# 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?

## 1 Answer

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.

More details can be found in an example here.

• Sorry, I don't understand how I can use this for this problem though. For partial fraction expansion, doesn't the denominator need to be a product, not a summation? – Jacob Parnacov Nov 7 '16 at 0:04
• 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