# Poles of 2nd order passive filter

Thread Starter

#### ntetlow

Joined Jul 12, 2019
44
I've been trying to calculate the two poles of the passive low pass filter with 2 resistors and 2 capacitors.
The answers would seem to be the roots of the quadratic equation formed by the denominator of the equation
G(s)= 395553973339.66 /
s2+1264150.9433962s+395553973339.66
The answers given are p1= -90575.686508642[Hz] and p2 = -110620.18494715[Hz]

However, my solution is different from the above. Can anyone tell me where I am going wrong?

#### SamR

Joined Mar 19, 2019
3,484
fc1=1/(2π√R1R2C1C2) fc2=1/(2π√R3R4C3C4) fr=√fc1*fc2 for bandpass, for just the HI/LO filter only use the first equation.

#### LvW

Joined Jun 13, 2013
1,299
ntetlow - tell us the values of the parts.
Are you aware that the roots of the equation are angular frequencies in 1/rad (and not f in Hz)?

Thread Starter

#### ntetlow

Joined Jul 12, 2019
44
ntetlow - tell us the values of the parts.
Are you aware that the roots of the equation are angular frequencies in 1/rad (and not f in Hz)?
No I wasn't aware of that, thank you. The two resistors are 1k and 100k. The two caps are 1.59n and 15.9p.

#### LvW

Joined Jun 13, 2013
1,299
For the combination 1k/1.59n and 100k/15.9p (without a buffer between both RC combinations the two real poles are:
p1=-6.95 E5 1/rad
p2=-5.691 E5 1/rad.

Thread Starter

#### ntetlow

Joined Jul 12, 2019
44
For the combination 1k/1.59n and 100k/15.9p (without a buffer between both RC combinations the two real poles are:
p1=-6.95 E5 1/rad
p2=-5.691 E5 1/rad.
Yes, thank you. It is because I hadn't converted the rps to Hz.

#### Papabravo

Joined Feb 24, 2006
16,775
Something is bothering me. You have what you claim is a second order system, but the roots are not complex conjugate pairs. How can this be? What you have is cascaded 1st order sections. If the roots of the quadratic equation are not complex conjugate pairs then they must lie along the negative real axis and they don't correspond to any frequency at all. They are just decaying exponentials.
Expressed another way:
$e^{-x+j\omega 0}=e^{-x}e^0=e^{-x}\cdot 1= e^{-x}$

There are literally no frequencies involved period full stop.

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