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Chapter 3: Laplace Transform Chapter 3.1: Definition : ∞ L{f(t)} = F(s) = ∫ f (t)e dt −st f(t) : original function F(s) : Laplace transform Chapter [2] Example1: Find Laplace Transform ? Unit Step Function u(t) or 1(t) : 1 ↔ for : t > u (t ) = 0 ↔ for : t < ∞ −st ∞ e L{u(t)} = ∫ e dt = = −s s Chapter −st [3] Example2: Find Laplace Transform ? Impulse or Delta or Dirac Function δ(t) : ∞ ↔ for : t = δ (t ) = 0 ↔ for : t ≠ ∞ δ(t) L{δ (t)} = ∫δ (t)e dt = e =1 −st 0 Chapter [4] 3.2: Properties of Laplace Transform : 1) Multiplication by a constant : If: L{f(t)} = F(s) Then: Chapter L{k.f(t)} = k.F(s) [5] 2) Addition / Subtraction: If: L{f1 (t)} = F1 (s) and L{f (t)} = F2 (s) Then: Chapter L{f1 (t) ± f (t) } = F1 (s) ± F2 (s) [6] 3) Translation in Time-domain: If: L{f(t)} = F(s) Then: Chapter L{f(t − t ).u(t − t ) } = F(s).e − st [7] 4) Translation in frequency-domain: If: L{f(t)} = F(s) Then: Chapter L{f(t).e − at } = F(s + a) [8] 5) Scale Changing: If: L{f(t)} = F(s) Then: Chapter s L{f(at) } = F a a [9] 6) Differentiation: If: L{f(t)} = F(s) Then: df (t) L = sF ( s ) − f (0) dt d f (t) ' L = s F ( s ) − s.f (0) − f (0) dt L {f Chapter (n ) (t)} = s F ( s ) − s f (0) − − f n n −1 (n −1) (0) [10] 10) Periodic signal : f(t) = f1 (t) + f (t) + f (t) + = f1 (t) + f1 (t − T)u(t − T) + f1 (t − 2T)u(t − 2T) + F(s) = F1 (s) + F1 (s).e − sT + F1 (s).e − s2T + Then: F1 (s) F(s) = − sT 1− e F1(s) = Laplace Transform over first period only Chapter f(t) t T 2T 3T A periodic function f1(t) t T f2(t) t T 2T f3(t) t 2T 3T Decomposition of f(t) [14] 11) Initial Value Theorem : If: L{f(t)} = F(s) Then: Chapter f (0 ) = lim+ f (t) = lim ( s.F(s) ) + t →0 s →∞ [15] 12) Final Value Theorem : If: L{f(t)} = F(s) Then: f (∞) = lim f (t) = lim ( s.F(s) ) t →∞ s →0 Attention: all poles of F(s) must be located in the left half of the s-plane, except s = Chapter [16] 3.3: Laplace Transform of Fundamental functions : 1) Unit Step Function u(t) or 1(t) : 1 ↔ for : t > u (t ) = 0 ↔ for : t < Chapter £{u(t)} = s [17] 2) Translated unit step u(t – t0) : 1↔ for :t >t0 u(t −t0) = 0 ↔ for :t