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4.6 Laplace transform • What else did he do? ∞ X ( s ) = Lx(t ) = ∫ e − st x(t )dt 0 • Main advantage: converting differential equation to algebraic ones ∞ ∞ dx ∞ − st dx dt = e − st x 0 + s ∫ e− st xdt L = ∫e dt 0 dt 0 = sX ( s ) − x(0) Pierre-Simon Laplace 1749 ( Beaumont-enAuge, Normandy) to 1827 ( Paris) Generalized impedance • Using the transforms of d 2x L 2 = s 2 X ( s ) − sx(0) − x (0) second derivative dt • Write spring-mass(ms 2 + cs + k ) X ( s ) = dashpot equation as F ( s ) + (ms + c) x(0) + mx (0) • Without initial F (s) = ms 2 + cs + k Z (s) = X (s) conditions get 1 X (s) generalized impedance = 2 G(s) = F ( s ) ms + cs + k and transfer function Laplace transform of delta and step functions • For delta function ∞ ∆( s ) = ∫ e − stδ (t )dt = e − st 0 • Since X ( s ) = G ( s ) F ( s ) • Then x(t ) = L-1G ( s ) F ( s ) Then g (t ) = L G ( s )∆ ( s ) = L G ( s ) -1 • Similarly -1 ∞ ∞ 1 U ( s ) = ∫ e u (t )dt = ∫ e dt = s 0 0 − st − st G(s) s(t ) = L G ( s )U ( s ) = L s -1 t =0 =1 -1 Laplace transform tables • Appendix B.8 • Boyce and Diprima, Elementary differential equations and boundary value problems. Example • Use Laplace transform to find solution of x + x = sin 2t x(0) = 2 x (0) = 1 • Laplace transform of equation (see table for sine) s 2 X ( s ) − sx(0) − x (0) + X ( s ) = 2 /( s 2 + 4) • Solution in s-plane 2 s + s + 8s + 6 X ( s) = 2 2 ( s + 1)( s + 4) 3 2 Solution in time domain • Partial fraction expansion 2s 5/3 2/3 X (s) = 2 + 2 − 2 s +1 s +1 s + 4 • Back-transform using table 5 1 x(t ) = 2 cos t + sin t − sin 2t 3 3 • Checks? Example 4.7 by table • Find response of undamped single degree of freedom system to the sawtooth pulse F (t ) = F0 [tu (t ) − (t − T0 )u (t − T0 ) − u (t − T0 ) ] 1 Ltu (t ) = 2 s (#3) e − sT0 L (t − T0 )u (t − T0 ) = 2 (#13) s e − sT0 Lu (t − T ) = (#12) s Example 4.7 - continued ⎡ T0 e − sT0 1 − sT0 ⎤ + 2 1− e ⎢− ⎥ s s ⎣ ⎦ 1 1 = G(s) = 2 ms + k m( s 2 + ωn2 ) F F (s) = 0 T0 • Solution in s-plane ( ) X ( s ) = F ( s )G ( s ) • From Table B.8 1 1 L = 2 (1 − cos ωnt ) 2 2 s ( s + ωn ) ωn -1 1 1 L 2 2 = 2 (ωnt − sin ωnt ) 2 s ( s + ωn ) ωn -1 Solution • With some algebra Checks? 4.7 General system response • Applying Laplace transform to mx + cx + kx = f (t ) x(0) = x0 x (0) = v0 • Get X ( s) = s + 2ζωn F (s) 1 x v + + 2 2 2 2 0 2 2 0 m( s + 2ζωn s + ωn ) s + 2ζωn s + ωn s + 2ζωn s + ωn • See development in textbook to get 1 x(t ) = mωd + ∫ t 0 F (τ )e −ζωn (t −τ ) sin ωd (t − τ )dτ x0ωn ωd e −ζωn t cos(ωd t −ψ ) + v0 ωd e −ζωnt sin ωd t Reading assignment Sections 4.7-4.9 Source: www.library.veryhelpful.co.uk/ Page11.htm