0
$\begingroup$

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 ?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Matt L. Jan 12 at 15:14
1
$\begingroup$

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.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Hi Lucas! Rational transfer functions lead to differential equations (or difference equations in discrete time) which are very common and useful in practice. $\endgroup$ – GKH Jan 12 at 14:59
  • $\begingroup$ 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. $\endgroup$ – Lukas Jan 12 at 15:03
  • $\begingroup$ That is true. Probably because they want to keep the content relevant to real systems (however, it depends on the book's audience). $\endgroup$ – GKH Jan 12 at 15:13
  • $\begingroup$ 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. $\endgroup$ – Lukas Jan 12 at 15:20
  • $\begingroup$ 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. $\endgroup$ – GKH Jan 12 at 15:34
1
$\begingroup$

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...

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.