ME_446_-_Lec_04-Laplace_Transformation

# ME_446_-_Lec_04-Laplace_Transformation - Laplace...

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Unformatted text preview: Laplace Transformation - I 446-4 446- Prof. Neil A. Duffie University of Wisconsin-Madison Wisconsin- Neil A. Duffie, 1996 All rights reserved. Utility of Laplace Transformation Substitutes relatively more easily solved algebraic equations for relatively more difficult to solve differential equations Apply to linear differential equations Apply to physically realizable signals Sequence of application: - obtain differential equations - transform to new equations - solve equations or analyze further Laplace Transformation Given a time variable, t and a function of time, f(t) Define a new variable, s s is a complex variable s = + j j Transform f(t) into L[f(t)] = F(s) Definition of Laplace Transformation L[f(t)] = 0 f(t)e -st dt s is a new, complex variable s = + j To simplify notation F(s) = L[f(t)] Time Function Multiplied by Constant L[kf(t)] = 0 kf(t)e -st dt L[kf(t)] = k 0 f(t)e -st dt L[kf(t)] = k L[f(t)] Sum of Two Time Functions L[x(t)+y(t)] = 0 (x(t)+y(t))e -st dt L[x(t)+y(t)] = 0 (x(t)e -st + y (t)e -st )dt L[x(t)+y(t)] = 0 x(t)e -st dt + 0 y(t)e -st dt L[x(t)+y(t)] = L[x(t)] + L[y(t)] First Derivative of a Time Function L [ ] df(t) dt = 0 df(t) -st e dt dt Integrate by parts du (t) df(t) = dt dt u(t) = f (t) v(t) = e -st dv(t) = -se -st dt du (t) v(t)dt = u(t)v(t) dt u(t) dv(t) dt dt First Derivative of a Time Function L L [ ] [ ] [ ] L df(t) dt df(t) dt df(t) dt = 0 df(t) -st e dt dt = f(t)e -st 0 0 f(t)(-se -st )dt = f()e-s - f(0+)e-s0 "+" for discontinuity in f(t) at t = 0 + 0 sf(t)e -st dt Derivative of a Time Function df(t) L dt L [ ] [ ] df(t) dt = s 0 f(t)e-st dt - f(0+) = sL[f(t)] - f(0+) Second Derivative of a Time Function L [ ] df2(t) dt2 = s2L[f(t)] - sf(0 +) - df(0 +) dt Higher Derivative of a Time Function L [ ] dfi(t) dti - si-2 = siL[f(t)] - si-1f(0+) df(0 +) dt -...- dfi-1(0+) dti-1 Often: Assume initial conditions = 0 function its derivatives L [ ] df i(t) dt i = s iL[f(t)] Time Delay f(t) f(t-D) f(t0 0 for t < D: f(t-D) = 0 f(tt D 0 L[f(t-D)] = 0 f (t-D)e -stdt = D f (t-D)e -stdt Define = t-D t = +D Time Delay f(t) f(t-D) f(t0 0 D 0 for t < D: f(t-D) = 0 f(tt L[f(t-D)] = 0 f()e -s(+D)d (+D ) L[f(t-D)] = e -sD 0 f()e -sd Time Delay f(t) f(t-D) f(t0 0 D 0 for t < D: f(t-D) = 0 f(tt L[f(t-D)] = e -sD L[f()] But L[f()] = L[f(t)] L[f(t-D )] = e -sD L[f(t)] Thermal System Fluid flow rate: Q e(t) Tank o(t) Heater e(t) Insulation resistance: R Heat input: e(t) Temperature difference: (t) = o(t) - e(t) Thermal capacitance: Ct Specific heat of fluid: Sf Thermal System e(t) Tank (t) = o(t) - e(t) o(t) Energy balance: Heater e(t) d (t) Ct = e (t) - 1 (t) - Q S f (t) dt R Laplace transformation: C t(s(s) - (0+)) = E (s) - 1 (s) - Q S f(s) R (C t s + 1 + Q S f )(s) = E (s) + Ct(0+) R Spring-Mass-Damper y(t) [m] Spring- MassAt time = 0: y(0) = 0 y(0) = 0 damper c [N-s/m] [Nx(t) [m] mass M [kg] spring k [N/m] 2 1 dy (t) + 2 dy(t) + y(t) = x(t) n dt n2 dt2 n = k [r/s] M = c 2 kM 1 s2Y(s) + 2 sY(s) + Y (s) = X (s) n n2 Mixing Valve and Pipe valve position x(t) [%] water mixture concentration c(t) [%] concentrated solution length dp [m] flow rate qp [m/s] Delay process d D = p [s] qp c(t) = x(t - D ) C(s) = e-sDX (s) [%] ...
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## This note was uploaded on 04/27/2010 for the course ME 446 taught by Professor Nd during the Spring '10 term at Universität für Bodenkultur Wien.

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