8.3: The Partial Fraction Expansion of the Resolvent

Partial Fraction Expansion of the Transfer Function

The Gauss-Jordan method informs us that \(R\) will be a matrix of rational functions with a common denominator. In keeping with the notation of the previous chapters, we assume the denominator to have the \(h\) distinct roots, \(\{\lambda_{j} | j = \{1, \cdots, h\}\}\) with associated multiplicities \(\{m_{j} | j = \{1, \cdots, h\}\}\)

Now, assembling the partial fraction expansions of each element of \(R\) we arrive at

\[R(s) = \sum_{j = 1}^{h} \sum_{k=1}^{m_{j}} \frac{R_{j,k}}{(s-\lambda_{j})^k} \nonumber\]

where, recalling the equation from Cauchy's Theorem, the matrix \(R_{j,k}\) equals the following:

\[R_{j,k} = \frac{1}{2\pi j} \int R(z)(z-\lambda_{j})^{k-1} dz \nonumber\]

Example \(\PageIndex{1}\)

As we look at this example in the introduction, we find

\[\begin{array}{ccc} {R_{1,1} = \begin{pmatrix} {1}&{0}&{0}\\ {0}&{1}&{0}\\ {0}&{0}&{0} \end{pmatrix}}&{R_{1,1} = \begin{pmatrix} {0}&{0}&{0}\\ {1}&{0}&{0}\\ {0}&{0}&{0} \end{pmatrix}}&{R_{2,1} = \begin{pmatrix} {0}&{0}&{0}\\ {0}&{0}&{0}\\ {0}&{0}&{1} \end{pmatrix}} \end{array} \nonumber\]

One notes immediately that these matrices enjoy some amazing properties. For example

\[\begin{array}{ccccc} {R_{1,1}^{2} = R_{1,1}}&{R_{1,2}^{2} = R_{1,2}}&{R_{1,1} R_{2,1} = 0}&{and}&{R_{2,1}^{2} = R_{2,1}} \end{array} \nonumber\]

Below we will now show that this is no accident. As a consequence of Equation and the first resolvent identity, we shall find that these results are true in general.

\(R_{j,1}^{2} = R_{j,1}\) as seen above.

Recall that the \(C_{j}\) appearing in Equation is any circle about \(\lambda_{j}\) that neither touches nor encircles any other root. Suppose that \(C_{j}\) and \(C_{j}'\) are two such circles and \(C_{j}'\) encloses \(C_{j}\). Now,

\[R_{j,1} = \frac{1}{2\pi j} \int R(z) dz = \frac{1}{2\pi j} \int R(z) dz \nonumber\]

and so

\[R_{j,1}^2 = \frac{1}{(2\pi j)^2} \int R(z) dz = \frac{1}{2\pi j} \int R(w) dw \nonumber\]

\[R_{j,1}^2 = \frac{1}{(2\pi j)^2} \int \int R(z) R(w) dw dz \nonumber\]

\[R_{j,1}^2 = \frac{1}{(2\pi j)^2} \int \int \frac{R(z)-R(w)}{w-z} dw dz \nonumber\]

\[R_{j,1}^2 = \frac{1}{(2\pi j)^2} (\int R(z)- \int \frac{1}{w-z} dw dz - \int R(w)- \int \frac{1}{w-z} dz dw) \nonumber\]

\[R_{j,1}^2 = \frac{1}{2\pi i} \int R(z) dz = R_{j,1} \nonumber\]

We used the first resolvent identity, This Transfer Function equation, in moving from the second to the third line. In moving from the fourth to the fifth we used only

\[\int \frac{1}{w-z} dw = 2 \pi i \nonumber\]

and

\[\int \frac{1}{w-z} dz = 0 \nonumber\]

The latter integrates to zero because \(C_{j}\) does not encircle ww

From the definition of orthogonal projections, which states that matrices that equal their squares are projections, we adopt the abbreviation

\[P_{j} \equiv R_{j,1} \nonumber\]

With respect to the product \(P_{j}P_{k}\), for \(j \ne k\), the calculation runs along the same lines. The difference comes in Equation where, as \(C_{j}\) lies completely outside of \(C_{k}\), both integrals are zero. Hence,

If \(j \ne k\) then \(P_{j}P_{k} = 0\)

Along the same lines we define

\[D_{j} \equiv R_{j,2} \nonumber\]

and prove

If \(1 \le k \le m_{j}-1\) then \(D_{j}^{k} = R_{j,k+1}\) \cdot D_{j}^{m_{j}} = 0\)

For \(k\) and \(l\) greater than or equal to one,

\[R_{j,k+1} R_{j,l+1} = \frac{1}{(2\pi i)^2} \int R(z)(z-\lambda_{j})^{k} dz \int R(w)(w-\lambda_{j})^{l} dw \nonumber\]

\[R_{j,k+1} R_{j,l+1} = \frac{1}{(2\pi i)^2} \int \int R(z) R(w)(z-\lambda_{j})^{k} (w-\lambda_{j})^{l} dw dz \nonumber\]

\[R_{j,k+1} R_{j,l+1} = \frac{1}{(2\pi i)^2} \int \int \frac{R(z)-R(w)}{w-z} (z-\lambda_{j})^{k} (w-\lambda_{j})^{l} dw dz \nonumber\]

\[R_{j,k+1} R_{j,l+1} = \frac{1}{(2\pi i)^2} \int R(z) (z-\lambda_{j})^{k} \int \frac{(w-\lambda_{j})^{l}}{w-z} dw dz-\frac{1}{(2\pi i)^2} \int R(w) (w-\lambda_{j})^{k} \int \frac{(z-\lambda_{j})^{k}}{w-z} dz dw \nonumber\]

\[R_{j,k+1} R_{j,l+1} = \frac{1}{2\pi i} \int R(z) (z-\lambda_{j})^{k+l} dz = R_{j,k+l+1} \nonumber\]

because

\[\int \frac{(w-\lambda_{j})^{l}}{w-z} dw = 2 \pi i (z-\lambda_{j})^{l} \nonumber\]

and

\[\int \frac{(z-\lambda_{j})^{k}}{w-z} dw = 0 \nonumber\]

With \(k = l = 1\) we have shown \(R_{j,2}^{2} = R_{j,3}\) i.e., \(D_{j}^{2} = R_{j,3}\). Similarly, with \(k = 1\) and \(l = 2\) we find \(R_{j,2} R_{j,3} = R_{j,4}\) i.e., \(D_{j}^{3} = R_{j,4}\). Continuing in this fashion we find \(R_{j,k} R_{j,k+1} = R_{j,k+2} = j\), or \(D_{j}^{k+1} = R_{j,k+2}\). Finally, at \(k = m_{j-1}\) this becomes

\[D_{j}^{m_{j}} = R_{j, m_{j}+1} = \frac{1}{2\pi i} \int R(z)(z-\lambda_{j})^{m_{j}} dz = 0 \nonumber\]

by Cauchy's Theorem.

With this we now have the sought after expansion

\[R(z) = \sum_{j = 1}^{h} \frac{1}{z-\lambda_{j}} P_{j}+\sum_{k = 1}^{m_{j-1}} \frac{1}{(z-\lambda_{j})^{k+1}} D_{j}^{k} \nonumber\]

along with the verification of a number of the properties laid out in Complex Integration Equations.