Inverse laplace transform

In summary: AxisLabel=x&yAxisLabel=y&grid=on&thick=1&plotcolor=%2300FF00&bgcolor=%23FFFFFF&lang=enIn summary, the inverse Laplace transform of the given function is f(t) = 3 - 4u(t-1)(t-1) + 5u(t-2)(t
  • #1
gtfitzpatrick
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0

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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  • #2
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?
 

1. What is an inverse Laplace transform?

An inverse Laplace transform is a mathematical operation that takes a function in the Laplace domain and returns the original function in the time domain. It is the opposite of the Laplace transform, which converts a function in the time domain to the Laplace domain.

2. Why is the inverse Laplace transform important?

The inverse Laplace transform is important because it allows us to solve differential equations in the time domain by transforming them into simpler algebraic equations in the Laplace domain. This makes it a powerful tool in various fields such as engineering, physics, and mathematics.

3. How is the inverse Laplace transform calculated?

The inverse Laplace transform can be calculated using various methods, such as partial fraction decomposition, convolution, and residue theorem. The specific method used depends on the complexity of the function in the Laplace domain and the desired accuracy of the result.

4. Can any function be inverted using the Laplace transform?

No, not all functions can be inverted using the Laplace transform. The function must meet certain conditions, such as being piecewise continuous and having a finite number of discontinuities and poles. Functions that do not meet these conditions cannot be inverted using the Laplace transform.

5. What are some real-world applications of the inverse Laplace transform?

The inverse Laplace transform has many real-world applications, including analyzing electrical circuits, solving heat transfer problems, and predicting the behavior of mechanical systems. It is also used in signal processing, control systems, and in the study of fluid dynamics.

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