1 Inverse Laplace Transform Note that most of the transforms are written as fractions

2 P(s)Q(s) sa o, a 1,..., a n b o, b 1,..., b m F(s) where P(s) and Q(s) are polynomials in the complex frequency variables s and the coefficients a o, a 1,..., a n, b o, b 1,..., b m are real numbers. A rational function is specified completely by the two sets of real coefficients which define the numerator and denominator polynomials. On the other hand, the polynomials can also be expressed in the factored form in terms of their zeros. Thus, an alternate representation of F(s) is given Partial-fraction Expansion

3 The first step in the partial-fraction expansion is to put the rational proper F(s) P(s) Q(s),P(s)Q(s) function into a proper form. We say that a rational function is proper if the degree of the numerator polynomial is less than the degree of the denominator polynomial. If the given rational function F(s) is not proper, i.e.., if the degree of P(s) is greater than or equal to that of Q(s), we divide (long division) P(s) by Q(s) and obtain R(s)R(s) Q(s), R(s)/Q(s) ,  (1),  (2), R(s)/Q(s) The quotient, is a polynomial and R(s) is the remainder; therefore, R(s) has a degree less than that of Q(s), and the new rational function R(s)/Q(s) is proper. Since is a polynomial, the corresponding time function is a linear combination of ,  (1),  (2), etc., and can be determined directly using Tables. We therefore go ahead with the new rational function R(s)/Q(s) which is proper. In the remaining part of this section we assume that all rational functions are proper.

4 Second Order Polynomial If the damping ratio >1 (roots are real & distinct ‘simple’) If the damping ratio <1 (roots are complex conjugates If the damping ratio =1 (roots are real & equal ‘repeated’) If the damping ratio >1 (roots are real & distinct ‘simple’) If the damping ratio <1 (roots are complex conjugates If the damping ratio =1 (roots are real & equal ‘repeated’)

5 We start with a simple example as follows: K 1, K 2 K 3 We claim that there are constants K 1, K 2, and K 3 such that Heaviside’s Expansion Compare Heaviside’s Expansion with solving for K or substituting convenient values for s. Case 1 Simple poles

6 K j p j and the residue K j of the pole p j is given by

7 Find the partial-fraction expansion of s =p 1 =-1n 1 =3 s = p 2 = 0n 2 = 2 The function has two multiple poles at s =p 1 =-1 (third order, n 1 =3 ) and at s = p 2 = 0 (second order, n 2 = 2 ). Thus, the partial fraction expansion is of the form K 11, K 12, K 13 F(s)(s +1) 3 To calculate K 11, K 12, and K 13, we first multiply F(s) by (s +1) 3 to obtain we find Case 2 Multiple poles

8 K 21 K 22 F(s)s 2 Similarly, to calculate K 21 and K 22, we first multiply F(s) by s 2 to obtain Using, we find Therefore, the partial-fraction expansion is The corresponding time function is

9 Case 3 Complex poles The two cases presented above are valid for poles which are either real or complex. However, if complex poles are present, the coefficients in the partial-fraction expansion are, in general, complex, and further simplification is possible. Using formula for simple poles, we obtain

10 F(s)sreal K2K1 Since F(s) is a rational function of s with real coefficients, it follows that K2 is the complex conjugate of K1. Once K and K* are determined, the corresponding two complex terms cn be combined as follows: Suppose K=a+jb. Then K*=a-jb, abd Once, we find a and b, given,, using the Inverse Laplace transform, we get

11 This formula, which gives the corresponding time function for a pair of terms due to the complex conjugate poles, is extremely useful. There are good Matlab examples utilizing the command residue Practice the examples