Why does the \$ Q\$-factor depend on the resonant frequency?

Question

The \$Q\$-factor is defined as $$Q=\frac{\omega_{0}}{\Delta \omega},$$

where \$ \Delta \omega\$ is the bandwidth around the resonant frequency \$ \omega _{0}\$. For a series RLC circuit, $$\Delta \omega = \frac{R}{L}$$

and $$\omega_{0}=\frac{1}{\sqrt{LC}}.$$

\$Q\$ is supposed to measure how sharp the frequency response is, so a higher \$Q\$ means a smaller \$\Delta \omega \$. We can clearly see above that for a series RLC circuit, the \$Q\$-factor can be increased just by increasing \$\omega_{0}\$, alone, by decreasing the value of \$C\$, without even changing how narrow the response curve is, i.e. without even changing \$\Delta \omega \$.

Why does a higher resonant frequency imply a higher \$Q\$-factor, which is supposed to measure how sharp the frequency response of our resonant circuit is? In other words, why should the resonant frequency have anything to do with the the \$Q\$-factor?

Answer 1

So why a higher resonant frequency implies a higher Q-factor, which is supposed to measure how sharp the frequency response of our resonant circuit is?

It measures how sharp the resonance is, relative to the resonance frequency.

This might make resonant circuit topologies or designs easier to compare for certain purposes if they're designed at different resonant frequencies.

If you want a measure of how sharp the resonance is that isn't dependent on the center frequency, you can always just use \$\Delta\omega\$.

Answer 2

The definition that makes more sense to me, is via the relative power loss per period:

$$Q = 2 \pi \frac{E}{|\Delta E|}$$

For a driven damped harmonic oscillator this boils down to $$Q\approx\frac{\omega_0}{\Delta \omega}$$

for small damping. See also the definitions in wikipedia.

Also, to me the second equation was not intuitive but seeing it comes from this more general definition it makes more sense.

Also with that: increasing frequency gives shorter periods, hence less is lost per cycle (although it could be that more is lost per time, I guess.)

Answer 3

The Q relationship makes spectrum and filter shapes a constant on a log scale.

Q applies to audio filters. And human perception of audio bandwidth (perceivable pitch differences, bass and treble bands, etc.) is more similar to a log scale than a linear scale. Constant ratio rather than a constant delta Hertz.

Same with the physical component tolerances required to create a musical instrument of RF circuit with a given Q.

Answer 4

Why does a higher resonant frequency imply a higher Q-factor

It only does so in the particular scenario you're citing. If you were to vary the resonant frequency of your circuit by varying the inductance, then the Q factor would go down with increasing resonant frequency.

In practical RF circuits the Q-factor is a function of everything. I'm used to lumped-impedance design (i.e. discrete R, L and C rather than transmission line or other resonators); In such circuits the Q factor of an unloaded circuit is usually a function of the inductor, and the inductor Q is usually going to go down with increasing frequency. Moreover, the Q factor of a loaded circuit depends on a combination of the inductor and the surrounding amplifying elements, and the amount that an amplifying element loads a circuit for the amount of gain it has also tends to negatively effect Q.

So the trend that you think you have noticed is, in general, not there in reality.

Use Q factor to understand how your circuit is going to work -- but if you want to predict the Q factor of a given actual circuit made out of actual physical components, be sure to take the component's characteristics into account before you try to make any generalizations.