Definition of Fourier Transform MCQ Quiz - Objective Question with Answer for Definition of Fourier Transform - Download Free PDF

Concept:

The Fourier transform for a discrete time sequence y[n] is given by:

$Y\left(e^{j\omega }\right)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}y\left(n\right)e^{-j\omega n}$

Analysis:

$x\left(n\right)={\left(\frac{1}{2}\right)}^{n}u\left(n\right)$

$y\left(n\right)=x^2\left(n\right)={\left(\frac{1}{2}\right)}^{2n}{u}^2\left(n\right)$

$={\left[{\left(\frac{1}{2}\right)}^2\right]}^n{u}^2\left(n\right)$

$={\left(\frac{1}{4}\right)}^{n}u\left(n\right)$

The Fourier transform of the above signal will now be:

$Y(e^{j\omega })=\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{1}{4}\right)}^n{e}^{-j\omega n}$

$Y\left(e^{j0}\right)=\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{1}{4}\right)}^n$

$=1+\frac{1}{4}+{\left(\frac{1}{4}\right)}^2+\dots$

$=\frac{1}{1-\frac{1}{4}}=\frac{4}{3}$

By time reversal

$x\left(-t\right)\stackrel{F}{\leftrightarrow }X\left(-\omega \right)$

For real functions $X\left(-\omega \right)=X^{\ast }\left(\omega \right)$

$\begin{array}{l}\Rightarrow x\left(-t\right)\stackrel{F}{\leftrightarrow }{X}^{\ast }\left(\omega \right)\\ \Rightarrow \left[x\left(t\right)-x\left(-t\right)\right]\stackrel{F}{\leftrightarrow }X\left(\omega \right)-X^{\ast }\left(\omega \right)\end{array}$

X(ω) – x*(ω) is complex conjugate subtracted from real function which will be pure imaginary.

Concept:

The Fourier Transform of a continuous-time signal x(t) is given as:

$X\left(\omega \right)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t)e^{-j\omega t}dt$

Analysis:

Given:

x(t) = e^j10t  defined from t = -1 to 1. 

$X\left(\omega \right)=\underset{-1}{\overset{1}{\int }}{e}^{j10t}.e^{-j\omega t}dt=\underset{-1}{\overset{1}{\int }}{e}^{j\left(10-\omega \right)t}dt$

$X(\omega )={\frac{e^{j\left(10-\omega \right)t}}{j\left(10-\omega \right)}|}_{-1}^1=\frac{2\sin \left(\omega -10\right)}{\left(\omega -10\right)}$

$\begin{array}{l}H\left(j\omega \right)=\frac{\left(2\cos \omega \right)\left(\sin 2\omega \right)}{\omega }\\ =\frac{\sin 3\omega }{\omega }+\frac{sin\omega }{\omega }\end{array}$

We know that the inverse Fourier transform of sinc function is a rectangular function.

$Arect(\frac{t}{\tau })\leftrightarrow A\tau Sa(\frac{\omega \tau }{2})$

Now,

$Sa(\frac{\omega \tau }{2})=\frac{sin(\frac{\omega \tau }{2})}{\frac{\omega \tau }{2}}$

1)

$\frac{sin3\omega }{\omega }=3\frac{sin(3\omega )}{3\omega }$

By comparing, we get

$\tau /2=3,A\tau =3$

hence, A = 0.5

2) 

$\frac{sin\omega }{\omega }=1\times \frac{sin(\omega \times 1)}{\omega \times 1}$

By comparing, we get

$\tau /2=1,A\tau =1$

hence, A = 0.5

So, inverse Fourier transform of H (jω)

h(t) = h_1­(t) + h₂(t)

h(0) = h₁(0) + h_2­­(0)

$\frac{1}{2}+\frac{1}{2}=1$

For a real valued signal $\mathrm{x}(\mathrm{t})={\mathrm{x}}^{\ast }(\mathrm{t})$

Taking its fourier transform, we have

$\mathrm{X}\left(\omega \right)={\mathrm{X}}^{\ast }(-\omega )$

Thus, Fourier transform of a real valued time signal has conjugate symmetry.

Concept:

The Fourier Transform of a continuous-time signal x(t) is given as:

$X\left(\omega \right)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t)e^{-j\omega t}dt$

Analysis:

Given:

x(t) = e^j10t  defined from t = -1 to 1. 

$X\left(\omega \right)=\underset{-1}{\overset{1}{\int }}{e}^{j10t}.e^{-j\omega t}dt=\underset{-1}{\overset{1}{\int }}{e}^{j\left(10-\omega \right)t}dt$

$X(\omega )={\frac{e^{j\left(10-\omega \right)t}}{j\left(10-\omega \right)}|}_{-1}^1=\frac{2\sin \left(\omega -10\right)}{\left(\omega -10\right)}$

$\begin{array}{l}H\left(j\omega \right)=\frac{\left(2\cos \omega \right)\left(\sin 2\omega \right)}{\omega }\\ =\frac{\sin 3\omega }{\omega }+\frac{sin\omega }{\omega }\end{array}$

We know that the inverse Fourier transform of sinc function is a rectangular function.

$Arect(\frac{t}{\tau })\leftrightarrow A\tau Sa(\frac{\omega \tau }{2})$

Now,

$Sa(\frac{\omega \tau }{2})=\frac{sin(\frac{\omega \tau }{2})}{\frac{\omega \tau }{2}}$

1)

$\frac{sin3\omega }{\omega }=3\frac{sin(3\omega )}{3\omega }$

By comparing, we get

$\tau /2=3,A\tau =3$

hence, A = 0.5

2) 

$\frac{sin\omega }{\omega }=1\times \frac{sin(\omega \times 1)}{\omega \times 1}$

By comparing, we get

$\tau /2=1,A\tau =1$

hence, A = 0.5

So, inverse Fourier transform of H (jω)

h(t) = h_1­(t) + h₂(t)

h(0) = h₁(0) + h_2­­(0)

$\frac{1}{2}+\frac{1}{2}=1$

Concept:

The Fourier transform for a discrete time sequence y[n] is given by:

$Y\left(e^{j\omega }\right)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}y\left(n\right)e^{-j\omega n}$

Analysis:

$x\left(n\right)={\left(\frac{1}{2}\right)}^{n}u\left(n\right)$

$y\left(n\right)=x^2\left(n\right)={\left(\frac{1}{2}\right)}^{2n}{u}^2\left(n\right)$

$={\left[{\left(\frac{1}{2}\right)}^2\right]}^n{u}^2\left(n\right)$

$={\left(\frac{1}{4}\right)}^{n}u\left(n\right)$

The Fourier transform of the above signal will now be:

$Y(e^{j\omega })=\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{1}{4}\right)}^n{e}^{-j\omega n}$

$Y\left(e^{j0}\right)=\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{1}{4}\right)}^n$

$=1+\frac{1}{4}+{\left(\frac{1}{4}\right)}^2+\dots$

$=\frac{1}{1-\frac{1}{4}}=\frac{4}{3}$

Concept:

The Fourier Transform of a continuous-time signal x(t) is given as:

$X\left(\omega \right)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(t)e^{-j\omega t}dt$

Analysis:

Given:

x(t) = e^j10t  defined from t = -1 to 1. 

$X\left(\omega \right)=\underset{-1}{\overset{1}{\int }}{e}^{j10t}.e^{-j\omega t}dt=\underset{-1}{\overset{1}{\int }}{e}^{j\left(10-\omega \right)t}dt$

$X(\omega )={\frac{e^{j\left(10-\omega \right)t}}{j\left(10-\omega \right)}|}_{-1}^1=\frac{2\sin \left(\omega -10\right)}{\left(\omega -10\right)}$

$\begin{array}{l}H\left(j\omega \right)=\frac{\left(2\cos \omega \right)\left(\sin 2\omega \right)}{\omega }\\ =\frac{\sin 3\omega }{\omega }+\frac{sin\omega }{\omega }\end{array}$

We know that the inverse Fourier transform of sinc function is a rectangular function.

$Arect(\frac{t}{\tau })\leftrightarrow A\tau Sa(\frac{\omega \tau }{2})$

Now,

$Sa(\frac{\omega \tau }{2})=\frac{sin(\frac{\omega \tau }{2})}{\frac{\omega \tau }{2}}$

1)

$\frac{sin3\omega }{\omega }=3\frac{sin(3\omega )}{3\omega }$

By comparing, we get

$\tau /2=3,A\tau =3$

hence, A = 0.5

2) 

$\frac{sin\omega }{\omega }=1\times \frac{sin(\omega \times 1)}{\omega \times 1}$

By comparing, we get

$\tau /2=1,A\tau =1$

hence, A = 0.5

So, inverse Fourier transform of H (jω)

h(t) = h_1­(t) + h₂(t)

h(0) = h₁(0) + h_2­­(0)

$\frac{1}{2}+\frac{1}{2}=1$

$x\left(t\right)=\{\begin{array}{cc}1for& -1\le t\le 1\\ 0& otherwise\end{array}$

Fourier transform:

$X\left(j\omega \right)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-j\omega t}x\left(t\right)dt$

$=\underset{-1}{\overset{1}{\int }}{e}^{-j\omega t}(1)dt$

$=\frac{1}{-j\omega }{\left[e^{-j\omega }\right]}_{-1}^1$

$X(j\omega )=\frac{1}{-j\omega }\left(e^{-j\omega }-e^{j\omega }\right)$

X(jω) is zero at ω = π and ω = 2π.

Concept:

The Fourier transform for a discrete time sequence y[n] is given by:

$Y\left(e^{j\omega }\right)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}y\left(n\right)e^{-j\omega n}$

Analysis:

$x\left(n\right)={\left(\frac{1}{2}\right)}^{n}u\left(n\right)$

$y\left(n\right)=x^2\left(n\right)={\left(\frac{1}{2}\right)}^{2n}{u}^2\left(n\right)$

$={\left[{\left(\frac{1}{2}\right)}^2\right]}^n{u}^2\left(n\right)$

$={\left(\frac{1}{4}\right)}^{n}u\left(n\right)$

The Fourier transform of the above signal will now be:

$Y(e^{j\omega })=\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{1}{4}\right)}^n{e}^{-j\omega n}$

$Y\left(e^{j0}\right)=\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{1}{4}\right)}^n$

$=1+\frac{1}{4}+{\left(\frac{1}{4}\right)}^2+\dots$

$=\frac{1}{1-\frac{1}{4}}=\frac{4}{3}$

By time reversal

$x\left(-t\right)\stackrel{F}{\leftrightarrow }X\left(-\omega \right)$

For real functions $X\left(-\omega \right)=X^{\ast }\left(\omega \right)$

$\begin{array}{l}\Rightarrow x\left(-t\right)\stackrel{F}{\leftrightarrow }{X}^{\ast }\left(\omega \right)\\ \Rightarrow \left[x\left(t\right)-x\left(-t\right)\right]\stackrel{F}{\leftrightarrow }X\left(\omega \right)-X^{\ast }\left(\omega \right)\end{array}$

X(ω) – x*(ω) is complex conjugate subtracted from real function which will be pure imaginary.

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

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

$\Rightarrow \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}X\left(j\omega \right)e^{j3\omega }d\omega =2\pi \times \left(-2\right)$

= -4π

≃ -12.56

Concept:

Frequency domain representation of the signal is its Fourier transform

$F.T\left[x\left(t\right)\right]\to \underset{-\mathrm{\infty }}{\overset{+\mathrm{\infty }}{\int }}x\left(t\right)e^{-j\omega t}dt$

Calculation:

The signal x(t) can be mathematically written as:

x(t) = u(t + 2) – 2u(t) +u(t – 2)

Taking the Fourier Transform

$X\left(j\omega \right)=\underset{-2}{\overset{0}{\int }}1.e_{}^{-j\omega t}dt+\underset{0}{\overset{2}{\int }}\left(-1\right)e^{-j\omega t}dt$

$\Rightarrow {\left[\frac{e^{-j\omega t}}{-j\omega }\right]}_{-2}^0-{\left[\frac{e^{-j\omega t}}{-j\omega }\right]}_0^2$

$\frac{-1}{j\omega }\left[e^0-e^{j2\omega }\right]+\frac{1}{j\omega }\left[e^{-j2\omega }-e^0\right]$

$\Rightarrow \frac{e^{j2\omega }-1+e^{-j2\omega }-1}{j\omega }$

$\Rightarrow \frac{2\left(\frac{e^{j2\omega }+e^{-j2\omega }}{2}\right)-1}{j\omega }$