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While studying Laplace transform, I also some questions which I want to understand:

a) We used to say that Laplace transform include both real and imaginary part whereas in Fourier transform we only have imaginary part. But when we have to say about convergence we also choose Real part to be either >0 or <0 . I want to know why we ignore imaginary part?

b) If we have any function x(t) how do we determine that we have to take bilateral integral or unilateral integral. In the above case we have u(t) with the function so our limits are changed. But if we don't have $u(t)$ function with the exponential like $$ x(t)=e^{at}$$ than how can we select bilateral or unilateral

Q2: what will be the Laplace Transform of $$f(t) = e^{at}$$

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The Fourier transform is the Laplace transform along the imaginary axis in the complex plane.

The convergence of the Laplace transform ignores the complex part, as the imaginary part breaks down the signal into sinusoids: which are bounded, and so have no effect on the convergence.

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  • $\begingroup$ can you please tell me about If we have any function $x(t)$ how do we determine that we have to take bilateral integral or unilateral integral. In the above case we have $u(t)$ with the function so our limits are changed. But if we don't have $u(t)$ function with the exponential like $x(t)=e^{at}$ than how can we select the integral bilateral or unilateral? $\endgroup$
    – Aadnan Farooq A
    Nov 19, 2015 at 15:51
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    $\begingroup$ Well, if my signal isn't defined for all time, why would I use a transform defined for all time? In terms of filter design: do you have feedforward coefficients as well as feedback coeffecients? $\endgroup$
    – Tom Kealy
    Nov 19, 2015 at 16:39
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    $\begingroup$ Yes. For example, you would use the unilateral transform for a pulse signal at t=0. $\endgroup$
    – Tom Kealy
    Nov 19, 2015 at 16:44
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    $\begingroup$ Yup I was also guessing that as the $t>0$ it should be $Re(s−a)>0$.. right? $\endgroup$
    – Aadnan Farooq A
    Nov 19, 2015 at 17:10
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    $\begingroup$ Yep! See, it wasn't so hard. $\endgroup$
    – Tom Kealy
    Nov 20, 2015 at 10:27

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