4
$\begingroup$

I'm learning about $\mathcal Z$-transforms in DSP and I have a transfer function of the following form:

$$H(z)=\frac{2-3z^{-1}}{1-1.6z^{-1}+0.8z^{-2}}$$

When I calculate zeros and poles of this system by hand, I get these poles from the equation

\begin{align} \frac{z^2-1.6z+0.8}{z^2} &= 0\\ p_1&=0.8-0.4i\\ p_2&=0.8+0.4i \end{align}

And a single zero from the equation

\begin{align} \frac{2z-3}{z}&=0\\ z_1&=1.5+0i \end{align}

However, when I use Wolfram Alpha to compute the poles and zeros, it also lists a second zero positioned at origin. Or in MATLAB:

Plots and zeros on z-plane

The question is, where does that second zero come from?

$\endgroup$
3
$\begingroup$

Maybe this way it is simpler to understand:

$$1-1.6z^{-1}+0.8z^{-2}=\frac{z^2-1.6z+0.8}{z^2}$$

Hence,

$$\frac{2-3z^{-1}}{1-1.6z^{-1}+0.8z^{-2}}=\frac{2-3z^{-1}}{\frac{z^2-1.6z+0.8}{z^2}}=\frac{z^2(2-3z^{-1})}{z^2-1.6z+0.8}$$

whose numerator is $2z^2-3z$.

$\endgroup$
  • $\begingroup$ Looks good, however I still can't wrap my head around that the original H(z) cannot be evaluated at z=0. So can z=0 be really considered a solution? $\endgroup$ – feek Nov 12 '16 at 16:27
  • $\begingroup$ By solution, do you mean a root of numerator? This is still the same as the original $H(z)$, but in a more tractable algebraic form. $\endgroup$ – msm Nov 12 '16 at 21:01
  • $\begingroup$ I mean that when I do H(0), for the original you get NaN (division by zero), but for the last tractable algebraic form it is equal to zero. So the expressions are the same for all numbers, with the exception of z=0, right? $\endgroup$ – feek Nov 13 '16 at 13:00
  • $\begingroup$ That happens when you evaluate the original expression on a computer, since the order of operations is first (1/z), then other operations. Hence, division by zero occurs right at the beginning (no matter what your expression is). $\endgroup$ – msm Nov 13 '16 at 21:55
1
$\begingroup$

You cannot multiply nominator and denominator by a different power of z. Instead, you have two treat nominator and demoninator equally.

$$ H(z) = \frac{2-3z^{-1}}{1-1.6z^{-1}-0.8z^{-2}}=\frac{z(2z-3)}{z^2-1.6z-0.8} $$

Now, you can see that the nominator has 2 zeros, the denominator also.

$\endgroup$

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.