# Laplace transform with delta function

1. Jun 14, 2005

### EvLer

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: Jun 14, 2005
2. Jun 15, 2005

### Staff: Mentor

Remember,

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

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

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.

3. Jun 15, 2005

### SGT

Since $$\delta (t-2)$$ exists only at t = 2, your f(t) is:
$$f(t) = cos(2\pi + \theta)*\delta (t-2) = cos(\theta)*\delta (t-2)$$ and
$$F(s) = cos(\theta)*e^{-2s}$$.

4. Jun 15, 2005

### EvLer

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