Partial fractions expansion

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, 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 ... 

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 ... 

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. 

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

If we do the partial fraction expansion and inverse transform, we should find, after some hard work, the time domain solution ... 

Since the model is stable, all the roots of P(s), whether they be real or complex, have negative real parts and their corresponding time domain terms will decay away exponentially. Thus if we are only interested in the time domain response at sufficiently large times, we only need to consider the partial fraction expansion of the two purely sinusoidal terms associated with the input ... 

Construct a polynomial from its roots Partial fraction expansion Find the roots to a polynomial Transfer function to zero-pole form conversion Zero-pole form to transfer function conversion... 

The partial fraction expansion of the four stage transform... 

The most common inversion method is called partial fractions expansion. The function to be inverted, F,), is merely rearranged into a series of simple functions ... 

Example 9.1. Given the F(,j below, find its inverseby partial fractions expansion. 

Using partial fractions expansion to invert (see Example 9.1) gives... 

B. PARTIAL-FRACTIONS EXPANSION. The linearity theorem [Eq. (18.36)] permits us to expand the function into a sum of simple terms and invert each individually. This is completely analogous to Laplace-transformation inversion. Let F, be a ratio of polynomials in z, Mth-order in the numerator and iVth-order in the denominator. We factor the denominator into its N roots pi, P2, Ps,... 

These are, of course, exactly the same results we found by partial fractions expansion in Example 18.4. 

Find the outputs X2, T,) of the two systems of Example 18.7 for a unit step input in m, . Use partial fractions expansion and long division. 

First, we assume that there is no delay in the current amplifier and the feedback circuit that is, t = 0 and RC = 0. From the partial fraction expansion of Eq. (11.35), we find the poles are at... 

The coefficients Q and the exponents w, can be obtained by making a partial fraction expansion of gi (s). 

For an infinitely thin sample (Z—>0), no light will be reflected (po >0). With increasing thickness (Z—>oo), the reflectance will approach a final value called p. Integration is achieved via partial fraction expansion and the reflectance at infinite sample thickness is p, (no transmission) ... 

The coefficients of the corresponding terms in the partial fractions expansion will also be complex conjugates of each other. 

If the polynomial P(s) has multiple roots, the denominator of x(s) has a term (s - P()m, where p, is the multiple root which is repeated m times. In such case the partial-fractions expansion produces terms such as... 

The expressions above can now be inverted using the partial-fractions expansion, as it was described in Section 8.2, to find the unknown solution x,(i) and X2(0-... 

In this case the inversion of eq. (11.6) by partial-fractions expansion yields... 

This method is completely parallel to the partial-fractions expansion methodology used to invert Laplace transforms and proceeds as follows ... 

Inverse Laplace transforms, 144, 145-52 computation of (see Partial fractions expansion) table of, 146 Inverse response, 216-20 references, 223... 

Pad6 approximations, 215-16 Pairing variables, 467-84, 494-503, 538 Parallel transmission of signals, 561 Partial fractions expansion ... 

As pointed out above, the critical point in finding the solution to a differential equation using Laplace transforms is the inversion of the Laplace transforms. In this section we will study a method developed by Heaviside for the inversion of Laplace transforms known as Heaviside or partial-fractions expansion. 

How does the procedure to compute the constants of the terms resulting from the partial fractions expansion vary in the presence of multiple roots ... 

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