Question

The Fourier transform of a continuos time signal x(t) is given by $X\left(\omega \right)=\frac{1}{(10+j\omega {)}^2},-\infty <\omega <\infty$, where $j=\sqrt{-1}$ and $\left(\omega \right)$ denotes frequency. Then the value of $|lnx\left(t\right)|$ at t = 1 is (upto decimal place). (ln denotes the logarithm to base e).

Answer 1

$x\left(t\right)\rightleftharpoons X\left(j\omega \right)=\frac{1}{(10+j\omega {)}^2}$
By taking inverse Fourier transform,
$x\left(t\right)=t{e}^{-10t}u\left(t\right)$

$Now,x\left(t\right){|}_{t=1}=1\times e^{-10}\times 1=e^{-10}$
Thus,$|ln\left\{x\right(t\left)\right\}|=|ln\left(e^{-10}\right)|=|-10|=10$