# How do filters really work?

#### antcollinet

##### Master Contributor
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.
Sure - but they store and release energy at the frequency of the input signal. So the averaging period is only 1 cycle of the lowest applied frequency.

OP

#### Curvature

##### Addicted to Fun and Learning
The pesky math has the upside of giving a quantitative description alongside a qualitative one.
I agree. And no harm in sharing all the quant stuff alongside the description. What's hard is understanding/communicating the physical thing happening in all the detail if you never verbalize it.

The idea that energy has to go somewhere, come from somewhere, is very compelling. Redistribution rather than destruction.

#### voodooless

##### Master Contributor
Forum Donor
The idea that energy has to go somewhere, come from somewhere, is very compelling. Redistribution rather than destruction.
There is more. As others have explained it’s also about not generating the energy in the first place

##### Active Member
I agree. And no harm in sharing all the quant stuff alongside the description. What's hard is understanding/communicating the physical thing happening in all the detail if you never verbalize it.

The idea that energy has to go somewhere, come from somewhere, is very compelling. Redistribution rather than destruction.
I'll just repeat what the peeps have been saying....in a slightly different way.

Energy must be conserved, yes. But in the case of passive filters their impedance goes high during the transition to the stop-band. Thus, the power amplifier does not supply any energy into the system in that instance. Thus, no dissipation.

A simple example:
If you put a 10 ohm 1 watt resistor on your power amplifier output, it would get warm.
If you put a 1000 ohm 1 watt resistor on your power amplifier output, it would not.

Power amplifiers are very close to what's called a "voltage-source." It will supply the voltage to a load......but only supplies current relative to the resistance/impedance load.

#### MCH

##### Major Contributor
regarding digital filters, this, and the following in the series:

#### antcollinet

##### Master Contributor
I agree. And no harm in sharing all the quant stuff alongside the description. What's hard is understanding/communicating the physical thing happening in all the detail if you never verbalize it.

The idea that energy has to go somewhere, come from somewhere, is very compelling. Redistribution rather than destruction.
Imagine an amp driving a speaker at full power.

Imagine a switch in series with the speaker. The switch when open is effectively an infinite resistance/impedance.

When you open the switch All the energy stops. It doesn't get absorbed, it doesn't get redirected, it doesn't get reflected. It just stops.

This is the equivalent of when a filter blocks a particular frequency band. For those frequencies, the filter is high impedance - so the current is reduced and energy for those frequencies doesn't go to the load. It doesn't need to be redistributed.

#### dfuller

##### Major Contributor
Is attenuation a kind of redistribution of energy?
Easiest to understand this on passive filters. Generally it's just impedance increasing as you get further above/below the passband, so less and less current flows. It's a little different in active filters as they're just adding more feedback.
Where does the extra energy come from when boosting?
Given boosting is only possible with active filters, it's coming from the power supply of the EQ. Generally active filters rely on varying passively filtered loop gain.

As far as DSP, that's just numbers.

#### Cbdb2

##### Major Contributor
Easiest to understand this on passive filters. Generally it's just impedance increasing as you get further above/below the passband, so less and less current flows. It's a little different in active filters as they're just adding more feedback.
Sometimes. In a simple series resistor capacitor filter the impedance the source sees always goes down with frequency (current increases), the filtering is done because of the voltage division. If you take the voltage across the capacitor you have a low pass filter, if you take the voltage across the resistor you have a high pass filter.

#### dfuller

##### Major Contributor
Sometimes. In a simple series resistor capacitor filter the impedance the source sees always goes down with frequency (current increases), the filtering is done because of the voltage division. If you take the voltage across the capacitor you have a low pass filter, if you take the voltage across the resistor you have a high pass filter.
Yes, that's the other option is a voltage divider.

#### Cbdb2

##### Major Contributor
Aren't most passive filters voltage dividers?

#### solderdude

##### Grand Contributor
Is attenuation a kind of redistribution of energy?
It just increases the impedance so for the same voltage the current decreases. Assuming a series filter in series with the load, so voltage division due to the increased impedance of the filter (and thus decreased current) which thus leads to lower voltage in the load resistance and a larger voltage drop over the filter.
For LRC the impedance increases at (near) the resonance frequency of a parallel filter.
In a series filter (which can be in parallel to a source) energy is dissipated in heat in the series resistor or when not present in the components and draws power near (at) the resonance frequency.

Where does the extra energy come from when boosting?
When there is a boost this is a resonance. There is no extra energy there. The voltage can swing higher but the current that can be drawn is lower as the impedance is higher. When you load the circuit there peak voltage will drop because of the load.

Digital filtering is quite different even though, in the end, the effect could be the same.

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#### Cbdb2

##### Major Contributor
It just increases the impedance so for the same voltage the current decreases. Assuming a series filter in series with the load, so voltage division due to the increased impedance of the filter (and thus decreased current) which thus leads to lower voltage in the load resistance and a larger voltage drop over the filter.
Not necessarily. A series resistor cap filter will decrease in impedance with freq. so the current increases. The filter works by shifting the voltage from the cap to the resistor as the freq. goes up.

#### solderdude

##### Grand Contributor
Yep as I also explained...

In a series filter (which can be in parallel to a source) energy is dissipated in heat in the series resistor or when not present in the components and draws power near (at) the resonance frequency.

#### Mnyb

##### Major Contributor
Forum Donor
You must get a grip of the maths of it . To get the intuition for it.

It was a weird revelation in school when you realised that for a complex impedance ( one that’s not a pure resistor but have L and C ).

Voltage and current does not happen at the same time , if you apply an sinusoidal voltage the corresponding current drawn is offset by the phase angle.

Then L and C vary with frequency.

As we speak of energy get a grip on active power apparent power and reactive power .
Hint for heat and mechanical movements it’s the active part that counts.

OP

#### Curvature

##### Addicted to Fun and Learning
You must get a grip of the maths of it . To get the intuition for it.

It was a weird revelation in school when you realised that for a complex impedance ( one that’s not a pure resistor but have L and C ).

Voltage and current does not happen at the same time , if you apply an sinusoidal voltage the corresponding current drawn is offset by the phase angle.

Then L and C vary with frequency.

As we speak of energy get a grip on active power apparent power and reactive power .
Hint for heat and mechanical movements it’s the active part that counts.
I can do the equations. I don't have a handle on what is happening physically, deep down low.

OP

#### Curvature

##### Addicted to Fun and Learning
When there is a boost this is a resonance. There is no extra energy there.
Resonance implies storage and release. Is that correct here?

#### IPunchCholla

##### Addicted to Fun and Learning
Forum Donor
I can do the equations. I don't have a handle on what is happening physically, deep down low.
As someone whose math is from 32 years ago, I can do some of the math, but not fast or well enough to get an intuitive feel for it. What has helped immensely are programs like iCircuit, where you can visualize the flow of current and see changes in voltage. LTSpice has also been nice for taking the bits I have gleaned from iCircuit and building things where I can test my understanding. I don’t know that either of these will help with understanding what is happening deep down, but I do feel I am slowly developing an intuitive sense of simple circuits.

#### solderdude

##### Grand Contributor
Resonance implies storage and release. Is that correct here?
Yes, depending on the load and the drive method (voltage or current) you can get substantial rises in voltage with parallel circuits.
Inductors can store current (phase shift in one direction) and capacitors can store voltage (phase shift in the other direction).

You can mimic inductors with opamps + capacitors by the way (called gyrators) and is often used in equalizers. Inductors have the nasty habit of picking up magnetic fields (depends on construction and shielding) and can be bulky and have a high resistance while gyrators don't have these disadvantages but they usually are noisy.

In audio, filters usually are voltage driven and loaded so in most cases there is no peaking with voltages above the input voltage, just attenuation in the load.
It all has to do with phase of the current vs voltage.

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