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 February 19th, 2017, 11:37 PM #1 Newbie   Joined: Feb 2017 From: malaysia Posts: 1 Thanks: 0 Finding laplace transform of unit step function? The initial function is: f(t) = 0 for 02 In function notation this function would be f(t)=U(t)-U(t-2). However the actual question is for us to find f(t-1). So what would this mean? Would it mean: f(t) = -1 for 02 or something else? February 20th, 2017, 02:23 AM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,834 Thanks: 649 Math Focus: Yet to find out. If $f(t) = U(t) - U(t - 2)$, then, $f(t - 1) = U(t - 1) - U((t - 1) - 2) = U(t - 1) - U(t - 3)$. Hence, $\displaystyle f(t - 1) =$$\displaystyle \left\{ \begin{array}{rl} 1, &\mbox{1$\displaystyle \leq$t < 3} \\ 0, &\mbox{elsewhere.} \end{array} \right.$ The Laplace Transform of $f(t - 1)$ should be straight forward to compute using tables or otherwise. Tags finding, function, laplace, step, transform, unit Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post KaiL Calculus 1 January 19th, 2017 09:28 AM jakejakejake Differential Equations 0 May 17th, 2014 07:25 PM gelatine1 Algebra 2 November 4th, 2012 12:48 PM FreaKariDunk Calculus 21 April 20th, 2012 02:52 AM PSBlueprint54 Real Analysis 0 November 15th, 2011 07:08 AM

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