What are the uses for fourth order active filters? - Electronic Circuits Projects Diagrams Free

alexander3133 · 15th October 2007, 03:11 PM

Any idea anyone?
Is it for speaker design?
Or equalizer design?
Or something else?

HiTech · 15th October 2007, 03:25 PM

Loudspeaker crossovers can make use of this type of filter. Most often they are made by connecting two 2nd order Butterworth filters. The attenuation should be twice that of a 2nd order filter, typically 24db/ octave, and the high-pass and low-pass outputs are always in phase with each other.

alexander3133 · 15th October 2007, 03:35 PM

Quote:

Originally Posted by HiTech

Loudspeaker crossovers can make use of this type of filter. Most often they are made by connecting two 2nd order Butterworth filters. The attenuation should be twice that of a 2nd order filter, typically 24db/ octave, and the high-pass and low-pass outputs are always in phase with each other.

Thanks for the reply,
Thats mean inside the loudspeaker, tweeter is connected to the output of high pass active filter, bass is connected to the output of low pass active filter?

What is the difference between 24dB / octave and 24dB / decade?
What I know for a fourth order active filters have -80dB / decade.

ronsimpson · 15th October 2007, 07:19 PM

An octave is 2:1 in frequency. 1khz and 2khz
A decade is 10:1 in frequency. 1khz to 10khz
If your filter slope is 80db/decade then it is much less than 80db/ octave. I bet you can find the answer with out a calculator.

audioguru · 15th October 2007, 09:27 PM

Quote:

Originally Posted by HiTech

the high-pass and low-pass outputs are always in phase with each other.

Nope.
Even-order (2nd, 4th, 6th etc) Butterworth and Linkwitz-Riley highpass and lowpass filters have outputs that are out-of-phase and make a notch at the crossover frequency.

The wires of the tweeter are usually reversed to correct the phase but then a Butterworth filter causes +3dB at the crossover frequency.

HiTech · 16th October 2007, 02:30 PM

Well, I thought to create a Linkwitz Riley type crossover, two 2nd order Butterworth sections are cascaded (now a 4th order filter) and from that we get an in-phase output that doesn't require reversing the driver's polarity connections. That's where I was coming from. A Butterworth filter by itself would have an out of phase output, thus requiring reversing the connections to the driver. Am I right or wrong?

speakerguy79 · 16th October 2007, 02:32 PM

2nd, 6th, and 10th order L-R filters are out of phase. 4th and 8th are in phase. It continues like that, but no one really uses more than 6th order acoustic slopes unless they are doing an elliptical crossover or something. Also note that audio designers will frequently use the inherent 2nd order falling response of a driver cascaded with a 2nd order electrical filter to get a 4th order acoustic.

audioguru · 16th October 2007, 02:55 PM

Hi HiTech and Speakerguy,
Thanks, I was mistaken.
Rane company confirms some of what you say about a 4th-order Linkwitz-Riley crossover having zero phase shift.

HiTech · 16th October 2007, 03:57 PM

Yep sound right to me after reading that bit. Point number 1 is something else I remembered from school but I didn't bother to mention that since it seemed to delv away from the topic somewhat.

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15th October 2007, 03:11 PM	(permalink)
alexander3133	What are the uses for fourth order active filters? Any idea anyone? Is it for speaker design? Or equalizer design? Or something else?

15th October 2007, 03:25 PM	(permalink)
HiTech	Loudspeaker crossovers can make use of this type of filter. Most often they are made by connecting two 2nd order Butterworth filters. The attenuation should be twice that of a 2nd order filter, typically 24db/ octave, and the high-pass and low-pass outputs are always in phase with each other. __________________ Don't make me reach through this monitor to slap you a good one!

15th October 2007, 07:19 PM	(permalink)
ronsimpson	An octave is 2:1 in frequency. 1khz and 2khz A decade is 10:1 in frequency. 1khz to 10khz If your filter slope is 80db/decade then it is much less than 80db/ octave. I bet you can find the answer with out a calculator.

16th October 2007, 02:30 PM	(permalink)
HiTech	Well, I thought to create a Linkwitz Riley type crossover, two 2nd order Butterworth sections are cascaded (now a 4th order filter) and from that we get an in-phase output that doesn't require reversing the driver's polarity connections. That's where I was coming from. A Butterworth filter by itself would have an out of phase output, thus requiring reversing the connections to the driver. Am I right or wrong? __________________ Don't make me reach through this monitor to slap you a good one! Last edited by HiTech; 16th October 2007 at 02:33 PM.

16th October 2007, 02:32 PM	(permalink)
speakerguy79	2nd, 6th, and 10th order L-R filters are out of phase. 4th and 8th are in phase. It continues like that, but no one really uses more than 6th order acoustic slopes unless they are doing an elliptical crossover or something. Also note that audio designers will frequently use the inherent 2nd order falling response of a driver cascaded with a 2nd order electrical filter to get a 4th order acoustic. Last edited by speakerguy79; 16th October 2007 at 02:34 PM.

16th October 2007, 03:57 PM	(permalink)
HiTech	Yep sound right to me after reading that bit. Point number 1 is something else I remembered from school but I didn't bother to mention that since it seemed to delv away from the topic somewhat. __________________ Don't make me reach through this monitor to slap you a good one!