# Q <= 0.5 with passive filters

#### Fredje

Joined Jun 1, 2022
15
A disadvantage of using a passive filter is that Q cant become higher than 0.5. But why it can't go over 0.5 and why would it be a bad thing?

#### ericgibbs

Joined Jan 29, 2010
16,393
Hi Fredje,
Is this a college or homework assignment.?
E

#### Fredje

Joined Jun 1, 2022
15
Hi Fredje,
Is this a college or homework assignment.?
E
Hello, I am currently learning for my exam, in the textbook it says that this is a disadvantage of a passive filter and I am just curious why this would be. It's not that this is a task or anything

#### Papabravo

Joined Feb 24, 2006
19,309
I don't think this is exclusively a problem with passive filters. It has to do with the nature of the polynomials that make up a transfer function. The polynomial in the denominator has a form which involves a natural frequency denoted by
$$\omega_0$$
And the parameter
$$Q$$
When you try various forms for that polynomial you can understand why that is the case.
For any odd order filter, we have a pole on the negative real axis, at an angle ψ = 0. The Q of that pole is:
$$Q\;=\;\cfrac{1}{2cos(\psi)}\;=\;0.5$$
For any even order filter, the poles occur in complex conjugate pairs and have a Q that uses the same formula as the first order pole. So for example in the case of a 3rd order Butterworth filter the same formula would give:
$$Q\;=\;\cfrac{1}{2cos(45^\circ)}\;=\;0.707$$

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#### Fredje

Joined Jun 1, 2022
15
It has to do with the nature of the polynomials that make up a transfer function. The polynomial in the denominator has a form which involves a natural frequency denoted by
$$\omega_0$$
And the parameter
$$Q$$
When you try various forms for that polynomial you can understand why that is the case.
So it would have to do something about the stability of the system? When Q becomes higher than 0.5 the roots of the denominator becomes positive?
and with an active second order filter you can get higher values of Q because $$\omega_0$$ can be higher or something like that

#### Papabravo

Joined Feb 24, 2006
19,309
So it would have to do something about the stability of the system? When Q becomes higher than 0.5 the roots of the denominator becomes positive?
and with an active second order filter you can get higher values of Q because $$\omega_0$$ can be higher or something like that
I amended my answer in post #4. The type of filter, active or passive, has nothing to do with the Q. Q is related to the angle the pole makes with the negative real axis, and is denoted by the Greek letter ψ. A real pole can only have a Q of 0.5, nothing more and nothing less. Poles at a small angle with respect to the negative real axis will have a Q of 0.5 just like the pole on the negative real axis. Poles at large angles to the negative real axis will have a Q that can be quite large.

Do the Gedankenexpriment of placing a complex conjugate pair of poles on the unit circle in the left half-plane. Then compute the Q, by the formula given above for ψ ∈ {1°, 45°, 89°} and see what answers you come up with.

As for the question in the thread title, there is NFW for a filter of any type to have a Q < 0.5. At least not in this universe.

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#### Fredje

Joined Jun 1, 2022
15
I amended my answer in post #4. The type of filter, active or passive, has nothing to do with the Q. Q is related to the angle the pole makes with the negative real axis, and is denoted by the Greek letter ψ. A real pole can only have a Q of 0.5, nothing more and nothing less. Poles at a small angle with respect to the negative real axis will have a Q of 0.5 just like the pole on the negative real axis. Poles at large angles to the negative real axis will have a Q that can be quite large.

Do the Gedankenexpriment of placing a complex conjugate pair of poles on the unit circle in the left half-plane. Then compute the Q, by the formula given above for ψ ∈ {1°, 45°, 89°} and see what answers you come up with.

As for the question in the thread title, there is NFW for a filter of any type to have a Q < 0.5. At least not in this universe.
If I understand this correctly, Q is related to the angle of the pole, so when Q becomes higher the poles will change position and so the response of the system will change aswell. Kind off the same as ζ in control systems if you know where I am talking about?

#### Papabravo

Joined Feb 24, 2006
19,309
If I understand this correctly, Q is related to the angle of the pole, so when Q becomes higher the poles will change position and so the response of the system will change aswell. Kind off the same as ζ in control systems if you know where I am talking about?
If you look at 2nd order polynomials that are the characteristic equations of 2nd order differential equations you can see an obvious to way to show how ζ and Q are related. the same equations represent identical physics.

$$2\zeta\;=\;\cfrac{1}{Q}$$