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You CAN Ace Calculus

17calculus > laplace transforms > shifting theorems

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Laplace Transform Shifting Theorems

On this page, we discuss two important theorems related to Laplace Transforms. They are rather cleverly named the First Shifting Theorem and the Second Shifting Theorem.

 First Shifting Theorem $$\displaystyle{ \mathcal{L}\{ e^{-at} f(t) \} = F(s+a) }$$ Second Shifting Theorem $$\displaystyle{ \mathcal{L}\{ u(t-c)g(t-c) \} = e^{-cs}G(s) }$$

First Shifting Theorem

The first shifting theorem says that in the t-domain, if we multiply a function by $$e^{-at}$$, this results in a shift in the s-domain a units. In your Laplace Transforms table you probably see the line that looks like $$\displaystyle{ \mathcal{L}\{ e^{-at} f(t) \} = F(s+a) }$$
This idea looks easy and watching other people using it may look easy but when you have to do it, it is not so easy. So it is important to get your hands dirty by working some problems on your own. After you watch this first video, you can work some practice problems to help you understand this better.

Let's watch a video that explains how this works. This video contains a very good explanation, a couple of examples and the proof of this theorem.

 Dr Chris Tisdell - First shifting theorem of Laplace transforms: a how to

Use the First Shifting Theorem on these practice problems.
Instructions - - Unless otherwise instructed,
- if $$f(t)$$ is given, find the Laplace transform $$F(s) = \mathcal{L}\{ f(t) \}$$ using a shifting theorem
- if $$F(s)$$ is given, find the inverse Laplace transform $$f(t) = \mathcal{L}^{-1} \{ F(s) \}$$

 Basic Problems

Practice 1

$$f(t)=e^{4t}t^6$$

solution

Practice 2

$$f(t)=(t+1)^2e^t$$

solution

Practice 3

$$f(t)=e^{3t}t^2$$

solution

Practice 4

$$f(t)=e^t\sin(t)$$

solution

Practice 5

$$\displaystyle{F(s)=\frac{1}{(s+1)^2}}$$

solution

Practice 6

$$\displaystyle{F(s)=\frac{\pi}{(s+\pi)^2}}$$

solution

 Intermediate Problems

Practice 7

$$f(t)=e^{2t}\cos(3t)$$

solution

Practice 8

$$\displaystyle{F(s)=\frac{s-2}{s^2-4s+5}}$$

solution

Practice 9

$$\displaystyle{F(s)=\frac{1}{s^2-4s+5}}$$

solution

Second Shifting Theorem

The second shifting theorem looks similar to the first but the results are quite different. In the t-domain we have the unit step function (Heaviside function) which translates to the exponential function in the s-domain. Your Laplace Transforms table probably has a row that looks like $$\displaystyle{ \mathcal{L}\{ u(t-c)g(t-c) \} = e^{-cs}G(s) }$$

Okay, let's watch a video that explains this very well and contains a couple of examples and discusses the big picture.

 Dr Chris Tisdell - Second shifting theorem of Laplace transforms

This next video clip contains a lot of the same explanation as the previous video but, toward the end, shows why the theorem holds. It is a good video to watch after the previous one to get some repetition and more detailed explanation. Then, work some practice problems to hone your skills.

 Dr Chris Tisdell - Second shifting theorem: Laplace transforms

Use the Second Shifting Theorem on these practice problems.
Instructions - - Unless otherwise instructed,
- if $$f(t)$$ is given, find the Laplace transform $$F(s) = \mathcal{L}\{ f(t) \}$$ using a shifting theorem
- if $$F(s)$$ is given, find the inverse Laplace transform $$f(t) = \mathcal{L}^{-1} \{ F(s) \}$$

 Basic Problems

Practice 10

$$f(t)=u(t-4)[t-4]$$

solution

Practice 11

$$u(t-4)[t-4]^2$$

solution

Practice 12

$$u(t-2\pi)\sin(t-2\pi)$$

solution

Practice 13

$$\displaystyle{F(s)=\frac{1-e^{-2s}}{s^2}}$$

solution

 Intermediate Problems

Practice 14

$$\cos[3(t-1)]u(t-1)$$

solution

Practice 15

$$f(t) = \left\{ \begin{array}{lr} t & 0 \leq 4 \\ 2 & t = 4 \\ 0 & t > 4 \end{array} \right.$$

solution