GATE 2018 EE – Question 50

Signals and Systems · Applications of Fourier Transform for continuous and discrete time signals, Laplace Transform and Z transform · 2 marks · Numerical answer

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

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Correct answer: 10

Explanation

$\frac{1}{(10+j\omega)^2}$ is the transform of $x(t)=te^{-10t}u(t)$. At $t=1$, $x(1)=e^{-10}$, so $|\ln x(1)|=10$.