Can Laplace Transform be understood as “Area under the curve”?












3












$begingroup$


Since Laplace Transform is basically a definite integral of multiplication of two functions $f(t)$ and $e^{-st}$. Can we interpret Laplace Transform as the area under the curve $f(t)e^{-st}$ from $-infty$ to $infty$ ?



For example, we know Laplace Transform of $f(t)=t$ for $t>0$ is equal to $F(s)=frac{1}{s^2}$. Can we interpret this graphically?






integration fourier-analysis laplace-transform harmonic-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you sure about your stated interval from $-infty$ to $infty$? The Laplace Transform is defined as the interal over $$ the interval $[0,infty)$!
    $endgroup$
    – mrtaurho
    Dec 12 '18 at 18:57












  • $begingroup$
    @mrtaurho there are two types of Laplace Transforms, Unilateral and Bilateral.. Even if you consider from $0$ to $infty$ can you interpret it graphically?
    $endgroup$
    – Simba
    Dec 12 '18 at 19:00












  • $begingroup$
    Well obviously I have forgot about this one. Anyway I would say the unilateral version is more common thus it confused me that you stated the interval $(-infty,infty)$.
    $endgroup$
    – mrtaurho
    Dec 12 '18 at 19:02










  • $begingroup$
    I am almost certain that there is no interest (even heuristical) to try to give an interpretation of Laplace Transform using areas.
    $endgroup$
    – Jean Marie
    Dec 12 '18 at 19:15










  • $begingroup$
    @JeanMarie Do you mean "no pratical interest"?
    $endgroup$
    – rafa11111
    Dec 12 '18 at 19:34
















3












$begingroup$


Since Laplace Transform is basically a definite integral of multiplication of two functions $f(t)$ and $e^{-st}$. Can we interpret Laplace Transform as the area under the curve $f(t)e^{-st}$ from $-infty$ to $infty$ ?



For example, we know Laplace Transform of $f(t)=t$ for $t>0$ is equal to $F(s)=frac{1}{s^2}$. Can we interpret this graphically?






integration fourier-analysis laplace-transform harmonic-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you sure about your stated interval from $-infty$ to $infty$? The Laplace Transform is defined as the interal over $$ the interval $[0,infty)$!
    $endgroup$
    – mrtaurho
    Dec 12 '18 at 18:57












  • $begingroup$
    @mrtaurho there are two types of Laplace Transforms, Unilateral and Bilateral.. Even if you consider from $0$ to $infty$ can you interpret it graphically?
    $endgroup$
    – Simba
    Dec 12 '18 at 19:00












  • $begingroup$
    Well obviously I have forgot about this one. Anyway I would say the unilateral version is more common thus it confused me that you stated the interval $(-infty,infty)$.
    $endgroup$
    – mrtaurho
    Dec 12 '18 at 19:02










  • $begingroup$
    I am almost certain that there is no interest (even heuristical) to try to give an interpretation of Laplace Transform using areas.
    $endgroup$
    – Jean Marie
    Dec 12 '18 at 19:15










  • $begingroup$
    @JeanMarie Do you mean "no pratical interest"?
    $endgroup$
    – rafa11111
    Dec 12 '18 at 19:34














3












3








3





$begingroup$


Since Laplace Transform is basically a definite integral of multiplication of two functions $f(t)$ and $e^{-st}$. Can we interpret Laplace Transform as the area under the curve $f(t)e^{-st}$ from $-infty$ to $infty$ ?



For example, we know Laplace Transform of $f(t)=t$ for $t>0$ is equal to $F(s)=frac{1}{s^2}$. Can we interpret this graphically?






integration fourier-analysis laplace-transform harmonic-analysis







share|cite|improve this question











$endgroup$




Since Laplace Transform is basically a definite integral of multiplication of two functions $f(t)$ and $e^{-st}$. Can we interpret Laplace Transform as the area under the curve $f(t)e^{-st}$ from $-infty$ to $infty$ ?



For example, we know Laplace Transform of $f(t)=t$ for $t>0$ is equal to $F(s)=frac{1}{s^2}$. Can we interpret this graphically?






integration fourier-analysis laplace-transform harmonic-analysis




integration fourier-analysis laplace-transform harmonic-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 18:58







Simba

















asked Dec 12 '18 at 18:50









SimbaSimba

1163




1163












  • $begingroup$
    Are you sure about your stated interval from $-infty$ to $infty$? The Laplace Transform is defined as the interal over $$ the interval $[0,infty)$!
    $endgroup$
    – mrtaurho
    Dec 12 '18 at 18:57












  • $begingroup$
    @mrtaurho there are two types of Laplace Transforms, Unilateral and Bilateral.. Even if you consider from $0$ to $infty$ can you interpret it graphically?
    $endgroup$
    – Simba
    Dec 12 '18 at 19:00












  • $begingroup$
    Well obviously I have forgot about this one. Anyway I would say the unilateral version is more common thus it confused me that you stated the interval $(-infty,infty)$.
    $endgroup$
    – mrtaurho
    Dec 12 '18 at 19:02










  • $begingroup$
    I am almost certain that there is no interest (even heuristical) to try to give an interpretation of Laplace Transform using areas.
    $endgroup$
    – Jean Marie
    Dec 12 '18 at 19:15










  • $begingroup$
    @JeanMarie Do you mean "no pratical interest"?
    $endgroup$
    – rafa11111
    Dec 12 '18 at 19:34


















  • $begingroup$
    Are you sure about your stated interval from $-infty$ to $infty$? The Laplace Transform is defined as the interal over $$ the interval $[0,infty)$!
    $endgroup$
    – mrtaurho
    Dec 12 '18 at 18:57












  • $begingroup$
    @mrtaurho there are two types of Laplace Transforms, Unilateral and Bilateral.. Even if you consider from $0$ to $infty$ can you interpret it graphically?
    $endgroup$
    – Simba
    Dec 12 '18 at 19:00












  • $begingroup$
    Well obviously I have forgot about this one. Anyway I would say the unilateral version is more common thus it confused me that you stated the interval $(-infty,infty)$.
    $endgroup$
    – mrtaurho
    Dec 12 '18 at 19:02










  • $begingroup$
    I am almost certain that there is no interest (even heuristical) to try to give an interpretation of Laplace Transform using areas.
    $endgroup$
    – Jean Marie
    Dec 12 '18 at 19:15










  • $begingroup$
    @JeanMarie Do you mean "no pratical interest"?
    $endgroup$
    – rafa11111
    Dec 12 '18 at 19:34
















$begingroup$
Are you sure about your stated interval from $-infty$ to $infty$? The Laplace Transform is defined as the interal over $$ the interval $[0,infty)$!
$endgroup$
– mrtaurho
Dec 12 '18 at 18:57






$begingroup$
Are you sure about your stated interval from $-infty$ to $infty$? The Laplace Transform is defined as the interal over $$ the interval $[0,infty)$!
$endgroup$
– mrtaurho
Dec 12 '18 at 18:57














$begingroup$
@mrtaurho there are two types of Laplace Transforms, Unilateral and Bilateral.. Even if you consider from $0$ to $infty$ can you interpret it graphically?
$endgroup$
– Simba
Dec 12 '18 at 19:00






$begingroup$
@mrtaurho there are two types of Laplace Transforms, Unilateral and Bilateral.. Even if you consider from $0$ to $infty$ can you interpret it graphically?
$endgroup$
– Simba
Dec 12 '18 at 19:00














$begingroup$
Well obviously I have forgot about this one. Anyway I would say the unilateral version is more common thus it confused me that you stated the interval $(-infty,infty)$.
$endgroup$
– mrtaurho
Dec 12 '18 at 19:02




$begingroup$
Well obviously I have forgot about this one. Anyway I would say the unilateral version is more common thus it confused me that you stated the interval $(-infty,infty)$.
$endgroup$
– mrtaurho
Dec 12 '18 at 19:02












$begingroup$
I am almost certain that there is no interest (even heuristical) to try to give an interpretation of Laplace Transform using areas.
$endgroup$
– Jean Marie
Dec 12 '18 at 19:15




$begingroup$
I am almost certain that there is no interest (even heuristical) to try to give an interpretation of Laplace Transform using areas.
$endgroup$
– Jean Marie
Dec 12 '18 at 19:15












$begingroup$
@JeanMarie Do you mean "no pratical interest"?
$endgroup$
– rafa11111
Dec 12 '18 at 19:34




$begingroup$
@JeanMarie Do you mean "no pratical interest"?
$endgroup$
– rafa11111
Dec 12 '18 at 19:34










1 Answer
1






active

oldest

votes


















2












$begingroup$

I think the best way is to think in analogy to Fourier transforms. In Fourier transforms you think as your time dependent signal as a superposition of simple sinusoidal functions with different frequencies. This works well for periodic signals. The Laplace transformation is used to describe transitions. You turn on something at $t=0$ and wait to equilibrate, at $t=infty$. You can write your time dependence as a superposition of exponential functions with different decay rates.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I am sorry but you certainly want to speak about Fourier series not Fourier transform as you write "This works well for periodic signals". Fourier transform does not assume any periodicity of the signals.
    $endgroup$
    – Jean Marie
    Dec 12 '18 at 19:21












  • $begingroup$
    @JeanMarie Indeed. I was thinking of the series when I've made the connection with periodicity.
    $endgroup$
    – Andrei
    Dec 12 '18 at 19:46











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

I think the best way is to think in analogy to Fourier transforms. In Fourier transforms you think as your time dependent signal as a superposition of simple sinusoidal functions with different frequencies. This works well for periodic signals. The Laplace transformation is used to describe transitions. You turn on something at $t=0$ and wait to equilibrate, at $t=infty$. You can write your time dependence as a superposition of exponential functions with different decay rates.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I am sorry but you certainly want to speak about Fourier series not Fourier transform as you write "This works well for periodic signals". Fourier transform does not assume any periodicity of the signals.
    $endgroup$
    – Jean Marie
    Dec 12 '18 at 19:21












  • $begingroup$
    @JeanMarie Indeed. I was thinking of the series when I've made the connection with periodicity.
    $endgroup$
    – Andrei
    Dec 12 '18 at 19:46
















2












$begingroup$

I think the best way is to think in analogy to Fourier transforms. In Fourier transforms you think as your time dependent signal as a superposition of simple sinusoidal functions with different frequencies. This works well for periodic signals. The Laplace transformation is used to describe transitions. You turn on something at $t=0$ and wait to equilibrate, at $t=infty$. You can write your time dependence as a superposition of exponential functions with different decay rates.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I am sorry but you certainly want to speak about Fourier series not Fourier transform as you write "This works well for periodic signals". Fourier transform does not assume any periodicity of the signals.
    $endgroup$
    – Jean Marie
    Dec 12 '18 at 19:21












  • $begingroup$
    @JeanMarie Indeed. I was thinking of the series when I've made the connection with periodicity.
    $endgroup$
    – Andrei
    Dec 12 '18 at 19:46














2












2








2





$begingroup$

I think the best way is to think in analogy to Fourier transforms. In Fourier transforms you think as your time dependent signal as a superposition of simple sinusoidal functions with different frequencies. This works well for periodic signals. The Laplace transformation is used to describe transitions. You turn on something at $t=0$ and wait to equilibrate, at $t=infty$. You can write your time dependence as a superposition of exponential functions with different decay rates.






share|cite|improve this answer









$endgroup$



I think the best way is to think in analogy to Fourier transforms. In Fourier transforms you think as your time dependent signal as a superposition of simple sinusoidal functions with different frequencies. This works well for periodic signals. The Laplace transformation is used to describe transitions. You turn on something at $t=0$ and wait to equilibrate, at $t=infty$. You can write your time dependence as a superposition of exponential functions with different decay rates.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 12 '18 at 19:02









AndreiAndrei

12.4k21128




12.4k21128












  • $begingroup$
    I am sorry but you certainly want to speak about Fourier series not Fourier transform as you write "This works well for periodic signals". Fourier transform does not assume any periodicity of the signals.
    $endgroup$
    – Jean Marie
    Dec 12 '18 at 19:21












  • $begingroup$
    @JeanMarie Indeed. I was thinking of the series when I've made the connection with periodicity.
    $endgroup$
    – Andrei
    Dec 12 '18 at 19:46


















  • $begingroup$
    I am sorry but you certainly want to speak about Fourier series not Fourier transform as you write "This works well for periodic signals". Fourier transform does not assume any periodicity of the signals.
    $endgroup$
    – Jean Marie
    Dec 12 '18 at 19:21












  • $begingroup$
    @JeanMarie Indeed. I was thinking of the series when I've made the connection with periodicity.
    $endgroup$
    – Andrei
    Dec 12 '18 at 19:46
















$begingroup$
I am sorry but you certainly want to speak about Fourier series not Fourier transform as you write "This works well for periodic signals". Fourier transform does not assume any periodicity of the signals.
$endgroup$
– Jean Marie
Dec 12 '18 at 19:21






$begingroup$
I am sorry but you certainly want to speak about Fourier series not Fourier transform as you write "This works well for periodic signals". Fourier transform does not assume any periodicity of the signals.
$endgroup$
– Jean Marie
Dec 12 '18 at 19:21














$begingroup$
@JeanMarie Indeed. I was thinking of the series when I've made the connection with periodicity.
$endgroup$
– Andrei
Dec 12 '18 at 19:46




$begingroup$
@JeanMarie Indeed. I was thinking of the series when I've made the connection with periodicity.
$endgroup$
– Andrei
Dec 12 '18 at 19:46


















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