Laplace Transform of 5{\delta (t-3)}+4{u(t-2)}+3{e^{-5(t-3)}u(t-4)}

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SUMMARY

The Laplace Transform of the expression 5{\delta (t-3)}+4{u(t-2)}+3{e^{-5(t-3)}u(t-4)} can be computed using established formulas. The first term transforms to 5e^{-3t}, and the second term transforms to 4/s e^{-2t}. The third term requires the application of the formula L{e^{-\alpha t}u(t)} = 1/(s + α) and the shifting theorem L{f(t - t_0)u(t - t_0)} = e^{-st_0}F(s). The challenge lies in determining F(s) for the third term, which can be resolved by breaking down the exponential function algebraically.

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Homework Statement


Find the Laplace Transform of the following:

[tex]5{\delta (t-3)}+4{u(t-2)}+3{e^{-5(t-3)}u(t-4)}[/tex]


The Attempt at a Solution


I know how to do the first and second terms, but I'm having trouble with the third.

[tex]5e^{-3t}+4\frac{1}{s}e^{-2t} + ?[/tex]

I think I should be using either

[tex]L\{ f(t-t_0)u(t-t_0) \} = e^{-st_0}F(s)[/tex]

or

[tex]L\{ e^{-\alpha t}u(t) \} = \frac{1}{s+\alpha}[/tex]

or both.

The problem is, in the first one, I don't know what to put for [tex]F(s)[/tex].

Any help would be appreciated.
 
Last edited:
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Unfortunately, you can't use your first equation because [tex]t_0[/tex] in f and u are not the same.

Try breaking up the exponential using algebra first.
 

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