# Why are we still using Continuous Time Fourier Transform when we have Laplace Transform? [duplicate]

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 ?

• What you heard is not entirely correct. Read this answer. There are also many related posts over at math.SE. – Matt L. Jan 5 '17 at 8:49
• 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. – Matt L. Jan 5 '17 at 8:53