# Need Help Solving This Really Difficult Quadratic Equation!

• Mar 24th 2010, 10:57 AM
TomMUFC
Need Help Solving This Really Difficult Quadratic Equation!
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:

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

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

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

a=1
b= $2rcos(\omega_cT)$
c= $r^2$

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.

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

all divided by 2

I can change the $b^2$ in the brackets to become = $4r^2cos^2(\omega_cT)$ 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).
• Mar 24th 2010, 11:49 AM
Hello TomMUFC
Quote:

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:

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

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

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

a=1
b= $2rcos(\omega_cT)$
c= $r^2$

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.

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

all divided by 2

I can change the $b^2$ in the brackets to become = $4r^2cos^2(\omega_cT)$ 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:
$z^2-2r\cos(\omega_cT)z +r^2=0$
In which case, you're on the right track:
$z = \frac{2r\cos(\omega_cT)\pm\sqrt{[-2r\cos(\omega_cT)]^2-4r^2}}{2}$
$= \Big(\cos(\omega_cT)\pm\sqrt{\cos^2(\omega_cT)-1}\Big)r$
Now $\cos^2(\omega_cT)-1 = -\sin^2(\omega_cT)$

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

So we get:
$z = \Big(\cos(\omega_cT)\pm i\sin(\omega_cT)\Big)r$
If you know that you can take the positive sign, then this will reduce to:
$z = re^{i\omega_cT}$
Does that make sense?

• Mar 24th 2010, 12:31 PM
TomMUFC
Yes you are correct. That is fantastic Grandad.

Now see this, I don't know what you know about transfer functions (regarding DSP) but the original question was to "find the range of (real) values of r for which the filter is stable?"

Now a filter is stable if the poles all lie within the "unit circle" of the plane (i.e. <1).

To obtain pole positions you would need to obtain the roots of the demominator (which I gave). Normally answers come in a complex format (real/imaginary) I think so for the filter to be stable the complex numbers need to be less than 1. But now I'm confused. You don't happen to know anything regarding this do you?
• Mar 24th 2010, 01:07 PM
Hello TomMUFC
Quote:

Originally Posted by TomMUFC
Yes you are correct. That is fantastic Grandad.

Now see this, I don't know what you know about transfer functions (regarding DSP) but the original question was to "find the range of (real) values of r for which the filter is stable?"

Now a filter is stable if the poles all lie within the "unit circle" of the plane (i.e. <1).

To obtain pole positions you would need to obtain the roots of the demominator (which I gave). Normally answers come in a complex format (real/imaginary) I think so for the filter to be stable the complex numbers need to be less than 1. But now I'm confused. You don't happen to know anything regarding this do you?

I'm afraid you're on your own as far as the electronics is concerned. (I took A level Physics in 1961!) But $z = re^{i\omega_cT}\Rightarrow |z| = r$, so if you are sure that $|z| < 1$, then $-1 < r < 1$. That's the best I can do.