# Partial fraction expansion - handling multiple roots

I am trying to solve the partial fraction expansion of the following:

$$X(z) = \dfrac{3-15z}{9z^2-6z+1}$$.

So I calculated the roots of $$9z^2 - 6z + 1$$.

This gives $$n_1 = n_2 = \frac{1}{3}$$ and $$9z^2-6z+1 = 9(z-\frac{1}{3})(z-\frac{1}{3})$$.

Now I don't know how to deal with it.

If there were no factor $$9$$, I would take the approach $$X(z) = \frac{A}{z-\frac{1}{3}} + \frac{B}{(z-\frac{1}{3})^2}$$.

And if there were two different roots, $$z_1$$ and $$z_2$$, I'd take the approach $$X(z) = \frac{A}{z-z_1} + \frac{B}{z-z_2}$$.

But since neither of these two cases is what we have here, I don't know what to do.

• Forget about the 9, then divide everything by it at the end? Sep 18, 2019 at 21:56
• or just use $\dfrac A {3z-1} + \dfrac B {(3z-1)^2}$ Sep 18, 2019 at 21:57
• Factor out the $9$! We write $\frac{3-15z}{9(z-1/3)(z-1/3)} = \frac{3-15z}{9}\cdot \frac{1}{(z-1/3)(z-1/3)}$. And you know how to write that last factor (it's exactly as you have written). Thus you would end up with $\frac{3-15z}{9} \left(\frac{A}{z-1/3} + \frac{B}{(z-1/3)^2}\right) = \frac{3-15z}{9z^2-6z+1}$ Sep 18, 2019 at 22:06
• Thanks to all of you! All your suggestions are helpful! :) Sep 19, 2019 at 12:43

You could absolutely do it your way: \begin{align} X(z)=\frac{3-15z}{9z^2-6z+1}&=\frac{3-15z}{9(z-\frac13)^2}\\ &=\frac19\bigg(\frac{3-15z}{(z-\frac13)^2}\bigg)\\ &=\frac19\bigg(\frac{A}{z-\frac13}+\frac{B}{(z-\frac13)^2}\bigg) \end{align}

Not impossible, but not very nice.

Alternatively: \begin{align} X(z)=\frac{3-15z}{9z^2-6z+1}&=\frac{3-15z}{(3z-1)^2}\\ &=\frac{A}{3z-1}+\frac{B}{(3z-1)^2}\\ &=\frac{A(3z-1)+B}{(3z-1)^2}\\ &=\frac{3Az-A+B}{(3z-1)^2}\Rightarrow \left\{\begin{array} & 3A &=-15\\ -A+B&=3\end{array} \right. \Rightarrow (A,B)=(-5,-2)\\ &=-\frac{5}{3z-1}-\frac{2}{(3z-1)^2} \end{align}

• Thank you for listing both options! So now I also know how to deal with the factor if it hasn't a square root in the natural numbers. This was very helpful. :) Sep 19, 2019 at 12:45

You could just use $$\dfrac A {3z-1} + \dfrac B {(3z-1)^2}=\dfrac{3-15z}{(3z-1)^2}$$

whence $$A(3z-1)+B=3-15z$$

so $$B-A=3$$ and $$3A=-15$$. Can you take it from here?

• Hi J. W. Tanner, thanks for your help! I didn't realize I could factor the 9. Now it's obvious. :) Sep 19, 2019 at 12:44