Consider the signal x(t) = e-|t|. Let X(jω) = \(\mathop \smallint \limits_{ - \infty }^\infty x\left( t \right){e^{ - j\omega t}}dt\) be the Fourier transform of x(t). The value of X(j0) is
x(t) = e-|t|
\(x\left( t \right) = {e^{ - \left| t \right|}}\;;\;\left\{ {\begin{array}{*{20}{c}} {{e^{ - t}}\;;\;\;t > 0}\\ {1;t = 0}\\ {{e^t}\;\;;t < 0\;} \end{array}} \right.\)
x(t) = e-tu(t) + et u(-t)
\(X\left( {j\omega } \right) = \;\mathop \smallint \limits_{ - \infty }^{ - \infty } d\left( t \right){e^{ - j\omega t}}dt\)
\( = \mathop \smallint \limits_{ - \infty }^0 {e^{\left( {1 - j\omega } \right)t}}dt + \;\mathop \smallint \limits_0^\infty {e^{ - \left( {1 + j\omega } \right)}}dt\)
\( = \left. {\frac{{{e^{\left( {1 - j\omega } \right)t}}}}{{\left( {1 - j\omega } \right)}}} \right|_{ - \infty }^0 + \left. {\frac{{{e^{ - \left( {1 + j\omega } \right)}}}}{{ - \left( {1 + j\omega } \right)}}} \right|_0^\infty \)
\(X\left( {j\omega } \right) = \frac{1}{{1 - j\omega }}\left[ {{e^{\left( {1 - j\omega } \right)0}} - {e^{\left( {1 - j\omega } \right)\left( { - \infty \;} \right)}}} \right] - \frac{1}{{1 + j\omega }}\;\left[ {\;{e^{ - \left( {1 - j\omega } \right)\infty }} - {e^{ - \left( {1 - j\omega } \right)0}}} \right]\)
\( = \frac{1}{{1 - j\omega }}\left[ {1 - {e^{\left( {1 - j\omega } \right)\left( { - \infty } \right)}}} \right] - \frac{1}{{1 + j\omega }}\;\left[ {{e^{ - \left( {1 - j\omega } \right)\left( \infty \right)}} - 1} \right]\)
\(X\left( {j\omega } \right) = \frac{1}{{1 - j\omega }}\left[ {1 - 0} \right] - \frac{1}{{1 + j\omega }}\left[ {0 - 1} \right]\)
\( = \frac{1}{{1 - j\omega }} + \frac{1}{{1 + j\omega }}\)
\( = \frac{{1 + j\omega + 1 - j\omega }}{{\left( {1 - j\omega } \right)\left( {1 + j\omega } \right)}}\)
\(X\left( {j\omega } \right) = \frac{2}{{1 + {\omega ^2}}}\)
Put ω = 0
\(X\left( {j0} \right) = \frac{2}{{1 + {0^2}}} = 2,\;\;\;X\left( {j0} \right) = 2\)
Important Questions from Properties of Fourier Transform
The given mathematical representation belongs to:
y(t) = x(t - T)
Which type of property is shown by the following function.
L{K f(t)} = K F(s)
The energy of the signal \(x(t) = \frac{{{\rm{sin}}\left( {4{\rm{\pi t}}} \right)}}{{4{\rm{\pi t}}}}\) is______
A real-valued signal 𝑥(𝑡) limited to the frequency band \(\left| f \right| \le \frac{W}{2}\) is passed through a linear time-invariant system whose frequency response is
\(H\left( f \right) = \left\{ {\begin{array}{*{20}{c}} {{e^{ - j4\pi f,\;\;\;\left| f \right| \le \frac{W}{2}}}}\\ {0,\;\;\;\;\left| f \right| > \frac{W}{2}} \end{array}} \right.\)
The output of the system isConsider 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 _________
