First Order Butterworth

[SoundRight]

I have a particularly dumb question regarding First Order Butterworth filters/shelves!

I am trying to establish the group delay for single inductor 1st order LP shelf chosen to give me a turn over point at around 200Hz (-3dB). There is no HP section

I understand the first order has a phase shift of 45deg at crossover (?) so presumably 90deg at 400Hz and so on?

Of course when I put this into a group delay calculator, it gives a constant group delay regardless of frequency (around .6ms).

I was under the impression that the GD should vary with frequency?

If this filter was implemented in DSP I am guessing the GD would still be the same as a passive component for IIR filters as they still have to implement this phase shift to achieve the "crossover"?

Layman language answers only please, I am not a mathematician!

Thanks for any help...

 

[AndrewT]

A first order filter has a F-3dB roll-off frequency defined by F-3dB = 1 / 2PiRC for capacitance, or F-3dB = R/2PiLfor Inductance

At F-3dB the phase has changed by 45degrees.
At infinite frequency the phase will have changed by 90degrees (an extra 45°, ref F-3db). It is not 45degrees per octave.

 

[SoundRight]

Hi Andrew, thanks, that's brilliant. So if I read that right the change of phase angle starts at 45 deg at -3dB and is very gradual ending in 90deg at "infinite" frequency. Would that mean at 800Hz for instance, at approx -12dB the phase angle is still around 45 deg. This would mean the gd (calculated at .6ms approx at 45deg) would be fairly constant still at 800Hz?

Very grateful for any response to this. I am guessing this holds true for IIR filters in dsp which would be configured to replicate this filter....?

 

[Dave Zan]

...First Order Butterworth filters/shelves!

There is only one sort of First Order filter.
So the 1st order Butterworth reduces to exactly the same filter as the 1st order version of any sort of filter.

...to establish the group delay

Why? There are a few subtleties, it would help to know what you want to achieve.

I understand the first order has a phase shift of 45deg at crossover (?) so presumably 90deg at 400Hz and so on?

As Andrew has pointed out, not "and so on".

Of course when I put this into a group delay calculator, it gives a constant group delay

The mistaken extrapolation causes this, the correct values will produce a GD that varies with frequency.
L. Paarmann has a book on Filter Analysis that is downloadable and covers this in all the detail you can stand, but plenty of plots to let you see the broad behavior even if you don't follow too much of the maths.
You can also download LTSpice for free and play with your planned circuit.
It has nice ways to plot phase and "delay", (in quotes because of some of the subtleties that I already mentioned.)

Best wishes
David

 

[krivium]

I am guessing this holds true for IIR filters in dsp which would be configured to replicate this filter....?

Yes. You could counteract the phase using FIR (rephase software can do that for example) but it will be at cost of latency and use of computer.

 

[Dave Zan]

.... Would that mean at 800Hz for instance, at approx -12dB the phase angle is still around 45 deg. This would mean the gd (calculated at .6ms approx at 45deg) would be fairly constant still at 800Hz?

No. The GD depends on the slope of the phase, not its actual value.
The GD for a 1st order LP filter is maximum at low frequencies, approaches a limit as frequency tends to 0.
Or to put it the other way round, decreases as frequency rises, slowly at first.
By 4 times the crossover frequency the GD will be quite small, less than 1/4 the value at the crossover frequency.
It will tend to 0 as frequency increases.

...holds true for IIR filters in dsp which would be configured to replicate this filter....?

I assume so too but have not studied DSP much, yet.
But DSP has true delay, no quotes, so this would need to be considered and allowed for.
There's some smart people in the Digital Line Level forum who could answer that part of the question better than I can.

Best wishes
David

The latency mentioned by Krivium is my "true delay"

 

[phase_accurate]

The pase of a first-order Lowpass can be calculated by

Phase= - arctan (f / f3)

If you use the phase calculated like that and enter it to your group-delay calculator you can easily see the behaviour as described by Dave.

Regards

Charles

 

[AndrewT]

Another reminder of how much of my simple school maths that I have forgotten!

Thanks for reminding me how old I am getting.

 

[SoundRight]

Hi...thanks. Some very helpful answers. I will go through them and expand my knowledge further!

 

[eriksquires]

Hi...thanks. Some very helpful answers. I will go through them and expand my knowledge further!

XSim is a fantastic learning environment and laboratory to try these ideas out. The FR plot will give you amplitude and phase, as will the impedance plot. This will help you play with and graphically experience what's being discussed. There are also group delay plots.

You can use XSim without any real data. Without input files XSim will create "ideal" drivers which is great for you to learn with. Then if you want to play with some real data you can go to Parts-Express and grab data files for any Dayton drivers.

Best,

Erik

 

[SoundRight]

Mine to Erik! Thanks for this. Will check it out. Had good success with Winisd as well...cheers

 

[eriksquires]

Mine to Erik! Thanks for this. Will check it out. Had good success with Winisd as well...cheers

WinISD is super, but different. it's more about cabinets. XSim is about electro-acoustics.

Plop a driver and a few caps down and examine the charts and you'll gain experience in leaps and bounds.

Best,

Erik

 

[Dave Zan]

...Phase= - arctan (f / f3)...

The GD is the slope of this curve, so it must be differentiated to produce the result he requested.
Very educational but a bit hard to make him do this himself
The answer is 1/((f/f3)^2 + 1) if we normalize it.
So at 0 Hz the delay is unity - about 0.8 msec for a 200 Hz crossover.
At crossover it's reduced to half - 0.4 msec
At 800 Hz it's 1/(4^2 + 1) = 1/(16 + 1) - under 0.05 msec.
But by 800 Hz the result is quite far down, so why does it matter?

Best wishes
David