For band pass and band stop filters, Q tells how sharp the curve is at the centre frequency. I guess in this way it is required to roll-off.

However, low pass and high pass filters do not have centre frequency. So, what meaning does Q factor have for them? Does it matter of it is less than 0.5 or more?

Looking at picture of frequency response, it seems that the high Q filter has a type of hump as it approaches the cut off frequency. Isn't this a bad thing since ripple in pass band is not desired.

  • 1
    \$\begingroup\$ In general terms, the Q factor will still refer to the steepness of the gain slope, no matter your selected filter type. With that said, as you have noted, many "real world" filters have imperfect reactions & frequency-response "humps" that can be exaggerated as a consequence of the added steepness of the response curve due to a higher Q. \$\endgroup\$ Mar 10, 2016 at 14:38

3 Answers 3


Here's a picture (I drag out now and then) that explains the effect of Q on a 2nd order low pass filter: -

enter image description here

The top three pictures show you the effect of varying the Q-factor. Q-factor can also be reduced to make a maximally flat pass-band (aka a butterworth filter).

The picture goes on to explain where the pole zero diagram comes from and how you can relate natural resonant frequency (\$\omega_n\$) with zeta (\$\zeta\$). For your reference, zeta = 1/2Q.

You will also find that the shape of the curve reverses (with a hump) for 2nd order high pass filters: -

enter image description here

The high-pass filter picture came from here.

However, low pass and high pass filters do not have centre frequency.

They have the equivalent of a centre frequency known as the natural resonant frequency and if you think about a series L and C making a notch filter: -

enter image description here

This becomes a 2nd order high pass filter if the output is taken from the junction of the capacitor and inductor. Also if L and C swap places, it's still a notch filter but now if you take the output from across C it becomes a 2nd order low pass filter. Same resonant frequency and Q formulas all apply.

  • \$\begingroup\$ I rather think that the pole frequency wp (magnitude of the pointer/vector to the pole location) is an "equivalernt" to the centre frequency wo of a bandpass (remember: for a bandpass wo=wp). \$\endgroup\$
    – LvW
    Mar 10, 2016 at 19:48
  • \$\begingroup\$ Hmmm, Q factor of the components (my answer), or the loaded Q that the design operates the components into (your answer). Re-reading the question, you may be righter! \$\endgroup\$
    – Neil_UK
    Mar 10, 2016 at 21:01
  • \$\begingroup\$ Neil - I think the question concerns the "pole-Q" only and NOT the "quality factor" of a passive component. We have to discriminate between the Q factor of a transfer function (pole position) and the Q factor of a part. \$\endgroup\$
    – LvW
    Mar 11, 2016 at 7:19

Even with theoretically perfect components, so infinite Q, you can design a lowpass filter that has a flat passband, or a bumpy passband, or a round-shouldered passband, so high Q doesn't equate to ripples.

Having designed the filter shape, it can acquire or lose humps if the components you build it with don't have exactly the design values, or if the terminations it's working between don't have the design values.

Q matters. If you want to design a filter with a steep transition band, there will be a minimum Q that you need to use. The steeper the transition band, the higher the Q your components must have.

A common filter design technique is to ignore the fact that all the design tables and simple design programs assume perfect components, and then build it with components with a finite Q. The result will be a filter that is more round-shouldered at the edge of the passband than you expected. With a high enough Q, the effect will be small enough to be ignored.

If a filter has to work with such a low Q that the simple approach doesn't work, then there are tables and programs that take account of the finite Q, but this restricts the steepness of the filter response that can be designed.

Ripple in the passband isn't necessarily the worst problem that a filter can have. There is a tradeoff between the number of components, the passband flatness, and the transition band steepness. By accepting a little passband ripple, one can get a lot more steepness, a trade that's usually (but not always, it depends on the application) worth making.


For second-order lowpass and highpass filters it the Q-factor that determines the filter approximation (Butterworth, Chebyshev, Cauer, Bessel,...). Hence, it is a very important parameter (form of the transfer function in the region between passband and stopband).. For higher-order filters (series of second-order sections) it is very important to use the correct Q-factors which are available as tabulated values.

Definition: Q-factors are defined using the pole location in the complex s-plane; therefore, they are also called Qp ("Pole Q"): Qp=wp/2|sigma| with sigma=real part of the pole and wp=Magnitude of the pointer from the origin to the pole.

The same definition applies to a second-order bandpass. However, in this case we have the equality Qp=Q (center frequency/bandwidth).


  • 2nd-order Butterworth: Qp=0.7071
  • 2nd-order Chebyshev (ripple 1 dB): Qp=0.9565
  • 2nd-order Thomson-Bessel: Qp=0.5773
  • 4th-order Butterworth: Strage 1: Qp=0.5412; stage 2: Qp=1.3065
  • \$\begingroup\$ Hmm, so the Q value is already fixed when I use the table with pole positions to design the filter \$\endgroup\$
    – quantum231
    Mar 10, 2016 at 16:34
  • \$\begingroup\$ Yes - the pole position defines the pole-Q. \$\endgroup\$
    – LvW
    Mar 10, 2016 at 19:43

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