# What is the laplace transform of (delta(t−(pi)/6))u(t−(pi)/6)

What is the laplace transform of $\left(\delta \left(t-\pi /6\right)\right)u\left(t-\pi /6\right)$
I know that $L\left(f\left(t-a\right).u\left(t-a\right)\right)={e}^{-as}.F\left(s\right)$
Does that mean the laplace transform of the above equation is ${e}^{-2\pi s}$?
You can still ask an expert for help

## Want to know more about Laplace transform?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Alannah Hanson
It is equal to 0 since
$f\left(t\right)\delta \left(t-{t}_{0}\right)=f\left({t}_{0}\right)\delta \left(t-{t}_{0}\right)$
and
$u\left(\frac{\pi }{6}-\pi \right)=0$
Also the statement you wrote is wrong. We have
$L\left(f\left(t-a\right)\right)={e}^{-as}F\left(s\right)$
as long as the ROCs are treated carefully.