laplace transform of unit step function $u(t+1)$
what is the laplace transform of unit step function $u(t+1)$?According to my understanding it should be $ e^s/s$ but I am not sure about it.
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$\begingroup$When you have doubt , you can always use the integral definition of the Laplace Transform:$$f(s)=\int_0^\infty u(t+1)e^{-st}dt$$$$f(s)=\int_{-1}^\infty e^{-st}dt$$$$f(s)=\dfrac {e^{-st}}{-s}\bigg|_{-1}^\infty$$$$f(s)=\dfrac {e^{s}}{s}$$So yes you're right.
$\endgroup$ $\begingroup$It is indeed. More generally,
$$\mathcal{L}(u(t-a)) = \frac{e^{-as}}{s}$$
Take $a=-1$ for your desired transform. A number of identities such as this can be found on this page I have bookmarked.
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