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Why do we need Laplace Transforms?


Differential equations and Fourier and Laplace transformsCompare Fourier and Laplace transformNeed guidance with inverse laplace transformA discussion on fourier and laplace transforms and differential equations …?Contradiction between Fourier and Laplace transforms?Why Fourier transform is not sufficient and we have to use Laplace transform?Why is it that the Laplace transform of $f(t)$ evaluated at $s = jomega$ is not the Fourier transformLaplace transform of $f(t^2)$Fourier and Laplace transforms together, is this possible?Why can one ignore the real part of a complex number because of Fourier transform “only using the rotational part”?













2












$begingroup$


If Fourier transforms can depict all types of signals, why do need Laplace transforms?



Can't Fourier transform handle exponential signals (or damped oscillations)?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Not every signal has its fourier transform. There are two conditions a signal must satisfy in order to have reversible fourier transormation.
    $endgroup$
    – user499203
    Nov 24 '17 at 18:33










  • $begingroup$
    For the existence of Fourier transform, I think just the function should be absolutely integrable.
    $endgroup$
    – L.Shane John Paul Newton
    Nov 24 '17 at 18:36










  • $begingroup$
    Actually, either a signal or it's square must be absolutely integrable. See Dirichlet statements about this topic.
    $endgroup$
    – user499203
    Nov 24 '17 at 18:37










  • $begingroup$
    Doesn't those conditions apply to Laplace transforms as well?
    $endgroup$
    – L.Shane John Paul Newton
    Nov 24 '17 at 18:39










  • $begingroup$
    The Laplace transform simply avoids the use of Dirac distributions. Otherwise there isn't much difference. For example, initial conditions in differential equations are more easily taken care of with the Laplace transform.
    $endgroup$
    – Chrystomath
    Nov 24 '17 at 19:05
















2












$begingroup$


If Fourier transforms can depict all types of signals, why do need Laplace transforms?



Can't Fourier transform handle exponential signals (or damped oscillations)?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Not every signal has its fourier transform. There are two conditions a signal must satisfy in order to have reversible fourier transormation.
    $endgroup$
    – user499203
    Nov 24 '17 at 18:33










  • $begingroup$
    For the existence of Fourier transform, I think just the function should be absolutely integrable.
    $endgroup$
    – L.Shane John Paul Newton
    Nov 24 '17 at 18:36










  • $begingroup$
    Actually, either a signal or it's square must be absolutely integrable. See Dirichlet statements about this topic.
    $endgroup$
    – user499203
    Nov 24 '17 at 18:37










  • $begingroup$
    Doesn't those conditions apply to Laplace transforms as well?
    $endgroup$
    – L.Shane John Paul Newton
    Nov 24 '17 at 18:39










  • $begingroup$
    The Laplace transform simply avoids the use of Dirac distributions. Otherwise there isn't much difference. For example, initial conditions in differential equations are more easily taken care of with the Laplace transform.
    $endgroup$
    – Chrystomath
    Nov 24 '17 at 19:05














2












2








2


1



$begingroup$


If Fourier transforms can depict all types of signals, why do need Laplace transforms?



Can't Fourier transform handle exponential signals (or damped oscillations)?










share|cite|improve this question











$endgroup$




If Fourier transforms can depict all types of signals, why do need Laplace transforms?



Can't Fourier transform handle exponential signals (or damped oscillations)?







fourier-analysis laplace-transform fourier-transform






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









nbro

2,41563174




2,41563174










asked Nov 24 '17 at 18:27









L.Shane John Paul NewtonL.Shane John Paul Newton

182




182












  • $begingroup$
    Not every signal has its fourier transform. There are two conditions a signal must satisfy in order to have reversible fourier transormation.
    $endgroup$
    – user499203
    Nov 24 '17 at 18:33










  • $begingroup$
    For the existence of Fourier transform, I think just the function should be absolutely integrable.
    $endgroup$
    – L.Shane John Paul Newton
    Nov 24 '17 at 18:36










  • $begingroup$
    Actually, either a signal or it's square must be absolutely integrable. See Dirichlet statements about this topic.
    $endgroup$
    – user499203
    Nov 24 '17 at 18:37










  • $begingroup$
    Doesn't those conditions apply to Laplace transforms as well?
    $endgroup$
    – L.Shane John Paul Newton
    Nov 24 '17 at 18:39










  • $begingroup$
    The Laplace transform simply avoids the use of Dirac distributions. Otherwise there isn't much difference. For example, initial conditions in differential equations are more easily taken care of with the Laplace transform.
    $endgroup$
    – Chrystomath
    Nov 24 '17 at 19:05


















  • $begingroup$
    Not every signal has its fourier transform. There are two conditions a signal must satisfy in order to have reversible fourier transormation.
    $endgroup$
    – user499203
    Nov 24 '17 at 18:33










  • $begingroup$
    For the existence of Fourier transform, I think just the function should be absolutely integrable.
    $endgroup$
    – L.Shane John Paul Newton
    Nov 24 '17 at 18:36










  • $begingroup$
    Actually, either a signal or it's square must be absolutely integrable. See Dirichlet statements about this topic.
    $endgroup$
    – user499203
    Nov 24 '17 at 18:37










  • $begingroup$
    Doesn't those conditions apply to Laplace transforms as well?
    $endgroup$
    – L.Shane John Paul Newton
    Nov 24 '17 at 18:39










  • $begingroup$
    The Laplace transform simply avoids the use of Dirac distributions. Otherwise there isn't much difference. For example, initial conditions in differential equations are more easily taken care of with the Laplace transform.
    $endgroup$
    – Chrystomath
    Nov 24 '17 at 19:05
















$begingroup$
Not every signal has its fourier transform. There are two conditions a signal must satisfy in order to have reversible fourier transormation.
$endgroup$
– user499203
Nov 24 '17 at 18:33




$begingroup$
Not every signal has its fourier transform. There are two conditions a signal must satisfy in order to have reversible fourier transormation.
$endgroup$
– user499203
Nov 24 '17 at 18:33












$begingroup$
For the existence of Fourier transform, I think just the function should be absolutely integrable.
$endgroup$
– L.Shane John Paul Newton
Nov 24 '17 at 18:36




$begingroup$
For the existence of Fourier transform, I think just the function should be absolutely integrable.
$endgroup$
– L.Shane John Paul Newton
Nov 24 '17 at 18:36












$begingroup$
Actually, either a signal or it's square must be absolutely integrable. See Dirichlet statements about this topic.
$endgroup$
– user499203
Nov 24 '17 at 18:37




$begingroup$
Actually, either a signal or it's square must be absolutely integrable. See Dirichlet statements about this topic.
$endgroup$
– user499203
Nov 24 '17 at 18:37












$begingroup$
Doesn't those conditions apply to Laplace transforms as well?
$endgroup$
– L.Shane John Paul Newton
Nov 24 '17 at 18:39




$begingroup$
Doesn't those conditions apply to Laplace transforms as well?
$endgroup$
– L.Shane John Paul Newton
Nov 24 '17 at 18:39












$begingroup$
The Laplace transform simply avoids the use of Dirac distributions. Otherwise there isn't much difference. For example, initial conditions in differential equations are more easily taken care of with the Laplace transform.
$endgroup$
– Chrystomath
Nov 24 '17 at 19:05




$begingroup$
The Laplace transform simply avoids the use of Dirac distributions. Otherwise there isn't much difference. For example, initial conditions in differential equations are more easily taken care of with the Laplace transform.
$endgroup$
– Chrystomath
Nov 24 '17 at 19:05










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