## 17Calculus - Laplace Transform Second Shifting Theorem

##### 17Calculus

On this page, we discuss the second of 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 $$\displaystyle{ \mathcal{L}\{ u(t-c)g(t-c) \} = e^{-cs}G(s) }$$

### Resources

Topics You Need To Understand For This Page

laplace transforms

Related Topics and Links

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 [10min-37secs]

video by Dr Chris Tisdell

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 [10min-36secs]

video by Dr Chris Tisdell

Okay, use the second shifting theorem to solve these problems.

Practice

Unless otherwise instructed,
- if $$f(t)$$ is given, find the Laplace transform $$F(s) = \mathcal{L}\{ f(t) \}$$ using the second shifting theorem
- if $$F(s)$$ is given, find the inverse Laplace transform $$f(t) = \mathcal{L}^{-1} \{ F(s) \}$$
Give your answers in exact, completely factored form.

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

Problem Statement

For $$f(t) = (t+1)^2 e^t$$, find the Laplace transform $$F(s) = \mathcal{L}\{ f(t) \}$$ using a shifting theorem.

Solution

### Dr Chris Tisdell - 657 video solution

video by Dr Chris Tisdell

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

Problem Statement

For $$f(t) = e^{3t} t^2$$, find the Laplace transform $$F(s) = \mathcal{L}\{ f(t) \}$$ using a shifting theorem.

Solution

### Dr Chris Tisdell - 661 video solution

video by Dr Chris Tisdell

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

Problem Statement

For $$f(t) = e^t \sin(t)$$, find the Laplace transform $$F(s) = \mathcal{L}\{ f(t) \}$$ using a shifting theorem.

Solution

### Dr Chris Tisdell - 663 video solution

video by Dr Chris Tisdell

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

Problem Statement

For $$\displaystyle{ F(s) = \frac{1}{(s+1)^2} }$$, find the inverse Laplace transform $$f(t) = \mathcal{L}^{-1} \{ F(s) \}$$ using a shifting theorem.

Solution

### Dr Chris Tisdell - 658 video solution

video by Dr Chris Tisdell

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

Problem Statement

For $$\displaystyle{ F(s) = \frac{\pi}{(s+\pi)^2} }$$, find the inverse Laplace transform $$f(t) = \mathcal{L}^{-1} \{ F(s) \}$$ using a shifting theorem.

Solution

### Dr Chris Tisdell - 664 video solution

video by Dr Chris Tisdell

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

Problem Statement

For $$f(t) = e^{2t} \cos(3t)$$, find the Laplace transform $$F(s) = \mathcal{L}\{ f(t) \}$$ using a shifting theorem.

Solution

### Dr Chris Tisdell - 660 video solution

video by Dr Chris Tisdell

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

Problem Statement

For $$\displaystyle{ F(s) = \frac{s-2}{s^2-4s+5} }$$, find the inverse Laplace transform $$f(t) = \mathcal{L}^{-1} \{ F(s) \}$$ using a shifting theorem.

Solution

### Dr Chris Tisdell - 659 video solution

video by Dr Chris Tisdell

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

Problem Statement

For $$\displaystyle{ F(s) = \frac{1}{s^2-4s+5} }$$, find the inverse Laplace transform $$f(t) = \mathcal{L}^{-1} \{ F(s) \}$$ using a shifting theorem.

Solution

### Dr Chris Tisdell - 662 video solution

video by Dr Chris Tisdell

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