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I'm a bit new to Laplace transforms and have come across the following question:

$$f(t) = \frac{\sin(2t)}{e^{2t}}+t{\cdot}u(t-4)$$

I've worked out the laplace transform for the first part, pretty standard, but I'm having trouble with what to do next! Thanks in advance.

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  • $\begingroup$ Hint: $\int_0^{\infty} u(t-a)f(t)= \int_a^{\infty}f(t)$ $\endgroup$ – tired Nov 30 '16 at 22:46
  • $\begingroup$ Apologies, I said $F(s)$ when I meant $f(t)$ $\endgroup$ – Tom Nov 30 '16 at 22:50
  • $\begingroup$ Does the $*$ mean convolution or it is the regular product? $\endgroup$ – polfosol Nov 30 '16 at 23:11
  • $\begingroup$ Regular product. $\endgroup$ – Tom Nov 30 '16 at 23:15
  • $\begingroup$ Google Second shifting theorem and rewrite $t=(t-4)+4$ $\endgroup$ – b00n heT Nov 30 '16 at 23:28
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$$\text{F}\left(\text{s}\right)=\mathcal{L}_t\left[\frac{\sin\left(2t\right)}{e^{2t}}+t\theta\left(t-4\right)\right]_{\left(\text{s}\right)}:=\int_0^\infty e^{-\text{s}t}\left(\frac{\sin\left(2t\right)}{e^{2t}}+t\theta\left(t-4\right)\right)\space\text{d}t$$

Where $\theta\left(x\right)$ is the Heaviside step function.

Now:

$$\text{F}\left(\text{s}\right)=\int_0^\infty e^{-\text{s}t}\cdot\frac{\sin\left(2t\right)}{e^{2t}}\space\text{d}t+\int_0^\infty te^{-\text{s}t}\theta\left(t-4\right)\space\text{d}t=$$ $$\int_0^\infty e^{-t\left(2+\text{s}\right)}\sin\left(2t\right)\space\text{d}t+\int_4^\infty te^{-\text{s}t}\space\text{d}t$$

So, we get:

  1. When $\Re\left[\text{s}\right]>-2$: $$\int_0^\infty e^{-t\left(2+\text{s}\right)}\sin\left(2t\right)\space\text{d}t=\frac{2}{\text{s}^2+4\left(2+\text{s}\right)}$$
  2. When $\Re\left[\text{s}\right]>0$: $$\int_4^\infty te^{-\text{s}t}\space\text{d}t=\frac{e^{-4\text{s}}\left(1+4\text{s}\right)}{\text{s}^2}$$
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You can use the formula for finding the laplace transform of secondary part-

L {f (t)u (t-a)}=L {f (t+a)}×e^(-as)/s

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  • $\begingroup$ Please use MathJax to format your answers. $\endgroup$ – Glorfindel May 9 '17 at 13:18

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