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Visualizing the Laplace Transform (for beginners)
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I'm taking a differential equations course and have just been introduced to the concept of the Laplace Transform. I know its usefulness as a tool for solving IVPs, but I'm having trouble visualizing what the transform actually is. How would one explain the physical interpretation of the Laplace transform to a dummy? If possible, please refrain from using too much real/complex analysis (I only have experience in Linear Algebra, Multivariable Calculus, and Diff Eqs with some dabbling in function analysis).
ordinary-differential-equations laplace-transform
asked Jan 20 at 17:54
HyperionHyperion
636110
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add a comment |
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I'm taking a differential equations course and have just been introduced to the concept of the Laplace Transform. I know its usefulness as a tool for solving IVPs, but I'm having trouble visualizing what the transform actually is. How would one explain the physical interpretation of the Laplace transform to a dummy? If possible, please refrain from using too much real/complex analysis (I only have experience in Linear Algebra, Multivariable Calculus, and Diff Eqs with some dabbling in function analysis).
ordinary-differential-equations laplace-transform
asked Jan 20 at 17:54
HyperionHyperion
636110
$endgroup$
3
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I very highly recommend to watch this lecture given by Arthur Mattuck - youtube.com/watch?v=sZ2qulI6GEk. It explains the Laplace transform in a beautiful manner, and shows the intuition behind it.
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– Kolja
Jan 20 at 18:00
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@Kolja Thank you so much! That video helped a ton with my understanding.
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– Hyperion
Jan 21 at 16:34
add a comment |
$begingroup$
I'm taking a differential equations course and have just been introduced to the concept of the Laplace Transform. I know its usefulness as a tool for solving IVPs, but I'm having trouble visualizing what the transform actually is. How would one explain the physical interpretation of the Laplace transform to a dummy? If possible, please refrain from using too much real/complex analysis (I only have experience in Linear Algebra, Multivariable Calculus, and Diff Eqs with some dabbling in function analysis).
ordinary-differential-equations laplace-transform
asked Jan 20 at 17:54
HyperionHyperion
636110
$endgroup$
I'm taking a differential equations course and have just been introduced to the concept of the Laplace Transform. I know its usefulness as a tool for solving IVPs, but I'm having trouble visualizing what the transform actually is. How would one explain the physical interpretation of the Laplace transform to a dummy? If possible, please refrain from using too much real/complex analysis (I only have experience in Linear Algebra, Multivariable Calculus, and Diff Eqs with some dabbling in function analysis).
ordinary-differential-equations laplace-transform
ordinary-differential-equations laplace-transform
asked Jan 20 at 17:54
HyperionHyperion
636110
asked Jan 20 at 17:54
HyperionHyperion
636110
asked Jan 20 at 17:54
HyperionHyperion
636110
asked Jan 20 at 17:54
HyperionHyperion
636110
asked Jan 20 at 17:54
HyperionHyperion
636110
636110
3
$begingroup$
I very highly recommend to watch this lecture given by Arthur Mattuck - youtube.com/watch?v=sZ2qulI6GEk. It explains the Laplace transform in a beautiful manner, and shows the intuition behind it.
$endgroup$
– Kolja
Jan 20 at 18:00
$begingroup$
@Kolja Thank you so much! That video helped a ton with my understanding.
$endgroup$
– Hyperion
Jan 21 at 16:34
add a comment |
3
$begingroup$
I very highly recommend to watch this lecture given by Arthur Mattuck - youtube.com/watch?v=sZ2qulI6GEk. It explains the Laplace transform in a beautiful manner, and shows the intuition behind it.
$endgroup$
– Kolja
Jan 20 at 18:00
$begingroup$
@Kolja Thank you so much! That video helped a ton with my understanding.
$endgroup$
– Hyperion
Jan 21 at 16:34
3
3
$begingroup$
I very highly recommend to watch this lecture given by Arthur Mattuck - youtube.com/watch?v=sZ2qulI6GEk. It explains the Laplace transform in a beautiful manner, and shows the intuition behind it.
$endgroup$
– Kolja
Jan 20 at 18:00
$begingroup$
I very highly recommend to watch this lecture given by Arthur Mattuck - youtube.com/watch?v=sZ2qulI6GEk. It explains the Laplace transform in a beautiful manner, and shows the intuition behind it.
$endgroup$
– Kolja
Jan 20 at 18:00
$begingroup$
@Kolja Thank you so much! That video helped a ton with my understanding.
$endgroup$
– Hyperion
Jan 21 at 16:34
$begingroup$
@Kolja Thank you so much! That video helped a ton with my understanding.
$endgroup$
– Hyperion
Jan 21 at 16:34
add a comment |
1 Answer
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The Laplace transform and the Fourier transform are closely related. One way to view them is to analyse what happens in convolution of a function with anexponential functions. Recall that convolution is defined as
$$ (f*g)(x) := int_{-infty}^infty f(t),g(x-t) dt. tag{1}$$
If we let $,g_s(x) := e^{s,x},,$ a nice test function, then
$$ (f*g_s)(x) = int_{-infty}^infty f(t),e^{s,(x-t)} dt =
e^{s, x}, (f*g_s)(0).tag{2}$$
Now notice that $,(f*g_s)(0),$ is just the bilateral Laplace transform of $,f(t).,$ A similar situation arises for the Fourier transform. One interpretation of this is that the linear transformation $, g to f*g,$ has exponential eigenvectors $,g_s,$ and the associated eigenvalue is the Laplace transform $,F(s) = (f*g_s)(0).$
As for visualization, you can consult the Wikipedia article on convolution for illustrative examples.
answered Jan 20 at 21:46
SomosSomos
14.4k11336
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1 Answer
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$begingroup$
The Laplace transform and the Fourier transform are closely related. One way to view them is to analyse what happens in convolution of a function with anexponential functions. Recall that convolution is defined as
$$ (f*g)(x) := int_{-infty}^infty f(t),g(x-t) dt. tag{1}$$
If we let $,g_s(x) := e^{s,x},,$ a nice test function, then
$$ (f*g_s)(x) = int_{-infty}^infty f(t),e^{s,(x-t)} dt =
e^{s, x}, (f*g_s)(0).tag{2}$$
Now notice that $,(f*g_s)(0),$ is just the bilateral Laplace transform of $,f(t).,$ A similar situation arises for the Fourier transform. One interpretation of this is that the linear transformation $, g to f*g,$ has exponential eigenvectors $,g_s,$ and the associated eigenvalue is the Laplace transform $,F(s) = (f*g_s)(0).$
As for visualization, you can consult the Wikipedia article on convolution for illustrative examples.
answered Jan 20 at 21:46
SomosSomos
14.4k11336
$endgroup$
add a comment |
$begingroup$
The Laplace transform and the Fourier transform are closely related. One way to view them is to analyse what happens in convolution of a function with anexponential functions. Recall that convolution is defined as
$$ (f*g)(x) := int_{-infty}^infty f(t),g(x-t) dt. tag{1}$$
If we let $,g_s(x) := e^{s,x},,$ a nice test function, then
$$ (f*g_s)(x) = int_{-infty}^infty f(t),e^{s,(x-t)} dt =
e^{s, x}, (f*g_s)(0).tag{2}$$
Now notice that $,(f*g_s)(0),$ is just the bilateral Laplace transform of $,f(t).,$ A similar situation arises for the Fourier transform. One interpretation of this is that the linear transformation $, g to f*g,$ has exponential eigenvectors $,g_s,$ and the associated eigenvalue is the Laplace transform $,F(s) = (f*g_s)(0).$
As for visualization, you can consult the Wikipedia article on convolution for illustrative examples.
answered Jan 20 at 21:46
SomosSomos
14.4k11336
$endgroup$
add a comment |
$begingroup$
The Laplace transform and the Fourier transform are closely related. One way to view them is to analyse what happens in convolution of a function with anexponential functions. Recall that convolution is defined as
$$ (f*g)(x) := int_{-infty}^infty f(t),g(x-t) dt. tag{1}$$
If we let $,g_s(x) := e^{s,x},,$ a nice test function, then
$$ (f*g_s)(x) = int_{-infty}^infty f(t),e^{s,(x-t)} dt =
e^{s, x}, (f*g_s)(0).tag{2}$$
Now notice that $,(f*g_s)(0),$ is just the bilateral Laplace transform of $,f(t).,$ A similar situation arises for the Fourier transform. One interpretation of this is that the linear transformation $, g to f*g,$ has exponential eigenvectors $,g_s,$ and the associated eigenvalue is the Laplace transform $,F(s) = (f*g_s)(0).$
As for visualization, you can consult the Wikipedia article on convolution for illustrative examples.
answered Jan 20 at 21:46
SomosSomos
14.4k11336
$endgroup$
The Laplace transform and the Fourier transform are closely related. One way to view them is to analyse what happens in convolution of a function with anexponential functions. Recall that convolution is defined as
$$ (f*g)(x) := int_{-infty}^infty f(t),g(x-t) dt. tag{1}$$
If we let $,g_s(x) := e^{s,x},,$ a nice test function, then
$$ (f*g_s)(x) = int_{-infty}^infty f(t),e^{s,(x-t)} dt =
e^{s, x}, (f*g_s)(0).tag{2}$$
Now notice that $,(f*g_s)(0),$ is just the bilateral Laplace transform of $,f(t).,$ A similar situation arises for the Fourier transform. One interpretation of this is that the linear transformation $, g to f*g,$ has exponential eigenvectors $,g_s,$ and the associated eigenvalue is the Laplace transform $,F(s) = (f*g_s)(0).$
As for visualization, you can consult the Wikipedia article on convolution for illustrative examples.
answered Jan 20 at 21:46
SomosSomos
14.4k11336
answered Jan 20 at 21:46
SomosSomos
14.4k11336
answered Jan 20 at 21:46
SomosSomos
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answered Jan 20 at 21:46
SomosSomos
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I very highly recommend to watch this lecture given by Arthur Mattuck - youtube.com/watch?v=sZ2qulI6GEk. It explains the Laplace transform in a beautiful manner, and shows the intuition behind it.
$endgroup$
– Kolja
Jan 20 at 18:00
$begingroup$
@Kolja Thank you so much! That video helped a ton with my understanding.
$endgroup$
– Hyperion
Jan 21 at 16:34