Definition of H(0) when inversing Laplace transform results in heaviside step function The...
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Definition of H(0) when inversing Laplace transform results in heaviside step function
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We have: $ L^{-1}{e^{-cs}F(s)} = H(t - c)f(t-c) $, with $ H $ is a heaviside function.
In many documents, $ H(t - c) $ is defined as:
$ H(t - c) = left{begin{matrix}
0 &, t < c \
1 &, t geq c
end{matrix}right. $
However, the Wikipedia seems to hint that the value of $ H(t - c) $ at $ t = c $ is actually of our choice. I know Wikipedia is not a reliable source but it is true that the definition of the heaviside function in the discrete form does vary at $ H(0) $.
dlmf.nist.gov defines $ H(0) $ to be $ 0 $ whereas uea.ac.uk does not define $ H(0) $.
So when we inverse the Laplace transform of a function and get the heaviside function as the result, we can define $ H(0) $ at our choice?
laplace-transform inverselaplace
asked Mar 17 at 5:55
ntvy95ntvy95
123
$endgroup$
add a comment |
$begingroup$
We have: $ L^{-1}{e^{-cs}F(s)} = H(t - c)f(t-c) $, with $ H $ is a heaviside function.
In many documents, $ H(t - c) $ is defined as:
$ H(t - c) = left{begin{matrix}
0 &, t < c \
1 &, t geq c
end{matrix}right. $
However, the Wikipedia seems to hint that the value of $ H(t - c) $ at $ t = c $ is actually of our choice. I know Wikipedia is not a reliable source but it is true that the definition of the heaviside function in the discrete form does vary at $ H(0) $.
dlmf.nist.gov defines $ H(0) $ to be $ 0 $ whereas uea.ac.uk does not define $ H(0) $.
So when we inverse the Laplace transform of a function and get the heaviside function as the result, we can define $ H(0) $ at our choice?
laplace-transform inverselaplace
asked Mar 17 at 5:55
ntvy95ntvy95
123
$endgroup$
add a comment |
$begingroup$
We have: $ L^{-1}{e^{-cs}F(s)} = H(t - c)f(t-c) $, with $ H $ is a heaviside function.
In many documents, $ H(t - c) $ is defined as:
$ H(t - c) = left{begin{matrix}
0 &, t < c \
1 &, t geq c
end{matrix}right. $
However, the Wikipedia seems to hint that the value of $ H(t - c) $ at $ t = c $ is actually of our choice. I know Wikipedia is not a reliable source but it is true that the definition of the heaviside function in the discrete form does vary at $ H(0) $.
dlmf.nist.gov defines $ H(0) $ to be $ 0 $ whereas uea.ac.uk does not define $ H(0) $.
So when we inverse the Laplace transform of a function and get the heaviside function as the result, we can define $ H(0) $ at our choice?
laplace-transform inverselaplace
asked Mar 17 at 5:55
ntvy95ntvy95
123
$endgroup$
We have: $ L^{-1}{e^{-cs}F(s)} = H(t - c)f(t-c) $, with $ H $ is a heaviside function.
In many documents, $ H(t - c) $ is defined as:
$ H(t - c) = left{begin{matrix}
0 &, t < c \
1 &, t geq c
end{matrix}right. $
However, the Wikipedia seems to hint that the value of $ H(t - c) $ at $ t = c $ is actually of our choice. I know Wikipedia is not a reliable source but it is true that the definition of the heaviside function in the discrete form does vary at $ H(0) $.
dlmf.nist.gov defines $ H(0) $ to be $ 0 $ whereas uea.ac.uk does not define $ H(0) $.
So when we inverse the Laplace transform of a function and get the heaviside function as the result, we can define $ H(0) $ at our choice?
laplace-transform inverselaplace
laplace-transform inverselaplace
asked Mar 17 at 5:55
ntvy95ntvy95
123
asked Mar 17 at 5:55
ntvy95ntvy95
123
asked Mar 17 at 5:55
ntvy95ntvy95
123
asked Mar 17 at 5:55
ntvy95ntvy95
123
asked Mar 17 at 5:55
ntvy95ntvy95
123
123
add a comment |
add a comment |
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