# Calculate the Laplace transform Lleft{sin(t-k) cdot H(t-k)right}

Calculate the Laplace transform
$L\left\{\mathrm{sin}\left(t-k\right)\cdot H\left(t-k\right)\right\}$
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

Aubree Mcintyre

Step 1
To find the Laplace transform of
$L\left\{\mathrm{sin}\left(t-k\right)\cdot H\left(t-k\right)\right\}$
Step 2
Since,
For the Heaviside function (or unit step function)
$\left({H}_{a}\left(t\right)=H\left(t-a\right)=\left\{\begin{array}{ll}0& ta\end{array}\right)$The second shifting theorem,

First we need to find the laplace transform of $\mathrm{sin}\left(t\right)$.
Step 3

$L\left\{f\left(t\right)\right\}=L\mathrm{sin}\left(t\right)$
$\text{Since ,}L\left\{f\left(t\right)\right\}=\frac{a}{\left({s}^{2}+{a}^{2}\right)}$
$L\left\{f\left(t\right)\right\}=\frac{1}{\left({s}^{2}+1\right)}$
$\text{Thus ,}F\left(s\right)=\frac{1}{\left({s}^{2}+1\right)}$
Step 4
By second shifting theorem,

Thus,
$L\left\{H\left(t-k\right)f\left(t-a\right)\right\}={e}^{-ks}F\left(s\right)$
$={e}^{-ks}\left(\frac{1}{\left({s}^{2}+1\right)}\right)$
$=\frac{{e}^{-ks}}{\left({s}^{2}+1\right)}$
Step 5
Therefore, $L\left\{\mathrm{sin}\left(t-k\right)\cdot H\left(t-k\right)\right\}=\frac{{e}^{-ks}}{\left({s}^{2}+1\right)}$