Need help with evaluating H(t-8t)2cosh(t-8)

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SUMMARY

The discussion centers on evaluating the expression H(t-8)2cosh(t-8) using the Laplace Transform. The correct interpretation involves applying the second shift theorem to the function H(t-8) multiplied by 2cosh(t-8). The Laplace Transform is expressed as $\mathcal{L}\left\{ H\left( t - 8 \right) \cdot 2\cosh{ \left( t - 8 \right) } \right\} = 2\,\mathrm{e}^{-2\,s}\left( \frac{s}{s^2 - 1} \right)$, confirming the necessity of substituting the correct function and parameters.

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need help with evaluating H(t-8t)2cosh(t-8)
 
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dean2203 said:
need help with evaluating H(t-8t)2cosh(t-8)

Hi dean,
What do you want to evaluate exactly?
Did you try substituting t=0?
Do you know what the function H is?
Btw, is there perchance a typo? Can it be that the first factor should be H(t-8) instead?
 
dean2203 said:
need help with evaluating H(t-8t)2cosh(t-8)

Hi Dean, this is Hayden, I'm assuming that you want to do a Laplace Transform of this. In future please post the ENTIRE question.

I also expect it was $\displaystyle \mathcal{L}\left\{ H \left( t - 8 \right) \cdot 2\cosh{ \left( t - 8 \right) } \right\} $.

Use the second shift theorem.

$\displaystyle \begin{align*} \mathcal{L}\left\{ H\left( t - 8 \right) \cdot 2\cosh{ \left( t - 8 \right) } \right\} &= 2\,\mathrm{e}^{-2\,s} \,\mathcal{L} \left\{ \cosh{ \left( t \right) } \right\} \\ &= 2\,\mathrm{e}^{-2\,s}\left( \frac{s}{s^2 - 1} \right) \end{align*}$
 

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