Fraction. Partial

It is interesting to note that the BET equation is equivalent to the difference between the upper branches of two rectangular hyperbolae, as may be seen by breaking up the right-hand side of Equation (2.12) into partial fractions ... 

Partial Fractions Rational functions are of the type f x)/g x) where /x) and g(x) are polynomial expressions of degrees m and n respectively. If the degree of/is higher than g, perform the algebraic division—the remainder will then be at least one degree less than the denominator. Consider the following types ... 

Equation (8-14) shows that starts from 0 and builds up exponentially to a final concentration of Kcj. Note that to get Eq. (8-14), it was only necessaiy to solve the algebraic Eq. (8-12) and then find the inverse of C (s) in Table 8-1. The original differential equation was not solved directly. In general, techniques such as partial fraction expansion must be used to solve higher order differential equations with Laplace transforms. 

In practice, inverse transformation is most easily achieved by using partial fractions to break down solutions into standard components, and then use tables of Laplace transform pairs, as given in Table 3.1. 

Fig. 3.18 Step response of a generalized second-order system for C < 1-Expanding equation (3.52) using partial fractions...

Note This result could also have been obtained at equation (7.44) by using z-trans-form number 7 in Table 7.1, but the solution demonstrates the use of partial fractions. 

Taking the partial fractions of Equation 3-62 with a = [Cgg (1/2)Cao] gives the following ... 

Converting Equation 8-22 into partial fraction and solving yields... 

Applying the general partial fraction theorem, Eq. (3-72), to Eq. (3-96) and then taking inverse transforms gives Eq. (3-99). 

Teilunga-brucb, m. Math.) partial fraction, -ebene, /. plane of division, -flacbe, /. (Geol.) division plane, -gesetz, n. law of partition, -koeffizient, m. partition coefficient, distribution coefficient, -zabl, /. dividend. -zeicben, n. mark of division hyphen, -zustand, m. state of division. Teil-verflUssigung,/. partial liquefaction, -vor-gang, m. partial process, -wand, /. division wall. 

Integration by Partial Fractions is of assistance in the integration of rational fractions. If... 

These equations can be solved (although not easily) by integration by the method of partial fractions, by matrices, or by Laplace transforms. For the case where [I]o = [P]0 = 0, the concentrations are... 

Since we are doing inverse transform using a look-up table, we need to break down any given transfer functions into smaller parts which match what the table has—what is called partial fractions. The time-domain function is the sum of the inverse transform of the individual terms, making use of the fact that Laplace transform is a linear operator. 

The linear property is one very important reason why we can do partial fractions and inverse transform using a look-up table. This is also how we analyze more complex, but linearized, systems. Even though a text may not state this property explicitly, we rely heavily on it in classical control. 

Since we rely on a look-up table to do reverse Laplace transform, we need the skill to reduce a complex function down to simpler parts that match our table. In theory, we should be able to "break up" a ratio of two polynomials in 5 into simpler partial fractions. If the polynomial in the denominator, p(s), is of an order higher than the numerator, q(s), we can derive 1... 

The next question is how to find the partial fractions in Eq. (2-25). One of the techniques is the so-called Heaviside expansion, a fairly straightforward algebraic method. We will illustrate three important cases with respect to the roots of the polynomial in the denominator (1) distinct real roots, (2) complex conjugate roots, and (3) multiple (or repeated) roots. In a given problem, we can have a combination of any of the above. Yes, we need to know how to do them all. 

If the order of q(s) is higher, we need first carry out "long division" until we are left with a partial fraction "residue." Thus the coefficients ai are also called residues. We then expand this... 

In practice, we seldom do the partial fraction expansion of a pair of complex roots. Instead, we rearrange the polynomial p(s) by noting that we can complete the squares ... 

The polynomial p(s) has the roots -1 repeated three times, and -2. To keep the numerator of each partial fraction a simple constant, we will have to expand to... 

Recall from the examples of partial fraction expansion that the polynomial Q(s) in the numerator, or the zeros, affects only the coefficients of the solution y(t), but not the time dependent functions. That is why for qualitative discussions, we focus only on the poles. 

We now put one and one together. The key is that we can "read" the poles—telling what the form of the time-domain function is. We should have a pretty good idea from our exercises in partial fractions. Here, we provide the results one more time in general notation. Suppose we have taken a characteristic polynomial, found its roots and completed the partial fraction expansion, this is what we expect in the time-domain for each of the terms ... 

It is cumbersome to write the partial fraction with complex numbers. With complex conjugate poles, we commonly combine the two first order terms into a second order term. With notations that we will introduce formally in Chapter 3, we can write the second order term as... 

In establishing the relationship between time-domain and Laplace-domain, we use only first and second order differential equations. That s because we are working strictly with linearized systems. As we have seen in partial fraction expansion, any function can be "broken up" into first order terms. Terms of complex roots can be combined together to form a second order term. 

Take note (again ) that the characteristic polynomials in the denominators of both transfer functions are identical. The roots of the characteristic polynomial (the poles) are independent of the inputs. It is obvious since they come from the same differential equation (same process or system). The poles tell us what the time-domain solution, y(t), generally would "look" like. A final reminder no matter how high the order of n may be in Eq. (3-4), we can always use partial fractions to break up the transfer functions into first and second order terms. 

Let say we have a high order transfer function that has been factored into partial fractions. If there is a large enough difference in the time constants of individual terms, we may try to throw away the small time scale terms and retain the ones with dominant poles (large time constants). This is our reduced-order model approximation. From Fig. E3.3, we also need to add a time delay in this approximation. The extreme of this idea is to use a first order with dead time function. It obviously cannot do an adequate job in many circumstances. Nevertheless, this simple... 

For the case with a unit step input such that X = 1/s, we have, after partial fraction expansion,... 

This is really an algebraic exercise in partial fractions. The answer hides in Table 2.1. 

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