Hello TomMUFC Originally Posted by

**TomMUFC** Hi all this relates to a Digital Signal Processing question, but it's the maths I can't understand when attempting to solve the quadratic equation.

This was originally part of a transfer function (denominator) and basically reads:

$\displaystyle 1-2rCos(\omega_cT)z^-1 + r^-2$ $\displaystyle \color{blue}\text{ Grandad says: should this be: }1-2r\cos(\omega_cT)z^-1 + r^2z^{-2}=0\text{ ?}$

Multiplying this line to positive values of z, to give the quadratic the NOW equation. I'm attempting to solve for r:

$\displaystyle z^2-2rCos(\omega_cT)z^1 +r^2$

a=1

b=$\displaystyle 2rcos(\omega_cT)$

c=$\displaystyle r^2$

But this where I fall on, and where I need help, because I cannot for the life of me solve with another unknown in it.

$\displaystyle (2rcos(\omega_cT) +/- [sqrt](-(2rcos(\omega_cT))^2-4r^2)$

all divided by 2

I can change the $\displaystyle b^2$ in the brackets to become = $\displaystyle 4r^2cos^2(\omega_cT)$ but then I've become snookered. Can anyone help? I can't go any further unfortunatley.

Apologies my Mathtags are terrible! (Can't do the quadratic properly).

Assuming the correction I've suggested, then you need to solve:$\displaystyle z^2-2r\cos(\omega_cT)z +r^2=0$

In which case, you're on the right track:$\displaystyle z = \frac{2r\cos(\omega_cT)\pm\sqrt{[-2r\cos(\omega_cT)]^2-4r^2}}{2}$$\displaystyle = \Big(\cos(\omega_cT)\pm\sqrt{\cos^2(\omega_cT)-1}\Big)r$

Now $\displaystyle \cos^2(\omega_cT)-1 = -\sin^2(\omega_cT)$

$\displaystyle \Rightarrow \sqrt{\cos^2(\omega_cT)-1} = i\sin(\omega_cT)$

So we get:$\displaystyle z = \Big(\cos(\omega_cT)\pm i\sin(\omega_cT)\Big)r$

If you know that you can take the positive sign, then this will reduce to:$\displaystyle z = re^{i\omega_cT}$

Does that make sense?

Grandad