First, some background on what I'm actually trying to achieve: I have a reflectance filter (a discrete IIR) for use in an FDTD boundary condition:


As you can see, both the numerator and denominator polynomials are univariate, of the same order, and their powers change by 1 per coefficient. However, I need to turn it into an impedance filter (also represented by an IIR of the same type), according to Kowalczyk 2008 it is done by:


So my plan was to simply divide $B_{(z)}$ by $A_{(z)}$ to get a single polynomial, and then adjust the first coefficient by $\pm1$.

Only my plan fell at the first hurdle because every polynomial division implementation I've found returns a quotient and remainder. Why are there no implementations that return non-integer (floating point in computer parlance) coefficients? Is it fundamentally impossible for a polynomial division result to be represented in any form other than quotient and remainder?

Or have I just gotten completely confused and the solution to my real problem is far simpler?

  • $\begingroup$ What are you trying to do? Express $\xi$ in terms of $A$ and $B$? $\endgroup$ – user37238 Jan 27 '16 at 11:47
  • $\begingroup$ @user37238 To put it as simply as possible, I have $R_{(z)}$ and I want $\xi_{(z)}$, but $\xi_{(z)}$ must be in the form of a polynomial ratio. I don't know how to get from one to the other. $\endgroup$ – cmannett85 Jan 27 '16 at 12:01

Yes it's fundamentally impossible, well at least if you require the quotient to be a polynomial.

Let's look at what (long) division would get you if you divide the polynomials $\phi$ and $\psi$. It's a expression on the form:

$$\phi(x) = q(x)\psi(x) + r(x)$$

this is the general form that the long division algorithm processes. $q$ is the quotient and $r$ is the remainder.

The problem is that we can't generally find a quotient $q$ such that $\phi(x) = q(x)\psi(x)$ (without remainder).

To see an example and why it fails we could assume that $\psi(x)$ is a polynomial of first degree, more concretely let $\psi(x) = x-a$, now we see that it's inevidable that $q(x)\psi(x)$ will have a zero at $x=a$, which means that if we found a quotient without a remainder the polynomial $\psi(x)$ would be required to have a zero for $x=a$ as well - that's not generally true.


What about$$\xi=\frac{1+R}{1-R}=\frac{1+\frac BA}{1-\frac BA}=\frac{A+B}{A-B}$$

So $$\xi = \dfrac{(a_0+b_0)+(a_1+b_1)z^{-1}+(a_2+b_2)z^{-2}} {(a_0-b_0)+(a_1-b_1)z^{-1}+(a_2-b_2)z^{-2}}$$


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