|The Crossover Design Cookbook|
Chapter 2: How Components Work
by Mark Lawrence
What are Crossovers?
1st order Crossover
2nd order Crossover
How Crossovers Work
Final Watt-V Crossover
What We've Learned
I recommend FireFox
Well, having plowed (or skipped) through that, here's how to read this stuff. "s" is shorthand for "i 2π f", where "i" is the square root of -1, "π" is that unusual number 3.14159, and "f" is frequency. Don't worry about the "i" or the "2π"; again, they are only important if you're writing computer software. Basically, you can think of "s" as "f".
Engineers are addicted to these graphs called "Bode plots", named after Dr. Bode. You're already familiar with bode plots - the X-axis is frequency on a logarithmic scale, and the Y- axis is dB. You can draw bode plots for all of the above equations about as fast as you can sketch, knowing only three rules. First, every time you see an "s" in the numerator, you turn upwards by 6dB per octave. Second, every time you see an "s" in the denominator, you turn downwards by 6dB per octave. Third, the turning point happens at the critical frequency, which is chosen from the following table:
|C s||0 (DC)|
|L s||0 (DC)|
|1 + R C s||1 / (2π R C)|
|1 + L s / R||R / (2π L)|
|1 + Qs + L C s²||1 / (2π √LC ) [turn two times] [Q greater than 1]|
|1 + Qs + L C s²||Q / (2π √LC ) and 1 / (2π Q √LC ) [Q less than 1]|
These are all the terms that ever come up.
We call "s" terms in the numerator "zeros", and we call "s" terms in the denominator "poles". An equation has a zero or a pole for each power of "s". If the numerator's highest power of "s" is 1, the equation has one zero. If the denominator's highest power of "s" is 2 (s²), the equation has two poles. "Poles" and "zeros" is just a naming convention which make this stuff easier to talk about. If we didn't have these names, we'd have to say "numerator" and "denominator" all the time. This would be cumbersome, and make us too easy to understand.
There is no such thing in the universe as a physical system which has more zeros than it has poles. There are always more poles than there are zeros. Even resistors have poles: due to self-inductance, and eventually quantum mechanics, resistors stop conducting electricity at extremely high frequencies. However, these effects do not concern us at audio frequencies, or even radio frequencies, so we ignore them.
Let's practice making Bode plots. The very first circuit, which is also the easiest, has the equation Z = (RCs + 1) / Cs. Looking at our table above, we see we have a pole at DC, and a zero at the frequency 1 / (2π R C). So, we start the graph sloping downwards at 6 dB per octave (the pole at DC), and turn upwards 6dB per octave at the frequency 1 / (2π R C). Since we started out sloping downwards at 6dB per octave, the turn upwards cancels the downwards slope and makes us go horizontal. Our graph looks like this:
Next, we'll try a second-order system. The first L-C circuit has an impedance
Z = (LCs² + 1) / Cs
We see from the table above that there are two zeros at the frequency 1 / (2π √LC). Also, there's a pole at DC. So, due to the pole at DC, the graph must start out sloping downwards at 6dB per octave. The two zeros slope us upwards by two 6dB per octave increments. The first of these increments cancels out the downwards slope from the pole, leaving us moving horizontally. The second of these turns us upwards at 6dB per octave. However, since the two zeros are right on top of each other, we never actually get to see the horizontal portion. So, our graph must look like a "V":
The graph above has more zeros than poles, so it can't be the whole story for a real physical system. Graphs for real physical systems always wind up sloping downwards. In this system, the capacitor has resistance and inductance in the leads which give us a couple more poles at very high frequencies. If they are decent capacitors, these poles are well above audio frequencies so we can ignore the in our graph.
Finally, we'll graph the voltage on the second-order low pass filter. The voltage response is V = 1 / ( 1 + Ls/R + LCs2 ). We'll assume that Q is .7, which makes this a Butterworth cross over. From our table above, we have no zeros, a pole at .7 / (2π √LC ), and a pole at 1 / (.7 2π √LC ). So, the first pole is at about 70% of the cross over frequency, and the second pole is at about 140% of the cross over frequency (1 / .7 = 1.4 ). Thus, our graph starts out horizontal, turns downwards at 6dB per octave at .7 f, and turns downwards again another 6dB per octave to a total of 12dB per octave at 1.4 f. The graph looks like this:
These graphs we are making are approximations. The lines we are drawing are asymptotes. The actual curves would look just like these, except with the corners rounded. For example, the Butterworth cross over above would actually be a smooth curve which just barely touched the 6dB per octave slope at the exact center. However, it's much easier to draw these graphs using only straight lines and the simple rules above, so that's how engineers almost always do it.
We can learn something else from these curves: if the curve is sloping downwards, the circuit looks mostly like a capacitor at that frequency. If the curve is sloping upwards, the circuit looks mostly like an inductor at that frequency. If the curve is exactly horizontal, then the circuit looks like a pure resistor at that frequency.
Copyright © 2002-2019 Mark Lawrence. All rights reserved. Reproduction is strictly prohibited.
Email me, firstname.lastname@example.org, with suggestions, additions, broken links.
Revised Thursday, 15-Aug-2019 09:30:53 CDT