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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 $$f(t) = e^{at}$$

Aadnan Farooq A
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  • This Question may help ( http://dsp.stackexchange.com/questions/25489/bilateral-mathcal-z-transform-of-exponential ) – spectre Aug 17 '16 at 05:35
  • I attempted to provide an intuitive explanation of what it means "to be" in the Laplace s-domain in relation to the Fourier domain.. https://dsp.stackexchange.com/a/40506/26009 – Envidia Jun 12 '17 at 16:29
  • 1/(s-a) R of real>real of a R – Mohamed Awad Dec 11 '20 at 08:33

1 Answers1

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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.

Tom Kealy
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  • 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? – Aadnan Farooq A Nov 19 '15 at 15:51
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    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? – Tom Kealy Nov 19 '15 at 16:39
  • means we have to check whether the signal is defined for all the time or not. if it defined for all the time then we will use bilateral .. right? – Aadnan Farooq A Nov 19 '15 at 16:43
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    Yes. For example, you would use the unilateral transform for a pulse signal at t=0. – Tom Kealy Nov 19 '15 at 16:44
  • I have edited the question can you please check that – Aadnan Farooq A Nov 19 '15 at 16:48
  • A2: by a straightforward integral (or an educated guess), and to check http://tutorial.math.lamar.edu/Classes/DE/Laplace_Table.aspx – Tom Kealy Nov 19 '15 at 16:50
  • exactly, but how can i prove that? can you provide the details as answer – Aadnan Farooq A Nov 19 '15 at 16:53
  • This one is for you to think about! – Tom Kealy Nov 19 '15 at 16:57
  • I have done some part. like we can not use bilateral LT as if we apply the $-\infty$ then the exponential goes to infinity and it will not converge. so we will use unilateral LT and then the expression will be $\int_{0}^{+\infty} e^{-(s-a)t}$ i am confuse about check the convergence, wheather it will $Re(s-a)>0$ or $Re(s-a)<0$ – Aadnan Farooq A Nov 19 '15 at 17:06
  • Hint: The integral isn't valid for t<0. – Tom Kealy Nov 19 '15 at 17:09
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    Yup I was also guessing that as the $t>0$ it should be $Re(s−a)>0$.. right? – Aadnan Farooq A Nov 19 '15 at 17:10
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    Yep! See, it wasn't so hard. – Tom Kealy Nov 20 '15 at 10:27
  • Complex numbers have a real part and an imaginary part. What is the complex part supposed to be? Based on context, I'm guessing it is the imaginary part...? – hops Jan 13 '17 at 07:09