Inverse Laplace Transforms (ILT)

Idea

The basic idea of ILT is to decompose Y(s) into a linear combination of a set of s-domain functions (or signals) whose ILT can be found from the Laplace transformation table.

Let’s say: (ILT.1)

in which ILT of is known, i.e., can be found from s-table. Then ILT of Y(s) can be obtained as: (ILT.2)

Partial Fraction Expansion (PFE)

PFE is a way to accomplish (14.1) for a s-function having the form of , where C(s) and

D(s) are polynomial of s, i.e.,

where p1, p2, , pn are roots of D(s).

For m < n, Y(s) can be decomposed into the following forms:

Case 1. If all p i are different , i.e., p1 p2 pn,

can be factorized as

(ILT.3)

where the coefficients can be found from the residual formula given by

With this factorization, the ILT of is

(ILT.4)

Ex 1: Given , it can be decomposed as

where Ai can be obtained by the comparing coefficient method or the residual formula (14.3) as:

Thus, the inverse Laplace transform of is

Case 2. If all p i are equal, i.e., repeated roots p1 = p2 = = pn

can be factorized as

(ILT.5)

where Ai can be found by the comparing coefficients method or the residue formula given by:

With this factorization, the ILT of is

Ex 2. Given , it can be decomposed to

where

Case 3. In general, where p1 p2 pl and ni is the multiplicity of the root pi , i = 1, , l. Then, can be decomposed to

(ILT.6)

where the coefficients Aji can be found by the comparing coefficient method or the residue formula:

The inverse Laplace transform of is then obtained as

(ILT.7)

Note: From (14.7), one can see that if all roots pi have negative real parts, then y(t) 0 as t .

Ex 3. Given , it can be decomposed to

where

Order of the Numerator C(s) = Order of the denominator D(s) : n = m

For n = m, first factorize Y(s) into the form

where the order of Cn-1(s) is less than the order of D(s). The inverse Laplace transform of is

where (t) is the unit impulse function and the inverse Laplace transform of can be obtained by

the partial fraction expansion (PFE) procedure discussed before.

Ex 4.

Since

Residue Command in MATLAB

The roots and the coefficients in the partial fraction expansion (PFE) in (ILT.6) can be obtained by the residue command in MATLAB as follows:

[A, P, K] = residue (num, den)where num and den are the vectors of the coefficients of the numerator and the denominator of Y(s) in descending order of s respectively. The return value A is the vector of the coefficients in PFE, i.e.,

and P is the vector of the roots, i.e.,

Ex: Given , using the MATLAB command:

>> [A, P, K] = residue ( [1, 0, 0, 2] , [1, 2, 1, 0] )

will return the following values: A = [ -4, -1, 2]T , P = [ -1, -1, 0] , K = 1

which means that

From which we can obtain the inverse Laplace transform of as

Complex Roots

In principle, the PFE procedure presented before applies to complex roots also as illustrated by the following example.

Ex C.1: Given

where

Useful Formula:

Note that the resulting coefficients for the complex roots using residual formula are complex numbers. So if we use the above PRE expansion to solve ILT, we have to work with operation of complex numbers as shown in the above example. As an alternative, we can also use the basic idea of ILT--decompose Y(s) into a linear combination of some simple functions whose ILT are known--to avoid the problem of dealing with complex numbers as shown below.

Since complex roots will appear in a complex conjugate form (for a polynomial with real coefficients), the complex roots of D(s) can be represented by pairs like and . Noting the fact that a pair of complex roots and correspond to a factor (s – pi)(s – pi+1) = (s - )2 +2 in the denominator D(s), instead of decomposing Y(s) into the forms like

(ILT.8)

where Ai and Ai+1 are complex numbers, we can decompose Y(s) into the following form

(ILT.9)

where Bi and Bi+1 are real numbers and can be obtained by the method of comparing coefficients as shown in the following example. With this PFE, the ILT of Y(s) is thus given by

(ILT.10)

Ex C.2: Given the same Y(s) as in Ex C.1, if we use (ILT.9) to do PFE, we have

where = -1 and =1. Multiplying both sides by (s+ 1)2 + 1, we obtain

Comparing coefficients:

Ex C.3 Given , by using PFE (ILT.9),

Multiplying both sides by :

Comparing coefficients:

Note:

Complex roots Oscillating response since cosine and sine terms appear !