\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus - Laplace Transform Shifting Theorems

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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 [12min-6secs]

video by Dr Chris Tisdell

Practice - First Shifting Theorem

Use the First Shifting Theorem on these practice problems.

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 \( f(t) = \mathcal{L}^{-1} \{ F(s) \} \)
Give your answers in exact, completely factored form.

Basic

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

Problem Statement

For \( f(t) = e^{4t}t^6 \), find the Laplace transform \( F(s) = \mathcal{L}\{ f(t) \} \) using a shifting theorem.

Solution

656 video

video by Dr Chris Tisdell

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

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

657 video

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

661 video

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

663 video

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

658 video

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

664 video

video by Dr Chris Tisdell

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Intermediate

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

Problem Statement

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

Solution

660 video

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

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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

662 video

video by Dr Chris Tisdell

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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

Practice - Second Shifting Theorem

Use the Second Shifting Theorem on these practice problems.

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 \( f(t) = \mathcal{L}^{-1} \{ F(s) \} \)
Give your answers in exact, completely factored form.

Basic

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

Problem Statement

For \( f(t) = u(t-4)[t-4] \), find the Laplace transform \( F(s) = \mathcal{L}\{ f(t) \} \) using a shifting theorem.

Solution

665 video

video by Dr Chris Tisdell

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\( f(t) = u(t-4)[t-4]^2 \)

Problem Statement

For \( f(t) = u(t-4)[t-4]^2 \), find the Laplace transform \( F(s) = \mathcal{L}\{ f(t) \} \) using a shifting theorem.

Solution

667 video

video by Dr Chris Tisdell

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\( f(t) = u(t-2\pi) \sin(t-2\pi) \)

Problem Statement

For \( f(t) = u(t-2\pi) \sin(t-2\pi) \), find the Laplace transform \( F(s) = \mathcal{L}\{ f(t) \} \) using a shifting theorem.

Solution

668 video

video by Dr Chris Tisdell

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

Problem Statement

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

Solution

666 video

video by Dr Chris Tisdell

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Intermediate

\( f(t) = \cos[3(t-1)] u(t-1) \)

Problem Statement

For \( f(t) = \cos[3(t-1)] u(t-1) \) find the Laplace transform \( F(s) = \mathcal{L}\{ f(t) \} \) using a shifting theorem.

Solution

670 video

video by Dr Chris Tisdell

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

Problem Statement

For \( f(t) = \left\{\begin{array}{lr} t & 0 \leq 4 \\ 2 & t = 4 \\ 0 & t > 4 \end{array} \right. \) find the Laplace transform \( F(s) = \mathcal{L}\{ f(t) \} \) using a shifting theorem.

Solution

669 video

video by Dr Chris Tisdell

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You CAN Ace Calculus

Topics You Need To Understand For This Page

Related Topics and Links

external links you may find helpful

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Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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Practice 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 \( f(t) = \mathcal{L}^{-1} \{ F(s) \} \)
Give your answers in exact, completely factored form.

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