# On the Heaviside Cover-up Method

Engineering students take introductory courses in ordinary differential equations where they encounter a barrage of Laplace transforms they must invert. The formal inverse is a contour integral over the region of convergence of the transform in the $$s$$-domain but this is difficult to do efficiently. Since engineering courses primarily discuss finite-dimensional linear ordinary differential equations there is a specific, restricted class of transforms they observe. All the Laplace transforms an engineer will see take the form $F(s) = \frac{a_{n-1} s^{n-1} + \cdots + a_1 s + a_0}{s^n + b_{n-1} s^{n-1} + \cdots + b_1 s + b_0}$ where $$a_i, b_i \in \mathbb{R}.$$ The fundamental theorem of algebra allows us to rewrite the denominator as $F(s) = \frac{a_{n-1} s^{n-1} + \cdots + a_1 s + a_0}{\prod (s - r_i)^{\kappa_i}}$ where $$r_i \in \mathbb{C}$$ are the unique roots of the polynomial with multiplicity $$\kappa_i \in \mathbb{N}.$$ Observe that the right hand-side, by algebra, can be written as a finite sum of rational maps $F(s) = \sum_i \sum_{k = 1}^{\kappa_i} \frac{A_{i, k}}{(s - r_i)^k}$ It is then a matter of collapsing this summation into one fraction and matching the unknown coefficients $$A_{i, k}$$ with those that form the numerator polynomial $$a_{n-1},\ldots, a_0$$; this results in a large system of linear equations. Solving a linear system of equations is how a computer would do it. But how would a practicing engineer do it without a computer?

## Heaviside Cover-up Method

The Heaviside Cover-up Method is a method named after engineer Oliver Heaviside to discover the coefficients $$A_{i, k}$$ through direct evaluation. Fix a root $$r_i.$$ There are $$\kappa_i$$ coefficients $$A_{i, 1}$$ up to $$A_{i, \kappa_i}$$ associated to this root. Observe that if we multiply both sides of the partial fraction expansion by $$(s - r_i)^{\kappa_i}$$ we arrive at $(s - r_i)^{\kappa_i} F(s) = A_{i, \kappa_i} + \sum_{k=1}^{\kappa_i - 1} (s - r_i)^{\kappa_i - k} A_{i, k} + \sum_{j\neq i} \sum_{k = 1}^{\kappa_i} A_{i, k} \frac{(s - r_i)^{\kappa_i}}{(s - r_i)^k}.$ Now consider the limit of both sides of this expression as $$s\to r_i.$$ Necessarily the latter two summations must vanish, as they have zeros at $$r_i$$ but no poles (infinite discontinuities at $$r_i$$). This results in the rule $A_{i, \kappa_i} = \lim_{s \to r_i} (s - r_i)^{\kappa_i} F(s).$ In simpler terms, to produce the coefficient $$A_{i, \kappa_i}$$ evaluate $$F(s)$$ at $$r_i$$ but exclude the denominator factors of $$F(s)$$ that will blowup at $$r_i.$$ This is all pretty well explained in the Wikipedia article linked. However there is a piece that is left out of the Wikipedia article as it stands. How do you solve for all the other coefficients associated to the lower powers of $$r_i$$? Wikipedia gives an answer that puts me off, quite frankly:

Since the equation of the numerators... is true for all values of $$x,$$ pick a value for $$x$$ and use it to solve for [$$A_{i, \kappa_i - 1}$$].

This is obviously absurd. It works for second order roots, but what about if the root was repeated 5 times over? You will need to evaluate at 4 points! Moreover, it begs the question: why bother with heaviside cover-up and just use this point evaluation technique on the original $$F(s)$$ at points away from its discontinuities. That too determines the coefficients of the partial fraction expansion!

No. We will not take this approach. Because it is actually far simpler than this. Since we know the coefficient $$A_{i, \kappa_i}$$ we can define a new Laplace transform $G(s) = F(s) - \frac{A_{i, \kappa_i}}{(s - r_i)^{\kappa_i}}.$ Facts:

1. $$F(s)$$ has a partial fraction expansion with root $$r_i$$ appearing with order $$\kappa_i$$,

2. the latter term cancels this higher order term in the partial fraction expansion of $$F(s),$$

3. $$G(s)$$ is now expressed as the partial fraction expansion of $$F(s)$$ with root $$r_i$$ appearing at an order less than, or equal to, $$\kappa_i - 1,$$

4. $$G(s)$$ has a pole at $$r_i$$ with order at most, but possibly less than, $$\kappa_i - 1,$$

5. the coefficients of the partial fraction expansion of $$G(s)$$ are equal to that of $$F(s)$$ with the exception of the $$A_{i, \kappa_i}$$ that does not appear.

So now we can repeat the algorithm described above for $$G(s)$$ to solve for the lower order coefficients. In particular, $A_{i, \kappa_i - 1} = \lim_{s \to r_i} (s - r_i)^{\kappa_i - 1} G(s) = \lim_{s \to r_i} \frac{(s - r_i)^{\kappa_i} F(s) - A_{i, \kappa_i}}{s - r_i}.$ Repeat until all the coefficients associated to the root $$r_i$$ are resolved. And yes, since the other coefficients remain unmutated by this process of producing new transforms, you can use the reduced order transform to compute the other coefficients as well thereby simplifying the computations you perform at every step!

## Example

It is best to demonstrate by way of example. Consider the transform $F(s) = \frac{s - 3}{(s + 3)^3 (s - 1)}.$ We know we can write this function in expanded form as $F(s) = \frac{A_{1,1}}{s + 3} + \frac{A_{1,2}}{(s + 3)^2} + \frac{A_{1,3}}{(s + 3)^3} + \frac{A_{2,1}}{s - 1}.$ Coefficient $$A_{1, 3}$$ is, by our formula, $A_{1, 3} = \lim_{s \to -3} (s + 3)^3 F(s) = \lim_{s \to -3} \frac{s - 3}{s - 1} = \frac{3}{2}.$ Substituting this number into our partial fraction expansion $F(s) = \frac{A_{1,1}}{s + 3} + \frac{A_{1,2}}{(s + 3)^2} + \frac{3/2}{(s + 3)^3} + \frac{A_{2,1}}{s - 1}.$ Define $$G(s) = F(s) - (3/2)1/(s + 3)^3.$$ Simplify this to obtain an expression for $$G(s),$$ $G(s) = \frac{s - 3}{(s + 3)^3 (s - 1)} - \frac{(3/2)(s - 1)}{(s + 3)^3 (s - 1)} = \frac{-\frac{1}{2}s - \frac{3}{2}}{(s + 3)^3 (s - 1)} = -\frac{1}{2} \frac{1}{(s+3)^2 (s - 1)}.$ Moreover, from our partial fraction expansion for $$F(s)$$ we see that $$G(s)$$ must also take the form $G(s) = \frac{A_{1,1}}{s + 3} + \frac{A_{1,2}}{(s + 3)^2} + \frac{A_{2,1}}{s - 1},$ where the $$A_{i, k}$$ are the same variables as before. As a result, we can compute $$A_{1, 2}$$ using our formula again $A_{1, 2} = \lim_{s \to -3} (s + 3)^2 G(s) = \lim_{s \to -3} -\frac{1}{2}\frac{1}{s - 1} = \frac{1}{8}.$ So we have that $G(s) = \frac{A_{1,1}}{s + 3} + \frac{1/8}{(s + 3)^2} + \frac{A_{2,1}}{s - 1}.$ Once again, define a new function $$H(s) = G(s) - (1/8) 1/(s+3)^2$$ and simplify to arrive at the expression \begin{aligned} H(s) &= -\frac{1}{2} \frac{1}{(s+3)^2 (s - 1)} - \frac{1}{8} \frac{s - 1}{(s+3)^2 (s-1)}\\ &= -\frac{1}{2} \left[ \frac{\frac{1}{4} s + \frac{3}{4}}{(s+3)^2 (s-1)} \right]\\ &= -\frac{1}{8} \frac{1}{(s+3)(s-1)}. \end{aligned} The partial fraction expansion of $$H(s)$$ must take the form $H(s) = \frac{A_{1,1}}{s + 3} + \frac{A_{2,1}}{s - 1}.$ At this point it should be clear the advantages of employing this strategy. Of course these final coefficients can be found by employing the same formulas: $A_{1, 1} = \frac{1}{32},\quad A_{2,1} = -\frac{1}{32}.$ This gives the complete partial fraction expansion for $$F(s)$$ $F(s) = \frac{1/32}{s + 3} + \frac{1/8}{(s + 3)^2} + \frac{3/2}{(s + 3)^3} + \frac{-1/32}{s - 1}.$