How do I find the laplace transformation of i(t)=(t)(e^t)(sinkt)?

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To find the Laplace transformation of i(t) = (t)(e^t)(sin(kt)), the integral Y(s) = ∫_{0}^{∞} e^{-st}i(t)dt can be quite complex. The integral simplifies by using integration by parts to eliminate the t-factor, and employing the complex exponential form of sin(kt) can further streamline the process. This approach not only simplifies calculations but also yields two Laplace transforms simultaneously. The discussion emphasizes the importance of understanding convolution and integration techniques in solving such problems. Overall, using these strategies can make the Laplace transformation more manageable.
mak_wilson
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please help me with this question

Find the laplace transformation of this function

i(t)=(t)(e^t)(sinkt)

i really don't know how to do!
 
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The Laplace Transform is defined as:

Y(s) = \int_{0}^{\infty} e^{-st}y(t)dt

where y(t) is the function you wish to find the Laplacian of.

In this example, the integral would be:

\int_{0}^{\infty} te^{-st}e^tsin(kt)dt

...which is unbelievably ugly.

Have you learned about convolution yet? This is a pretty nasty problem, unless I'm missing something, which it seems probable that I am.
 
Last edited:
thz

You didnt miss anything, i can do up to this stage, but it contain 3 t in it, I don't really know how to solve it!
 
1. Since the integral of the e^((1-s)t)*sin(kt) will "rotate" during integration by parts (i.e. you will gain back a multiple of what you began integrating), evaluating the integral of this function alone should pose no problems.
(Assuming s>1, that is)

2. You can now go back to the original problem, using integration by parts to eliminate the t-factor.

3. Alternatively, you might use the complex exponential as a simplifying measure.
 
arildno said:
3. Alternatively, you might use the complex exponential as a simplifying measure.

That's what I would do, too. The beautiful thing about that is that, not only is it a lot easier to calculate, but it also gives you TWO Laplace transforms simultaneously.

mak_wilson, I would recommend that you take this suggestion. Make the replacement:

sin(kt)--->eikt

and take the imaginary part at the end.
 
solution

Here is a solution,
Max.
 

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thank You~~
 

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