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Old 10th June 2006, 03:04 AM   #1
Clipped is offline Clipped  Thailand
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Default how are crossovers and eq's designed?

how are crossovers and eq's designed to,...cutoff-boost-cut... a specific frequency? and can someone explain how caps and resistors work in conjunction to determine a specific frequency?

if i would want to modify my eq or crossover for a different frequency range anyone know what i would have to look for?

thanx
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Old 10th June 2006, 06:07 AM   #2
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A couple of electronics classes and a differential equations class wouldn't hurt

I don't know where to begin without some kind of background...
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Old 10th June 2006, 08:39 AM   #3
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thank you for your 2 cents...

now if anyone could help to explain, id really appreciate it...
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Old 10th June 2006, 09:19 AM   #4
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im no expert but if its a normal crossover with op amps, the cutoff freq. is determined by a couple of resistors and caps. the values and number of caps and resistors depends a little on the slope of the filter and if its high or low pass. here is a nice url that explains it a little bit better than i can.

http://www.linkwitzlab.com/images/protos/LR2.gif

12db/oct (or 2nd order) is the most common in low price car audio. In the picture you have the formula to create your own freq if you have one fixed component value in your filter. If you have a filter with adjustable crossover freq. you most likely have fixed cap values and a big potentiometer with 2 or maybe 4 gangs. (that means its like 4 seperate potentiometers all adjusted by one common axle.

there are some more good information on the linkwitz site about delays and phase correction and god knows what. my knowledge is fairly basic when it comes to these things so this is all i can help you with.
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Old 10th June 2006, 09:30 AM   #5
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Search the net for Sallen-Key filters and State-Variable filters. Notice how they're designed (in general) and determine which your crossover uses (likely S-K). Study the filter design notes. Then you can modify your circuit. It's likely that you'll change the cap values to change the range of frequencies. You'll have to change two caps per channel. The ratio between the two caps for the filter is important so you need to keep it constant. If you don't maintain the proper ratio, you'll likely introduce ripple (non-flat frequency response) in the output of the crossover.
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Old 10th June 2006, 03:51 PM   #6
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Quote:
Originally posted by Clipped
thank you for your 2 cents...

now if anyone could help to explain, id really appreciate it...
Alright, you asked for it....


The three basic components in a passive crossover are the inductor, capacitor, and resistor. They all behave differently to AC signals, and by combining them together you can achieve some pretty cool things.

First things first, terminology
Impedance - Z- an electrical device's opposition to the flow of AC current
Reactance - X - The imaginary part of the device's opposition to the flow of AC current, and its value is normally frequency-dependent
Resistance - R - The real part of the device's opposition to the flow of AC current, this is basically the DC component of impedance, it's frequency-independent

Impedance is simply the combination of both reactance and resistance, Z=sqrt(X^2 + R^2)

A resistor has a resistance but no reactance, that means that it provides the same impedance for all frequencies (the X in the above equation is 0). Useful for attenuation, but not filtering (since there's no frequency component in its value).

A capacitor has a reactance but [ideally] no resistance, that means that its impedance to the flow of AC current is entirely imaginary (the R in the above equation is 0). The equation for the current through a capacitor is C*dV/dt, or in the Laplacian domain, the impedance of a capacitor is simply 1/(C*s), where 's' is 2*pi*frequency*sqrt(-1). As you can see, the magnitude of the impedance for a capacitor lowers as frequency rises (the frequency term is on the bottom of the equation for impedance, as frequency rises, impedance lowers), and the magnitude of the impedance rises as the frequency lowers. At DC, a capacitor has an infinite impedance, and at high frequencies it behaves like a wire.

An inductor also has a reactance but [ideally] no resistance, just like a capacitor, but its equation is 1/L multiplied by the integral of V, or in the Laplacian domain, the impedance of an inductor is simply L*s, where 's' is the same as above. As you can see, the magnitude of the impedance for an inductor rises as frequency rises (the frequency term is on the top of the equation for impedance), and the magnitude of the impedance lowers as the frequency lowers. At DC, an inductor is just a wire, and at high frequencies it has ideally an infinite impedance.


Now, take a speaker, it's essentially an inductor, but as it moves, the magnetic field is constantly changing, and as a result its impedance vs frequency curve is much more complicated than a simple inductor. The manufacturer tells you it's a "4 ohm" speaker, but in reality its impedance can be anywhere between ~3.3 ohms (the DC resistance) to 30-50 ohms or higher. This link is a decent example of a typical speaker's impedance curve (it's the curve with the spike at ~31hz). For simplicity in this example, we'll just assume the speaker is a resistor, with a resistance equal to the nominal impedance (2 ohm, 4 ohm, 8 ohm, etc).

At this point, we should go into some basic circuit analysis. When you wire two impedances in series, they add. In addition to this, the voltage that is "dropped" across each device depends on the ratio of their impedances, and they each get the same current. The voltage dropped across device 1 is equal to V*Z1/(Z1+Z2), and for device 2 it's V*Z2/(Z1+Z2). As you can see, the device with the majority of the impedance receives the majority of the voltage.

When you wire two impedances in parallel, the final impedance is equal to 1/(1/Z1 + 1/Z2). The voltage across each device is the same, and the current through each device depends on the ratio of their impedances. The current through device 1 is equal to I*Z2/(Z1+Z2), and the current through device 2 is equal to I*Z1/(Z1+Z2). As you can see, the device with the lowest impedance receives the majority of the current.

Now that that's out of the way, assuming a speaker just has a resistance and no reactance, what happens when you put a capacitor in series with it? At high frequencies, the capacitor has almost no impedance, so it receives almost none of the voltage, and the speaker plays like there's no capacitor there at all. As the frequency lowers, the impedance of the capacitor rises and it starts "taking" more and more of the voltage. The speaker receives less of the voltage and its output drops. At DC, the capacitor has an infinite impedance and the speaker gets none of the voltage. This is what's called a 1st order highpass filter. It has a cutoff slope of 6dB/oct (pretty shallow) and is easy to implement. Given the above equations you could figure out the cutoff frequency for a specific capacitance and resistance (the cutoff frequency is simply where the speaker has fallen to -3dB, or the voltage has dropped to 70.7% of its value at high frequencies (these two definitions are the same thing, just different ways of saying it)). In the real world, this means that the speaker receives the normal signal at high frequencies, but low frequencies are attenuated.

Conversely, if you wire an inductor in series with a speaker, the impedance of the inductor is [ideally] 0 at DC and infinite at high frequencies. As the frequency rises, the inductor starts taking more and more of the voltage, and eventually the speaker is receiving nothing. This is called a 1st order lowpass filter, and follows the same rules as the highpass filter. In the real world, this means that the speaker receives the normal signal at low frequencies, but high frequencies are attenuated.

These filters are called "1st order" because when you solve for their transfer function in the Laplacian domain, they have a single zero/pole (you can essentially count the number of 's's in your equation, one capacitor or inductor is 1 s, so 1st order, one of each is 2nd order, 2 of 1 and one of the other is 3rd order, 2 of each is 4th order, etc). When you get to higher order filters things get more complicated. For example, with a 2nd order highpass filter, you have an inductor wired in parallel with the speaker, and then a capacitor in series with the both of them. At high frequencies the inductor has an infinite impedance (the speaker/inductor combination has an impedance equal to the speaker's), and the capacitor has an impedance of 0 (the speaker gets all of the voltage from the series combination). So basically, at high frequencies the speaker gets everything. As the frequency lowers, the inductor's impedance lowers, which means the impedance of the speaker/inductor lowers. Now you have the capacitor impedance rising, and the speaker/inductor impedance lowering, so the capacitor pulls the voltage away from the speaker at a much higher rate, which means this type of filter has a higher cutoff slope (12dB/oct). 3rd order is 18dB/oct, 4th is 24dB/oct, etc.

The other guys gave you some ideas to check out for determining the specific values needed for different cutoff frequencies and different types of filters, but you had asked how caps, resistors, etc work in conjunction with each other to "filter" the signal going to the speaker, and I hope I answered it above.
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Old 10th June 2006, 04:20 PM   #7
Clipped is offline Clipped  Thailand
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thanx fellas...

sr20demon...umm i meant active not passive, but on a side note, how do you measure a millihenrie (sp) ?

what takes the place of that big inductor in an active filter?

ok , i'll search now....
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Old 10th June 2006, 05:06 PM   #8
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The same equations and properties still apply in active filters, you just go about implementing it differently.

http://www.swarthmore.edu/NatSci/ech...d/Filters.html

About halfway down the page you can see the 1st order highpass filter. With a typical inverting amplifier implementation, the circuit looks like it does there but without the capacitor. The gain is simply the ratio of the resistances, R1/R2. When you add the capacitor, the same thing happens as I described before. At high frequencies the capacitor has an impedance of 0, so it basically doesn't exist and you get your normal R1/R2 gain. As the frequency lowers, your effective R2 (which is just the series combination of R2 and your capacitor) starts increasing, so your gain lowers. At dc, your "R2" is infinite, so the op-amp's gain is 0, and there is no signal leaving.

Above the highpass is the lowpass, you can see just by looking at it that the same basic thing is happening, only this time at DC you have the normal R1/R2 gain, and as the frequency rises the effective "R1" lowers, and as a result your gain lowers.

You can find the cutoff frequency for either just like before with some simple circuit analysis. If you wanted to change the cutoff frequency, the easiest thing would be to change the cap, but first you need to find out what type of filter is being used so you know what size capacitor would give you the change you want.

There are several ways you can measure inductance, but you shouldn't normally have to. Any inductor you buy should have the inductance labelled, and any speaker should have it listed as the Le in the T/S parameters. A quick google search should turn up some handy procedures you can use to measure inductance if you're so inclined.
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