Inverse Laplace transform for 1/(350+s) * X(s)

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SUMMARY

The discussion focuses on finding the inverse Laplace transform of the expression 1/(350+s) * X(s). The incorrect approach of inverting the components separately and multiplying them is highlighted. Instead, the correct method involves using convolution, as the product of two Laplace transforms corresponds to the convolution of their respective time-domain functions. The convolution integral is defined as (f * g)(x) = ∫₀^∞ f(y) g(x-y) dy.

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DinaZhang1
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Hi, everyone, the question is as below:

Find the inverse Laplace transform to 1/(350+s) * X(s). 's' is the Laplace variable and 'X(s)' is also a variable.

I inverted 1/(350+s) and X(s) separately and multiplied them together directly. But this seems not giving me the correct answer. Could anyone help me on this?

Thank you.
 
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DinaZhang1 said:
Hi, everyone, the question is as below:

Find the inverse Laplace transform to 1/(350+s) * X(s). 's' is the Laplace variable and 'X(s)' is also a variable.


I inverted 1/(350+s) and X(s) separately and multiplied them together directly. But this seems not giving me the correct answer. Could anyone help me on this?

Thank you.

Multiplication won't work; you need convolution.

If ##f(t) \leftrightarrow F(s)## and ##g(t) \leftrightarrow G(s)##, then ##F(s) G(s)## is the Laplace transform of the convolution ##f*g##, where
$$(f * g)(x) = \int_0^{\infty} f(y) g(x-y) \, dy = \int_0^{\infty} f(x-y) g(y) \, dy .$$
 

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