Fourier Transform of Standard Signals MCQ Quiz in বাংলা - Objective Question with Answer for Fourier Transform of Standard Signals - বিনামূল্যে ডাউনলোড করুন [PDF]

$s\left(j\omega \right)=\frac{2\sin \omega }{\omega }\stackrel{FT}{\leftrightarrow }s\left(t\right)=\{\begin{array}{ccc}1& ,& \left|t\right|\le 1\\ 0& ,& otherwise\end{array}$ 

x(t) = s(t) * s(t)

$\frac{1}{j\omega }+\pi \delta \left(j\omega \right)\stackrel{FT}{\leftrightarrow }u\left(t\right)$

$\frac{1}{2+j\omega }\stackrel{FT}{\leftrightarrow }{e}^{-2t}u\left(t\right)$

$2\pi \delta \left(\omega \right)\stackrel{FT}{\leftrightarrow }1$

$X\left(j\omega \right)=-0.5\frac{1}{\left(j\omega +2\right)}+0.5\frac{1}{j\omega }+0.5\pi \delta \left(\omega \right)-1.5\pi \delta \left(\omega \right)$

$X\left(j\omega \right)\stackrel{FT}{\leftrightarrow }x\left(t\right)=-0.5e^{-2t}u\left(t\right)+0.5u\left(t\right)-\frac{3}{4}$

At t = 0

$x\left(t\right)=a\left(t\right)\ast b\left(t\right)\stackrel{FT}{\leftrightarrow }X\left(j\omega \right)=A\left(j\omega \right)B\left(j\omega \right)$ 

$\frac{\mathrm{s}\mathrm{i}\mathrm{n}\left(Wt\right)}{\pi t}\stackrel{FT}{\leftrightarrow }\{\begin{array}{ccc}1& ,& \omega \le W\\ 0& ,& otherwise\end{array}$ 

$\frac{d}{dt}s\left(t\right)\stackrel{FT}{\leftrightarrow }j\omega s\left(\omega \right)$ 

$x\left(t\right)=\left[2\sin \left(\frac{3\pi t}{\pi t}\right)\right]\left[\frac{\sin \left(2\pi t\right)}{\pi t}\right]$ 

$\frac{\sin Wt}{\pi t}\leftrightarrow \{\begin{array}{cc}1,& \omega \le W\\ 0,& otherewise\end{array}$ 

$s_1\left(t\right)s_2\left(t\right)\leftrightarrow \frac{1}{2\pi }{s}_1\left(j\omega \right)\ast s_2\left(j\omega \right)$ 

$x\left(t\right)=\frac{j}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}{e}^{j\frac{\omega t}{4}}{e}^{j\omega t}d\omega$ 

$=\frac{\sin \left(\pi \left(t+0.25\right)\right)}{\pi \left(t+0.25\right)}$ 

$x\left(t\right)=\frac{j}{2\pi }\underset{-2}{\overset{0}{\int }}{e}^{j\omega t}d\omega -\frac{j}{2\pi }\underset{0}{\overset{2}{\int }}{e}^{j\omega t}d\omega$

$=\frac{1-\cos \left(2t\right)}{\pi t}$

At t = π,

$x\left(t\right)=\frac{\sin \left(\pi t\right)}{\pi t}\stackrel{FT}{\leftrightarrow }X\left(j\omega \right)=\{\begin{array}{ccc}1& ,& \left|\omega \right|\le \pi \\ 0& ,& otherwise\end{array}$

Multiply the denominator and the numerator by π

$\pi \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\left(\frac{\sin \left(\pi t\right)}{\pi t}\right)}^{2}dt=\frac{1}{2}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\left|X\left(j\omega \right)\right|}^{2}d\omega$

$x\left(t\right)=\frac{1}{2\pi }\underset{-3}{\overset{3}{\int }}\frac{2}{3}j\omega e^{j\omega t}d\omega$

$x\left(t\right)=\frac{2\cos \left(3t\right)}{2\pi t}-\frac{2\sin \left(3t\right)}{3\pi t^2}$

Hilbert transform of signal

$\begin{array}{l}h\left(t\right)=y\left(t\right)\ast \frac{1}{\pi t}\\ h\left(t\right)=\frac{1}{\pi t}\ast \frac{1}{\pi t}\end{array}$  

H(jω) = (-j sgn ω)²

H(jω) = -1

The signal x(t) is liner combination of rectangular triangular function

x(t) = x₁(t) - x₂(t)

x(ω) = x₁(ω) - x₂(ω)

to find x₁(ω)

$\frac{d{x}_1\left(t\right)}{d\left(t\right)}=2\left(\delta +2\right)-2\delta \left(t-2\right)$

$\Rightarrow j\omega x_1\left(\omega \right)=2\left[e^{j2\omega }-e^{-j2\omega }\right]$

$x_1\left(\omega \right)=\frac{4}{\omega }\left[\frac{e^{j2\omega }-e^{-j2\omega }}{2j}\right]=\frac{4\sin 2\omega }{\omega }$

To find x₂(ω)

$\frac{d^2{x}_2\left(t\right)}{d{t}^2}=2\delta \left(t+2\right)-4\delta \left(t\right)+2\delta \left(t-2\right)$

$x_2\left(\omega \right)=\frac{4\left(1-\cos 2\omega \right)}{{\omega }^2}$

$\therefore x\left(\omega \right)=\frac{4\sin 2\omega }{\omega }-\frac{4\left(1-\cos 2\omega \right)}{{\omega }^2}$