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 anythingHi Fredje,
Is this a college or homework assignment.?
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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?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.
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.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
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?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 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.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?
by Jake Hertz
by Jake Hertz
by Duane Benson