S. Boyd EE102Table of Laplace TransformsRemember that we consider all functions (signals) as defined only on t ≥ 0.Generalf(t) F (s) =Z∞0f(t)e−stdtf + g F + Gαf (α ∈ R) αFdfdtsF (s) − f(0)dkfdtkskF (s) − sk−1f(0) − sk−2dfdt(0) − · · · −dk−1fdtk−1(0)g(t) =Zt0f(τ) dτ G(s) =F (s)sf(αt), α > 01αF (s/α)eatf(t) F (s − a)tf(t) −dFdstkf(t) (−1)kdkF (s)dskf(t)tZ∞sF (s) dsg(t) =(0 0 ≤ t < Tf(t − T ) t ≥ T, T ≥ 0 G(s) = e−sTF (s)1Specific11sδ 1δ(k)skt1s2tkk!, k ≥ 01sk+1eat1s − acos ωtss2+ ω2=1/2s − jω+1/2s + jωsin ωtωs2+ ω2=1/2js − jω−1/2js + jωcos(ωt + φ)s cos φ − ω sin φs2+ ω2e−atcos ωts + a(s + a)2+ ω2e−atsin ωtω(s + a)2+ ω22Notes on the derivative formula at t = 0The formula L(f0) = sF (s) − f(0−) must be interpreted very carefully when f has a discon-tinuity at t = 0. We’ll give two examples of the correct interpretation.First, suppose that f is the constant 1, and has no discontinuity at t = 0. In other words,f is the constant function with value 1. Then we have f0= 0, and f(0−) = 1 (since there isno jump in f at t = 0). Now let’s apply the derivative formula above. We have F (s) = 1/s,so the formula readsL(f0) = 0 = sF (s) − 1which is correct.Now, let’s suppose that g is a unit step function, i.e., g(t) = 1 for t > 0, and g(0) = 0.In contrast to f above, g has a jump at t = 0. In this case, g0= δ, 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 first case, f has no jump at t = 0, while in the secondcase g does. As a result, f0has 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