I'm learning about Z-transforms and transfer functions in DSP. Suppose I have the following transfer function:

  1. $H(z) = \frac{1}{z-0.5}$

I want to find the corresponding $h(n)$. If I factor out $z^{-1}$ from the numerator, I'll be creating a new zero and a pole in $z=0$.

  1. $H(z) = z^{-1}(\frac{z}{z-0.5})$

Which its inverse is

$h(n) = 0.5^{(n-1)}u(n-1) $

However, my assignment says it only has one pole in $z=0.5$, with which I do not fully agree since from (2) we keep the system equivalent in terms of stability.

Is it wrong to say that the TF in (2) is also valid as it has two poles and one zero?

Also, can I say that there can be an unlimited amount of zeros and poles from any transfer function in any other value $z=a$ if present in both numerator and denominator? Or is it properly considered a pole and a zero only after I cancel out the equivalent factors from the TF?

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    $\begingroup$ Why make it more complicated than it already is?$Z^{-1}(\frac{1}{z-0.5}) = e^{0.5n}u(n)$ $\endgroup$
    – Volpina
    Commented Apr 18, 2023 at 8:21
  • $\begingroup$ @Volpina Good point! But I was rooting for $Z^{-1}(z^{-100}\frac{z^{100}}{z-0.5} )= e^{0.5(n-100)}u(n-100)$, not to be more complicated but just to give us more time. But I guess that's a subterfuge since they are all still identically $e^{0.5n}u(n)$ $\endgroup$ Commented Apr 20, 2023 at 14:08

1 Answer 1


There is nothing wrong per say with adding a pole and zero in the same location, but what should be clear is it will have no effect on the resulting characteristics of that transfer function (such as the frequency response). It is just an equation not described in its simplest form And therefore not representative of systems of that order.

If the pole completely cancels the zero, as you have done- then we would say it doesn’t exist (when describing the system as a black box- but see RBJ’s good comment below of when we may want to consider the pole and zero before they annihilate each other). This is similar to me having one dollar to my name but borrowing a million dollars so that I have a million dollar asset and a million dollar liability: I still have one dollar to my name. A more mathematical analogy is asking if we would call this a 102nd order polynomial or not:

$$y = \frac{x^{102}+ x^{100}}{x^{100}}$$

The number of poles and zeros is given by the order of the polynomials. I suggest letting the forces of the poles and zeros do their battle- annihilating each other when they match, and then once the dust settles, the victors can hold there place in declaring the true number of occupants. In other words, attempt to put the transfer function in its minimum reduced form, and then the order as given will be consistent with what we would expect from a system of such order.

I will also point out that the system given by $H(z) = \frac{1}{z-.5}$ has one pole but it also has one zero: when $z= \infty$, $H(z)= 0$ and thus infinity is indeed a zero. With that all linear system transfer functions that can be described by a ratio of polynomials will always have the same number of poles and zeros. We should be explicit and say “finite poles” and “finite zeros” if we do not wish to acknowledge the infinites.

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    $\begingroup$ //"If the pole completely cancels the zero, as you have done- then we would say it doesn’t exist."// ----- I would agree if we're just discussing the input-output relationship with a black box and assume that the system is completely observable and completely controllable. But it's certainly possible to realize the pole and the zero separately (which increases the order of the system by 1) where they actually exist in the internal signal path separately. $\endgroup$ Commented Apr 17, 2023 at 2:55
  • $\begingroup$ @robertbristow-johnson yes very good point. And a subsystem could be of higher order where poles and zeros get cancelled due to its interconnection with other systems - we might hold back in that case from saying those poles and zeros don’t exist (unless we concern ourselves with the black box only as you say). Good point $\endgroup$ Commented Apr 17, 2023 at 3:04
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    $\begingroup$ When I took Advanced Control Theory in grad school 4 decades ago, we created and simulated (in DINAP, i think it was called) a system that was fine at the terminals of the black box, but with a state inside that was going to hell. And we couldn't really see it going to hell (at the output terminals) until numerical limitations forced the issue. $\endgroup$ Commented Apr 17, 2023 at 15:43
  • $\begingroup$ One does have to watch out when the pole/zero is unstable. $\endgroup$
    – fibonatic
    Commented Apr 18, 2023 at 10:38
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    $\begingroup$ @testing_22 thanks for your upvote and check! Looking forward to your future questions and answers, welcome to DSP.SE. $\endgroup$ Commented Apr 21, 2023 at 16:42

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