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I realized someone asked a question about the exact same section of a book, which can be found here: Question about zeroes of simple 2nd order FIR filter.

While the answer to this question looks useful to me I have a question about the what the author did to find the zeros. Particularly this part:

$0=1+\alpha_1z^{-1}+\alpha_2z^{-2}$

We know that the zeroes need to be complex conjugates, so can change it to this:

$0=(1-Z_1z^{-1})(1-Z_2z^{-1})$

where $Z_1=Re^{j\theta}$ and $Z_2=Re^{-j\theta}$

The author says that $Z_1$ and $Z_2$ are the zeros, but I don't understand how this can be. Wouldn't $Z_1$ or $Z_2$ have to $=z$ for this to be true?

Steve M
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Below is a polynomial in $z$ :

$$ P(z) = 1+\alpha_1z^{-1}+\alpha_2z^{-2} $$

Zeros of this polynomial is also known as its roots; i.e., those $z_r$ for which $P(z_r)=0$.

Solving for the roots of a general $n$-th order polynomial is not possible. They are numerically found.

However, one alternative way to define a polynomial is to write it as the product of factors of the form $1 - d_k z^{-1}$, where each $d_k$ is a zero/root of $P(z)$; i.e., $(1 - d_k d_k^{-1}) = 1-1 = 0$.

Then your second order polynomial $P(z)$ can be written as $P(z) = (1-Z_1z^{-1})(1-Z_2z^{-1})$

where the complex valued $Z_1=Re^{j\theta}$ and $Z_2=Re^{-j\theta}$ are the roots of the polynomial. You should solve the second order equation $P(z)=0$ to find out what those zeros are.

Fat32
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  • if $Z_1z^{-1}$ =1 and $z=e^{j\theta}$ wouldn't $R=1$ always? – Steve M Oct 03 '20 at 19:43
  • if $Z_1z^{-1}=1$ then $z=Z_1$, whatever that $Z_1$ is. Note, a second order polynomial with real coefficients can have two real (distinct) roots, double real roots, or a pair of complex conjugate roots. You have to solve it to see which is the case. $R=1$ means the zeros are on the unit-circle ; no they don't have to be there. But if it's said that zeros are on the unit cirle then you can set $R=1$. – Fat32 Oct 03 '20 at 19:48
  • Ok, I think where I was getting confused is the book stated that the z substitution was $z=e^{j\theta}$, but really a function of z is over the whole imaginary plane and not just a function of the angle (where the magnitude always =1). Does that sound right? – Steve M Oct 04 '20 at 02:18