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.

$F(s)={\mathcal {L}}\left\{f(t)\right\}=\int _{0}^{\infty }e^{-st}f(t)\,dt$

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 ${\mathcal {L}}\left\{\right\}$.

$F(s)={\mathcal {L}}\left\{1\right\}=\int _{0}^{\infty }e^{-st}\,dt=$ ${\frac {1}{s}}$

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

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

$F(s)={\mathcal {L}}\left\{sin(\omega t)\right\}=\int _{0}^{\infty }e^{-st}sin(\omega t)\,dt=$ ${\frac {\omega }{s^{2}+\omega ^{2}}}$

$F(s)={\mathcal {L}}\left\{cos(\omega t)\right\}=\int _{0}^{\infty }e^{-st}cos(\omega t)\,dt=$ ${\frac {s}{s^{2}+\omega ^{2}}}$

$F(s)={\mathcal {L}}\left\{t^{n}g(t)\right\}=\int _{0}^{\infty }e^{-st}t^{n}g(t)\,dt=$ ${\frac {(-1)^{n}d^{n}G(s)}{ds^{n}}}{\mbox{ for}}~n\ {\mbox{= 1,2,...}}$

$F(s)={\mathcal {L}}\left\{tsin(\omega t)\right\}=\int _{0}^{\infty }e^{-st}tsin(\omega t)\,dt=$ ${\frac {2\omega s}{(s^{2}+\omega ^{2})^{2}}}$

$F(s)={\mathcal {L}}\left\{tcos(\omega t)\right\}=\int _{0}^{\infty }e^{-st}tcos(\omega t)\,dt=$ ${\frac {s^{2}-\omega ^{2}}{(s^{2}+\omega ^{2})^{2}}}$

$F(s)={\mathcal {L}}\left\{g(at)\right\}=\int _{0}^{\infty }e^{-st}g(at)\,dt=$ ${\frac {1}{a}}G\left({\frac {s}{a}}\right),{\mbox{ for}}~a>0.$

$F(s)={\mathcal {L}}\left\{e^{at}g(t)\right\}=\int _{0}^{\infty }e^{-st}e^{at}g(t)\,dt=G(s-a)$

$F(s)={\mathcal {L}}\left\{e^{at}t^{n}\right\}=\int _{0}^{\infty }e^{-st}e^{at}t^{n}\,dt=$ ${\frac {n!}{(s-a)^{n+1}}}{\mbox{ for}}~n\ {\mbox{= 1,2,...}}$

$F(s)={\mathcal {L}}\left\{te^{-t}\right\}=\int _{0}^{\infty }e^{-st}te^{-t}\,dt=$ ${\frac {1}{(s+1)^{2}}}$

$F(s)={\mathcal {L}}\left\{1-e^{-t/T}\right\}=\int _{0}^{\infty }e^{-st}(1-e^{-t/T})\,dt=$ ${\frac {1}{s(1+Ts)}}$

$F(s)={\mathcal {L}}\left\{e^{at}sin(\omega t)\right\}=\int _{0}^{\infty }e^{-st}e^{at}sin(\omega t)\,dt=$ ${\frac {\omega }{(s-a)^{2}+\omega ^{2}}}$

$F(s)={\mathcal {L}}\left\{e^{at}cos(\omega t)\right\}=\int _{0}^{\infty }e^{-st}e^{at}cos(\omega t)\,dt=$ ${\frac {s-a}{(s-a)^{2}+\omega ^{2}}}$

$F(s)={\mathcal {L}}\left\{u(t)\right\}=\int _{0}^{\infty }e^{-st}u(t)\,dt=$ ${\frac {1}{s}}$

$F(s)={\mathcal {L}}\left\{u(t-a)\right\}=\int _{0}^{\infty }e^{-st}u(t-a)\,dt=$ ${\frac {e^{-as}}{s}}$

$F(s)={\mathcal {L}}\left\{u(t-a)g(t-a)\right\}=\int _{0}^{\infty }e^{-st}u(t-a)g(t-a)\,dt=e^{-as}G(s)$

$F(s)={\mathcal {L}}\left\{g'(t)\right\}=\int _{0}^{\infty }e^{-st}g'(t)\,dt=sG(s)-g(0)$

$F(s)={\mathcal {L}}\left\{g''(t)\right\}=\int _{0}^{\infty }e^{-st}g''(t)\,dt=s^{2}\cdot G(s)-s\cdot g(0)-g'(0)$

$F(s)={\mathcal {L}}\left\{g^{(n)}(t)\right\}=\int _{0}^{\infty }e^{-st}g^{(n)}(t)\,dt=s^{n}\cdot G(s)-s^{n-1}\cdot g(0)-s^{n-2}\cdot g'(0)-...-g^{(n-1)}(0)$

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

$H(s)={\frac {{\mathcal {L}}(response~signal)}{{\mathcal {L}}(input~signal)}}.$

Impedances and admittances are special cases of transfer functions.

Example

Solve the differential equation:

$y''-2y'-15y=6\qquad y(0)=1\qquad y'(0)=3$

We start by taking the Laplace transform of each term.

${\mathcal {L}}\left\{y''\right\}-2{\mathcal {L}}\left\{y'\right\}-15{\mathcal {L}}\left\{y\right\}={\mathcal {L}}\left\{6\right\}$

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

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

Using association, the equation is rearranged:

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

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

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

$A(s+3)(s-5)+Bs(s-5)+Cs(s+3)=-s^{2}+s+6\,$

$A+B+C=-1\,$

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

$-15A=6\,$

$A={\frac {-2}{5}}\qquad B={\frac {-1}{4}}\qquad C={\frac {-7}{20}}$

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

${\mathcal {L}}\left\{y\right\}={\frac {\frac {-2}{5}}{s}}+{\frac {\frac {-1}{4}}{s+3}}+{\frac {\frac {-7}{20}}{s-5}}$

Applying anti-Laplace transforms, we get the equation:

$y={\mathcal {L}}^{-1}\left\{{\frac {\frac {-2}{5}}{s}}\right\}+{\mathcal {L}}^{-1}\left\{{\frac {\frac {-1}{4}}{s+3}}\right\}+{\mathcal {L}}^{-1}\left\{{\frac {\frac {-7}{20}}{s-5}}\right\}$

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

$y(t)={\frac {-2}{5}}+{\frac {-1}{4}}e^{-3t}+{\frac {-7}{20}}e^{5t}$

References

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

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: I believe that instead of

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

you wanted to write

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