I have few confusions while starting Laplace Transform. So far I have studied, Fourier series and Fourier Transform. The basic difference which I found from different books is Fourier Transform is only considered the imaginary part whereas the Laplace transform considers both real and imaginary for general values.
i) I want to ask that, is it only difference we have in Laplace and Fourier Transform? Then i saw the two different equations of Laplace transform which is bilateral Laplace Transform $$ X(s) = \int _{-\infty}^{+\infty} x(t) e^{-st} dt$$ Whereas in the second equation of Laplace Transform is called unilateral Laplace Transform and it defined as: $$ X(s) = \int _{0}^{+\infty} x(t) e^{-st}dt $$ It omit the negative part and only have for $t>0$

ii) Here I want to ask that what is the reason of omitting the $t<0$ part?

And Lastly, there was an example $$ x(t)=e^{-at}u(t)$$ After applying Laplace transform it was written The transform exists only if $Re(s+a)$ is possitive.
iii) Now here is am confuse that why it took for possitive?

  • $\begingroup$ Good to know that someone is also working on understanding Laplace Transform from basics. 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 ar $\endgroup$ Commented Nov 19, 2015 at 13:31
  • $\begingroup$ That's also interesting thing if someone could answer. @MattL $\endgroup$ Commented Nov 19, 2015 at 13:38
  • $\begingroup$ Please ask another question. This is NOT an answer to the original question. $\endgroup$
    – Peter K.
    Commented Nov 19, 2015 at 14:42

1 Answer 1


The unilateral Laplace transform is used for analyzing causal linear time-invariant systems, which have an impulse response $h(t)$ that is zero for $t<0$. The unilateral Laplace transform can be used to solve initial value problems, due to the correspondence

$$x'(t)\Longleftrightarrow sX(s)-x(0)$$

where $x(0)$ is a given initial value for the function $x(t)$. Note that for the bilateral Laplace transform the equivalent correspondence is simply $x'(t)\Longleftrightarrow sX(s)$.

Concerning the signal $x(t)=e^{-at}u(t)$, note that its Laplace transform is


This integral only converges if the exponential decays, which results in the condition $\text{Re}(a+s)>0$. This condition defines the region of convergence (ROC) of the Laplace transform. $X(s)$ only exists for values of the complex variable $s$ satisfying $\text{Re}(s)>-\text{Re}(a)$.

  • $\begingroup$ 1)That means unilateral Laplace Transform is only existed for causal linear time-invariant systems, whereas, Bilateral Laplace transform existed for causal and non-causal systems?? 2) Here, integral always not converges for exponential increasing ?? $\endgroup$ Commented Nov 19, 2015 at 11:51
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    $\begingroup$ @AadnanFarooqA: (1) The unilateral transform is only useful for causal systems. It does exist for other systems/signals too, but then you throw away everything for $t<0$, which is usually not what you want. (2) The integral $\int_0^{\infty}e^{at}dt$ is infinite for $a>0$ because the integrand increases exponentially. $\endgroup$
    – Matt L.
    Commented Nov 19, 2015 at 11:55
  • $\begingroup$ I have understood the condition for unilateral Laplace transform. but what will be the case for bilateral Laplace transform? 2) Ok I understood the part you have explained. But in the next example i have $x(t)=-e^{-at}u(-t)$ Its Laplace transform is $X(s)=\frac{1}{s+a}$ and convergence is defined as $Re(s+a)<0$ $\endgroup$ Commented Nov 19, 2015 at 12:02
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    $\begingroup$ @AadnanFarooqA: The bilateral transform exists if the corresponding integral converges. The signal you describe is simply a unilateral signal which is only non-zero for $t<0$, so it's completely analogous with the causal case in your original question. The integral converges if the exponential decays towards the negative $t$-axis. $\endgroup$
    – Matt L.
    Commented Nov 19, 2015 at 12:05
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    $\begingroup$ @AadnanFarooqA: Well, $u(t)$ is non-zero for $t>0$ and $u(-t)$ is non-zero for $t<0$. $\endgroup$
    – Matt L.
    Commented Nov 19, 2015 at 12:07

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