The input 4sinc(2t) is fed to a Hilbert transformer to obtain y(t), as shown in the figure below:

Here \({\rm{sinc}}\left( {\rm{x}} \right) = \frac{{\sin \left( {{\rm{\pi x}}} \right)}}{{{\rm{\pi x}}}}\) The value (accurate to two decimal places) \(\mathop \smallint \limits_{ - \infty }^\infty {\left| {{\rm{y}}\left( {\rm{t}} \right)} \right|^2}{\rm{dt\;}}\) is ________.

This question was previously asked in

GATE EC 2018 Official Paper

CT 1: Ratio and Proportion

2846

10 Questions
16 Marks
30 Mins

__Concept__:

The impulse response of a Hilbert transformer is given as:

\({\rm{h}}\left( {\rm{t}} \right) = \frac{1}{{{\rm{\pi t}}}}\)

The Fourier transform representation is:

H(f) = -j sgn(f)

|H(f)| = 1

__Application__:

Fourier transform of a standard rectangular pulse is given as:

\({\rm{A\;rect\;}}\left( {\frac{{\rm{t}}}{{\rm{T}}}} \right)\mathop \to \limits^{{\rm{F.T.}}} {\rm{ATsinc\;}}\left( {{\rm{FT}}} \right)\)

\({\rm{ATsin\;c}}\left( {{\rm{tT}}} \right)\mathop \to \limits^{{\rm{F}}.{\rm{T}}} {\rm{Arect}}\left( {\frac{{ - {\rm{f}}}}{{\rm{T}}}} \right) = {\rm{Arect}}\left( {\frac{{\rm{f}}}{{\rm{T}}}} \right)\)

For the given input signal, the Fourier representation will be:

\(4\sin {\rm{c}}\left( {2{\rm{t}}} \right)\mathop \to \limits^{{\rm{F}}.{\rm{T}}} 2{\rm{rect}}\left( {\frac{{\rm{f}}}{2}} \right)\)

Here A = 2, T = 2

Using Parseval's theorem, the energy is calculated as:

\(E= \mathop \smallint \limits_{ - \infty }^\infty {\left| {{\rm{y}}\left( {\rm{f}} \right)} \right|^2}{\rm{dF}}\)

\(E = \mathop \smallint \limits_{ - \infty }^\infty {\left| {2{\rm{rect}}\left( {\frac{{\rm{f}}}{2}} \right)} \right|^2}{\rm{df}} = 4 \times 2 = 8\)