Question

find laplace transform of f(t)=d(t-4)*t*sin(10t+0.2)*e^2t where d(t) is an impulse function..

Answer 1

what is d(t-4) it does'nt make any sense

Answer 2

\[f(t)=d(t-4)*t*\sin(10t+0.2)*e^{2t}\]
is this the function?
@JK24

Answer 3

yes @hysenberg

Answer 4

what is d(t-4) ??

Answer 5

d(t) is the delta of the DIRAC DELTA FUNCTION...impulse function

Answer 6

omg if u dont mention it who will understand :(
now it make sense u should at least mention it

Answer 7

i have mentioned it...
"impulse function" in my question

Answer 8

\[\begin{align*}
\mathcal{L}\left\{f(t)\right\}(s)&=\mathcal{L}\left\{\delta(t-4)~t~\sin(10t+0.2)e^{2t}\right\}(s)
\end{align*}\]
First thing we can remove is the \(e^{2t}\) factor using the rule,
\[\large e^{ct}f(t)\overbrace{\iff}^{\large\mathcal{L\{f(t)\}(s)}}F(s-c)\]
So whatever the transform of \(g(t)=\delta(t-4)t\sin(10t+0.2)\) is, we will shift \(s\) by 2, i.e. replace any \(s\) with \(s-2\).
\[\begin{align*}
\mathcal{L}\left\{e^{2t}g(t)\right\}(s)&=\mathcal{L}\left\{\delta(t-4)~t~\sin(10t+0.2)\right\}(s-2)
\end{align*}\]
Now let's remove the Dirac Delta factor using the rule,
\[\large \delta(t-c)f(t)\overbrace{\iff}^{\large\mathcal{L\{f(t)\}(s)}}e^{-cs}f(c)\]
\[\large\begin{align*}
\mathcal{L}\left\{e^{2t}\delta(t-4)h(t)\right\}(s)&=e^{-4(s-2)}h(4)=e^{-4s}e^8(4\sin40.2)
\end{align*}\]
where \(h(t)=t\sin(10t+0.2)\), and \(h(4)\approx7127.59\) (if using radians) or \(h(4)\approx7696.33\) (if using degrees).

Answer 9

good explanation... thanks @SithsAndGiggles