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:

\(Y\left( {{e^{j\omega }}} \right) = \;\mathop \sum \limits_{n = - \infty }^\infty y\left( n \right){e^{ - j\omega n}}\)

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:

\(Y(e^{j\omega })= \mathop \sum \limits_{n\; = \;0}^\infty {\left( {\frac{1}{4}} \right)^n}{e^{ - j\omega n}}\)

\(Y\left( {{e^{j0}}} \right) = \mathop \sum \limits_{n = 0}^\infty {\left( {\frac{1}{4}} \right)^n} \)

\(= 1 + \frac{1}{4} + {\left( {\frac{1}{4}} \right)^2} + \ldots \; \)

\(= \frac{1}{{1 - \frac{1}{4}}} = \frac{4}{3}\)

By time reversal

\(x\left( { - t} \right)\mathop \leftrightarrow \limits^F X\left( { - \omega } \right)\)

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

\(\begin{array}{l} \Rightarrow x\left( { - t} \right)\mathop \leftrightarrow \limits^F {X^*}\left( \omega \right)\\ \Rightarrow \left[ {x\left( t \right) - x\left( { - t} \right)} \right]\mathop \leftrightarrow \limits^F X\left( \omega \right) - {X^*}\left( \omega \right) \end{array}\)

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:

\(X\left( \omega \right) = \mathop \smallint \limits_{ - \infty}^{\infty} x(t) ~{e^{ - j\omega t}}~dt \)

Analysis:

Given:

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

\( X\left( \omega \right) = \mathop \smallint \limits_{ - 1}^1 {e^{j10t}}.{e^{ - j\omega t}}dt = \mathop \smallint \limits_{ - 1}^1 {e^{j\left( {10 - \omega } \right)t}}dt\)

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

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

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

\(A~rect(\frac{t}{\tau}) \leftrightarrow A \tau ~Sa(\frac{\omega \tau}{2})\)

Now,

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

1)

\(\frac{sin ~3\omega}{\omega} = 3 \frac{sin~(3\omega)}{3\omega}\)

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 \(\rm{x(t)=x^*(t)}\)

Taking its fourier transform, we have

\(\rm{X\left(\omega\right)=X^*\left(-\omega\right)}\)

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:

\(X\left( \omega \right) = \mathop \smallint \limits_{ - \infty}^{\infty} x(t) ~{e^{ - j\omega t}}~dt \)

Analysis:

Given:

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

\( X\left( \omega \right) = \mathop \smallint \limits_{ - 1}^1 {e^{j10t}}.{e^{ - j\omega t}}dt = \mathop \smallint \limits_{ - 1}^1 {e^{j\left( {10 - \omega } \right)t}}dt\)

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

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

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

\(A~rect(\frac{t}{\tau}) \leftrightarrow A \tau ~Sa(\frac{\omega \tau}{2})\)

Now,

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

1)

\(\frac{sin ~3\omega}{\omega} = 3 \frac{sin~(3\omega)}{3\omega}\)

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:

\(Y\left( {{e^{j\omega }}} \right) = \;\mathop \sum \limits_{n = - \infty }^\infty y\left( n \right){e^{ - j\omega n}}\)

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:

\(Y(e^{j\omega })= \mathop \sum \limits_{n\; = \;0}^\infty {\left( {\frac{1}{4}} \right)^n}{e^{ - j\omega n}}\)

\(Y\left( {{e^{j0}}} \right) = \mathop \sum \limits_{n = 0}^\infty {\left( {\frac{1}{4}} \right)^n} \)

\(= 1 + \frac{1}{4} + {\left( {\frac{1}{4}} \right)^2} + \ldots \; \)

\(= \frac{1}{{1 - \frac{1}{4}}} = \frac{4}{3}\)

Concept:

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

\(X\left( \omega \right) = \mathop \smallint \limits_{ - \infty}^{\infty} x(t) ~{e^{ - j\omega t}}~dt \)

Analysis:

Given:

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

\( X\left( \omega \right) = \mathop \smallint \limits_{ - 1}^1 {e^{j10t}}.{e^{ - j\omega t}}dt = \mathop \smallint \limits_{ - 1}^1 {e^{j\left( {10 - \omega } \right)t}}dt\)

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

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

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

\(A~rect(\frac{t}{\tau}) \leftrightarrow A \tau ~Sa(\frac{\omega \tau}{2})\)

Now,

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

1)

\(\frac{sin ~3\omega}{\omega} = 3 \frac{sin~(3\omega)}{3\omega}\)

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)

\(\frac{1}{2}+\frac{1}{2} = 1\)

\(x\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {1\;for}&{ - 1 \le t \le 1}\\ 0&{otherwise} \end{array}} \right.\)

Fourier transform:

\(X\left( {j\omega } \right) = \mathop \smallint \limits_{ - \infty }^\infty {e^{ - j\omega t}}x\left( t \right)dt\)

\(=\mathop \smallint \limits_{ - 1 }^1 {e^{ - j\omega t}}(1)dt\)

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

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

X(jω) is zero at ω = π and ω = 2π.

Concept:

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

\(Y\left( {{e^{j\omega }}} \right) = \;\mathop \sum \limits_{n = - \infty }^\infty y\left( n \right){e^{ - j\omega n}}\)

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:

\(Y(e^{j\omega })= \mathop \sum \limits_{n\; = \;0}^\infty {\left( {\frac{1}{4}} \right)^n}{e^{ - j\omega n}}\)

\(Y\left( {{e^{j0}}} \right) = \mathop \sum \limits_{n = 0}^\infty {\left( {\frac{1}{4}} \right)^n} \)

\(= 1 + \frac{1}{4} + {\left( {\frac{1}{4}} \right)^2} + \ldots \; \)

\(= \frac{1}{{1 - \frac{1}{4}}} = \frac{4}{3}\)

By time reversal

\(x\left( { - t} \right)\mathop \leftrightarrow \limits^F X\left( { - \omega } \right)\)

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

\(\begin{array}{l} \Rightarrow x\left( { - t} \right)\mathop \leftrightarrow \limits^F {X^*}\left( \omega \right)\\ \Rightarrow \left[ {x\left( t \right) - x\left( { - t} \right)} \right]\mathop \leftrightarrow \limits^F X\left( \omega \right) - {X^*}\left( \omega \right) \end{array}\)

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

\(x\left( t \right) = \frac{1}{{2\pi }}\mathop \smallint \limits_{ - \infty }^\infty X\left( {j\omega } \right){e^{j\omega t}}d\omega \)

\(\Rightarrow \frac{1}{{2\pi }}\mathop \smallint \limits_{ - \infty }^\infty X\left( {j\omega } \right){e^{j\omega \left( 3 \right)}}d\omega = x\left( 3 \right)\)

\(\Rightarrow \mathop \smallint \limits_{ - \infty }^\infty X\left( {j\omega } \right){e^{j3\omega }}d\omega = 2\pi \times \left( { - 2} \right)\)

= -4π

≃ -12.56

Concept:

Frequency domain representation of the signal is its Fourier transform

\(F.T\left[ {x\left( t \right)} \right] \to \mathop \smallint \limits_{ - \infty }^{ + \infty } x\left( t \right){e^{ - j\omega t}}dt\)

Calculation:

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

x(t) = u(t + 2) – 2u(t) +u(t – 2)

Taking the Fourier Transform

\(X\left( {j\omega } \right) = \mathop \smallint \limits_{ - 2}^0 1.e_{}^{ - j\omega t}dt + \mathop \smallint \limits_0^2 \left( { - 1} \right){e^{ - j\omega t}}dt\)

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

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

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

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