f(t) = (e^t) , 0 <t<3
0 , t>3
the formula of is L { f(t-a) H(t-a) = (e^-as)(F (s) )
why there's no need to transform (e^t)(H(t-3) ) into e^(t-3+3) H(t-3) ) ?
express the function in terms of step unit function and find laplace transform-dsc_0004.jpg