Question

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

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

2020-11-02

Step 1
To find the Laplace transform of
$$L\left\{\sin(t-k) \cdot H(t-k)\right\}$$
Step 2
Since,
For the Heaviside function (or unit step function)
$$(H_a(t)=H(t-a)=\begin{cases}0 & ta\end{cases})$$The second shifting theorem,
$$\text{If } L{f(t)}=F(s) \text{ then } L{H(t-k)f(t-a)}=e^{-as}F(s)$$
First we need to find the laplace transform of $$\sin(t)$$.
Step 3
$$\text{Let }, f(t)=\sin(t)$$
$$L\left\{f(t)\right\}=L{\sin(t)}$$
$$\text{Since ,} L\left\{f(t)\right\}=\frac{a}{(s^2+a^2)}$$
$$L\left\{f(t)\right\}=\frac{1}{(s^2+1)}$$
$$\text{Thus ,} F(s)=\frac{1}{(s^2+1)}$$
Step 4
By second shifting theorem,
$$\text{If } L{f(t)}=F(s) \text{then } L{H(t-k)f(t-a)}=e^{-as}F(s)$$
Thus,
$$L\left\{H(t-k)f(t-a)\right\}=e^{-ks}F(s)$$
$$=e^{-ks}\left(\frac{1}{(s^2+1)}\right)$$
$$=\frac{e^{-ks}}{(s^2+1)}$$
Step 5
Therefore, $$L\left\{\sin(t-k) \cdot H(t-k)\right\} = \frac{e^{-ks}}{(s^2+1)}$$