Question

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

Laplace transform
ANSWERED
asked 2020-11-01
Calculate the Laplace transform
\(L\left\{\sin(t-k) \cdot H(t-k)\right\}\)

Answers (1)

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, \(L\left\{\sin(t-k) \cdot H(t-k)\right\} = \frac{e^{-ks}}{(s^2+1)}\)

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