Question #277182

fine laplace transforms of : (t-1) u(t-1)


Expert's answer

Time-displacement theorem:

g(ta)u(ta)easG(s)g(t-a) u(t-a)\to e^{-as}G(s)

so:

(t1)u(t1)ses(t-1) u(t-1)\to se^{-s}



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