# Bandpass Filter Attenuation?

I'm trying to split an audio signal into 5 channels, letting 5 different ranges through each filter (low, mids, highs, etc..). What I'm finding is that each bandpass filter has different levels of attenuation, which I don't understand.

I expect there to be some attenuation, but my "passed" frequencies for some of the filters are at best -6dB. This doesn't make sense to me because shouldnt the corner frequencies be defined as -3dB and they bandpass should allow signals through between these at an attenuation of less than -3dB.

Here's the schematic I'm working with: Here's a magnitude/phase plot of the circuit: I don't know much about filter design but I chose my capacitor value to be 0.1uF and used the following the following equations to solve for a resistor value to match the cutoff frequencies I want.

Wc1 = 1 / (2piRC)

Wc2 = 1 / (2piRC)

I did this for each bandpass filter diving the ranges of 20Hz - 20kHz into five regions. Perhaps the design I'm using itself is a flawed design? I've looked around a bit and see that a lot of designs include some sort of amplifier between the low pass and high pass portion of the bandpass, could this be the reason?

• 1 In your two equations are "R" and "C" meant to refer to the same thing? 2. sim.okawa-denshi.jp/en/RLCbpkeisan.htm – The Photon Feb 27 '17 at 2:32
• 1. No, they refer to the resistor and the capacitor. I chose C = 0.1uF and chose a Wc then solved for the resistor R I would need to make the filter work. 2. Interesting website I'll spend some time looking/playing with it to see if it helps. – Zearia Feb 27 '17 at 2:35
• 3. When cascading filters, you need to account for the 2nd stage loading the first stage. – The Photon Feb 27 '17 at 2:35
• 4. Bandpass filters, when the pass band is less than an octave, are often designed by their own special rules because of this. – The Photon Feb 27 '17 at 2:37
• Cascade means the output of one filter goes in to the input of the next filter. You have 5 sets of cascaded low-pass and high-pass filters. – The Photon Feb 27 '17 at 2:42

Wc1 = 1 / (2piRC)

Wc2 = 1 / (2piRC)

You haven't designed a bandpass filter; you've designed a low-pass filter and a high-pass filter and connected one to the output of the other.

The equation for the low-pass only gives the cut-off frequency when the load on the output has a very high impedance. Your 2nd stage (the high-pass part) is loading the first stage (low-pass part) and changing its cut-off frequency.

Similarly the high-pass equation assumes the source feeding it has very low impedance, which is again not valid once you connect the low-pass filter as the input source.

I expect there to be some attenuation, but my "passed" frequencies for some of the filters are at best -6dB.

The best case scenario at the midpoint of the passband is that C1 looks like an open circuit while C2 looks like a short circuit. In that case you'd have an equivalent circuit given by the resistor divider formed by C1 and C2. For your "bass" filter, for example, this would be about -3 dB because the two resistor values are roughly equal.

But things won't even be that good, because you're using equal capacitor values for the two filter stages, so the capacitors will have equal impedance at any given frequency. I won't try to work out the effect mathematically since you've already run the simulation and seen what it is.