Inverse Laplace transform of 1/s

In summary, the conversation discusses the inverse Laplace transform of 1/s, which is given as u(t) in the data sheet and 1 in Wolfram Alpha. Both are correct, but the data sheet result needs to be multiplied by a unit step for t > 0. The unit step, u(t), cannot take on values other than 0 or 1, but it can be scaled by a constant.
  • #1
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Homework Statement


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Homework Equations


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The Attempt at a Solution


The left hand side (red box) is the data sheet provided to us in the exam. The right hand side (blue box) is Wolfram Alpha. The data sheet says that the inverse Laplace transform of 1/s is equal to u(t) (i.e. the unit step) whereas Wolfram Alpha says that it is equal to 1. Which one is correct? Are they both correct? If so, how can that be since the unit inpulse, u(t), can take more than one value. For instance you can have u(t) =5? So shouldn't the data sheet also be 1 rather than u(t)?

Thanks
 
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  • #2
u(t) is the unit step, not unit impulse.

Both are correct. You cannot have u(t) = 5, since the step size of a unit step is unity: 1. What you can do is scale the unit step by multiplying it by a constant. So you might have 5u(t) = 5, for t > 0.

The Wolfram result makes the implied assumption that the result applies for t > 0. That is, the result should be multiplied by a unit step: ##1⋅u(t)##.
 

FAQ: Inverse Laplace transform of 1/s

What is the inverse Laplace transform of 1/s?

The inverse Laplace transform of 1/s is the function f(t) = 1, where t is the time variable. This means that the function does not change over time and remains at a constant value of 1.

How is the inverse Laplace transform of 1/s calculated?

The inverse Laplace transform of 1/s is calculated using the formula f(t) = limT→∞c-iTc+iT F(s)est ds, where F(s) = 1/s is the Laplace transform of the function f(t) = 1. This formula involves taking the limit of an integral over a complex plane and can be solved using various methods such as partial fraction decomposition or contour integration.

Are there any special cases for the inverse Laplace transform of 1/s?

Yes, there are two special cases for the inverse Laplace transform of 1/s. If the constant c in the formula is equal to 0, then the inverse Laplace transform is f(t) = u(t), where u(t) is the unit step function. If c is equal to -∞, then the inverse Laplace transform is f(t) = δ(t), where δ(t) is the Dirac delta function.

What is the physical interpretation of the inverse Laplace transform of 1/s?

The physical interpretation of the inverse Laplace transform of 1/s is the response of a system to a unit impulse input. This means that when an impulse is applied to a system, the output will be a constant value of 1 over time. This can be seen in various fields such as electrical circuits and control systems.

Are there any real-life applications of the inverse Laplace transform of 1/s?

Yes, the inverse Laplace transform of 1/s has various real-life applications. One example is in circuit analysis, where it is used to calculate the response of a circuit to an impulse input. It is also used in control systems to analyze the stability and performance of a system. In addition, it has applications in fields such as fluid dynamics, heat transfer, and signal processing.

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