Multiple Feedback bandpass filter: inverting or not?

Started by earthtonesaudio, May 04, 2009, 01:49:35 PM

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earthtonesaudio

Saw this thread:
http://www.diystompboxes.com/smfforum/index.php?topic=75384.0
and this picture:
http://www.geofex.com/article_folders/wahpedl/opampwah1.gif
and noted that the caption on this picture says the output is inverted.  But I immediately thought of the twin-T notch filter, which inverts the phase 180 degrees at the notch frequency, and how if you put that filter in a neg. feedback loop, it inverts again (and you have a twin-t oscillator).  Does this filter work the same way?  Is the bandpass frequency inverted (nearly) twice, or just once?  I don't quite understand it at the moment (probably a time domain vs. frequency domain mental exercise) so if someone could help clear it up for me, that would be wonderful!

Just trying to make sense of this.  Thanks!

R.G.

Talking about a filter as inverting or not only has meaning far from the filter frequency.

Near the filter frequency, say from 1/10 to 10x the filter frequency, the phase will be between 0 and 360 degrees, not simply in phase or out of phase.

If you think this makes designing crossovers for hifi speaker complicated, you're right.
R.G.

In response to the questions in the forum - PCB Layout for Musical Effects is available from The Book Patch. Search "PCB Layout" and it ought to appear.

Cliff Schecht

The amplifier is acting as a simple gain block to add some resonance to the second order filter. The filter will definitely cause some phase shift, but there are other requirements that need to be met before oscillation will occur. The 1 meg resistor in the circuit acts as a Q dampener, where the smaller it is less Q you have. If you make the 1 meg resistor larger, you decrease the amount of negative feedback and start getting more resonance. Eventually, of course, it will self-oscillate.

If you look at something like this on a pole-zero plot (showing you the location of your poles and zeroes on a real vs. imaginary graph), as the resonance increases you will see your poles start to get closer and closer to crossing the imaginary axis. As the poles get really close, the filter gets very ringy and starts to sound like a dampened sinusoid after an excitation (like an impulse function). The second those poles cross the imaginary axis, the filter WILL become an oscillator and will resonate at the center frequency. I'm not the best at explaining this stuff, but it helps sometimes to look at your system on a pole-zero plot, just looking at frequency/phase plots won't always tell you everything about your system. It's important to understand that frequency domain plots are time invariant and can't always be used to predict system stability.

chilecocula

#3
Quote from: Cliff Schecht on May 04, 2009, 02:18:48 PM
The amplifier is acting as a simple gain block to add some resonance to the second order filter. The filter will definitely cause some phase shift, but there are other requirements that need to be met before oscillation will occur. The 1 meg resistor in the circuit acts as a Q dampener, where the smaller it is less Q you have. If you make the 1 meg resistor larger, you decrease the amount of negative feedback and start getting more resonance. Eventually, of course, it will self-oscillate.

If you look at something like this on a pole-zero plot (showing you the location of your poles and zeroes on a real vs. imaginary graph), as the resonance increases you will see your poles start to get closer and closer to crossing the imaginary axis. As the poles get really close, the filter gets very ringy and starts to sound like a dampened sinusoid after an excitation (like an impulse function). The second those poles cross the imaginary axis, the filter WILL become an oscillator and will resonate at the center frequency. I'm not the best at explaining this stuff, but it helps sometimes to look at your system on a pole-zero plot, just looking at frequency/phase plots won't always tell you everything about your system. It's important to understand that frequency domain plots are time invariant and can't always be used to predict system stability.

I don't think that pole-zero and frequency/phase plots are common DIY knowledge (at least in a system control context)
in conservative stompboxes, tone is neither created nor destroyed, but transformed

Cliff Schecht

Frequency plots are common to see in the DIY arena, but yeah the other stuff isn't. But hey, just because people don't know doesn't mean we can't teach them! Anybody with a copy of Matlab and a transfer function can look at a pole-zero plot and anybody with a piece of simulation software can do time and frequency based signal analysis. While these aren't essential to understanding circuit design, having some basic knowledge about the topics can sure help one out quite a bit. Making a thread to explain this stuff wouldn't work too well though because, honestly, some of the information is very hard to explain without a solid math background.

Also, a big problem here is that your typical guitar pedal is a non-linear system, so actual modeling and circuit analysis can be a bit more tricky. You can usually separate out different components of a circuit and do analysis separately on each, but this becomes tedious pretty quickly and when you put it all back together the results can be unpredictable. It's not uncommon to design parts of a schematic separately and put them together after testing the individual sections, but then ones knowledge of impedance matching/bridging, biasing and filter design (like the high-pass filter formed by EVERY DC blocking cap) becomes more and more important.

earthtonesaudio

So perhaps a way to rephrase the question would be something like:

"If you summed together an MFB bandpass filter's output with a dry signal, would the result at the exact center frequency of the filter be added to, or subtracted from, the dry signal?"

...In the hopes of getting an answer that doesn't require too much math.   ::)

Cliff Schecht

#6
With just additive mixing, all your going to do is mix in the filtered signal with the unfiltered one. In essence, whatever passes through the filter will be added at the mixer output and everything else that is still left in the dry signal won't change. You'll get some slight variation in signal timbre because of the associated signal phase shift, but this is characteristic of any filter (even "linear-phase" filters aren't perfect :)).  If you're trying to get the thing to resonate, then you have to decrease the amount of negative feedback (the 1 meg resistor changes this).

So to make sure I'm clear on the answer: At the exact center frequency the filtered signal will get added to the dry signal.

R.G.

Quote from: earthtonesaudio on May 04, 2009, 03:22:15 PM
"If you summed together an MFB bandpass filter's output with a dry signal, would the result at the exact center frequency of the filter be added to, or subtracted from, the dry signal?"
Google search "active crossover design" for the answer you want.

The short answer is that it depends on the exact filter and it's sensitivity to components; if the components are exactly what the filter equations call for it's not exactly what the equations predict the phase response will be. However, it will not be either 0 or N times 360 degrees and also not 180 or (1+2N)*180 degrees for any MFB filter I've seen, nor for most of the active filters of any type. I would say for no active filter, but I'm not all that widely experienced in active filters, only a working hack's knowledge. Cauer and elliptical filters use phase shifts of 180 degrees just outside the passband to put an infinite rejection notch just outside the pass band to get greater falloff, but that's not at the center frequency.

So you'll get some phase shifted version of the signal at the center frequency, neither "in phase" nor "out of phase" the way those terms are usually used. Beyond that, you're into learning active filters.
R.G.

In response to the questions in the forum - PCB Layout for Musical Effects is available from The Book Patch. Search "PCB Layout" and it ought to appear.

Cliff Schecht

Quote from: R.G. on May 04, 2009, 09:02:27 PM


The short answer is that it depends on the exact filter and it's sensitivity to components; if the components are exactly what the filter equations call for it's not exactly what the equations predict the phase response will be. However, it will not be either 0 or N times 360 degrees and also not 180 or (1+2N)*180 degrees for any MFB filter I've seen, nor for most of the active filters of any type. I would say for no active filter, but I'm not all that widely experienced in active filters, only a working hack's knowledge. Cauer and elliptical filters use phase shifts of 180 degrees just outside the passband to put an infinite rejection notch just outside the pass band to get greater falloff, but that's not at the center frequency.

So you'll get some phase shifted version of the signal at the center frequency, neither "in phase" nor "out of phase" the way those terms are usually used. Beyond that, you're into learning active filters.

You would be correct in saying that NO active filter will have a completely predictable phase response. Even your "linear" Sallen-Key op-amp filters aren't all that linear when you start trying to actually apply superposition and get consistent results. There is too much variation from component to component, not to mention stray capacitance and inductance (plus the internal capacitors in an op amp), that somehow have an influence on the phase of your signal.

As far as whether a certain active filter can actually get to 180 degrees of phase shift, I would think the answer is yes but only at one specific frequency (or a few specific frequencies for a multi-pole filter).

earthtonesaudio

Thanks guys, this confirms real-life results on the breadboard.  A while ago I built a MFB filter with a parallel dry path, and both were summed together.  I didn't notice any drop in signal level at the passband, so I assumed the phase was approximately aligned to (but obviously not canceling) the dry signal.