The Fourier transform of a continuous-time signal x(t) is given by \(X\left( \omega \right) = \frac{1}{{{{\left( {10 + j\omega } \right)}^2}}}, - \infty < \omega < \infty ,\) where \(j = \sqrt { - 1} \) and ω denotes frequency. Then the value of |ln x(t)| at t = 1 is _____ (up to 1 decimal place). (In denotes the logarithm to base e)
\(e^{-at}\rm \overset{F.T.}{\leftrightarrow}\frac{1}{a+j\omega}\)
Frequency differentiation property
\(tx(t)\rm \overset{F.T.}{\leftrightarrow}j\dfrac{d}{d\omega}x(\omega)\)
Explanation:
\(X\left( \omega \right) = \frac{1}{{{{\left( {10 + j\omega } \right)}^2}}}\)
By applying inverse Fourier transform,
⇒ x(t) = t e-10t u(t)
\(x\left( t \right) = {e^{ - at}}u\left( t \right) \leftrightarrow \frac{1}{{a + j\omega }}\)
By using frequency differentiation property,
\(t\;{x_1}\left( t \right) = t{e^{ - at}}u\left( t \right) \leftrightarrow j\frac{d}{{d\omega }} \times \left( \omega \right) = \frac{1}{{{{\left( {a + j\omega } \right)}^2}}}\)
\(\Rightarrow t{e^{ - 10t}}u\left( t \right) \leftrightarrow \frac{1}{{{{\left( {10 + j\omega } \right)}^2}}}\)
x(t) = te-10t u(t)
Taking logarithm (to base e ) on both sides,
\(\left| {lnx\left( t \right)} \right| = \left| {ln\left[ {t{e^{ - 10t}}u\left( t \right)} \right]} \right|\)
\(at\;t = 1, = \left| {ln\left[ {{e^{ - 10}}} \right]} \right| = \left| { - 10} \right| = 10\)
Important Questions from Fourier Transform
- The FT of $x(t) = e^{4t} u(-t)$ is:
The function f(t) is a periodic function of period 2π. In the range (-π, π), it equals e-t. If f(t) = \(\sum\nolimits_{ - \infty }^\infty {{c_n}{e^{{\mathop{\rm int}} }}}\) denotes its Fourier series expansion, the sum \({\sum\nolimits_{ - \infty }^\infty {\left| {{c_n}} \right|} ^2}\) is
Fourier transform of the unit impulse δ(t) is
Differentiating a signal in the time domain corresponds to _________ its FT in the frequency domain by _________.
The function f(t) has a Fourier transform F(ω). The Fourier transform of F(t) is
