Properties of Discrete Fourier Transform MCQ Quiz in मराठी - Objective Question with Answer for Properties of Discrete Fourier Transform - मोफत PDF डाउनलोड करा

By definition

\(\rm \begin{array}{l} X\left[ k \right] = \mathop \sum \limits_{n = 1}^{N - 1} x\left[ n \right]{e^{ - jk\frac{{2\pi n}}{N}}}\\ \rm \Rightarrow X\left[ {N - k} \right] = \mathop \sum \limits_{n = 1}^{N - 1} x\left[ n \right].{e^{ - j\frac{{\left( {N - k} \right)2\pi n}}{N}}}\\ \rm = \mathop \sum \limits_{n = 1}^{N - 1} x\left[ n \right]{e^{jk\frac{{2\pi n}}{N}}}.{e^{ - j2\pi n}}\\ \rm = \mathop \sum \limits_{n = 1}^{N - 1} x\left[ n \right]{e^{\frac{{jk2\pi n}}{N}}}\\ \rm \Rightarrow X\left[ {N - k} \right] = {\left( {\mathop \sum \limits_{n = 1}^{N - 1} x\left[ n \right].{e^{ - jk\frac{{2\pi n}}{N}}}} \right)^*} = {X^*}\left[ k \right] \end{array}\)

\(\rm x[n]\)  is discrete and aperiodic \(\rm \Rightarrow X\left( {{e^{j\omega }}} \right)\)is periodic and continuous.

\(\rm X\left( {{e^{j\omega }}} \right)\)is periodic with \(\rm 2π\).

Thus \(\rm \mathop \smallint \limits_0^{6\pi } {\left| {X\left( {{e^{j\omega }}} \right)} \right|^2}d\omega = 3\left[ {\mathop \smallint \limits_0^{2\pi } {{\left| {X\left( {{e^{j\omega }}} \right)} \right|}^2}d\omega } \right]\)

Now from Parseval’s theorem.

\(\rm \frac{1}{{2\pi }}\mathop \smallint \limits_0^{2\pi } {\left| {X\left( {{e^{j\omega }}} \right)} \right|^2}d\omega = \mathop \sum \limits_{n = - \infty }^\infty {\left| {x\left[ n \right]} \right|^2}\)

\(\rm \Rightarrow \mathop \smallint \limits_0^{2\pi } {\left| {X\left({e^{j\omega }}\right)} \right|^2}d\omega = 2\pi \mathop \sum \limits_{n = - \infty }^\infty {\left| {x\left[ n \right]} \right|^2}\)

\(\rm = 2\pi \left[ {1 + 4 + 9 + 4 + 1} \right]\)

\(\rm = 2\pi .19\)

Now,

\(\rm \mathop \smallint \limits_0^{6\pi } {\left| {X\left( {{e^{j\omega }}} \right)} \right|^2}d\omega = 3\left[ {2\pi .19} \right] = 358.14\)