16.1 Partial Fractions

Partial fraction expansion simplifies fractions of the form

$$\frac{T(s)}{B(s)}$$

where $T(s)$ and $B(s)$ are polynomials in some variable $s$. This is of importance, for example, in integration of such fractions and in Laplace transformation. This section explains the procedure.

First, in case $T(s)$ has a degree equal or higher than $B(s)$, you should perform long division to take the ratio apart into powers of $s$ plus a remainder. That remainder is again a fraction of the form $T(s)/B(s)$, but the new $T(s)$ is of a degree less than $B(s)$.

Now the only thing you still have to know is now how to deal with a ratio $T(s)/B(s)$ when the degree of $T(s)$ is less than that of $B(s)$.

The first thing you will need to do is factor $B(s)$. Assume it is a polynomial of some degree $n$, in other words

$$B(s) = C (s^n + b_{n-1} s^{n-1} + \ldots b_2 s^2 +b_1 s + b_0)$$

where $C$ is some nonzero constant and the $b_i$ for $i$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n{-}1,\ldots,2,1,0$ are $n$ additional constants. You can simply take $C$ out of the entire ratio $T(s)/B(s)$ before proceeding. From now on, it will be assumed that you have done that, so that there is no longer a $C$ in the above expression to worry about. It is then known from complex variable theory that $B(s)$ can always be written in the form

$$B(s) = (s-s_1) (s-s_2) (s-s_3) \ldots (s-s_n)$$

Here the $n$ constants $s_i$, ($i$ $\vphantom0\raisebox{1.5pt}{$=$}$ $1,2,\dots,n$), are the locations, or roots, where $B(s)$ is zero. So $B(s_i)$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 for $i$ $\vphantom0\raisebox{1.5pt}{$=$}$ $1,2,\dots,n$. Note that some of these roots may coincide. In other words some of the $s_i$ may be equal to each other. For example

$$s^3 + s^2 - 5 s + 3 = (s-1)(s-1)(s+3)$$

has $s_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $s_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 and $s_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-3$.

There is a further complication. Some of the roots $s_i$ may be complex. For example, a quadratic with a negative discriminant has complex roots instead of real ones. However, assuming that the polynomial $B(s)$ is real, (for real $s$), you would probably not want to deal with complex numbers. You can avoid that because for real $B(s)$, the complex roots come in “complex-conjugate pairs.” That means that for every complex root $s_i$ there is a second root $s_j$ so that $(s-s_i)(s-s_j)$ is a real quadratic with roots $s_i$ and $s_j$. So a real $B(s)$ can always be written as a product of real factors linear in s and real factors quadratic in s.

For example, you might have a $B(s)$ that can be written as

$$B(s)=(s-4)(s-3)^3(s^2-2s+5)(s^2+4s+13)^2$$

That would be a polynomial of degree 10, ($1+3+2+4$ $\vphantom0\raisebox{1.5pt}{$=$}$ 10). For this $B(s)$, any desired ratio $T(s)/B(s)$ can be written as

$$\begin{eqnarray*} \frac{T(s)}{B(s)} & = & \frac{C_1}{s-4} + \\ && \frac{C_2... ... \frac{C_7s+C_8}{s^2+4s+13} + \frac{C_9s+C_{10}}{(s^2+4s+13)^2} \end{eqnarray*}$$

The right hand side is the partial fraction expansion of the ratio.

Note that the individual terms in the partial fraction expansion are much simpler than the original ratio. The original ratio has a polynomial of degree 10 in the bottom. And a polynomial of a degree up to 9 in the top. So if you want to do anything with the given ratio, it will become much easier if you use the right hand side above to do it. That is why you want to do partial fraction expansions.

That leaves two key questions to be answered:

1. In general, how do you know what terms there are in the right hand side?
2. How do you find the values of all these coefficients $C_1, C_2, C_3, \ldots$?

The answer to the first question is as follows;

1. For every factor $(s-s_i)$ that appears $k$ times in $B(s)$, there are terms in the right hand side of the form
   
   $$\frac{C_{i,1}}{s-s_i} + \frac{C_{i,2}}{(s-s_i)^2} + \ldots + \frac{C_{i,k}}{(s-s_i)^k}$$
   
   
   Check it out for the factors $(s-4)$ (single) and $(s-3)$ (triple) in the example given earlier. (What you want to call the constants is of course up to you, as long as each has a unique name.)
2. For every factor $(s^2+a_is+b_i)$ that appears $k$ times in $B(s)$, there are terms in the right hand side of the form
   
   $$\frac{C_{i,1}s+C_{i,2}}{s^2+a_is+b_i} + \frac{C_{i,3}s+C... ...} + \ldots + \frac{C_{i,2k-1}s+C_{i,2k}}{(s^2+a_is+b_i)^k}$$
   
   
   Check it out for the factors $(s^2-2s+5)$ (single) and $(s^2+4s+13)$ (double) in the example above.

The second question was how to find all these constants. The method you must use in this class is to crunch it out:

1. Bring all the terms in the partial fraction expansion over the common denominator $B(s)$.
2. Multiply out the top. This must equal the given $T(s)$. So the net coefficient of each power of $s$ must match the corresponding coefficient in $T(s)$. That gives you your equations for your unknown coefficients. Use Gaussian elimination to solve them.

For the example, its partial fraction expansion becomes, when brough over the common denominator $B(s)$:

$$\frac{\begin{array}{l} C_1(s-3)^3(s^2-2s+5)(s^2+4s+13)^2... ...s^2-2s+5) \end{array}}{(s-4)(s-3)^3(s^2-2s+5)(s^2+4s+13)^2}$$

Multiply out the top, (don't make any mistakes, of course), then equate the net coefficients of the $s^0$, $s$, $s^2$, ..., $s^8$ and $s^9$ powers to the corresponding coefficients in the given $T(s)$. That gives 10 equations for the 10 unknowns $C_1$, $C_2$, ... $C_{10}$. Solve using Gaussian elimination. (Cramer's rule is not recommended.) That will be fun!

Note: Of course, if $B(s)$ is not real, there is no point in using quadratics. Just expand in linear factors of the form $s-s_i$ (with $s_i$ now complex) only.

Note: there are more intelligent ways of finding the coefficients than crunching it out as we must do in this class. For example, consider once more the example:

$$\begin{eqnarray*} \frac{T(s)}{B(s)} & = & \frac{C_1}{s-4} + \\ && \frac{C_2... ... \frac{C_7s+C_8}{s^2+4s+13} + \frac{C_9s+C_{10}}{(s^2+4s+13)^2} \end{eqnarray*}$$

If you multiply the expansion in the right hand side by the single factor $(s-4)$ and then evaluate it at $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 4, you get $C_1$. (All the other terms are zero at $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 4 because of the multiplication by $(s-4)$.) The bottom line is therefore that if you multiply the left hand side by $(s-4)$ and then evaluate it at $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 4, you get $C_1$ too. And that is doable because the left hand side is given. (To evaluate at $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 4, you either need to divide out the common factor $(s-4)$ from top and bottom or use l’Hopital once. Dividing out the common factor from top and bottom is a simple matter of a long division of the bottom if you have it in unfactored form.) To find $C_4$, multiply by $(s-3)^3$ and evaluate at $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3. To find $C_3$, multiply by $(s-3)^3$, differentiate the result once, and evaluate at $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3. Etcetera. Especially if you want just a single coefficient, this can be much more convenient. Or you can use it to test the correctness of a few sample coefficients.