Laplace and inverse laplace transformation of unit step functions u(t).

Click For Summary
SUMMARY

The discussion centers on the Laplace transformation of the unit step function, u(t), where u(t) = 0 for t < 0 and u(t) = 1 for t > 0. The Laplace transform of both u(t) and the constant function 1 is represented as F(s) = 1/s. Since both functions are equivalent over the domain t ∈ [0, ∞), there is no ambiguity in using either function when computing the inverse Laplace transform of 1/s.

PREREQUISITES
  • Understanding of Laplace transforms
  • Familiarity with unit step functions
  • Knowledge of inverse Laplace transformations
  • Basic calculus concepts related to function domains
NEXT STEPS
  • Study the properties of Laplace transforms in detail
  • Learn about the application of unit step functions in control systems
  • Explore examples of inverse Laplace transformations
  • Review the implications of function equivalence in Laplace transforms
USEFUL FOR

Students and professionals in engineering, mathematics, and physics who are working with Laplace transforms and need clarity on the use of unit step functions in transformations.

thepatient
Messages
160
Reaction score
0

Homework Statement



So I have this laplace transformation chart and was a bit unsure about the laplace and inverse laplace of this.

The unit step function, where u(t) = 0 where t < 0, u(t) = 1 where t > 0.

The laplace transformation chart that I have has two columns, the column on the left shows f(t) and the column on the right shows inverse laplace{f(t)} = F(s) for the corresponding f(t) on the left. Under f(t), both u(t) and 1 are on the same row as F(s) = 1/s. Does this mean that if I take the inverse laplace of 1/s, I can choose between using 1 or u(t)?

Thanks.
 
Physics news on Phys.org
The Laplace transform involves the domain t\in [0,\infty). Both f(t)=1 and u(t) are the same on this domain, so there's no ambiguity in computing the inverse Laplace transform.
 
fzero said:
The Laplace transform involves the domain t\in [0,\infty). Both f(t)=1 and u(t) are the same on this domain, so there's no ambiguity in computing the inverse Laplace transform.

Thanks.

Thats what I thought.
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
2
Views
2K