Laplace transform with delta function

  • Thread starter 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:

Astronuc

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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.
 

SGT

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].
 
458
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Agree, I should have searched at other places on these forums.
Thanks everyone for explanation.
 

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