(-t) \;\;(from \;(i))\\ &= x_1(t)\cdot x_1^*(t)\\ &= |x_1(t)|^2=\text{real function} \end{aligned}

Q16GATE 2014MCQ

A signal is represented by

x(t)=\left\{\begin{matrix} 1 &|t| \lt 1 \\ 0& |t|\gt 1 \end{matrix}\right.

The Fourier transform of the convolved signal $y(t)= x(2t)* x(t/2)$ is

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📖 Explanation

Given signal can be drawn as Therefore,

\begin{aligned} &x(t)\leftrightarrow X(\omega )=2Sa(\omega ) \\ &\text{Now, } x(t)\leftrightarrow X(\omega )\\ &\text{then by time scaling,} \\ &x(at)\leftrightarrow \frac{1}{|a|}X(\omega /a) \\ &\therefore \; x(2t)\leftrightarrow Sa \left ( \frac{\omega }{2} \right )\;\;...(i) \\ &x\left ( \frac{t}{2} \right ) \leftrightarrow 4Sa(2\omega )\;\;...(ii)\\ &\text{Now, }y(t)=x(2t) \times x(t/2) \end{aligned}

Convolution in time domain multiplication in frequency domain

\begin{aligned} Y(\omega )&=4Sa\left ( \frac{\omega }{2}\right ) Sa(2\omega ) \\ Y(\omega )&=\frac{4\sin \left ( \frac{\omega }{2}\right ) }{\left ( \frac{\omega }{2}\right ) } \frac{\sin (2\omega )}{2\omega }\\ Y(\omega )&=\frac{4}{\omega ^2}\sin \left ( \frac{\omega }{2}\right ) \sin (2\omega ) \end{aligned}

Q20GATE 2008MCQ

Let $x(t)= rect(t-\frac{1}{2}$ ) (where $rect(x) = 1$ for $-\frac{1}{2}\leq x\leq \frac{1}{2}$ and zero otherwise. Then if $sinc(x)=\frac{sin(\pi x)}{\pi x}$ , the Fourier Transform of $x(t) + x(-t)$ will be given by

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📖 Explanation

\begin{aligned} x(t)&= \text{rect}[t-1/2]\\ x(t)&=1 \;\;\;0\leq t\leq 1\\ &=0 \;\;\; \text{otherwise}\\ F[x(t)]&=\int_{-\infty }^{\infty }x(t)e^{-j\omega t}dt\\ &=\int_{0}^{1}1e^{-j\omega t}dt=\frac{1}{j\omega }(1-e^{-j\omega })\\ &=\frac{2}{\omega }\left[ \frac{e^{j\omega t} -e^{-j\omega /2}}{2j} \right]\$xe^{-j\omega /2}\\ &=\frac{\sin \omega /2}{\omega /2}\cdot e^{-j\omega /2}\\ F[x(t)+x(-t)]&\overset{F.T.}{\rightarrow}X(\omega )+X(-\omega )\\ &=\frac{\sin \omega /2}{\omega /2}\$[ e^{-j\omega /2}+e^{j\omega /2}]\$\\ &=\frac{\sin \omega /2}{\omega /2} 2 \cos \frac{\omega }{2}\\ &=2 \sin c\left ( \frac{\omega }{2 \pi} \right ) \cos \left ( \frac{\omega }{2} \right ) \end{aligned}