How do filters really work?

[Curvature]

In an (1) electrical or (2) digital filter, what happens to the energy that is attenuated or boosted?

Is attenuation a kind of redistribution of energy? Where does the extra energy come from when boosting?

Does this work any different acoustically? I assume all of those phenomena we usually discuss can be thought of as physical filters.

Do the clear I'm not asking about what filters can do or the results. I'm asking about what happens to that energy that is removed or added to the signal.

E.g., in a porous absorber, the attenuated energy is converted to heat. What I'm not clear about is how that conversion actually works and how it relates to bandwidth.

 

[RayDunzl]

A digital filter changes the digits... I would rate this "no energy change" in the way you are asking.

Analog (electrical) filter can block or dissipate a portion of the signal energy. If dissipated, heat. If blocked, maybe less dissipation. Look for resistors, capacitors, and coils in this application.

 

[DVDdoug]

what happens to the energy that is attenuated or boosted?

Imagine a motor without a load, or an amplifier without a speaker, or a power outlet in your house with nothing connected. Or a speaker without a woofer.... Without a woofer, the crossover diverts the low-frequency voltage to the woofer connections, but with nothing connected no current flows and there is no low frequency energy/wattage.

Where does the extra energy come from when boosting?

With an active filter it comes from an amplifier, which is what makes it "active" and able to boost.

With digital EQ or filtering, amplification is simply numerical multiplication and any additional energy comes from the analog electronics.

 

[fpitas]

Analog (electrical) filter can block or dissipate a portion of the signal energy. If dissipated, heat. If blocked, maybe less dissipation. Look for resistors, capacitors, and coils in this application.

Generally, unless you have a diplexer, the energy is simply blocked without dissipation.

 

[Curvature]

A digital filter changes the digits... I would rate this "no energy change" in the way you are asking.

I would imagine the math says something different. A filter can't arbitrarily change values without introducing errors.

What I'm asking about is the actual operation. Is it, for example, convolution? Not that I know the details of how that works. I would guess it's something like noise-shaping, where you have the same amount of information/energy, but it is pushed to a different band.

The reason why I'm asking is because I want a better mental picture of what's happening.

Generally, unless you have a diplexer, the energy is simply blocked without dissipation.

What is the difference between blocked and dissipated?

Imagine a motor without a load, or an amplifier without a speaker, or a power outlet in your house with nothing connected. Or a speaker without a woofer.... Without a woofer, the crossover diverts the low-frequency voltage to the woofer connections, but with nothing connected no current flows and there is no low frequency energy/wattage.

I think you in this example you are skipping the part I'm actually asking about: what happens in the crossover? I know what a crossover "does" and how to operate it. What I don't understand is the itty-bitty physics of what's happening inside it.

 

[fpitas]

What is the difference between blocked and dissipated?

Blocked in this case means does not dissipate energy. Dissipated energy would take amplifier power and heat the filter.

 

[fpitas]

For example, the tweeter highpass filter in a speaker does not dissipate bass energy, but presents a high impedance.

 

[RayDunzl]

I would imagine the math says something different. A filter can't arbitrarily change values without introducing errors.

I'm sure someone with a higher pay scale than myself will happily disagree and explain more profusely.

A digital filter introduces error (as compared to the original signal) - more highs, more lows, more phase, or less... whatever.

By changing the digits of the source data.

The digits represent a waveform. change the waveform by changing the digits.

 

[Curvature]

Blocked in this case means does not dissipate energy. Dissipated energy would take amplifier power and heat the filter.

Bear with me.

AFAIK, high resistance implies dissipation. Is that accurate?

Impedance indicates that something frequency specific and involving phase is happening.

Does that then mean that "blocked" energy is either "reflected" or "canceled", given the involvement of phase?

 

[fpitas]

Bear with me.

AFAIK, high resistance implies dissipation. Is that accurate?

Impedance indicates that something frequency specific and involving phase is happening.

Does that then mean that "blocked" energy is either "reflected" or "canceled", given the involvement of phase?

Well...no. In the case of a speaker you would see a real impedance (meaning, resistive) of a few ohms if the network was dissipative. In reality the impedance is purely reactive, generally high, outside the passband.

Impedance simply includes real and imaginary parts; pure resistance has a phase angle of 0 degrees: https://en.wikipedia.org/wiki/Electrical_impedance

By blocked energy I simply meant you would see a voltage there, but no energy would be absorbed and dissipated.

 

[voodooless]

I would imagine the math says something different. A filter can't arbitrarily change values without introducing errors.

Errors is not energy. The concept of energy isn’t really applicable in the digital domain. The only energy consumed is the one needed for the calculations. There is some correlation between the type of filter and the energy needed for the operations, but it’s quite a fuzzy one, and not really relevant to the presumed energy of the audio signal

What I'm asking about is the actual operation. Is it, for example, convolution?

Convolution is one type, infinite impulse response is another. There are other types, but these are the main ones.

I would guess it's something like noise-shaping, where you have the same amount of information/energy, but it is pushed to a different band.

That’s not what is happening. These filters actually remove information content from the digital data. The data is gone, not just placed somewhere else.

 

[fpitas]

Not sure how far you want to pursue, but this is a practical yet inexpensive look at filters:

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[Geert]

Does that then mean that "blocked" energy is either "reflected" or "canceled", given the involvement of phase?

'Reflection' is only applicable to high frequencies where transmission line theory applies (wire lenght more than about 1/10 of a wavelength). Then reflection happens when filters cause an impedance mismatch.

 

[voodooless]

For example, the tweeter highpass filter in a speaker does not dissipate bass energy, but presents a high impedance.

An L-pad does however. For the layman: it attenuates the signal by a fixed ratio. Whatever energy is not transmitted by the tweeter is lost as heat in a resistor.

There are other complex schemes that do not filter anything in the frequency domain, but do burn energy. For instance a impedance linearization network may flatten the impedance of a speaker driver. This will not change the frequency response, but will make the amp put out more power in a limited frequency band, giving it a more constant load overall (tubes will be happy). These filters convert this energy to heat as well.

… now I have the feeling that I’ve only made it more complicated

 

[fpitas]

… now I have the feeling that I’ve only made it more complicated

It will never be straightforward to explain.

 

[Cbdb2]

Power transfer is from the current and the voltage which are related by the impedance, ohms law. So if the filters impedance changes with freq. so does the power.
Say you have a 1volt signal going into a pair of freq. dependant resistors (impedances) in series (the same current in both) and take the output between the two, a voltage divider. At 100hz one resistor is 1Ω and the other is 100Ω. If you use the signal across the 100Ω as output it will be close to 1volt.
Now we increase the freq. to 1khz where both "resistors" are 50Ω and now the output is only 1/2volt.
At 10khz the impedances are 100Ω and 1Ω the output would be close to .01volts. You have a low pass filter. If you took the output from the other resistor you would have a high pass filter.
Real filters use caps/inductors as the variable resistors (impedances), but these are reactive (they don't burn energy and they cause phase shifts) so things are more complicated than above, but thats the jist.
To really understand you need to do the math.

 

[Curvature]

I appreciate the efforts here.

The textbooks use math to describe the events. That's nice, although nicer would be the clumsy and longwinded translation into boring old regular language.

 

[AnalogSteph]

The pesky math has the upside of giving a quantitative description alongside a qualitative one.

Anyway, a passive filter can obviously never have more power going out than is coming in. It may be less if there are some resistive losses as usually the case (and for sure if there is attenuation involved). If you are getting a boost, there will be an impedance transformation of sorts involved. In other words, you may be getting more voltage out but this is afforded by presenting a lower-impedance load to the amplifier output which has to be delivering correspondingly more current instead. The voltage to current ratio is, guess what, impedance.

Impedance transformation is something that people tend to think of routinely in the RF world, not so much in the "basically DC" world. It is still feasible when LC circuits are involved though.

If you've got active filters, the active components may be pumping in energy from the power supplies when needed. And digital is a different beast altogether.... that's just math.

 

[voodooless]

Anyway, a passive filter can obviously never have more power going out than is coming in.

Only on average Inductors and capacitors are energy storage devices. They can store energy when available, and release it at some other point in time.

 

[AnalogSteph]

Only on average Inductors and capacitors are energy storage devices. They can store energy when available, and release it at some other point in time.

Which is why they are particularly useful for power supply applications when combined with switches.

Speaking of which, a filter could literally be containing a transformer and make the impedance transformation dead obvious right there. An MC step-up transformer (or same for a ribbon mic) would be a classic example.