# Filters and Oscillation

#### jp1390

Joined Aug 22, 2011
45
Hey, I'm wondering if anyone can set me in the correct direction to interpreting this concept. Let's say you have a second-order low-pass or high-pass filter with a reasonable Q factor. I am curious as to why the relaxation oscillation frequency 'w0' does not provide the peak value in magnitude, but rather that frequency is equal to: (low-pass)

$$w_{max} = w_{0}\sqrt{1 - \frac{1}{2Q^{2}}}$$ Slightly less than w0 for Q > 0.5.

I know that this is what the math works about to be for these filters, but in terms of physically what is going on, wouldn't it make more sense that the relaxation oscillation frequency would have the highest peak?

If you injected a signal at the same frequency as w0 (in phase), wouldn't it resonate much more effectively than if the injected frequency was at this so called 'wmax'?

Thanks!

#### jp1390

Joined Aug 22, 2011
45
bump to the top! #### t_n_k

Joined Mar 6, 2009
5,455
As an example consider the series LCR circuit configured as a LPF. That is to say, the output response is measured as the capacitor voltage when the driving input voltage is applied across the combined series LCR network. The R value can be varied to set the LPF Q.

Interestingly, as R varies, the peak series current always occurs when the driving frequency is ω0=1/√LC. That is, current resonance occurs at ω0. This must be true since the peak current occurs when the capacitive and inductive reactance cancel each other. However as you note in your post, the output peak voltage occurs at the 'damped' relaxation frequency which varies as the value of R [Q] varies.

Why this is so is hard to pin down without resorting to a mathematical 'description'. One might simply comment that this is a fundamental physical property of driven damped oscillatory systems. For the case in question, I suspect it may have something to do with the temporal energy distribution (storage and dissipation) in the circuit.

• jp1390