Given the following transfer function,

$$H(z) = \frac{6 + 4z^{-1}}{2 + 5z^{-1} - 3z^{-2}}$$

How do we find the zeros of the transfer function? We can write the above expression as

$$\frac{3(1+\frac{2}{3}z^{-1})}{(1-\frac{1}{2}z^{-1})(1 + 3z^{-1})}$$

As per the above expression, the system has one zero at $z = -\frac{2}{3}$. Alternatively, $H(z)$ can be written as

$$H(z) = \frac{z(6z + 4)}{2z^{2} + 5z - 3}$$

In this case, there is an additional zero at $z = 0$. Can anyone explain what is going on here and which of the two solutions is correct?

  • $\begingroup$ If you solve for the poles of both cases, what do you get? $\endgroup$ Dec 17, 2022 at 9:14
  • $\begingroup$ $z = \frac{1}{2}$, $z = -3$ in both cases (?) $\endgroup$
    – MaxFrost
    Dec 17, 2022 at 9:23
  • $\begingroup$ No, in the first case it's 2 and -1/3, and in the other it's 1/2 and -3. Can you spot the pattern? $\endgroup$ Dec 17, 2022 at 9:31
  • $\begingroup$ This is basically the same question, take a look. $\endgroup$
    – Matt L.
    Dec 17, 2022 at 11:44
  • $\begingroup$ @aconcernedcitizen: the OP is correctly finding the zeros in z in both cases. Just because you express something as a polynomial in $-^{-1}$ that doesn't you can't solve it for roots in $z$. $\endgroup$
    – TimWescott
    Dec 17, 2022 at 15:53

2 Answers 2


Your first solution is just incomplete because you apparently find it harder to see the zero at $z=0$ if the transfer function is written in powers of $z^{-1}$. But with a bit of practice you could see that as $z\to 0$, the $z^{-2}$ term in the denominator goes to infinity faster than the term with $z^{-1}$ in the numerator, making the whole expression approach zero for $z\to 0$.

Also note that due to the term $z^{-2}$ in the denominator, you have a second order transfer function, which must have two poles as well as two zeros. If only one of them is obvious to you, make sure you look at the cases $z\to 0$ and $z\to\infty$.


If you look at the part $f(z)=-3z^{-2}$ from your first formula and if that equals to $f(z)=\frac{-3}{z²}$ you got division by zero there for $z=0$. When you move z up from denominator to numerator in your last formula, where you changed it from $\frac{1}{z^{-x} }$ to $z^x$, your division by zero turns into an additional zero of your formula.

So your first formula is discontinuous at that point. I plotted your top and bottom functions and both are the same. Since both are equal, both are correct. Your graph plotter just might fool you about the division by zero.


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