Science:Math Exam Resources/Courses/MATH215/December 2011/Question 05 (a)

MATH215 December 2011

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Question 05 (a)
Given the function $f(t)={\begin{cases}0,&t<1\\2\pi ,&1\leq t<2\\\pi ,&t\geq 2\end{cases}}$ (a) Find the Laplace transform of f(t).

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Recall that the Laplace transform for a function f(t) is ${\mathcal {L}}(f(t))=\int _{0}^{\infty }{}f(t)\exp(-st)dt$

Hint 2
How do we integrate a piecewise function?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The Laplace transform, F(s), for a function f(t) is given by $F(s)={\mathcal {L}}(f(t))=\int _{0}^{\infty }{}f(t)\exp(-st)dt.$ Since our function is piecewise defined we can split up the integral and write ${\begin{aligned}F(s)&=\int _{0}^{1}f(t)dt+\int _{1}^{2}f(t)\exp(-st)dt+\int _{2}^{\infty }{}f(t)\exp(-st)dt\\F(s)&=\int _{0}^{1}0dt+2\pi \int _{1}^{2}\exp(-st)dt+\pi \int _{2}^{\infty }\exp(-st)dt.\end{aligned}}$ The first integral is just zero. For the second integral we get, $2\pi \int _{1}^{2}\exp(-st)dt=-\left.{\frac {2\pi }{s}}\exp(-st)\right|_{1}^{2}=-{\frac {2\pi }{s}}\exp(-2s)+{\frac {2\pi }{s}}\exp(-s).$ For the third integral we get $\pi \int _{2}^{\infty }\exp(-st)dt=-\left.{\frac {\pi }{s}}\exp(-st)\right|_{2}^{\infty }={\frac {\pi }{s}}\exp(-2s).$ Adding the two integrals together we get $F(s)=-{\frac {2\pi }{s}}\exp(-2s)+{\frac {2\pi }{s}}\exp(-s)+{\frac {\pi }{s}}\exp(-2s)={\frac {\pi }{s}}\left(2\exp(-s)-\exp(-2s)\right).$

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