The Laplace Transform is another form of integral transform which is used in applied mathematics to solve differential equations. Although it lacks some of the mathematical elegance of the Fourier Transform, it is still extremely useful. It is very popular in engineering, and unlike the Fourier Transform can be used without knowledge of complex numbers.


Let f(t) be a function which does not grow exponentially as t increases. Then the Laplace Transform is defined as

F(p) = integral(f(t)exp(-pt)dt, t=0...inf).

The Laplace Transform F(p) is often denoted as f with a circumflex ("^"). Since the Laplace Transform does not depend on the values of f for negative t, we usually assume f(t)=0 in this region.

Basic properties

Using elementary properties of integration we can prove the following useful results.

  1. Linearity. Let h(t) = f(t) + g(t). Then H(p) = F(p) + G(p).
  2. Translation. Let g(t) = f(t-a) for a>0. Then G(p) = exp(-ap) F(p).
  3. Frequency shift. Let g(t) = exp(lt) f(t). Then G(p) = F(p-l) for p>l.
  4. Scaling. Let g(t) = f(at) for a>0. Then G(p) = F(p/a)/a.
  5. Derivative. Let g(t) = f'(t). Then G(p) = pF(p) - f(0).
  6. Second derivative. Let g(t) = f''(t). Then G(p) = p2F(p)-pf(0)-f'(0).
  7. Factors of t. Let g(t) = tf(t). Then G(p) = - dF(p)/dp.

Many of these properties are similar to the Fourier Transform. Results 5 and 6 are critical in solving differential equations: see the Laplace Transformation node for more details.


Inversion is one place where the Laplace Transform lacks mathematical elegance, since it requires an ugly contour integral in the complex plane. Consequently most people who use Laplace Transforms for practical purposes look up Laplace inversions in tables.

If f(t) has Laplace Transform F(p), then we have the relation

F(t) = 1/(2πi) integral(F(p)exp(pt)dp, p=γ-i∞...γ+i∞)

where γ is a large real number, chosen so that the path of integration lies to the right of all the singularities of F(p). This is known as the Bromwich inversion integral; see this node for a full mathematical treatment of where this result comes from.

Laplace versus Fourier

Both Laplace Transforms and Fourier Transforms can be used to solve differential equations, so a natural question to ask is "which one is better?".

Because of the negative exponential term in the Laplace Transform integration, convergence is a lot stronger, and polynomial expressions which could not be handled with the Fourier Transform can be happily considered. However Fourier Transforms make up for this with a greater mathematical elegance, and tie in beautifully with many areas of applied mathematics, particularly quantum mechanics. The Fourier Transform also has many applications beyond the solution of differential equations.

The Laplace Transform picks out precisely the solution to a differential equation that obeys certain initial conditions at t=0. Hence it is ideally suited to initial value problems, like the one given above.

The Fourier Transform picks out precisely the solution to a differential equation which decays to zero at large distances. In many applied problems this is exactly the solution we want, since a function which grows at large distances would be considered unphysical.

Often a mixture of the two methods proves the most effective. We are often interested in functions that depend on time and space. The most physical solution to an equation is one which obeys initial conditions at t=0, and which tends to zero at large distances. This directly corresponds to Fourier transforming with respect space and Laplace transforming with respect to time.

The Laplace Transformation (after the French mathematician Pierre Simon Marquis de Laplace) is a method for solving differential equations, and the corresponding initial and boundary value problems.

Finding a solution of a differential equation using the Laplace Transformation consists of three steps:

  1. Transformation of the equation into a subsidiary equation (moving the equation from the so called t-space to s-space)
  2. Solving the subsidiary equation by simple algebraic manipulations
  3. Transforming the subsidiary equation back into t-space to obtain the solution of the given problem
The Laplace Transformation is widely used in engineering, especially in problems where the (mechanical/electrical) driving force has discontinuities. Another common use of the Laplace Transformation is in Process Control.

I will only describe the basic definition of the Laplace Transform, some basic properties, and a basic example on the use of this method.

Consider a function f(t), defined for all t≥0. Multiply this function by e-st, and integrate with respect to t from zero to infinity. The resulting integral F(s) is given by:

          F(s) = ∫ e-st f(t) dt
The function F(s) of the variable s is called the Laplace Transform, denoted by L(f). Furthermore, the original function f(t) is called the inverse transform, denoted by L-1(F):
          L(f) = ∫ e-st f(t) dt

          f(t) = L-1(F)
Some simple general transforms are given in the following table. More complex transforms can be found in books of mathematical tables.
  f(t)        L(f)
  ----        ----
  1           1/s
  t           1/s2
  tn          n!/(sn+1)    n=1, 2, ...
  eat         1/(s-a)

An important property of the Laplace Transform is its linearity:

          L{af(t) + bg(t)} = aL{f(t)} + bL{g(t)}
In order to solve differential equations using Laplace Transforms, the transform of the derivative of a function f(t) is needed. This is given by:
          L(f') = sL(f) - f(0)
This definition can be extended to transforms of derivatives of arbitrary higher orders:
          L(fn) = snL(f) - sn-1 f(0) - sn-2 f'(0) - ... - f(n-1)(0)

Example: solve the initial value problem:

          y" + 4y' +3y = 0    y(0) = 3, y'(0) = 1
  1. First set up the subsidiary equations, with Y(s) = L(y):
              L(y') = sY - y(0) = sY - 3
              L(y") = s2Y -sy(0) - y'(0) = s2Y - 3s - 1
              s2Y + 4sY +3Y = 3s + 1 + 4*3
  2. Solve the subsidiary equation for Y:
              (s + 3)(s + 1)Y = 3s + 13
                      3s + 13      -2     5
              Y = -------------- = --- + ---
                  (s + 3)(s + 1)   s+3   s+1
  3. Solve the given problem by using the inverse transform:
              L-1{1/(s + 3)} = e-3t
              L-1{1/(s + 1)} = e-t
              y(t) = -2e-3t + 5e-t
Of course, this differential equation would have been quite easy to solve using a substitution method. However, the Laplace Transformation allows for a systematic method to solve more complex differential equations, and also systems of differential equations.
As an additional note, there are a few different definitions of the Laplace transform. The one that seems to appear in mathematics classes is the one listed above; the range of integration is from 0 to infinity.

However, some sources, such as my book for the Electrical Engineering Signals and Systems course I took, define the Laplace transform to range from negative infinity to infinity. This is referred to as the bilateral transform, as opposed to the unilateral transform described above. The two have identical properties; which to use depends on one's application. Initial value problems require the unilateral transform; it seems the bilateral transform is more useful for signals and systems.

The Fourier transform and the Laplace transform are closely related; in a sense, the Fourier transform can be seen as a special case of the bilateral Laplace transform, where the complex variable s in the integral is restricted to be on the imaginary axis. Because of this, Laplace transforms apply to a larger set of functions than Fourier transforms, which can run into discontinuities along the imaginary axis that cause either the transform or its inverse to be divergent.

In control theory, a system is stable only if all the poles of the system function, which is the Laplace transform of the system's impulse response, are on the left side of the imaginary axis. This is a basic tool of Classical Control

Here's a table of some basic Laplace transforms.

            function            Laplace
1 1/s
eat 1/ (s-a)
tn n! / s (n + 1)
tp, p >-1 Γ(p + 1) / s (p + 1)
sin at a / (s 2 + a 2 )
cos at s / (s 2 + a 2 )
sinh at a / (s 2 - a 2 )
cosh at s / (s 2 - a 2 )
e at sin bt b / [ (s-a) 2 + b 2 ]
e at cos bt (s - a) / [ (s-a) 2 + b 2 ]
tneat n! / (s - a) (n + 1)
uc(t) e -cs / s
uc(t)f(t-c) e-csF(s)
ect f(t) F(s - c)
f(ct) 1/c F(s/c)
0tf(t-τ)g(τ)dτ F(s)*G(s)
δ(t - c) e-cs
f(n)(t) snF(s)-sn-1f(0)-...-f(n - 1)(0)
(-t)nf(t) F(n)(s)

For notation clarity, I used uc(t) in this table to denote the unit step function, or Heaviside function (this is often H(t-c) in some differential equations books).

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