CCC = Crossover Calculation Concept

[Jay]

In order to have an answer for whether putting a resistor before or after a 3rd order filter is equal, I tried to calculate from C=1/wC, R=R, and L=wL. But I forgot the basic concept or the target equation!!

You know, it is easier for engineers to forget the real concept to get things more... readable? In amplifiers, instead of thinking the real thing about voltage amplification, I tend to think that the music flow from input to output like waters. In crossovers, instead of thinking the real thing about impedance, I tend to think that capacitors can pass highs and inductors can pass lows and resistors can attenuate signals.

To answer the question, I didn't want to take the simple route, which is to try to compare by ears. If the sound is the same (no crossover changes) then it must be equal (putting the resistor before or after the filter).

2 professional speaker designers told me that it is equal, but I didn't think so (say intuitively), so I need to prove it.

[1] Is it equal?

[2] What is the target equation/function (to be solved) in the crossover design by math? I think it has something to do with the condition at frequency of angularity...

[Svante]

You want to know wether it gives the same result putting a series resistor before and after 3rd order a crossover filter and how to do the math?

Well it is not identical, but can give a similar result. For example if you use a 3rd order lowpass filter at 2 kHz applied to a bass-reflex box it can look like the picture below. The black curves are without a series resistance, the red curve has the resistor after the filter, the blue curve has the resistor before the filter. The filter components apart from the resistor are the same in all curves. In real life you would want to change the filter components in order to adapt it for the new load impedance for the red system.



To do the math on paper, you will need to learn about the jw-method and complex numbers. In short you assign an impedance of jwL to the coil and 1/jwC for the capacitor, where j^2 =-1.

[Jay]

Originally posted by Svante
In real life you would want to change the filter components in order to adapt it for the new load impedance for the red system.

And why is that? The blue one is better in my eyes....?

Originally posted by Svante
To do the math on paper, you will need to learn about the jw-method and complex numbers. In short you assign an impedance of jwL to the coil and 1/jwC for the capacitor, where j^2 =-1.

Yup, the question is why j^2=-1?

[Jay]

I read again about complex number (It has been uhmm 13 years since the last time I learnt about complex number), and I think I know what you were trying to explain.

Indeed, in a complex plane or complex RLC circuit, imaginary numbers is required to cope with variable phase delay. I just tried to find a simple way to prove that R in front and R at rear is not equal. The reason I believed so is first from practical experience and second because I thought crossovers are just ordinary linear circuits where Ohm law is the base for any equations involved. With R after the filter, the R is in parallel with the L so I thought it cannot be equal with R before the filter.

Unfortunately I couldn't find yet literature explaining the derivation of crossover formulas. But I looked into simple second order formula for hi pass (C=1/wR), and I think I understand why the target equation is Xc=R, but not 100% sure (Damm! I understood the whole concept back in high school ). Will soon go further with the formula for L...

[sr20dem0n]

Originally posted by Jay
And why is that? The blue one is better in my eyes....?

Because if you change the load impedance without changing the crossover components, you change the crossover frequency of the filter.

Originally posted by Jay
Yup, the question is why j^2=-1?

That's just the definition of 'j', mathematicians call it 'i'


The derivation of crossover forumlas isn't too bad, though the math can get a little tricky in higher ordered filters where you have to account for the Q of the filter to solve the system of equations.

With a second order highpass you have an inductor in parallel with the load, and a capacitor in series with them. Replace the inductor with jwL and the capacitor with 1/jwC, and then you just do simple circuit analysis until you get an equation for the voltage across the load/inductor (same voltage) as a function of j, w, R, C, and L. Solve for the magnitude of the equation and the j component goes away. You know R, you will know w once you plug in for the crossover frequency, you're just left with C and L. Now you have to deal with the Q of the filter to get your other equation (2 unknowns require 2 equations to solve). With a butterworth filter, the Q=.707, and if you do the math you see that wL has to be equal to 1/wC, which is your second equation. Solve for either L or C as a function of the other, plug it into the original equation, and you have your result. If you put the resistor aft of the filter, R will change, and if you don't change C and L accordingly then you change the cutoff frequency, which is pretty obvious once you solve the equations and do some examples.