# Improper to proper rational functions - does the order of division matter?

• Nov 11th 2012, 11:29 AM
Lancet
Improper to proper rational functions - does the order of division matter?
My question has to do with changing improper rational functions to the sum of a polynomial and proper rational function. If it matters, the context of this is inverse Z-transforms.

My textbook says to do this by doing polynomial long division with the polynomials written in **reverse** order. But I have assignments where the solutions do the long division in the original order, and if I tried doing it in reverse order, I get an answer with different signs that I can't easily factor.

Textbook example (reverse order):

$X(z)= \frac{1 + 3z^{-1} + \frac{11}{6} z^{-2} + \frac{1}{3} z^{-3}}{1 + \frac{5}{6} z^{-1} + \frac{1}{6} z^{-2}}$

$X(z) = 1 + 2 z^{-1} + \frac{\frac{1}{6} z^{-1}}{1 + \frac{5}{6} z^{-1} + \frac{1}{6} z^{-2}}$

Assignment example (normal order):

$H(z) = \frac{1 + z^{-1} + 6 z^{-2}}{1 - z^{-1} - 6 z^{-2}}$

$H(z) = -1 + \frac{2}{1 - z^{-1} - 6 z^{-2}} = -1 + \frac{2}{(1+2z^{-1})(1-3z^{-1})}$

Now, if I do the assignment example using the reverse order, I get:

$H(z) = \frac{1 + z^{-1} + 6 z^{-2}}{1 - z^{-1} - 6 z^{-2}}$

$H(z) = -1 + \frac{2}{1 + z^{-1} + 6 z^{-2}}$

...which I can't factor out the denominator in the nice way it was done in the normal order example.

So what's the deal here? Does the order I do the division not matter, and I should just choose the one that gives me the easier answer to work with? Or is one right and the other wrong?
• Nov 11th 2012, 12:58 PM
a tutor
Re: Improper to proper rational functions - does the order of division matter?
Quote:

Originally Posted by Lancet
Now, if I do the assignment example using the reverse order, I get:

$H(z) = \frac{1 + z^{-1} + 6 z^{-2}}{1 - z^{-1} - 6 z^{-2}}$

$H(z) = -1 + \frac{2}{1 + z^{-1} + 6 z^{-2}}$

...which I can't factor out the denominator in the nice way it was done in the normal order example.

$H(z) = \frac{1 + z^{-1} + 6 z^{-2}}{1 - z^{-1} - 6 z^{-2}}$

$= \frac{2 - 1 + z^{-1} + 6 z^{-2}}{1 - z^{-1} - 6 z^{-2}}$

$= \frac{2}{1 - z^{-1} - 6 z^{-2}} -1$
• Nov 11th 2012, 04:17 PM
Lancet
Re: Improper to proper rational functions - does the order of division matter?
So are you saying that the textbook is incorrect? Or is there a reason it says to do the division in the reverse order?
• Nov 11th 2012, 11:54 PM
a tutor
Re: Improper to proper rational functions - does the order of division matter?
No I'm not saying your textbook is incorrect.

I just avoided long division altogether which I think is a good idea for that particular example.

On reflection I think my post was unhelpful.

When you do need to use long division you can think of your variable as 1/z and write out your problems starting with the highest power of 1/z. Is this what you mean by reverse order?
• Nov 12th 2012, 03:43 AM
Lancet
Re: Improper to proper rational functions - does the order of division matter?
No. The book means instead of dividing the numerator by the denominator, divide the denominator by the numerator.
• Nov 12th 2012, 11:52 PM
hollywood
Re: Improper to proper rational functions - does the order of division matter?
I think you might have misunderstood the textbook example, because the denominator of the proper fraction in the result is the same as the original denominator. When you applied the method, that didn't happen.

- Hollywood