Laplace Transform

Laplace transforms are an adapted integral form of a differential equation (created and introduced by the French mathematician Pierre-Simon Laplace (1749-1827)) used to describe electrical circuits and physical processes. Adapted from previous notions given by other notable mathematicians and engineers like Joseph-Louis Lagrange (1736-1812) and Leonhard Euler (1707-1783), Laplace transforms are used to be a more efficient and easy-to-recognize form of a mathematical equation.

Standard Form

This is the standard form of a Laplace transform that a function will undergo.

$F(s)={ℒ}\{f(t)\}=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}f(t)dt$

Sample Functions

The following is a list of commonly seen functions of which the Laplace transform is taken. The start function is noted within the Laplace symbol ${ℒ}\left\{\right\}$.

$F(s)={ℒ}\left\{1\right\}=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}dt=$ $\frac{1}{s}$

$F(s)={ℒ}\left\{t^n\right\}=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}{t}^{n}dt=$ $\frac{n!}{s^{n+1}}$

$F(s)={ℒ}\left\{e^{at}\right\}=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}{e}^{at}dt=$ $\frac{1}{s-a}$

$F(s)={ℒ}\{sin(\omega t)\}=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}sin(\omega t)dt=$ $\frac{\omega }{s^2+{\omega }^2}$

$F(s)={ℒ}\{cos(\omega t)\}=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}cos(\omega t)dt=$ $\frac{s}{s^2+{\omega }^2}$

$F(s)={ℒ}\{t^{n}g(t)\}=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}{t}^{n}g(t)dt=$ $$\begin{gathered} \frac{(-1{)}^n{d}^{n}G(s)}{d{s}^n}\text{ for}n \\ \text{= 1,2,...} \end{gathered}$$

$F(s)={ℒ}\{tsin(\omega t)\}=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}tsin(\omega t)dt=$ $\frac{2\omega s}{(s^2+{\omega }^2{)}^2}$

$F(s)={ℒ}\{tcos(\omega t)\}=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}tcos(\omega t)dt=$ $\frac{s^2-{\omega }^2}{(s^2+{\omega }^2{)}^2}$

$F(s)={ℒ}\{g(at)\}=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}g(at)dt=$ $\frac{1}{a}G\left(\frac{s}{a}\right),\text{ for}a>0.$

$F(s)={ℒ}\{e^{at}g(t)\}=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}{e}^{at}g(t)dt=G(s-a)$

$F(s)={ℒ}\left\{e^{at}{t}^n\right\}=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}{e}^{at}{t}^{n}dt=$ $$\begin{gathered} \frac{n!}{(s-a{)}^{n+1}}\text{ for}n \\ \text{= 1,2,...} \end{gathered}$$

$F(s)={ℒ}\left\{t{e}^{-t}\right\}=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}t{e}^{-t}dt=$ $\frac{1}{(s+1{)}^2}$

$F(s)={ℒ}\{1-e^{-t/T}\}=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}(1-e^{-t/T})dt=$ $\frac{1}{s(1+Ts)}$

$F(s)={ℒ}\{e^{at}sin(\omega t)\}=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}{e}^{at}sin(\omega t)dt=$ $\frac{\omega }{(s-a{)}^2+{\omega }^2}$

$F(s)={ℒ}\{e^{at}cos(\omega t)\}=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}{e}^{at}cos(\omega t)dt=$ $\frac{s-a}{(s-a{)}^2+{\omega }^2}$

$F(s)={ℒ}\{u(t)\}=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}u(t)dt=$ $\frac{1}{s}$

$F(s)={ℒ}\{u(t-a)\}=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}u(t-a)dt=$ $\frac{e^{-as}}{s}$

$F(s)={ℒ}\{u(t-a)g(t-a)\}=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}u(t-a)g(t-a)dt=e^{-as}G(s)$

$F(s)={ℒ}\{g^{\prime }(t)\}=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}{g}^{\prime }(t)dt=sG(s)-g(0)$

$F(s)={ℒ}\{g^{″}(t)\}=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}{g}^{″}(t)dt=s^2\cdot G(s)-s\cdot g(0)-g^{\prime }(0)$

$F(s)={ℒ}\{g^{(n)}(t)\}=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}{g}^{(n)}(t)dt=s^n\cdot G(s)-s^{n-1}\cdot g(0)-s^{n-2}\cdot g^{\prime }(0)-...-g^{(n-1)}(0)$

Transfer Function

The Laplace transform of the impulse response of a circuit with no initial conditions is called the transfer function. If a single-input, single-output circuit has no internal stored energy and all the independent internal sources are zero, the transfer function is

$H(s)=\frac{{ℒ}(responsesignal)}{{ℒ}(inputsignal)}.$

Impedances and admittances are special cases of transfer functions.

Example

Solve the differential equation:

$y^{″}-2y^{\prime }-15y=6y(0)=1y^{\prime }(0)=3$

We start by taking the Laplace transform of each term.

${ℒ}\left\{y^{″}\right\}-2{ℒ}\left\{y^{\prime }\right\}-15{ℒ}\left\{y\right\}={ℒ}\left\{6\right\}$

The next step is to perform the respective Laplace transforms, using the information given above.

$(s^2{ℒ}\left\{y\right\}-s-3)-2(s{ℒ}\left\{y\right\}-1)-15{ℒ}\left\{y\right\}={ℒ}\left\{6\right\}$

Using association, the equation is rearranged:

$(s^2-2s-15){ℒ}\left\{y\right\}=\frac{6}{s}+s+3-2$

Continuing on using the method of partial fractions, the equation is progressed:

$\frac{s^2+s+6}{s(s+3)(s-5)}=\frac{A}{s}+\frac{B}{s+3}+\frac{C}{s-5}$

$A(s+3)(s-5)+Bs(s-5)+Cs(s+3)=s^2+s+6$

$A+B+C=-1$

$-2A-5B+3C=1$

$-15A=6$

$A=\frac{-2}{5}B=\frac{1}{2}C=\frac{9}{10}$

Plugging the above values back into the equation further up, we get:

${ℒ}\left\{y\right\}=\frac{\frac{-2}{5}}{s}+\frac{\frac{1}{2}}{s+3}+\frac{\frac{9}{10}}{s-5}$

Applying anti-Laplace transforms, we get the equation:

$y={{ℒ}}^{-1}\left\{\frac{\frac{-2}{5}}{s}\right\}+{{ℒ}}^{-1}\left\{\frac{\frac{1}{2}}{s+3}\right\}+{{ℒ}}^{-1}\left\{\frac{\frac{9}{10}}{s-5}\right\}$

Applying the Laplace transforms in reverse (as the above equation utilizes inverse Laplace transforms) for the above equation, we get the solution:

$y(t)=\frac{-2}{5}+\frac{1}{2}{e}^{-3t}+\frac{9}{10}{e}^{5t}$

References

DeCarlo, Raymond A.; Lin, Pen-Min (2001), Linear Circuit Analysis, Oxford University Press, ISBN 0-19-513666-7 .

Zill, Dennis G. (2005), A First Course in Differential Equations with Modeling Applications Ninth Edition, Brooks/Cole Cengage Learning, ISBN 0-495-10824-3 .

External links

• The Laplace Transform.

Authors

Colby Fullerton

Brian Roath

Reviewed By

David Robbins

Thomas Wooley

Read By

Jaymin Joseph

John Hawkins

Comments

John Hawkins:

• Nice list of transforms! Where did you find it? Inside the back cover of the textbook has a good list, but none including transforms of $g(t)$. I see your reference to the textbook. What page?

• • Oops! It seems I forgot to add a reference to the list, thanks for mentioning it. I got the others from a list in the back of the ODE textbook. I'll list the reference up above.

• I believe that

$\frac{6+s-s^2}{s(s+3)(s-5)}=\left(\frac{A}{s}\right)\left(\frac{B}{s+3}\right)\left(\frac{C}{s-5}\right)$

should be

$\frac{6+s-s^2}{s(s+3)(s-5)}=\frac{A}{s}+\frac{B}{s+3}+\frac{C}{s-5}$

• • Thank you for pointing that out. It should be fixed now.

• Also, the solution $y(t)=-\frac{2}{5}-\frac{1}{4}{e}^{-3t}-\frac{7}{20}{e}^{5t}$ does not match the initial condition of $y(0)=1$.

• • Yeah, you're right. A minor sign error was what messed it up. It should be fixed now.