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Laplace transform of the impulse δ(t) and step σ(t)

The impulse function is not a function in the sense of classical analysis, but a distribution (pseudo-function). Therefore without entering the theory of distributions the integral

(2.39)

is not defined. The singularity exactly matches with the lower integration limit. The impulse function can be approximately described by the limit

   

with the rectangular impulse function

(2.40)

Strictly speaking this representation of is not a distribution, as for is not arbitrarily often differentiable. Because of the simple description compared with other functions (e.g. Gaussian functions) this approach is preferred here. From Eq. (2.39) it follows that

(2.41)

As Eq. (2.40) can also be represented in the form

(2.42)

where is the unit step. Since the integration is independent of , the limit and integration can be permuted so that

   
   

By applying lHospital's rule one obtains

(2.43)

As the impulse has an area of unity it is also called unit impulse.

Example 2.6.1   Given the differential equation

   

Find solution .

Remark: The derivative theorem according to Eq. (2.6) is - as mentioned in section A.1.3 - valid only for classical functions. If, however, a signal consists of a function at , then the lower integration limit of Eq. (2.1) must be chosen equal to and also in Eq. (A.11) the left-hand initial condition to . According to the definition of Eq. (2.1) all left-hand initial conditions are always zero.

The solution can be determined in the following three steps:

Step 1:
The Laplace transform of the given differential equation is:

   with    

Step 2:
The solution of the algebraic equation is:

   

Step 3:
From the back transformation the solution follows as

   

where is the unit step function.



Next: Transfer functions Up: The Laplace transform Previous: Solving linear differential equations   Contents
Christian Schmid 2005-05-09