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$$ f(t):= \begin{cases} 3\cos(2t) & t<5 \\ 4\sin(3t) & 5 \leq t \leq 9 \\ 0 & t>9. \end{cases} $$

The above function I know how to solve using straight Laplace Transformation by Integration. However these integrals are not very quick to calculate by hand. I was told that there is a property involving Laplace Transformations that may make this problem quicker. Any suggestions which properties would help to find the Laplace transform in the most efficient manner?

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  • $\begingroup$ Do you mean in this situation add the Integrals of the first two functions together or is that only part of what you are getting at? $\endgroup$ Nov 14 '13 at 2:49
  • $\begingroup$ just saw your link sorry for the question $\endgroup$ Nov 14 '13 at 2:49
  • $\begingroup$ These tables are very informative and the examples are as well. My last conceptual question is what do I do with the constraints on t? $\endgroup$ Nov 14 '13 at 3:18
  • $\begingroup$ I am going to attempt to add it myself for you to review. $\endgroup$ Nov 14 '13 at 4:54
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We can rewrite your function using the Heaviside Unit Step function and then we can use the Shifting Property and Laplace Transform table. I would recommend practicing this with these examples.

Step 1: Lets plot your piece-wise function. We get:

enter image description here

Step 2: Lets rewrite the piece-wise function using the unit-step function as a single function.

We get:

$$f(t) = 3 \cos(2t) + (4 \sin(3t) - 3 \cos(2t)) u(t-5) + (0 - 4 \sin(3t)) u(t - 9)$$

Lets plot this function and it should be identical to the first one.

enter image description here

Step 3

Now, using the Laplace Table and the shifting property, we can easily write the Laplace Transform.

The final result will be:

$$\mathcal{L}\{f(t)\} = -\dfrac{3 e^{-5 s} (s \cos(10) - 2 \sin(10))}{s^2 + 4} + \dfrac{ 4 e^{-5 s} (3 \cos(15) + s \sin(15))}{s^2 + 9}$$

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