Laplace transform with delta function

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Discussion Overview

The discussion centers around the computation of the Laplace transform of a signal involving a delta function, specifically the function f(t) = cos(pi*t + theta)*delta(t-2). Participants are exploring the evaluation of the integral directly, with a focus on the properties of the delta function and its implications for the transform.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant expresses uncertainty about evaluating the Laplace transform integral and suggests a potential answer involving the exponential function.
  • Another participant provides the definition of the Laplace transform and emphasizes the role of the delta function in the evaluation.
  • A third participant clarifies that since the delta function is non-zero only at t = 2, the function simplifies to cos(2π + theta)*delta(t-2) and proposes that the Laplace transform results in cos(theta)*e^(-2s).
  • A later reply acknowledges the need to search for additional resources on the topic, indicating a willingness to learn from the discussion.

Areas of Agreement / Disagreement

Participants generally agree on the simplification of the function due to the delta function's properties, but there is no consensus on the initial evaluation of the Laplace transform integral.

Contextual Notes

Some assumptions about the properties of the delta function and the evaluation of the integral may not be fully articulated, and the discussion does not resolve all mathematical steps involved in the transform.

EvLer
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I am sort of stuck on this one:
compute Laplace trasnform of this signal (directly by evaluating the integral)

f(t) = cos(pi*t + theta)*delta(t-2);

I know what the LT integral looks like, but I don't think I'm evaluating it right.
Would the answer be just: cos(pi*t + theta)*e^(-2s) ?

Any help is much appreciated.
 
Last edited:
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Remember,

[tex]\mathcal{L}[\delta(t-c)]=\int_{0}^{\infty}e^{-st}\delta(t-c)dt[/tex]


[tex]\mathcal{L}[f(t)]=\int_{0}^{\infty}e^{-st}\,f(t)\,dt[/tex]

There are numerous examples of the Laplace Transform in the Mathematics > Calculus & Analysis forum, e.g. https://www.physicsforums.com/showthread.php?t=49248 , or just search the forum for "Lapalce Transform".

There is no time variable 't' in the transform.
 
Since [tex]\delta (t-2)[/tex] exists only at t = 2, your f(t) is:
[tex]f(t) = cos(2\pi + \theta)*\delta (t-2) = cos(\theta)*\delta (t-2)[/tex] and
[tex]F(s) = cos(\theta)*e^{-2s}[/tex].
 
Agree, I should have searched at other places on these forums.
Thanks everyone for explanation.
 

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