My Math Forum  

Go Back   My Math Forum > College Math Forum > Differential Equations

Differential Equations Ordinary and Partial Differential Equations Math Forum


Reply
 
LinkBack Thread Tools Display Modes
February 20th, 2017, 12:37 AM   #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 0<t<1
1 for 1<t<2
0 for >2

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 0<t<1
0 for 1<t<2
-1 for >2

or something else?
frostloh is offline  
 
February 20th, 2017, 03:23 AM   #2
Senior Member
 
Joined: Feb 2016
From: Australia

Posts: 1,164
Thanks: 382

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.
Joppy is offline  
Reply

  My Math Forum > College Math Forum > Differential Equations

Tags
finding, function, laplace, step, transform, unit



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Z-transform - Unit step KaiL Calculus 1 January 19th, 2017 10:28 AM
Laplace Transform, Finding solution: y′′+4y′+4y=f(t) jakejakejake Differential Equations 0 May 17th, 2014 07:25 PM
unit step and dirac delta function gelatine1 Algebra 2 November 4th, 2012 01:48 PM
Unit step function FreaKariDunk Calculus 21 April 20th, 2012 02:52 AM
Laplace transform for initial value from of step function PSBlueprint54 Real Analysis 0 November 15th, 2011 08:08 AM





Copyright © 2017 My Math Forum. All rights reserved.