Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
where P(s)and Q(s) are polynomials in the complex frequency variables s and the coefficients ao, a1, . . . , an, bo, b1, . . . , bm 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 ofF(s)is given
function into a proper form. We say that a rational function is proper ifthe 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
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.
We start with a simple example as follows:
We claim that there are constants K1, K2, and K3such that
Compare Heaviside’s Expansion with solving for K or substituting convenient values for s.
Find the partial-fraction expansion of
The function has two multiple poles at s =p1=-1 (third order, n1=3) and at s = p2 = 0 (second order,n2= 2). Thus, the partial fraction expansion is of the form
To calculate K11, K12, and K13, we first multiply F(s)by (s +1)3to obtain
Using (5-115) we find
Using (5-115),we find
Therefore, the partial-fraction expansion is
The corresponding time function is
First, we observe that F(s) is a ratio of polynomials with real coefficients; hence zeros and poles, if complex, must occur in complex conjugate pairs. More precisely, if p1 = 1 + j1is a pole, that is, Q1(p1)=0, then is also a pole; that is, . This is due to the fact that any polynomial Q(s)with real coefficients has the property that for all s.
Let us assume that the rational function has a simple pole at s = p1 =-+ j; then it must have another pole at s = p2 = =- - j. The partial-fraction expansion of F(s)must contain the following two terms:
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 formulafor simple poles, we obtain
Since F(s)is a rational function of s with realcoefficients, it follows that K2 is the complex conjugate of K1.
Once K and K* are determined, the corresponding two complex terms can be combined as follows: Suppose K=a+jb. Then K*=a-jb, and
Once, we find a and b, given, , using the Inverse Laplace transform, we get
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 in your text book utilizing the command residue
Practice the examples