Definition of Fourier Transform MCQ Quiz in తెలుగు - Objective Question with Answer for Definition of Fourier Transform - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Given, inverse Fourier of $\mathrm{X}\left({\mathrm{e}}^{\mathrm{j}\omega }\right)$ is $\mathrm{x}\left[\mathrm{n}\right]$ then,

${\mathrm{x}}_{\mathrm{e}}\left[\mathrm{n}\right]=\frac{\mathrm{x}\left[\mathrm{n}\right]+\mathrm{x}\left[-\mathrm{n}\right]}{2}\stackrel{\mathrm{F}\mathrm{T}}{\leftrightarrow }\mathrm{R}\mathrm{e}\left\{\mathrm{X}\left({\mathrm{e}}^{\mathrm{j}\omega }\right)\right\}=\mathrm{P}\left({\mathrm{e}}^{\mathrm{j}\omega }\right)$

and, ${\mathrm{x}}_{\mathrm{o}}\left[\mathrm{n}\right]=\frac{\mathrm{x}\left[\mathrm{n}\right]-\mathrm{x}\left[-\mathrm{n}\right]}{2}\stackrel{\mathrm{F}\mathrm{T}}{\leftrightarrow }\mathrm{j}\mathrm{Q}\left({\mathrm{e}}^{\mathrm{j}\omega }\right)$

Thus, inverse Fourier of $\mathrm{Q}\left({\mathrm{e}}^{\mathrm{j}\omega }\right)$ is $-\mathrm{j}{\mathrm{x}}_{\mathrm{o}}\left[\mathrm{n}\right]$

And, $\mathrm{P}\left({\mathrm{e}}^{\mathrm{j}\omega }\right){\mathrm{e}}^{-\mathrm{j}\omega }\stackrel{\mathrm{F}{\mathrm{T}}^{-1}}{\leftrightarrow }{\mathrm{x}}_{\mathrm{e}}\left[\mathrm{n}-1\right]$

Thus, inverse Fourier of $\mathrm{Y}\left({\mathrm{e}}^{\mathrm{j}\omega }\right),\mathrm{y}\left[\mathrm{n}\right]$is $\mathrm{y}\left[\mathrm{n}\right]={\mathrm{x}}_{\mathrm{e}}\left[\mathrm{n}-1\right]-\mathrm{j}{\mathrm{x}}_{\mathrm{o}}\left[\mathrm{n}\right]$.

we have $\mathrm{x}\left[\mathrm{n}\right]=\left\{\begin{array}{c}\begin{array}{ccc}-1& 1& 2\end{array}\\ \uparrow \end{array}\right\}$

${\mathrm{x}}_{\mathrm{e}}\left[\mathrm{n}\right]=\left\{\frac{1}{2},{}_{\uparrow }^1,\frac{1}{2}\right\}\Rightarrow {\mathrm{x}}_{\mathrm{e}}\left[\mathrm{n}-1\right]=\left\{\frac{1}{\begin{array}{c}2\\ \uparrow \end{array}},1,\frac{1}{2}\right\}$

and, ${\mathrm{x}}_{\mathrm{o}}\left[\mathrm{n}\right]=\left\{\frac{-3}{2}{,}_{\uparrow }^0,\frac{3}{2}\right\}\Rightarrow -\mathrm{j}{\mathrm{x}}_{\mathrm{o}}\left[\mathrm{n}\right]=\left\{\mathrm{j}\frac{3}{2}{,}_{\uparrow }^0,-\mathrm{j}\frac{3}{2}\right\}$

$\Rightarrow \mathrm{y}\left[\mathrm{n}\right]=\left\{\begin{array}{c}\begin{array}{ccc}\left(\mathrm{j}\frac{3}{2}\right),& \frac{1}{2},& \begin{array}{cc}\left(1-\mathrm{j}\frac{3}{2}\right),& \frac{1}{2}\end{array}\end{array}\\ \begin{array}{ccc}\uparrow & & \end{array}\end{array}\right\}$

We know the Fourier pair,  $\mathrm{x}\left(\mathrm{t}\right)=\frac{\sin \mathrm{t}}{\pi \mathrm{t}}\stackrel{\mathrm{F}\mathrm{T}}{\leftrightarrow }\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\left(\frac{\omega }{2}\right)$

Now, let $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{x}}^2\left(\mathrm{t}\right)\stackrel{\mathrm{F}\mathrm{T}}{\leftrightarrow }\mathrm{Y}\left(\mathrm{j}\omega \right)$

Then, $\mathrm{y}\left(\mathrm{t}\right)={\left(\frac{\sin \mathrm{t}}{\pi \mathrm{t}}\right)}^2\stackrel{\mathrm{F}\mathrm{T}}{\leftrightarrow }\mathrm{Y}\left(\mathrm{j}\omega \right)=\frac{1}{2\pi }\left[\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\left(\frac{\omega }{2}\right)\ast \mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\left(\frac{\omega }{2}\right)\right]$

Now, let $\mathrm{z}\left(\mathrm{t}\right)=\mathrm{t}\mathrm{y}\left(\mathrm{t}\right)$ then,

$\mathrm{t}{\left(\frac{\sin \mathrm{t}}{\pi \mathrm{t}}\right)}^2\stackrel{\mathrm{F}\mathrm{T}}{\leftrightarrow }\mathrm{j}\frac{\mathrm{d}\mathrm{Y}\left(\mathrm{j}\omega \right)}{\mathrm{d}\omega }=\mathrm{Z}\left(\mathrm{j}\omega \right)$

Now, using Parseval’s theorem, we have

$\begin{array}{l}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\left|\mathrm{z}\left(\mathrm{t}\right)\right|}^2\mathrm{d}\mathrm{t}=\frac{1}{2\pi }\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\left|\mathrm{Z}\left(\mathrm{j}\omega \right)\right|}^2\mathrm{d}\omega \\ \Rightarrow \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\mathrm{t}}^2{\left(\frac{\sin \mathrm{t}}{\pi \mathrm{t}}\right)}^4\mathrm{d}\mathrm{t}=\frac{1}{2\pi }\left[{\left(\frac{1}{2\pi }\right)}^2\times 4\right]=\frac{1}{2{\pi }^3}\end{array}$

Thus, $\Rightarrow \mathrm{A}=\frac{1}{2{\pi }^3}$.

We have inverse Fourier of$\frac{\sin \left(3\omega \right)}{\omega }$ given by,

Thus,${\mathrm{x}}_1\left(\mathrm{t}\right)=\frac{1}{2}\left[\mathrm{u}\left(\mathrm{t}+3\right)-\mathrm{u}\left(\mathrm{t}-3\right)\right]$

Using multiplication property we have

${\mathrm{x}}_2\left(\mathrm{t}\right)={\mathrm{x}}_1\left(\mathrm{t}\right)\ast {\mathrm{x}}_1\left(\mathrm{t}\right)\stackrel{\mathrm{F}\mathrm{T}}{\leftrightarrow }\frac{1}{2\pi }\left[{\mathrm{X}}_1\left(\mathrm{j}\omega \right){\mathrm{X}}_1\left(\mathrm{j}\omega \right)\right]=\frac{1}{2\pi }\cdot \frac{{\sin }^2\left(3\omega \right)}{{\omega }^2}$

Thus, inverse Fourier of ${\mathrm{X}}_2\left(\mathrm{j}\omega \right)=\frac{{\sin }^2\left(3\omega \right)}{{\omega }^2}$ is

$\begin{array}{l}\Rightarrow {\mathrm{x}}_2\left(\mathrm{t}\right)=\frac{1}{2}\left[\mathrm{u}\left(\mathrm{t}+3\right)-\mathrm{u}\left(\mathrm{t}-3\right)\right]\ast \frac{1}{2}\left[\mathrm{u}\left(\mathrm{t}+3\right)-\mathrm{u}\left(\mathrm{t}-3\right)\right]\\ \Rightarrow {\mathrm{x}}_2\left(\mathrm{t}\right)=\frac{1}{4}\left[\mathrm{r}\left(\mathrm{t}+6\right)-2\mathrm{r}\left(\mathrm{t}\right)+\mathrm{r}\left(\mathrm{t}-3\right)\right]\end{array}$

Now ${\mathrm{X}}_3\left(\mathrm{j}\omega \right)=\frac{2{\sin }^2\left(3\omega \right)\cos \omega }{{\omega }^2}=2\left[\frac{1}{2}{\mathrm{e}}^{\mathrm{j}\omega }{\left(\frac{\sin 3\omega }{\omega }\right)}^2+\frac{1}{2}{\mathrm{e}}^{-\mathrm{j}\omega }{\left(\frac{\sin 3\omega }{\omega }\right)}^2\right]$

$\begin{array}{l}\Rightarrow {\mathrm{x}}_3\left(\mathrm{t}\right)={\mathrm{x}}_2\left(\mathrm{t}+1\right)+{\mathrm{x}}_2\left(\mathrm{t}-1\right)\\ \Rightarrow {\mathrm{x}}_3\left(\mathrm{t}\right)=\frac{1}{4}\left[\mathrm{r}\left(\mathrm{t}+7\right)-2\mathrm{r}\left(\mathrm{t}+1\right)+\mathrm{r}\left(\mathrm{t}-5\right)\right]+\frac{1}{4}\left[\mathrm{r}\left(\mathrm{t}+5\right)-2\mathrm{r}\left(\mathrm{t}-1\right)+\mathrm{r}\left(\mathrm{t}-7\right)\right]\end{array}$

Net slope for $\mathrm{t}\in \left(-1,1\right)$ is $\left(-1\right)+\left(1\right)=0$

Thus $\mathrm{h}\left(\mathrm{t}\right)$ is constant for $-1<t<1$.

$\frac{\mathrm{x}\left(\mathrm{t}\right)+\mathrm{x}\left(-\mathrm{t}\right)}{2}\leftrightarrow \frac{\mathrm{X}\left(\mathrm{j}\omega \right)+\mathrm{X}\left(-\mathrm{j}\omega \right)}{2}$

$\Rightarrow \frac{\mathrm{x}\left(\mathrm{t}\right)+{\mathrm{x}}^{\ast }\left(-\mathrm{t}\right)}{2}\leftrightarrow \frac{\mathrm{X}\left(\mathrm{j}\omega \right)+{\mathrm{X}}^{\ast }\left(\mathrm{j}\omega \right)}{2}=\mathrm{R}\mathrm{e}\left\{\mathrm{X}\left(\mathrm{j}\omega \right)\right\}$         ($\therefore {\mathrm{x}}^{\ast }\left(-\mathrm{t}\right)=\mathrm{x}\left(-\mathrm{t}\right)$as$\mathrm{x}\left(\mathrm{t}\right)$ is real)

$\Rightarrow \mathrm{E}\mathrm{v}\mathrm{e}\mathrm{n}\left\{\mathrm{x}\left(\mathrm{t}\right)\right\}\leftrightarrow \mathrm{R}\mathrm{e}\left\{\mathrm{X}\left(\mathrm{j}\omega \right)\right\}$

Thus for a real signal, Fourier transform of even part of time domain signal is the real part of frequency domain signal.

$\mathrm{E}\mathrm{v}\mathrm{e}\mathrm{n}\left\{\mathrm{x}\left(\mathrm{t}\right)\right\}=\frac{\mathrm{x}\left(\mathrm{t}\right)+\mathrm{x}\left(-\mathrm{t}\right)}{2}\stackrel{\mathrm{F}\mathrm{T}}{\leftrightarrow }\mathrm{R}\mathrm{e}\left\{\mathrm{X}\left(\mathrm{j}\omega \right)\right\}$

Now, we are given $\mathrm{F}{\mathrm{T}}^{-1}\left\{\mathrm{R}\mathrm{e}\left\{\mathrm{X}\left(\mathrm{j}\omega \right)\right\}\right\}=\left|\mathrm{t}\right|{\mathrm{e}}^{-\left|\mathrm{t}\right|}$

Therefore, $\mathrm{E}\mathrm{v}\mathrm{e}\mathrm{n}\left\{\mathrm{x}\left(\mathrm{t}\right)\right\}=\frac{\mathrm{x}\left(\mathrm{t}\right)+\mathrm{x}\left(-\mathrm{t}\right)}{2}=\left|\mathrm{t}\right|{\mathrm{e}}^{-\left|\mathrm{t}\right|}$

Given, $\mathrm{x}\left(\mathrm{t}\right)=0$ for $\mathrm{t}\le 0$ (as x(t) is causal)

$\Rightarrow \mathrm{x}\left(-\mathrm{t}\right)=0$for$\mathrm{t}\ge 0$

Thus, we can write $\mathrm{x}\left(\mathrm{t}\right)=2\left|\mathrm{t}\right|{\mathrm{e}}^{-\left|\mathrm{t}\right|}$, $\mathrm{t}\ge 0$

$\Rightarrow \mathrm{x}\left(\mathrm{t}\right)=2\left|\mathrm{t}\right|{\mathrm{e}}^{-\left|\mathrm{t}\right|}\mathrm{u}\left(\mathrm{t}\right)$

$\Rightarrow \mathrm{x}\left(\mathrm{t}\right)=2\mathrm{t}{\mathrm{e}}^{-\mathrm{t}}\mathrm{u}\left(\mathrm{t}\right)$

Let there be function y(t) such that

$\frac{dy\left(t\right)}{dt}=x\left(t\right)$

And $x\left(t\right)\stackrel{F.T}{\leftrightarrow }X\left(\omega \right)$

Let $y\left(t\right)=\frac{-2}{\left(1+t^2\right)}$

Then $y\left(\omega \right)=\frac{-2}{\left(1+{\omega }^2\right)}$

Then inverse Fourier of y(ω) is $-e^{-\left|t\right|}\left(=w\left(t\right)let\right)$

Then Fourier transform of y(t) is 2π w(-ω)

Thus, $\Rightarrow 2\pi w\left(-\omega \right)=-2\pi e^{-\left|\omega \right|}$

Now Fourier transform of $\frac{dy\left(t\right)}{dt}$ is jω.FT{y(t)}

$\begin{array}{l}\Rightarrow FT\left\{\frac{dy\left(t\right)}{dt}\right\}=j\omega .FT\left\{y\left(t\right)\right\}=-j\omega .2\pi e^{-\left|\omega \right|}\\ \Rightarrow FT\left\{x\left(t\right)\right\}=-j2\pi \omega .e^{-\left|\omega \right|}\\ \Rightarrow X\left(\omega \right)=-j\omega 2\pi e^{-\left|\omega \right|}\end{array}$

Now $\left|X\left(\omega \right)\right|=\omega .2\pi e^{-\left|\omega \right|}$

Thus ${\left|X\left(\omega \right)\right|}_{\omega =0}=0$

Note we have used the duality property and differentiation in time domain to solve this problem.

We have $X\left(j\omega \right)=2\left[u\left(\omega +3\right)-u\left(\omega -3\right)\right]e^{j\left(\frac{-3\omega }{2}+\pi \right)}$

$\Rightarrow X\left(j\omega \right)=-2\left[u\left(\omega +3\right)-u\left(\omega -3\right)\right]e^{-j\frac{3\omega }{2}}$

Let $X_1\left(\mathrm{j}\omega \right)=-2\left[\mathrm{u}\left(\omega +3\right)-\mathrm{u}\left(\omega -3\right)\right]$

$\Rightarrow x_1\left(t\right)=-2\left(\frac{\sin 3t}{\pi t}\right)$

Now inverse Fourier of $X\left(j\omega \right)=X_1\left(j\omega \right)e^{\frac{-j3\omega }{2}}$ is $x_1\left(t-\frac{3}{2}\right)$

Thus,

$\begin{array}{l}x\left(t\right)=x_1\left(t-\frac{3}{2}\right)\\ \Rightarrow x\left(t\right)=-2\left[\frac{\sin 3\left(t-\frac{3}{2}\right)}{\pi \left(t-\frac{3}{2}\right)}\right]\end{array}$

Creating the tum $\frac{\sin \theta }{\theta }$ we have

Thus $x\left(t\right)att=\frac{3}{2}=\frac{-6}{\pi }=-1.9098$

Let $s\left(t\right)=\frac{dx\left(t\right)}{dt}=g\left(t\right)+f\left(t\right)$

$\Rightarrow S\left(j\omega \right)=\left(\frac{2sin\omega }{\omega }\right)-e^{j\omega }-e^{-j\omega }$  ________ (1)

Now

$\begin{array}{l}x\left(t\right)=\underset{-\mathrm{\infty }}{\overset{t}{\int }}\left[\frac{dx\left(t\right)}{dt}\right]dt=\underset{-\mathrm{\infty }}{\overset{t}{\int }}s\left(t\right)dt\\ \Rightarrow X\left(j\omega \right)=\frac{S\left(j\omega \right)}{j\omega }+\pi S\left(0\right)\delta \left(\omega \right)\end{array}$

from (1) $S\left(0\right)=\frac{2\sin \omega }{\omega }-e^{j\omega }-{e^{-j\omega }|}_{\omega =0}=0$

$\begin{array}{l}\Rightarrow X\left(j\omega \right)=\frac{S\left(j\omega \right)}{j\omega }=\frac{1}{j\omega }\left(G\left(j\omega \right)+F\left(j\omega \right)\right)\\ \left(s\left(t\right)=g\left(t\right)+f\left(t\right)\Rightarrow S\left(j\omega \right)=G\left(j\omega \right)+F\left(j\omega \right)\right)\end{array}$

Now $G\left(j\omega \right)=\frac{2sin\omega }{\omega }$ and $F\left(j\omega \right)=-e^{j\omega }-e^{-j\omega }=-2\mathrm{c}\mathrm{o}\mathrm{s}\omega$

Now $\frac{X\left(j\omega \right)}{G\left(j\omega \right)}=\frac{1}{j\omega }\left(1+\frac{F\left(j\omega \right)}{G\left(j\omega \right)}\right)=\frac{1}{j\omega }\left(1+\frac{-2\cos \omega }{2\left(\frac{\sin \omega }{\omega }\right)}\right)$

$\Rightarrow \frac{X\left(j\omega \right)}{G\left(j\omega \right)}=\frac{1}{j\omega }\left(1-\omega \cot \omega \right)$

On another approach we see that three options have $\frac{1}{j\omega }$ term in them, which prompts that x(t) is integral of g(t) (as people usually forget the impulses) and then when you try, you get $X\left(j\omega \right)=\frac{G\left(j\omega \right)}{j\omega }+\pi G\left(0\right)\delta \left(\omega \right)$ (remember you have missed the impulses) $\Rightarrow \frac{X\left(j\omega \right)}{G\left(j\omega \right)}=\frac{1}{j\omega }+\frac{\pi \cdot 2\cdot \delta \left(\omega \right)}{2\frac{\sin \omega }{\omega }}$

$\left(G(0\right)=2andG\left(j\omega \right)=\frac{2\sin \omega }{\omega })$ and then you realise that no option matches. After this you rethink, and if you can discover that

$\frac{dx\left(t\right)}{dt}\ne g\left(t\right)$ but $\frac{dx\left(t\right)}{dt}=g\left(t\right)+f\left(t\right)$

$\begin{array}{l}\Rightarrow x\left(t\right)=\underset{-\mathrm{\infty }}{\overset{t}{\int }}\left(g\left(t\right)+f\left(t\right)\right)dt\\ \Rightarrow X\left(j\omega \right)=\frac{1}{j\omega }\left(G\left(j\omega \right)+F\left(j\omega \right)\right)+\pi \left(G(0\right)+F\left(0\right)\delta \left(\omega \right)\\ G\left(j\omega \right)+F\left(j\omega \right)=\frac{2\sin \omega }{\omega }-e^{j\omega }-e^{-j\omega }=\frac{2\sin \omega }{\omega }-2\cos \omega \end{array}$

And $G\left(j0\right)+F\left(j0\right)=0$

$\Rightarrow \frac{X\left(j\omega \right)}{G\left(j\omega \right)}=\frac{1}{j\omega }\left(1+\frac{F\left(j\omega \right)}{G\left(j\omega \right)}\right)$

Here you see that a $\cot \omega$ term comes and you can mark option d without solving further.

$e^{-b\left|t\right|}\stackrel{FT}{\leftrightarrow }\frac{2b}{{\omega }^2+b^2}$

Using duality, $\frac{2b}{t^2+b^2}\stackrel{FT}{\leftrightarrow }2\pi e^{-b\left|-\omega \right|}$

$\Rightarrow \frac{1}{t^2+b^2}\stackrel{FT}{\leftrightarrow }\frac{\pi }{b}{e}^{-b\left|\omega \right|}$

We have the Nyquist rate for $\mathrm{x}\left(\mathrm{t}\right)$ is ${\omega }_{\mathrm{o}}$.

$\Rightarrow$ Highest frequency component of $\mathrm{x}\left(\mathrm{t}\right)$ is $\frac{{\omega }_{\mathrm{o}}}{2}$

$e^{-j{\omega }_{0}t}X\left(t\right)\to X\left(\omega +{\omega }_0\right)$

$X\left(t\right)\left[e^{-j{\omega }_{0}t}X\left(t\right)\right]\to \frac{1}{2\pi }\left[X\left(\omega \right)\times X\left(\omega +{\omega }_0\right)\right]$

Upper limit of convolution$=-\frac{{\omega }_0}{2}+\frac{{\omega }_0}{2}$=0

Lower limit of convolution =$=\frac{-3{\omega }_0}{2}+\frac{-{\omega }_0}{2}$=-2ω₀

Convolution of two same duration rectangular is triangular pulse

After sampling

Shifted sample position occurs at f_m ± nf_s

For centre frequency -ω₀

Shifted is at +ω₀

ω₀ = -ω₀ ± f_s

f_s = 2ω₀

Let

Now,

$\mathrm{P}(\mathrm{t})\star \mathrm{P}(\mathrm{t})=\mathrm{P}(\omega ).\mathrm{P}(\omega )$

So,

Now inverse Fourier of $\mathrm{X}(\omega )$ can be written as $\frac{\sin \mathrm{t}}{\pi \mathrm{t}}$Thus,

$\mathrm{t}\left[\frac{\sin \mathrm{t}}{\pi \mathrm{t}}\times \frac{\sin \mathrm{t}}{\pi \mathrm{t}}\right]=\mathrm{t}\left[\frac{\sin \mathrm{t}}{\pi \mathrm{t}}\right]=\frac{1}{\pi }\sin \mathrm{t}$