Partial fraction expansion - handling multiple roots

Question

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.

Answer 1

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}

Answer 2

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?