# Calculating Phase/Time Delay induced by a 2-pole Bandpass filter? ie. Group Delay?

If you have a two-pole Butterworth bandpass filter described by a particular Q value (Q>0.5) and a resonant center frequency "f", is there any equation that will tell you the phase/time delay induced by this filter at "f"?

Thanks

• As far as the group delay is concerned, just apply the definition of the group delay: Tg=d(phi)/dw. That means: Find the slope (first derivative) of the phase function vs. frequency at the corresponding frequency. – LvW Jun 4 '18 at 8:51
• A butterworth filter will have a Q of 0.7071 and not "Q>0.5". – Andy aka Jun 4 '18 at 9:17
• The filter has a variable Q to allow different bandwidth/resonance settings. Perhaps that makes it a modified Butterworth. But it's labelled a Butterworth filter in my system. – mike Jun 4 '18 at 14:54

The formulas for phase ($pd(\omega)$) and group ($gd(\omega)$) delays of a H(s) transfer function are related to the phase:

$$\Phi(\omega)=\arctan{\frac{\Im\left(H(s)\right)}{\Re\left(H(s)\right)}}$$ $$pd(\omega)=-\frac{\Phi(\omega)}{\omega}$$ $$gd(\omega)=-\frac{\text{d}\Phi(\omega)}{\text{d}\omega}$$

However, you'll find that these don't really hold up to the reality, when the most certain way to determine them is to simply measure the relevant delay of input vs the output. Also, the direct measurement will most likely tell you the phase delay, while the group delay will need special considerations, in that you'll have to amplitude modulate your input signal with an envelope of the desired frequency, then measure the delay of the whole group compared to the input. Quite often, reality will show inconsitencies with the calculations.

Not sure if this helps... 2nd Order Butterworth BPF, by empirical measurements;

The Phase delay at $$\f=f_o=180 \$$ deg. = π radians

The Group Delay $$\gd[Q,f_o]~ =\dfrac{log({Q)}^\pi)}{f_o}\$$ approx for Q>2

• @1kHz Q=1 , gd=0.34 ms
• @1kHz Q=10, gd=3.17 ms
• @1kHz Q=100,gd=31.7 ms
• .
• @1MHz Q=10, gd=3.17 μs
• @1MHz Q=100 gd=31.7 μs