17Calculus - Inverse Laplace Transforms

17Calculus

Since we use an integral to calculate the Laplace Transform, you might expect to use derivatives to go the other direction. Counterintuitively, this is not the case. Calculating the inverse Laplace Transform involves fewer actual computations. Most of the time, you will use formulas and tables that were derived from calculating the Laplace Transform.

Notation

The notation we used for the Laplace Transform, looked like this. $F(s) = \mathcal{L}\{ f(t) \}$ Intuitively, the notation for the Inverse Laplace Transform is written in a logical manner. $f(t) = \mathcal{L}^{-1} \{ F(s) \}$

Rules

Since the Laplace Transform uses integration, which is a linear operation, so too is the inverse Laplace Transform. This means that $\mathcal{L}^{-1} \{ F(s) + G(s) \} = \mathcal{L}^{-1} \{ F(s) \} + \mathcal{L}^{-1} \{ G(s) \}$ For the same reason, using the constant $$a$$, this equation also holds. $\mathcal{L}^{-1} \{ aF(s) \} = a \mathcal{L}^{-1} \{ F(s) \}$

Calculating the Inverse Laplace Transform

To actually calculate the inverse Laplace Transform, we use tables. However, you will need to make sure your partial fraction expansion skills are sharp.

Okay, here is the table you need to use for calculating inverse Laplace Transforms. The practice problems column contain links to practice problems using this table.

Laplace Transforms Table

Laplace Transforms

$$f(t)$$

$$\displaystyle{ F(s) }$$

Basic Functions

$$t^n, ~ n = 1, 2, 3, \ldots$$

$$\displaystyle{ \frac{n!}{s^{n+1}} }$$

$$e^{at}$$

$$\displaystyle{ \frac{1}{s-a} }$$

$$\sin(\alpha t)$$

$$\displaystyle{ \frac{\alpha}{s^2 + \alpha^2} }$$

$$\cos(at)$$

$$\displaystyle{ \frac{s}{s^2 + a^2} }$$

$$\sinh(at)$$

$$\displaystyle{ \frac{a}{s^2 - a^2} }$$

$$\cosh(at)$$

$$\displaystyle{ \frac{s}{s^2 - a^2} }$$

Special Functions

$$\delta(t)$$ unit impulse

$$1$$

$$\delta(t-\tau)$$ shifted unit impulse

$$e^{-\tau s}$$

$$u(t)$$ unit step

$$\displaystyle{ \frac{1}{s} }$$

$$u(t-\tau)$$ shifted unit step

$$\displaystyle{ \frac{1}{s} e^{-\tau s} }$$

Combined Functions

$$e^{at}\sin(\alpha t)$$

$$\displaystyle{ \frac{\alpha}{(s-a)^2 + \alpha^2} }$$

$$f(t)u(t-a)$$

$$\displaystyle{ e^{-sa} \mathcal{L}\{ f(t+a) \} }$$

$$t^n e^{at}, ~ n = 1, 2, 3, \ldots$$

$$\displaystyle{ \frac{n!}{(s-a)^{n+1}} \} }$$

Derivatives and Integrals

$$f'(t)$$

$$sF(s) - f(0)$$

$$f''(t)$$

$$s^2F(s) - sf(0) - f'(0)$$

$$\displaystyle{ f^{(n)}(t) }$$

$$\displaystyle{ s^nF(s) - s^{n-1}f(0) - }$$ $$\displaystyle{ s^{n-2}f'(0) - . . . - f^{(n-1)}(0) }$$

$$\int_0^t{ f(v)~dv }$$

$$\displaystyle{ \frac{F(s)}{s} }$$

Okay, so why do we need Laplace Transforms? Why are they useful? We use them to solve differential equations that cannot be solved otherwise, sometimes involving some special functions. These special functions also have a purpose. Some that you will run across are the unit step function, unit impulse function and the square wave.

Practice

Unless otherwise instructed, calculate the inverse Laplace transform $$f(t) = \mathcal{L}^{-1} \{ F(s) \}$$ using a table. Give your answer in exact, completely factored form.

Basic

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^4} \right\} }$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^4} \right\} }$$

$$t^3/6$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^4} \right\} }$$

Solution

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$$t^3/6$$

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Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{6s+3} \right\} }$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{6s+3} \right\} }$$

$$\displaystyle{ \frac{e^{-t/2}}{6} }$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{6s+3} \right\} }$$

Solution

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$$\displaystyle{ \frac{e^{-t/2}}{6} }$$

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Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{s+1}{s^2+2} \right\} }$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{s+1}{s^2+2} \right\} }$$

$$\displaystyle{ \cos(t\sqrt{2}) + \frac{\sin(t\sqrt{2})}{\sqrt{2}} }$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{s+1}{s^2+2} \right\} }$$

Solution

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$$\displaystyle{ \cos(t\sqrt{2}) + \frac{\sin(t\sqrt{2})}{\sqrt{2}} }$$

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Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^2+2s} \right\} }$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^2+2s} \right\} }$$

$$\displaystyle{ \frac{1}{2} ( 1-e^{-2t} ) }$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^2+2s} \right\} }$$

Solution

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$$\displaystyle{ \frac{1}{2} ( 1-e^{-2t} ) }$$

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Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{s}{(s+2)^2} \right\} }$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{s}{(s+2)^2} \right\} }$$

$$e^{-2t} - 2te^{-2t}$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{s}{(s+2)^2} \right\} }$$

Solution

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$$e^{-2t} - 2te^{-2t}$$

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Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{se^{-\pi s/2}}{s^2+1} \right\} }$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{se^{-\pi s/2}}{s^2+1} \right\} }$$

$$\cos(t-\pi/2)u(t-\pi/2)$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{se^{-\pi s/2}}{s^2+1} \right\} }$$

Solution

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$$\cos(t-\pi/2)u(t-\pi/2)$$

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Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{s}{s^2+2s+2} \right\} }$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{s}{s^2+2s+2} \right\} }$$

$$e^{-t}( \cos(t) - \sin(t) )$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{s}{s^2+2s+2} \right\} }$$

Solution

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$$e^{-t}( \cos(t) - \sin(t) )$$

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Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{(s+2)^5} \right\} }$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{(s+2)^5} \right\} }$$

$$\displaystyle{ \frac{t^4 e^{-2t}}{24} }$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{(s+2)^5} \right\} }$$

Solution

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$$\displaystyle{ \frac{t^4 e^{-2t}}{24} }$$

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Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{s+8}{s^2+4s+13} \right\} }$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{s+8}{s^2+4s+13} \right\} }$$

$$e^{-2t}( \cos(3t) + 2\sin(3t) )$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{s+8}{s^2+4s+13} \right\} }$$

Solution

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$$e^{-2t}( \cos(3t) + 2\sin(3t) )$$

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Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^4+5s^2+4} \right\} }$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^4+5s^2+4} \right\} }$$

$$\displaystyle{ \frac{1}{6} (2\sin(t) - \sin(2t)) }$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^4+5s^2+4} \right\} }$$

Solution

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$$\displaystyle{ \frac{1}{6} (2\sin(t) - \sin(2t)) }$$

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Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^4 e^{10s}} \right\} }$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^4 e^{10s}} \right\} }$$

$$\displaystyle{ \frac{(t-10)^3}{6} u(t-10) }$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^4 e^{10s}} \right\} }$$

Solution

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$$\displaystyle{ \frac{(t-10)^3}{6} u(t-10) }$$

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Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \ln \left[ \frac{s^2+9}{s^2+1} \right] \right\} }$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \ln \left[ \frac{s^2+9}{s^2+1} \right] \right\} }$$

$$\displaystyle{ \frac{2}{t}( \cos(t) -\cos(3t)) }$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \ln \left[ \frac{s^2+9}{s^2+1} \right] \right\} }$$

Solution

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$$\displaystyle{ \frac{2}{t}( \cos(t) -\cos(3t)) }$$

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Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^4-16} \right\} }$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^4-16} \right\} }$$

$$\frac{1}{32}( e^{2t} - e^{-2t} -2\sin(2t) )$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^4-16} \right\} }$$

Solution

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$$\frac{1}{32}( e^{2t} - e^{-2t} -2\sin(2t) )$$

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Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{s^3}{(s^4-16)^2} \right\} }$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{s^3}{(s^4-16)^2} \right\} }$$

Hint

Use the result from the previous problem to solve this one.

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{s^3}{(s^4-16)^2} \right\} }$$

Hint

Use the result from the previous problem to solve this one.

Solution

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$$\displaystyle{ F(s) = \frac{1}{s-3} - \frac{16}{s^2+9} }$$

Problem Statement

For $$\displaystyle{ F(s) = \frac{1}{s-3} - \frac{16}{s^2+9} }$$, find the inverse Laplace transform $$f(t) = \mathcal{L}^{-1} \{ F(s) \}$$.

Solution

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$$\displaystyle{ F(s) = \frac{s+3}{s^2+4s+13} }$$

Problem Statement

For $$\displaystyle{ F(s) = \frac{s+3}{s^2+4s+13} }$$, find the inverse Laplace transform $$f(t) = \mathcal{L}^{-1} \{ F(s) \}$$.

Solution

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Intermediate

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^3 (s^2+1)} \right\} }$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^3 (s^2+1)} \right\} }$$

$$\cos(t) + t^2/2 - 1$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^3 (s^2+1)} \right\} }$$

Solution

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$$\cos(t) + t^2/2 - 1$$

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Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \arctan(1/s) \right\} }$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \arctan(1/s) \right\} }$$

$$\displaystyle{ \frac{\sin(t)}{t} }$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \arctan(1/s) \right\} }$$

Solution

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$$\displaystyle{ \frac{\sin(t)}{t} }$$

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Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{\sqrt{s}} + \frac{1}{\sqrt{e^s}} \right\} }$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{\sqrt{s}} + \frac{1}{\sqrt{e^s}} \right\} }$$

$$\displaystyle{ \frac{1}{\sqrt{\pi t}} + \delta(t-1/2) }$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{\sqrt{s}} + \frac{1}{\sqrt{e^s}} \right\} }$$

Solution

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$$\displaystyle{ \frac{1}{\sqrt{\pi t}} + \delta(t-1/2) }$$

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Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^4+4s^2+4} \right\} }$$

Problem Statement

Evaluate $$\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^4+4s^2+4} \right\} }$$

Solution

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