GO Electrical
0 votes
The Fourier transform of a continuous-time signal $x(t)$ is given by $X(\omega) = \frac{1}{(10+j \omega)^2}, – \infty < \omega < \infty$, where $j = \sqrt{-1}$ and $\omega$ denoes frequency. Then the value of $\mid \text{ln } x(t) \mid$ at $t=1$ is _________ (up to $1$ decimal place). ($\text{ln}$ denotes the logarithm base $e$)
in new by (5.4k points)
retagged by

Please log in or register to answer this question.

Welcome to GATE Overflow, Electrical, where you can ask questions and receive answers from other members of the community.

847 questions
38 answers
26,458 users