How Do You Graph the Inverse Laplace Transform of a Piecewise Function?

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SUMMARY

The inverse Laplace transform of the function F(s) = 3/s - (4e^-s)/s^2 + (5e^-2s)/s^2 is determined to be f(t) = 3 - 4u(t-1)(t-1) + 5u(t-2)(t-2). The graph of f(t) is segmented into three intervals: for 0 < t < 1, f(t) is a constant 3; for 1 < t < 2, f(t) is 3 - 4(t-1); and for t > 2, f(t) combines all parts resulting in f(t) = 3 - 4(t-1) + 5(t-2). Users can verify their graph using the online tool at fooplot.com.

PREREQUISITES
  • Understanding of inverse Laplace transforms
  • Familiarity with piecewise functions
  • Knowledge of unit step functions (u(t))
  • Basic graphing skills for piecewise functions
NEXT STEPS
  • Learn how to use MATLAB for plotting piecewise functions
  • Study the properties of the Laplace transform
  • Explore advanced graphing techniques using Python's Matplotlib
  • Investigate the application of Laplace transforms in differential equations
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Students studying engineering or mathematics, particularly those focusing on control systems, differential equations, or signal processing, will benefit from this discussion.

gtfitzpatrick
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Homework Statement



determine the inverse laplace transform f(t) of the function

Homework Equations



F(s) = 3/s - (4e^-s)/s^2 + (5e^-2s)/s^2


The Attempt at a Solution



i broke it up into the 3 part and got them seperatly which gave me

f(t) = 3 - 4u(t-1)(t-1) + 5u(t-2)(t-2)

Am i right sofar ?

but I'm not sure how to graph the function?all tips welcome
 
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Yes, that appears to be correct. I got the same answer. To graph it, think of the graph of f(t) as being split into 3 parts. 0<t<1, 1<t<2, t>2.

For the first interval, notice that the unit step function has yet to be "activated", meaning to say that you consider only the part of f(t) which doesn't depend on the unit step function. That graph is f(t)=3. So for the first interval, it is just a straight horizontal line f(t)=3.

For the 2nd interval, 2 things into play; both the f(t)=3 from the first part of the interval as well as the part of the function which is multiplied to u(t-1). The 3rd part of f(t) isn't represented in that interval because it hasnt' been "switched on" yet (only when t>2). means the graph of the function during that interval is made up of 3 -4(t-1).

Finally for the final interval where t>2, all three parts of the function are active, so the final graph for t>2 is simply the graph of f(t)= 3- 4(t-1) + 5(t-2).

Combine all of the above into 1 graph and there you have it. To check your answer, you can graph the function online here:
http://fooplot.com/index.php?
 

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