11.2: Partial Fractions

Partial fractions are presented here, in the context of control systems, as the fundamental link between pole locations and stability. Solving linear time-invariant systems by the Laplace Transform method will generally create a signal containing the (factored) form

\[ Y(s) \, = \, \dfrac{K(s + z_1) (s + z_2) \cdots (s + z_m)}{(s + p_1)(s + p_2) \cdots (s + p_n)}. \]

Although for the moment we are discussing the signal \(Y(s)\), later we will see that dynamic systems are described in the same format: in that case we call the impulse response \(G(s)\) a transfer function. A system transfer function is identical to its impulse response, since \(L(\delta(t)) = 1\).

The constants \(-z_i\) are called the zeros of the transfer function or signal, and \(-p_i\) are the poles. Viewed in the complex plane, it is clear that the magnitude of \(Y(s)\) will go to zero at the zeros, and to infinity at the poles.

Partial fraction expansions alter the form of \(Y(s)\) so that the simple first- and second-order transform pairs can be used to find the time-domain output signals. We must have \(m<n\) for this procedure; if this is not the case, then we have to strip off extra powers of \(s\) to solve the problem, and then add them back on at the end.

11.2.1: Partial Fractions: Unique Poles

Under the condition \(m<n\), it is a fact that \(Y(s)\) is equivalent to \[ Y(s) \, = \, \dfrac{a_1}{s + p_1} + \dfrac{a_2}{s + p_2} + \cdots \dfrac{a_n}{s + p_n}, \]

in the special case that all of the poles are unique and real. The coefficient \(a_i\) is termed the residual associated with the \(i\)'th pole, and once all these are found it is a simple matter to go back to the transform table and look up the time-domain responses.

How to find \(a_i\)? A simple rule applies: multiply the right-hand sides of the two equations above by \( (s + p_i) \), evaluate them at \(s = -p_i ,\) and solve for \(a_i ,\) the only one left.

Example \(\PageIndex{1}\): Partial Fractions with Unique Real Poles

\[ G(s) \, = \, \dfrac{s(s+6)}{(s+4)(s-1)} e^{-2s}. \nonumber \]

Since we have a pure delay and \(m = n\), we can initially work with \(G(s) / se^{-2s}\). We have

\[ \dfrac{s+6}{(s+4)(s-1)} \, = \, \dfrac{a_1}{s+4} + \dfrac{a_2}{s-1}, \nonumber \]

giving

\[\begin{align*} a_1 \, &= \, \left[ \frac{(s+6)(s+4)}{(s+4)(s-1)} \right]_{s = -4} \\[4pt] &= \, - \dfrac{2}{5} \end{align*}\]

and

\[\begin{align*} a_2 \, &= \, \left[ \frac{(s+6)(s-1)}{(s+4)(s-1)} \right]_{s=1} \\[4pt] &= \,\, \dfrac{7}{5} \end{align*}\]

Thus

\begin{align*} L^{-1} (G(s)/se^{-2s}) \, &= \, - \dfrac{2}{5} e^{-4t} + \dfrac{7}{5}e^t \longrightarrow \\[4pt] g(t) \, &= \, \delta (t-2) + \dfrac{8}{5} e^{-4(t-2)} + \dfrac{7}{5} e^{t-2}. \end{align*}

The impulse response is needed to account for the step change at \(t=2\). Note that in this example, we were able to apply the derivative operator \(s\) after expanding the partial fractions. For cases where a second derivative must be taken, i.e., \(m \geq n+1,\) special care should be used when accounting for the signal slope discontinuity at \(t=0\). The more traditional method, exemplified by Ogata, may prove easier to work through.

The case of repeated real roots may be handled elegantly, but this condition rarely occurs in applications.

11.2.2: Partial Fractions: Complex-Conjugate Poles

A complex-conjugate pair of poles should be kept together, with the following procedure: employ the form

\[ Y(s) \, = \, \dfrac{b_1 s + b_2}{(s+p_1)(s+p_2)} + \dfrac{a_3}{s+p_3} + \cdots, \]

where \(p_1 = p_2^*\) (complex conjugate). As before, multiply through by \( (s+p_1)(s+p_2), \) and then evaluate at \(s = -p_1.\)

Example \(\PageIndex{2}\): Partial Fractions with Complex Poles

\[ G(s) \, = \, \dfrac{s+1}{s (s+j)(s-j)} \, = \, \dfrac{b_1 s + b_2}{(s+j)(s-j)} + \dfrac{a_3}{s}: \nonumber \]

\begin{align*} \left[ \dfrac{s+1}{s} \right]_{s=-j} \,\, & = \,\, \left[ b_1 s + b_2 \right]_{s=-j} \longrightarrow \\[4pt][5pt] 1 + j \,\, &= \,\, -b_1 j + b_2 \longrightarrow \\[4pt] b_1 \,\, &= \,\, -1 \\[4pt] b_2 \,\, &= \,\, 1; \text{ also} \\[4pt] \left[ \dfrac{s+1}{(s+j)(s-j)} \right]_{s=0} \,\, &= \,\, a_3 \, = 1. \end{align*}

Working out the inverse transforms from the table of pairs, we have simply (noting that \(\eta = 0\)):

\[ g(t) \, = \, - \cos t + \sin t + 1(t). \]