Extremely Basic Low Pass Filter Question

[Pano]

As Sofaspud says, don't worry about the coil in 1st order. It does not matter to the high pass filter. Really. Test it and see.

 

[macboy]

If you assume that the tweeter is just a perfect 8 ohm resistor (a false assumption of course but one you need to make for simplicity), then you can calculate the response at any given frequency like this:

First calculate the X_C, which is the reactance of the Cap at that frequency.
X_C = 1 / (2 * Pi * f * C)
Let's assume 5 kHz and 4.7 µF.
Then X_C = 6.77 Ω

The capacitor and the speaker (tweeter) form a voltage divider. You calculate the proportion of the voltage across only the tweeter, then convert that to dB.

Attenuation (dB) = 20 * log ( R / (R + X_C))
where R is resistance of speaker, and log is a base-10 logarithm. Let's assume R = 8 Ω.

Attenuation = -5.32 dB

You can plot for various values of f to get the curve.

 

[miragem3i]

Thanks! Your calcs were what I was looking for to verify and sanity check the 5551 and 6.1 figures and they do. When the XC of the cap meets the R of the tweeter is where the xo point is. Got it In the case of this xo it is 5551 hz.

So my DC resistance meter measurement is 6.1 ohms and I would plug that in to generate a curve of the tweeter attenuation, right? Seems easy enough. One spreadsheet and a chart and it's done.

I know that the tweeter won't have perfect resistance across the band but assumptions need to be made for simplicity.

The 6.1 Ohms point is reached at 5551 hz but the -6.0dB point is reached at 5577 hz

 

[Pano]

No. DC resistance of the tweeter and its AC impedance are not the same. Do you have a published impedance curve of the tweeter? Or just a single number?