# Taking the inverse $\mathcal Z$- transform with a summation in the denominator

I'm learning about z-transforms, and was going through some practice problems and I've been stuck on this one for a little bit. I'm trying to take the inverse z-transform of the following: $$\frac{1}{\sum_{k=0}^M c_kz^{-k}}$$

The only thing I can think of is that the denominator looks kind of like the definition of the z transform, but I don't think I'm going anywhere with that. Can anyone point me in the right direction?

The standard approach to deal with this is partial fraction expansion (PFE). Assume the numerator and denumerator are represented by $B(z)$ and $A(z)$, respectively. The objective of PFE is writing the expression as $$\frac{B(z)}{A(z)}=\frac{r_1}{1-p_1z^{-1}}+\frac{r_2}{1-p_2z^{-1}}+\cdots+\frac{r_N}{1-p_Nz^{-1}},$$ where $N$ is the degree of $A(z)$.
After this conversion, the inverse z-transform of each term is straightforward. A good beginning point would be finding the roots of $A(z)$ (the poles $p_1$, $p_2$,..) and writing $A(z)$ in form of $(1-p_1z^{-1})(1-p_2z^{-1})\cdots(1-p_Nz^{-1})$. Then you should find $r_i$ values.
• You can write a polynomial in form of products, when you know the roots. Example: $z^2-3z+2$ is written as $(z-1)(z-2)$. If you have a specific expression, add it to your question and I will show you the details. – msm Nov 7 '16 at 1:49
• The expression is actually the summation in the question. If you expand the summation you get $$\frac{1}{c_0z^{-0}+c_1z^{-1}...+c_Mz^{-M}}$$ and I have no idea what to do with that. – Jacob Parnacov Nov 7 '16 at 2:35