The energy of the signal \(x(t) = \frac{{{\rm{sin}}\left( {4{\rm{\pi t}}} \right)}}{{4{\rm{\pi t}}}}\) is______
We know,
\({\rm{x}}\left( {\rm{t}} \right) = \frac{{{\rm{sin}}4{\rm{\pi t}}}}{{4{\rm{\pi t}}}}\)
Integrating mod square of x(t) to find the energy will be a tedious process.
Alternate Method
Fourier transform of \(\frac{{{\rm{sin}}4{\rm{\pi t}}}}{{4{\rm{\pi t}}}}\)
From Parseval’s theorem
\(\smallint \limits_{ - \infty }^\infty {\left| {{\rm{x}}\left( {\rm{t}} \right)} \right|^2}{\rm{dt}} = \frac{1}{{2{\rm{\pi }}}}\smallint \limits_{ - \infty }^\infty {\left| {{\rm{X}}\left( {\rm{\omega }} \right)} \right|^2}{\rm{d\omega }}\)
we have,
\(\begin{array}{l} \frac{1}{{2{\rm{\pi }}}}\smallint \limits_{ - 4{\rm{\pi }}}^{4{\rm{\pi }}} {\left( {\frac{1}{4}} \right)^2}{\rm{d\omega }} = \frac{1}{{2{\rm{\pi }}}}\smallint \limits_{ - 4{\rm{\pi }}}^{4{\rm{\pi }}} \frac{1}{{16}}{\rm{d\omega }}\\ \Rightarrow = \left. {\frac{1}{{2{\rm{\pi }}}}.\frac{1}{{16}}{\rm{\omega }}} \right|_{ - 4{\rm{\pi }}}^{4{\rm{\pi }}}\\ = \frac{1}{{2{\rm{\pi }}}}.\frac{1}{{16}}.8{\rm{\pi }} = \frac{1}{4} \end{array}\)
Thus,
signal's energy \(\smallint \limits_{ - \infty }^\infty {\left| {{\rm{x}}\left( {\rm{t}} \right)} \right|^2}{\rm{dt}} = \frac{1}{4}\)
Important Questions from Properties of Fourier Transform
Which one of the following is the Fourier transform of the signal given in Fig. II, if the Fourier transform of the signal in Fig. I is \(=2\dfrac{\sin\omega T_1}{\omega}\) ?
The Hilbert transform of cos ω1t + sin ω2t is
Consider two continuous time signals $x(t)$ and $y(t)$ as shown below
If $X(f)$ denotes the Fourier transform of $x(t)$, then the Fourier transform of $y(t)$ is _________Let $X(\omega)$ be the Fourier transform of the signal
$x(t) = e^{-t^4} \cos t$, $-\infty < t < \infty$.
The value of the derivative of $X(\omega)$ at $\omega=0$ is ________ (rounded off to 1 decimal place).- Let $f(t)$ and $g(t)$ represent continuous-time real-valued signals. If $h(t)$ denotes cross-correlation between $f(t)$ and $g(-t)$, its continuous-time Fourier transform $H(j\omega)$ equals
Note: $F(j\omega)$ and $G(j\omega)$ denote the continuous-time Fourier transforms of $f(t)$ and $g(t)$, respectively.
