Coming from a guy with considerably less experience than some of the other gentlemen in your thread I'll stop short of saying box size isn't all that important ... it is ... but, you'd be hard pressed to actually tell the difference in driver response from Q =0.7 up to Q =1.0 I'd bet. Even Q =1.2 isn't like "holy crap, that's waaayyy too much bass for me" territory ... unless you can sense another 1db of gain down there
As somebody has pointed out, your room would have a bigger influence on the bass response than the 0.707 to 0.8/8.5 Q would. I wouldn't even sweat this.
I could hear quite a big difference between Q 1.2 and Q 0.85. It was more to do with quality than quantity though....
Fair enough ... quick question though, did you know it ahead of time or asked to tell them apart blindfolded?
Size of speaker box
Remember them two transistor sets? The battery was more expensive than the set and a tube set was virtually the same price working on mains. My first set that I got when I started to work was so expensive that I paid two years for a portable radio with ten transistors!!
Guys if the size is not a lot bigger I have not heard a difference yet that make such a huge difference but some of us live in postage stamps divided in rooms and maybe that is the reason why. when I was a youngster we put our pocket two transistor radios in a big box and the difference in soud quality was remarkable. So this is why I say the size must be considerable bigger.
Remember them two transistor sets? The battery was more expensive than the set and a tube set was virtually the same price working on mains. My first set that I got when I started to work was so expensive that I paid two years for a portable radio with ten transistors!!Hi Andrew,
Do you count a single negative real pole as Q=sqrt(2)/2? It is often counted as Q=1/2 (as omega_n/(-2*Re(p))=-Re(p)/(-2*Re(p))=1/2 for a negative real pole).
Anyway, you are right: the product of the pole pair Q's of an even-order Butterworth filter is sqrt(2)/2. For an odd-order Butterworth the product of the Q's of the complex pole pairs is 1, so with the real pole included it is whatever Q you attribute to a single negative real pole. Thanks, I never knew this!
Best regards,
Marcel
Do you count a single negative real pole as Q=sqrt(2)/2? It is often counted as Q=1/2 (as omega_n/(-2*Re(p))=-Re(p)/(-2*Re(p))=1/2 for a negative real pole).
Anyway, you are right: the product of the pole pair Q's of an even-order Butterworth filter is sqrt(2)/2. For an odd-order Butterworth the product of the Q's of the complex pole pairs is 1, so with the real pole included it is whatever Q you attribute to a single negative real pole. Thanks, I never knew this!
Best regards,
Marcel
Hi Andrew,
To make things simple, forget about loading effects for the time being.
You can then make an even-order high-pass filter by cascading a bunch of second-order high-passes. Each second-order high-pass has a Q, and when you design the total cascade to have a Butterworth response, the product of those Q's is indeed sqrt(2)/2~=0.7071068, like you wrote.
In post number 25, you do a similar calculation for an odd-order Butterworth high-pass. An odd-order high-pass must have a first-order section somewhere. What do you consider the Q of such a first-order section to be?
Best regards,
Marcel
To make things simple, forget about loading effects for the time being.
You can then make an even-order high-pass filter by cascading a bunch of second-order high-passes. Each second-order high-pass has a Q, and when you design the total cascade to have a Butterworth response, the product of those Q's is indeed sqrt(2)/2~=0.7071068, like you wrote.
In post number 25, you do a similar calculation for an odd-order Butterworth high-pass. An odd-order high-pass must have a first-order section somewhere. What do you consider the Q of such a first-order section to be?
Best regards,
Marcel
A first order filter can be named ANY characteristic, or NO characteristic at all (because there is only ONE possible response), and it has NO Q. Q is defined only and only for 2nd order filters (Highpass : H(s) = 1 / (1+w0/s/Q+(w0/s)^2), any multiplication of Q's of 2nd order stages for a "total Q" makes no sense and isn't used in transfer function theory.
The characteristics of orders above 1st are named for different effects (of lowpass filter!), has nothing to do with individual Q's
Butterworth is maximum flat magnitude before roll-off, Bessel is maximum flat group delay before group delay roll-off, Chebycheff is maximum allowed magnitude ripple, Linkwitz-Riley is Butterworth squared.
Highpasses are derived by substitution of s/w0 with w/s0 (inversion at the natural frequency).
You dial in your individual w0's and Q's of 2nd order stages (and w0 of a 1st order stage if odd order) to get to whatever target. Only for textbook filters all the w0's are identical. But the Q product is useless because there is infinite number of Q's to get there and most of them give different responses.
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for me here is the following rule: The smaller the value of Qec (i. e. higher value of Bl), the better the sound - even for large Vb. If you need a higher Value of Qtc, add a 20W resistor with the suited value in series.
At higher Qe/Qt speaker value suffers the sound more and more, because the power compression in your voice coil increases already by small SPL level.
Hi,
Sorry but that is all wrong. Q can only be discussed in terms of a classic
second order system. Butterworth is 0.71 for second order, for third
order you can cascade 1st order and Q=1 second order at the same
frequency for a Butterworth 3rd order, but Q is now meaningless.
A series combination of 2nd order Q=1 and Q=0.7 does not
yield a 4th order Butterworth response, you can't multiply Q's.
You can't ascribe a Q to a 1st order filter. You can't ascribe a Q to 3rd
order and higher filters. Practically they need to be described as a
cascade of 1st order frequencies and 2nd order Q and frequencies.
rgds, sreten.
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I thought I had picked up enough to have reached the conclusions the way I stated them.
I have to admit that your 2nd order theory is beyond my capabilities and as a result I cannot argue usefully one way or the other.
So for the time being I desist.
Sorry Sreten, but I won't accept your rejection of my statement/s.
See my previous post to Kstr, for an acceptance of his more reasoned argument.
Greets!
Yes, thanks for re-posting, I lost the links and its saved info.
GM
It's certainly true that a cascade of second-order filters with the product of the Q's equal to sqrt(2)/2 is not necessarily a Butterworth filter. But Andrew is right about the opposite: in an even-order Butterworth low-pass or high-pass filter the product of the Q's of the pole pairs equals sqrt(2)/2. In an odd-order Butterworth low-pass or high-pass filter the product of the Q's of the pole pairs equals 1. By the way, the Q of a pole pair is defined such that it equals the Q of a second-order filter section having the same poles.
It took me a while before I understood this. A well-known property of Butterworth high- and low-pass filters is that their poles are evenly distributed over a semi-circle in the left half plane. This implies that all second-order sections have the same natural frequency omega_0. Another well-known property is that the response has dropped to sqrt(2)/2 times the passband gain (-3.01029... dB point) at this frequency.
If you take a look at the equation for the transfer of a second-order section with unity passband gain, you see that its transfer is Q at the natural frequency omega_0. So the transfer of a cascade of them is the product of the Q's. Given the property that the transfer of a Butterworth filter is sqrt(2)/2 times the passband gain at omega_0, that means that the product of the Q's must indeed be sqrt(2)/2. For an odd-order filter it must be 1, because the first-order section already has a gain of sqrt(2)/2 at omega_0.
Coming back to loudspeaker design: Andrew's rule might be useful as a rule of thumb to get a reasonably flat overall response when you can't quite get the poles in Butterworth positions. Two examples:
High-pass with one pole pair with Q=sqrt(2) and another one with Q=1/2, both with equal omega_0:
-3.01 dB at 1 omega_0 (which makes sense)
-1.01 dB at 1.2 omega_0
-0.25112 dB at 2.42 omega_0
-0.02123 dB at 10 omega_0
Not bad at all, even though it is not as flat as a real Butterworth filter.
More extreme example: Q=2 and Q=sqrt(2)/4.
-3.01 dB at 1 omega_0 (which again makes sense)
-0.99955 dB at 3.76 omega_0
-0.25169 dB at 8.31 omega_0
-0.17723 dB at 10 omega_0
Not very good, but at least it does not peak and the -3.01 dB point is where you expect it to be.
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