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For what I'm learning, the bandwidth (of a BPF, for example) refers to the $\Delta$ between $f_L$ and $f_H$ at (usually) $-3\textrm{ db}$ from the $f_0$ (from wiki). For a peaking filter instead, $f_L$ and $f_H$ are situated (usually) at the peak $\textrm{dB Gain}/2$.

Q Factor become $f_0 /\Delta$. It's clear till here.

  • Now, if I take a lowpass filter, how is calculated $Q$ Factor?
  • I have $f_0$, but how would I calculate the $\Delta$ of a lowpass?

There is no a "peak" as reference, so $-3\textrm{ dB}$ (or $\textrm{dB Gain/2}$) make no sense as references points for the "created" bandwidth.

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    $\begingroup$ Can you please clarify if you are referring to low pass filters with resonance?. In that case, perhaps the points you are looking for are apparent from its frequency response (?) $\endgroup$
    – A_A
    Nov 21, 2016 at 11:37
  • $\begingroup$ actually paizza, the definition of Q for the peaking filter is more of a cookbook thing than it is "usually". we had a discussion about this not too long ago. $\endgroup$ Nov 22, 2016 at 17:06
  • $\begingroup$ Not sure if its really only a "cookbook" thing :) From this : The threshold value is often defined relative to the maximum value, and is most commonly the 3dB-point, that is the point where the spectral density is half its maximum value (or the spectral amplitude, in V or V/Hz, is more than 70.7% of its maximum). It also seems that "your" dbGain/2 band edges (for the Peaking filter) refers to Half power point (which is again -3db to the peak). It looks like all is related. $\endgroup$
    – markzzz
    Nov 22, 2016 at 17:48
  • $\begingroup$ Here (more or less) I'm asking if a Low Pass Filter's band edge are situated at -3db as for a Band Pass Filter (for the standard convention). I see this "-3db" only when talking about band pass (or notch)... $\endgroup$
    – markzzz
    Nov 22, 2016 at 17:49

2 Answers 2

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For basic lowpass and highpass filters, Q (quality) factor is not well defined.

Q factor is a useful metric for second-order bandpass frequency selective filters, as it shows their selectivity power. It can be defined as the ratio of the center frequency of passband to the bandwidth of passband, assuming that the system has only a single passband (which is indeed the case for a second-order system)

For a highpass filter, the passband width is infinite, hence this Q will be zero. And for a lowpass filter, Q will be 1/2 always. As a result, Q factor, as defined above, is not a useful merit for HighPass or Lowpass filters.

Note that there's also an identical ringing definition of Q, which focuses on the resonation power of those systems. Both are equally valid definitions but from different application perspectives.

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  • $\begingroup$ So you are saying that for low (and high) pass, Q become "resonance"? $\endgroup$
    – markzzz
    Nov 21, 2016 at 11:27
  • $\begingroup$ @paizza yes, kind of. Q is for Quality (of resonance)! From wikipedia's Q factor article's headline: Q factor is a dimensionless parameter that describes how under-damped an oscillator or resonator is, and characterizes a resonator's bandwidth relative to its center frequency. $\endgroup$ Nov 21, 2016 at 11:41
  • $\begingroup$ Yes, but "and characterizes a resonator's bandwidth relative". Since I don't specify a bandwidth for a Low Pass filter, it refers to what? i.e. Q (resonance) of 2.4 of a Low Pass filter at 1khz refers to the bandwidth created by the resonance of the filter itself around that cutoff? $\endgroup$
    – markzzz
    Nov 21, 2016 at 11:47
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    $\begingroup$ Oh ok, so also on Low/High Pass filter the bandwidth across the resonance refers to -3db to the peak (a peak that increase when resonance increase). $\endgroup$
    – markzzz
    Nov 21, 2016 at 14:24
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    $\begingroup$ god, he will kill me :D I stressed him a lot days ago... $\endgroup$
    – markzzz
    Nov 21, 2016 at 20:14
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enter image description here

So I once chaired a workshop about Q and I really wish we would nail down the definition for purposes of automation and consistent user experience.

So the transfer function is:

$$ H(s) = \frac{1}{\left(\frac{s}{\omega_0}\right)^2+\frac{1}{Q}\frac{s}{\omega_0}+1} $$

Where $\omega_0 = \frac{1}{\sqrt{L C}}$ and $Q=\frac{\omega_0 L}{R}$ .

That's the EE definition for a LPF.

Now, if you look at the impulse response shown above, and to define $RT60$ as the time for this output to decay to -60 dB, then you get

$$ \frac{\omega_0}{2Q} = \frac{2 \pi f_0}{2Q} = \ln(10)\frac{\frac{60}{20}}{RT60} $$

or

$$ Q = \frac{RT60}{2.1988} f_0 $$.

What this means is that for every unit of $Q$, the impulse response rings for 2.1988 cycles before it decays below -60 dB. It appears that this is an approximation since the actual frequency of the ringing is a little different than $f_0$. I forgot my thinking about this.

enter image description here

In case anyone is interested, it appears that I turned slides into a pdf.

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    $\begingroup$ So as you show the Q for the inductor is energy stored over energy dissipated which I believe is the EE definition of Q. Capacitors also have Q using the same relationship ($X_C/R$) but typically it is so much higher than that for the inductors that it can in most cases be ignored. $\endgroup$ Mar 23 at 23:36
  • $\begingroup$ @DanBoschen I'm just showing idealized components. For a parallel RLC filter it's the same because it's $$ Q = \frac{\omega_0 C}{1/R} = \frac{\omega_0 L}{R}$$ I think that's correct. Lemme look at this. $\endgroup$ Mar 24 at 0:55
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    $\begingroup$ I would think h(t) should be dimensionless when it is translating the input to the output in the same units (such as volts to volts). Also a LPF (to be applicable to this question specifically) would be series L shunt C (with possible R in series with the L and possible R in parallel with C). $\endgroup$ Mar 24 at 16:24
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    $\begingroup$ No, Dan. Any $h(t)$ for any continuous-time LTI that is voltage-in-voltage-out has dimension of $\frac{1}{\mathrm{time}}$. Look at the convolution integral and you should be able to see that right away. But I also fucked up with the exponent in the second $h(t)$ in the slide at top. This is embarrassing. $\endgroup$ Mar 24 at 16:36
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    $\begingroup$ I wanna make clear that this is fucked up. I'm pretty sure that the "$f_0$" in the exponent in the second $h(t)$ in the top slide does not belong there. I gotta deal with this, but there's fresh snow out there now, and I'm going skiing. Will fix this later. $\endgroup$ Mar 24 at 17:22

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