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I've read that Laplace Transform is more versatile and can cover a broad range of signals compared to Continuous Time Fourier Transform. Then why are we still using Continuous Time Fourier Transform ?

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  • $\begingroup$ What you heard is not entirely correct. Read this answer. There are also many related posts over at math.SE. $\endgroup$ – Matt L. Jan 5 '17 at 8:49
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    $\begingroup$ Note that the Laplace transform of $\sin(\omega_0t)$, $-\infty<t<\infty$ doesn't exist. Also the impulse responses of ideal brickwall filters (ideal low pass, high pass, etc.) have no Laplace transform, while they do have a Fourier transform. $\endgroup$ – Matt L. Jan 5 '17 at 8:53