Can Laplace Transform be understood as “Area under the curve”?
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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
asked Dec 12 '18 at 18:50
SimbaSimba
1163
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show 4 more comments
$begingroup$
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
asked Dec 12 '18 at 18:50
SimbaSimba
1163
$endgroup$
$begingroup$
Are you sure about your stated interval from $-infty$ to $infty$? The Laplace Transform is defined as the interal over $$ the interval $[0,infty)$!
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– mrtaurho
Dec 12 '18 at 18:57
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@mrtaurho there are two types of Laplace Transforms, Unilateral and Bilateral.. Even if you consider from $0$ to $infty$ can you interpret it graphically?
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– Simba
Dec 12 '18 at 19:00
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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)$.
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– mrtaurho
Dec 12 '18 at 19:02
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I am almost certain that there is no interest (even heuristical) to try to give an interpretation of Laplace Transform using areas.
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– Jean Marie
Dec 12 '18 at 19:15
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@JeanMarie Do you mean "no pratical interest"?
$endgroup$
– rafa11111
Dec 12 '18 at 19:34
|
show 4 more comments
$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
asked Dec 12 '18 at 18:50
SimbaSimba
1163
$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
asked Dec 12 '18 at 18:50
SimbaSimba
1163
asked Dec 12 '18 at 18:50
SimbaSimba
1163
edited Dec 12 '18 at 18:58
Simba
asked Dec 12 '18 at 18:50
SimbaSimba
1163
asked Dec 12 '18 at 18:50
SimbaSimba
1163
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
|
show 4 more comments
$begingroup$
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
|
show 4 more comments
1 Answer
1
active
oldest
votes
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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.
answered Dec 12 '18 at 19:02
AndreiAndrei
12.4k21128
$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
add a comment |
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1 Answer
1
active
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1 Answer
1
active
oldest
votes
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$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.
answered Dec 12 '18 at 19:02
AndreiAndrei
12.4k21128
$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
add a comment |
$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.
answered Dec 12 '18 at 19:02
AndreiAndrei
12.4k21128
$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
add a comment |
$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.
answered Dec 12 '18 at 19:02
AndreiAndrei
12.4k21128
$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.
answered Dec 12 '18 at 19:02
AndreiAndrei
12.4k21128
answered Dec 12 '18 at 19:02
AndreiAndrei
12.4k21128
answered Dec 12 '18 at 19:02
AndreiAndrei
12.4k21128
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
add a comment |
$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
add a comment |
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$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