S. Boyd EE102Table of Laplace TransformsRemember that we consider all functions (signals) as deflned only on t ‚ 0.Generalf(t) F(s) = Z01 f(t)e¡st dtf + g F + Gfif (fi 2 R) fiFdfdt sF(s) ¡ f(0)dkfdtk skF(s) ¡ sk¡1f(0) ¡ sk¡2ddt f (0) ¡ ¢ ¢ ¢ ¡ ddt k¡k¡1f1 (0)g(t) = Z0t f(¿) d¿ … Continue reading “Table of Laplace Transforms | My Assignment Tutor”
S. Boyd EE102Table of Laplace TransformsRemember that we consider all functions (signals) as deflned only on t ‚ 0.Generalf(t) F(s) = Z01 f(t)e¡st dtf + g F + Gfif (fi 2 R) fiFdfdt sF(s) ¡ f(0)dkfdtk skF(s) ¡ sk¡1f(0) ¡ sk¡2ddt f (0) ¡ ¢ ¢ ¢ ¡ ddt k¡k¡1f1 (0)g(t) = Z0t f(¿) d¿ G(s) = Fs(s)f(fit), fi > 0 1fiF(s=fi)eatf(t) F(s ¡ a)tf(t) ¡dFdstkf(t) (¡1)k dkF(s)dskf(t)t Zs1 F(s) dsg(t) = ( 0 f(t ¡ T) t0‚•Tt 0, and g(0) = 0.In contrast to f above, g has a jump at t = 0. In this case, g0 = –, and g(0¡) = 0. Now let’sapply the derivative formula above. We have G(s) = 1=s (exactly the same as F!), so theformula readsL(g0) = 1 = sG(s) ¡ 0which again is correct.In these two examples the functions f and g are the same except at t = 0, so they havethe same Laplace transform. In the flrst case, f has no jump at t = 0, while in the secondcase g does. As a result, f0 has no impulsive term at t = 0, whereas g does. As long as youkeep track of whether your function has, or doesn’t have, a jump at t = 0, and apply theformula consistently, everything will work out.3
