Why does this transfer function has a second zero

Question

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:

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

msm · Accepted Answer · 2016-11-12 13:25:08Z

3

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 $2z^2-3z$.

answered Nov 12, 2016 at 13:25

msm

4,3151 gold badge14 silver badges26 bronze badges

$\begingroup$ 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
Commented Nov 12, 2016 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
Commented Nov 12, 2016 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
Commented Nov 13, 2016 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
Commented Nov 13, 2016 at 21:55

Add a comment |

Maximilian Matthé · Accepted Answer · 2016-11-12 13:26:47Z

1

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.

answered Nov 12, 2016 at 12:56

6,2382 gold badges13 silver badges19 bronze badges

Add a comment |

2 Answers 2