Q factor question

I'm reading about the Q-factor from Wikipedia and here:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/serres.html

The Q-factor is defined as the resonant frequency divided by the bandwidth (the range of frequencies that oscillate with power greater than half of the power of the resonant frequency). Apparently, the definition is such, so that the amplitude against frequency plot resembles a sharper curve when the Q-factor is high.

But I don't understand why this is true. The numerator is the resonant frequency, not the amplitude of the resonant frequency. If the hold the bandwidth constant, why does the Q-factor improve simply by having the resonant frequency higher (moving the curve towards right/higher on the frequency domain plot)? If you look at the example pictures under the link I provided, you will see that the higher Q curve has a higher amplitude at the resonant frequency and the curve is less spread out. Naturally it would seem to me more intuitive to take the height of the curve and divide it by some number that characterises how spread out the curve is (here the bandwidth). Then a higher peak and less spread would give a higher Q/sharper "spike" on the plot.

It seems that there is something extremely simple that I'm misunderstanding but I'm just quite confused.

• Seems that one problem with defining a Q-factor according to your way would be that it would no longer be a dimensionless parameter. In fact, the dimensions would vary according to the application. The Q-factor would have the dimensions of amplitude (in either volts, or amps, or meters, etc.) divided by frequency. As the Q-factor is defined now, it has a simple interpretation: The Q-factor is $2\pi$ times the stored energy of the oscillator divided by the energy lost per cycle. That's dimensionless, easy to grasp intuitively, and can be used regardless of the type of oscillator. – Samuel Weir Jun 24 '17 at 19:15
• What you're missing is that filter responses are essentially in terms of ratios, e.g, a first order low-pass filter attenuates by a factor of (approximately) 10 at a frequency 10 times the cutoff frequency. So, for example, if the cut-off frequency is 10kHz, the response is down by a factor of 10 at 100kHz - a 90 kHz difference. However, if the cut-off frequency is 100kHz, the response is down by a factor of 10 at 1Mhz - a 900kHz difference. Now, apply this insight to a bandpass filter has both a high and low cut-off frequency, i.e, a bandwidth. – Alfred Centauri Jun 24 '17 at 19:48
• "...so that the amplitude against frequency plot resembles a sharper curve when the Q-factor is high." That is simply not true unless you scale the frequency axis as discussed in Alfred's answer. – DanielSank Jun 24 '17 at 20:18

Consider the following plots of the admittance magnitude versus frequency for a series RLC circuit: Image Credit

I chose this plot because the scales are normalized (I would have preferred a log horizontal axis too but this will do).

The crucial point is that the Q entirely determines the shape of the curve on these normalized scales.

So, for example, take a look at the curve for Q = 2 and let's look at the points where the curve intersects the 0.707 horizontal line. I'd eyeball those to be about 0.75 and about 1.25 for a normalized bandwidth of 0.5. This is consistent with a Q of 2 since

$$Q = \frac{\omega_0}{\Delta \omega} = \frac{1}{0.5} = 2$$

You ask in your question why the Q increases if the resonance frequency is increased while holding the bandwidth constant. You can see the answer here in this plot - the normalized bandwidth determines the Q

If you increase the (actual) resonance frequency while holding the (actual) bandwidth constant, the normalized bandwidth is decreasing thus increasing the Q.