# Q factor of a circuit.

1. Nov 16, 2012

### tomz

On my text book, there are several definition given for this Q factor.

1) 2 pi * maximum energy stored in reactive element / energy dissipated in a period

2)resonance frequency (in terms of ω) / band width

3)Q = 1/(2 * damping factor)

I have tried a couple of random circuits, and its seems not all of them are correct for arbitrary circuit. (Some o them may only true for simple RLC series of parallel).

May I ask which statement is always true?

My text book also says resonant frequency = natural frequency *(1-2*zeta^2) where zeta is damping factor

and undamped natural frequency is = natural frequency *(1-*zeta^2) where zeta is damping factor

Are these true for RLC series only??

Thank you!

2. Nov 16, 2012

### rbj

first of all, Q essentially has meaning only for 2nd-order circuits. if it's 4th or higher order, it will have more than one Q to talk about. now, for a 2nd-order circuit, you will get a transfer function that will look like:

\begin{align} H(s) & = \frac{b_0 + b_1 s^{-1} + b_2 s^{-2}}{a_0 + a_1 s^{-1} + a_2 s^{-2}} \\ \\ & = \frac{b_0 s^2 + b_1 s + b_2}{a_0 s^2 + a_1 s + a_2} \\ \\ & = \frac{(b_0/a_2) s^2 + (b_1/a_2) s + b_2/a_2}{(s/\omega_0)^2 + (1/Q) (s/\omega_0) + 1 } \\ \end{align}

put your 2nd-order transfer function in the form shown and then $\omega_0$ is your resonant frequency and the thing that multiplies your $s/\omega_0$ term is 1/Q. that is the definition from the POV of a transfer function.

Last edited: Nov 16, 2012
3. Nov 16, 2012

### Staff: Mentor

For me, they all ring a bell.

I'd say they are all correct (for, as rbj explains, a second-order underdamped system).

4. Nov 17, 2012

### Bassalisk

Quality factor is all 3.
Well I am familiar with first 2, but never used(yet) 3rd one.

I can connect physically first 2.

Quality of your RLC series combination is a measure how "good" your circuit filters one specific frequency.

If the circuit is ideal, with no resistance, it will have infinite quality factor meaning it will oscillate. The oscillation will be a sinusoid. Which means you will have a delta dirac function at your resonant frequency f0 given by $f0=\frac{1}{2\pi \sqrt{LC}}$. And the bandwidth would there for be 0.
(from $B=\frac{f0}{Q}\text{ Q ->}\infty \text{ B ->0}$

But if you have some resistance there, your delta will broaden and you will have a non-zero bandwidth.

So yes both of them are correct. Third one is beyond me, never used it.