# Non-polynomial Z-transform

My prof said that when a transfer function described by a z-transform is not polynomial, then i can't perform the anti-transformation. But, what does it means to be not polynomial ? Can you explain to me ?

• If your prof's statement was "only polynomials have an inverse Z-transform" then this is clearly wrong. If I misunderstood what you (or he) meant then please clarify your question. – Matt L. Jan 12 at 15:14

First of all - I myself am not a pro in Control theory, but a mathematician - so I write what I think might be what your prof means.

Part where i am quite certain:

Not to be polynomial means, that there exists NO polynomial which describes the transfer function. E.g. let $$H$$ be your transfer function, then there exists no $$n\in\mathbb{N}$$ and $$p\in P^n[X]$$ such that $$H(s) = \sum_{i = 0}^{n}p_is^i$$ holds. Examples for functions which are not polynomial are: exponential functions (e.g $$e^s$$) or trigonometric functions (e.g. $$sin(s)$$), but there are many more.

The next part is where I am not 100% sure, because it is not my main subject.

I assume you mean the inverse Z-Transform when saying anti transform. I do not know too much about the z-transform, but i think it is the discrete-time equivalent to the laplace-transform and therefore:

Let $$H$$ be a transfer function. Then $$H$$ is the Laplace-transform of an impulse response. Thus the inverse Laplace transform (i.e. $$h$$) exists. (The stability however is something different.)

But if you want to check for a given "transfer function" (an arbitrary function in the "s"-Domain) if there exists an impulse response corresponding to this "transfer function", then I think you need the inverse Laplace transform to converge, i.e. $$\frac{1}{2\pi i} \lim_{y\rightarrow\infty}\int_{\gamma-iy}^{\gamma+iy}e^{st}H(s)\,ds$$ to exist for some $$\gamma\in\mathbb{R}$$. The integral then is your impulse answer. There exist non-polynomial functions for which this integral exists. Thus, there are non-polynomial transfer functions which correspond to a system. Many many authors in literature however assume only rational valued transfer functions... i really do not know why.

If this answer is NOT correct (because for the discrete-time case there is something mayor different), or does not help, please say so and apology. As I said, this is not my main subject.

• Hi Lucas! Rational transfer functions lead to differential equations (or difference equations in discrete time) which are very common and useful in practice. – GKH Jan 12 at 14:59
• yes, i know. I just wanted to point out, that there exist non-rational (therefore non-polynomial) transfer functions. But still many authors just ignore the non-rational transfer functions, as if they did not exist at all. – Lukas Jan 12 at 15:03
• That is true. Probably because they want to keep the content relevant to real systems (however, it depends on the book's audience). – GKH Jan 12 at 15:13
• I dont know if you mean "real" like "applicable" systems, or real like $\mathbb{R}$ in terms of systems. In my masters thesis I have certain systems corresponding to partial differential equations which yield transcendental transfer functions. And those PDEs in fact have a real meaning. – Lukas Jan 12 at 15:20
• Yes, I meant applicable, "real-life" systems. Textbooks on DSP or signals & systems tend to bother only with rational transfer functions since the latter have been proved to be almost everywhere in analog or digital systems. – GKH Jan 12 at 15:34

Assuming that $$H(z) = A\frac{\prod_k (1-c_kz^{-1})}{\prod_l (1-d_l z^{-1})}, \: \: R_H$$ you can perform Partial Fraction Expansion (PFE) to quickly get your impulse response $$h[n]$$ (what you probably call anti-transformation) using Z-transform properties and tables of Z-transform pairs.

If your transfer function is not rational, such as $$H(z) = \mathrm{ln}(1+az^{-1})$$ or $$H(z) = \cos(z)$$ then you can not perform PFE but there are still methods to find the impulse response $$h[n]$$. For example, the first transfer function has an impulse response of $$h[n] = \frac{(-1)^{n+1}}{n}a^n u[n-1]$$ for $$|z| > |a|$$ while the second one's impulse response is $$h[n] = \frac{(-1)^{-n/2}}{(-n)!}$$ for $$n < 0$$ and even.

The majority of practical and useful discrete time systems can be easily implemented as difference equations. Difference equations can always be described as rational transfer functions in the Z domain.

However, I am not sure what your professor meant by his statement. Maybe he wanted to say that you cannot find the inverse Z-transform using tables and properties...