Laplace transform of ##f(t)=(u(t)-u(t-2\pi))\sin{t}##

  • #1
zenterix
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Homework Statement
Find the Laplace transform of ##f(t)=(u(t)-u(t-2\pi))\sin{t}## by use of the ##t##-shift rule.
Relevant Equations
$$\mathcal{L}(u(t-a)f(t-a))=e^{-as}\mathcal{L}(f(t))$$

$$\mathcal{L}(u(t-a)f(t))=e^{-as}\mathcal{L}(f(t+a))$$
I tried to solve this as follows

$$f(t)=(u(t)-u(t-2\pi))\sin{t}$$

$$=u(t)\sin{t}-u(t-2\pi)\sin{t}$$

$$\mathcal{L}(f(t))=e^{0\cdot s}\mathcal{sin{t}}-e^{-2pi s}\mathcal{L}(\sin{(t+2\pi)})$$

$$=\frac{1-e^{-2\pi s}}{s^2+1}$$

where I used the fact that ##\sin{(t+2\pi)}=\sin{t}##.

Then I looked at the solution and it has the following

1712756211131.png

I don't understand this solution.

First of all, in ##\mathcal{L}(u(t)-u(t-2\pi)\sin{(t-2\pi)})## why is there no ##\sin{t}## multiplying the ##u(t)##?
 
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  • #2
zenterix said:
First of all, in ##\mathcal{L}(u(t)-u(t-2\pi)\sin{(t-2\pi)})## why is there no ##\sin{t}## multiplying the ##u(t)##?
Because sometimes there are errors in the provided solutions.
 
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