High vs. low Q factor filters

[cjs94]

Hi,

Can someone please explain why you might choose a high or low Q factor high (or low) pass filter for a particular application?

I can understand why it's useful for band pass and notch filters, for selectivity, but I cannot see any reason for it in a low or high pass.

Thanks,
Chris

 

[berkeman]

Hi,

Can someone please explain why you might choose a high or low Q factor high (or low) pass filter for a particular application?

I can understand why it's useful for band pass and notch filters, for selectivity, but I cannot see any reason for it in a low or high pass.

Thanks,
Chris

What do you mean by the Q of LPF and HPF? Do you mean the number of poles and choice of polynomial?

 

[sophiecentaur]

What do you mean by the Q of LPF and HPF? Do you mean the number of poles and choice of polynomial?

Losses in the components of HP and LP filters will affect the sharpness of cut-off, which is equivalent to BP bandwidth.

 

[LvW]

Different Q factors (pole Q of a complex pole pair) are important, in particular, for higher filter orders (n>3) if you are using a cascade of several stages.
For example, a 6th-order Butterworth lowpass needs 3 stages - each with a different Q factor: Q1=0.518, Q2=0.7071, Q3=1.932.
A 4th-order Chebyshev lowpass (1dB ripple) needs a maximum value of Q=3.56.

 

[cjs94]

Thank you, but I don't think I explained myself well enough. Assume for example that I am designing a 3rd order high pass filter. I have a few choices: I could make the filter under damped (Q<1), critically damped (Q=1) or over damped (Q>1). Is there any reason, any possible application where I might choose an under- or over damped response rather than the critically damped response?

 

[LvW]

It is the main task of a filter to "filter" a mixture of different frequencies (either suppress or let pass a certain frequency range).
Hence, the filter specifications in 99% of all applications are expressed as requirements in the frequency domain (damping scheme)
Therefore, it happens not very often to ask for a specific behaviour in the time domain (under-/overdamped step response).
As a consequence, all filter requirements in the frequency domain result in a corresponding time behaviour which, normally, is accepted (mostly underdamped due to Q>0.5).

However, in some cases there are some specific requirements in the time domain (step response with small overshoot only). In these cases very often we require a constant group delay (phase response as linear as possible). These requirements can be met, for example, with a lowpass Thomson-Bessel approximation, which resembles a nearly "critically damped" (slightly underdamped) system.

 

[donpacino]

Thank you, but I don't think I explained myself well enough. Assume for example that I am designing a 3rd order high pass filter. I have a few choices: I could make the filter under damped (Q<1), critically damped (Q=1) or over damped (Q>1). Is there any reason, any possible application where I might choose an under- or over damped response rather than the critically damped response?

cost and design complexity. There are times when the components necessary to make a critically damped system vs an under-damped system will be too large.

In engineering there is a such thing as "close enough." if a design works and meets all requirements, in many cases it is wise to stop there, rather than add complexity.

 

[NascentOxygen]

Thank you, but I don't think I explained myself well enough. Assume for example that I am designing a 3rd order high pass filter. I have a few choices: I couldoutke the filter under damped (Q<1), critically damped (Q=1) or over damped (Q>1). Is there any reason, any possible application where I might choose an under- or over damped response rather than the critically damped response?

The higher Q filter gives a more rapid change of phase with frequency. When the different frequency components of a complex waveform (e.g., squarewave) undergo widely different phase changes, the filter output is very time-distorted. So a squarewave input to your filter can result in an output that looks nothing like you were expecting. A lower Q filter can be designed to have a phase shift that is proportional to frequency over a good part of its range, this means a squarewave when filtered will emerge looking like a squarewave, but rounded off a bit. The rapid attenuation of higher frequencies is not as pronounced in the lower Q filter, but this may be an acceptable tradeoff for keeping the squarewave at least resembling a squarewave and not distorted by wild oscillatory edges.