Inverse Laplace transform of 1/s

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SUMMARY

The inverse Laplace transform of 1/s is correctly identified as u(t), the unit step function, according to standard mathematical conventions. While Wolfram Alpha simplifies this to 1, it implicitly assumes the context of t > 0, where the unit step function is defined. Both interpretations are valid, but the unit step function cannot take arbitrary values like u(t) = 5 without scaling. The correct representation is 1·u(t) for t > 0.

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  • Understanding of Laplace transforms
  • Familiarity with unit step functions
  • Knowledge of mathematical notation and functions
  • Basic principles of signal processing
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  • Learn about scaling functions in signal processing
  • Explore the implications of the unit step function in control systems
  • Investigate the differences between unit step and unit impulse functions
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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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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)##.
 

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