# How to find zeros of a transfer function

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?

• If you solve for the poles of both cases, what do you get? Commented Dec 17, 2022 at 9:14
• $z = \frac{1}{2}$, $z = -3$ in both cases (?) Commented Dec 17, 2022 at 9:23
• 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? Commented Dec 17, 2022 at 9:31
• This is basically the same question, take a look. Commented Dec 17, 2022 at 11:44
• @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 $z$. Commented Dec 17, 2022 at 15:53

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.