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Laplace Transform

Laplace transform converts a time domain function to s-domain function by integration from zero to infinity

 of the time domain function, multiplied by e-st.

The Laplace transform is used to quickly find solutions for differential equations and integrals.

Derivation in the time domain is transformed to multiplication by s in the s-domain.

Integration in the time domain is transformed to division by s in the s-domain.

Laplace transform function

The Laplace transform is defined with the L{} operator:

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

Inverse Laplace transform

The inverse Laplace transform can be calculated directly.

Usually the inverse transform is given from the transforms table.

Laplace transform table

Function nameTime domain functionLaplace transform

f (t)

F(s) = L{f (t)}

Constant1\frac{1}{s}
Lineart\frac{1}{s^2}
Power

t n

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

Power

t a

Γ(a+1) · s -(a+1)

Exponent

e at

\frac{1}{s-a}

Sine

sin at

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

Cosine

cos at

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

Hyperbolic sine

sinh at

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

Hyperbolic cosine

cosh at

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

Growing sine

t sin at

\frac{2as}{(s^2+a^2)^2}

Growing cosine

t cos at

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

Decaying  sine

e -at sin ωt

\frac{\omega }{\left ( s+a \right )^2+\omega ^2}

Decaying cosine

e -at cos ωt

\frac{s+a }{\left ( s+a \right )^2+\omega ^2}

Delta function

δ(t)

1

Delayed delta

δ(t-a)

e-as

Laplace transform properties

Property nameTime domain functionLaplace transformComment
 

f (t)

F(s)

 
Linearitya f (t)+bg(t)aF(s) + bG(s)a,b are constant
Scale changef (at)\frac{1}{a}F\left ( \frac{s}{a} \right )a>0
Shifte-at f (t)F(s + a) 
Delayf (t-a)e-asF(s) 
Derivation\frac{df(t)}{dt}sF(s) - f (0) 
N-th derivation\frac{d^nf(t)}{dt^n}snf (s) - sn-1f (0) - sn-2f '(0)-...-f (n-1)(0) 
Powert n f (t)(-1)^n\frac{d^nF(s)}{ds^n} 
Integration\int_{0}^{t}f(x)dx\frac{1}{s}F(s) 
Reciprocal\frac{1}{t}f(t)\int_{s}^{\infty }F(x)dx 
Convolutionf (t) * g (t)F(s) · G(s)* is the convolution operator
Periodic functionf (t) = f (t+T)\frac{1}{1-e^{-sT}}\int_{0}^{T}e^{-sx}f(x)dx 

Laplace transform examples

Example #1

Find the transform of f(t):

f (t) = 3t + 2t2

Solution:

F(s) = 3/s2 + 4/s2

 

Example #2

Find the inverse transform of F(s):

F(s) = 3 / (s2 + s - 6)

Solution:

In order to find the inverse transform, we need to change the s domain function to a simpler form:

F(s) = 3 / (s2 + s - 6) = 3 / [(s-2)(s+3)] = a / (s-2) + b / (s+3)

[a(s+3) + b(s-2)] / [(s-2)(s+3)] = 3 / [(s-2)(s+3)]

a(s+3) + b(s-2) = 3

To find a and b, we get 2 equations - one of the s coefficients and second of the rest:

(a+b)s + 3a-2b = 3

a+b = 0 , 3a-2b = 3

a = 3 , b = -3

F(s) = 3 / (s-2) - 3 / (s+3)

Now F(s) can be transformed easily by using the transforms table for exponent function:

f (t) = 3e2t - 3e-3t

 


See also

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