13
$\begingroup$

What is it about a capacitor which allows it to filter frequencies?

I understand the construction of a high-pass RC filter, and the mathematics behind it, but I'm struggling to find an explanation of the physics behind the phenomenon.

In my mind I can picture the broad spectrum signal hitting the capacitor, but I feel like the "output" behaviour would be mush, not a controlled and predictable behaviour. I'm not a physicist, but I'd like to understand this problem better.

What is the physical behaviour which allows a capacitor to act as a high or low pass filter?

$\endgroup$
  • 4
    $\begingroup$ Imagine wiggling a springy stick with a weight at the end. If you wiggle slowly it will move more or less with your movements but if you wiggle very quickly your contributions will average out and have much less effect. $\endgroup$ – copper.hat Oct 28 at 20:34
  • $\begingroup$ Another great visual description, many thanks @copper.hat $\endgroup$ – gingerbreadboy Oct 31 at 9:42
  • $\begingroup$ From reading the answers below I'm starting to think a low pass filter is easier to visualise, even though they are kinda conceptually the same thing? Or am I crazy? $\endgroup$ – gingerbreadboy Oct 31 at 9:45

10 Answers 10

26
$\begingroup$

An capacitor has one intuitive property:

Its voltage can't change instantly since its voltage is dependent on the charge it has stored, and charge doesn't move at infinite speeds (there is always resistance somewhere), therefore you can't instantly charge up a capacitor without infinite current. More capacitance means less voltage for the same amount of charge, like how a wide bucket or a narrow bucket will have different water levels inside them for the same amount of water. It's also impossible to fill a bucket up instantly.

A corollary to this is that a capacitor cannot fully oppose a changing voltage being applied by a source. This is because if a voltage source is constantly changing, it's a moving target and the capacitor is always lagging behind. Because it is always lagging, the capacitor can never develop a voltage equal to the source to completely oppose it and stop all current from flowing. That means there is always some leftover voltage from the source not being opposed by the capacitor. That leftover voltage has to appear somewhere. That leftover voltage from the source must appear across other components in series with the capacitor and if a voltage appears across those components, then a current will flow.

But that is not to say it can't partially oppose the source voltage. It obviously does, and the fact that opposition is monotonic with frequency lets it behave as a high-pass filter.

  • If it were not monotonic, then it would be mush, but it would still be predictable mush and therefore useful (like a notch or comb filter or something).
  • If it charged up instantly, it would just always fully oppose the source voltage leaving no voltage drop across the other components in the circuit meaning no current flow meaning no signal which is just like an open circuit blocking the signal.
  • If it never charged up at all it would always have 0V across it which would allow the full source voltage to appear across the other components and just allow the signal to pass through, which would just be like a piece of wire.
| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Re "charge doesn't move at infinite speeds": But doesn't the model of the ideal capacitor break down if finite speed is assumed? $\endgroup$ – Peter Mortensen Oct 29 at 2:26
  • 1
    $\begingroup$ @PeterMortensen I see what you mean. What I said was a bit handwavey and I guess it does assume some resistance somewhere whether it be in the circuit or in the capacitor itself. Something to not allow infinite current without infinite applied voltage. $\endgroup$ – DKNguyen Oct 29 at 4:09
9
$\begingroup$

What is the physical behaviour which allows a capacitor to act as a high or low pass filter?

A capacitor alone cannot act as either.

To create a filter you need a combination of resistance and capacitance or inductance and capacitance (or RL). You need two immittances, at least one of which is reactive.

Let's take a practical example, an RC circuit.

enter image description here

This circuit has a certain frequency associated with it $\ \omega_c $ this is the frequency at which the impedance of the capacitor is the same as the impedance of the resistor, known as the "corner frequency", or the "crossover frequency" and by other names as well.

Above $\ \omega_c $ the impedance of the capacitor decreases at a rate which makes the magnitude of the frequency response of the transfer function of the circuit drop by 20 dB per dec.

Here is what it looks like simulated in LTSpice

enter image description here

I added a small arrow at the -3dB point. Notice that above this point the frequency response decreases at a rate of 20dB per. dec.

I can also demonstrate that you don't need a capacitor to create a filter, an inductor - resistor combination will work just as well. As stated in the beginning of my answer; You need at least two immittances, at least one of which is reactive.

enter image description here

The result is exactly the same.

But a capacitor, or any single immittance element alone, cannot act as a filter.

You can also look at it from a different perspective;

What is it usually that we do when implementing a digital filter? we integrate!

Looking at the equations for the RC filter, what does it do? well the resistor limits the current, and the capacitor integrates the current:

$\ V_{c}(t) = \frac{1}{C}\int I_c(t) \, dt $

It is hard to know what other people find intuitive, the two ways of thinking about it which I have explained above are the ones I find most intuitive.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Minor correction: "the frequency at which the impedance of the capacitor is the same as the impedance of the resistor" No it isn't. The magnitude of the impedance is the same, but there is that imaginary unit that makes a world of difference. If the impedance were equal, at wc the TF would be 1/2. It's 1/(1-j) and the absolute value is the well known 1/Sqrt[2]. $\endgroup$ – Peltio Oct 28 at 15:24
  • 1
    $\begingroup$ Oops, it's 1/(1-j) for the high pass filter, and 1/(1+j) for the low pass one, but the result is the same at wc and it's not what would happen with identical impedances. $\endgroup$ – Peltio Oct 28 at 19:02
  • $\begingroup$ Is the resistor really necessary? The wire itself will provide some resistance anyway. Seems like the capacitor is all you need. $\endgroup$ – Juan Perez Oct 31 at 19:49
  • $\begingroup$ @JuanPerez "The wire itself will provide some resistance anyway" Then you have also got a resistor, ie. you have got an RC filter. This concerns an ideal capacitor, if you could construct an ideal capacitor without any resistance and measure it without resistance, of course this is not possible in practice but it is certainly possible in theory and in equations. A capacitor alone cannot act as a filter. $\endgroup$ – Vinzent Oct 31 at 19:59
  • $\begingroup$ I thought that, as long as voltage changes much faster than the capacitor can charge to full capacity, there will always be considerable current. If the voltage changes so slowly that the capacitor can be (almost) fully charged at every instant, there will always be very little current. I don't see a resistor playing any role in that. Sorry if I'm making too many wrong assumptions $\endgroup$ – Juan Perez Oct 31 at 20:24
8
$\begingroup$

Imagine electricity as water in a pipe. The current can flow in either direction (direct current, DC) or one way then the other way (alternating current, AC). Now put a rubber membrane in the pipe. This is the capacitor. Now it will slow and then stop DC, but AC can still keep wobbling back and forth. In this way, capacitors block DC but enable AC.

Considering AC, if the frequency of oscillation of the water is low, the membrane has to stretch a fair way before it gets to return to its middle position and out the other way. So if the membrane is quite tight/inflexible, then it will inhibit low frequency oscillation. At higher frequency, the membrane doesn't have to move far before it gets to go back the other way. In this way, the tightness and flexibility provides varying resistive effect to different frequencies.

Not a perfect answer, but maybe helps with imagining what's going on.

| cite | improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ I love this analogy, it really helps me picture the behaviour. More so than graphs and equations. $\endgroup$ – gingerbreadboy Oct 28 at 18:43
  • $\begingroup$ @gingerbreadboy If that helps you picture the behaviour then you have got the wrong picture in your head. $\endgroup$ – Vinzent Oct 30 at 20:55
  • $\begingroup$ @vinzent Maybe, but is a more intuitive answer than some others. Whist I can believe that other answers are more robustly correct they kinda missed my point. As stated in the question, I understand the construction of the circuit and what the circuit does, I am looking for a way to picture this in my mind. Some answers here are quite dry and prescriptive, more like exam answers than a way to help a non-expert like myself. Thank you for taking the time to help though, all answers are appreciated. One day I hope I can get my head around the "hard" answers too. $\endgroup$ – gingerbreadboy Oct 31 at 9:30
  • $\begingroup$ @vinzent if you feel there is a better/more accurate analogy than this I would genuinely love to hear it, many thanks. $\endgroup$ – gingerbreadboy Oct 31 at 9:34
  • $\begingroup$ Interesting idea! I wonder how close this matches an RC circuit in equations? It probably has a bit of an inductive element to it as well because the membrane can oscillate. In particular there will be a resonant frequency in the fundamental drum mode. $\endgroup$ – Ryan Thorngren Oct 31 at 19:10
6
$\begingroup$

Sometimes for physical intuition, it's nice to think about the extreme cases. For instance, a zero frequency signal is just a DC voltage. If we send it through the RC high pass filter, the capacitor is just like a break in the circuit, and prevents any current from flowing.

Slightly more quantitatively, the capacitor equation $Q = CV$ implies that if we think about voltage as the driving force and current as the response, a capacitor acts as a differentiator: $I = dQ/dt = C dV/dt$ (while a resistor is a scalar and an inductor is an integrator). The physical content of the capacitor equation is just that the capacitor is in equilibrium, with the electrostatic charge repulsion exactly balancing the voltage.

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

Vinzent already explained about RC filter transfer charcteristics, which apply nicely to single frequency.

I will attempt to address the "broad spectrum signal" part of your question.

As you probably know, any kind of signal can be looked as a composition of single frequencies (see Fourier transform). Broad spectrum signal can (or actually any signal does) incorporate infinite amount of different frequencies - but this doesn't change anything. In RC (or any other) filter all single frequencies behave as described and resulting signal is just sum of same frequencies with modified amplitudes and phases. What does change is your signal waveform - you may get rectangular impulses rounded and similar effects.

This is not mush - this is controlled output, although on screen of oscilloscope it may not look like original signal at all :)

Depend on signal processing needs this waveform change can be irrelevant - or can be showstopper. If you process only frequency part(s) of your signal, then all is good. If you require that rectangular pulses remain more or less rectangular and so on, then you likely need another kind of filter than RC one.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Not all signals have an infinite amount of frequencies, if we consider only ones with non-zero amplitude. A pure sine wave has only that one frequency. A sum of two sine waves has only those two frequencies. Other shapes like a square wave do have harmonics all the way up to infinity if they're perfectly sharp. $\endgroup$ – Peter Cordes Oct 28 at 16:19
  • $\begingroup$ @PeterCordes Yes and no :) If we could have pure sine wave for infinite time, then yes, it would contain single frequency. Real waves tend to have finite duration, thereby they do include some other frequencies... Well, for any practical purposes such components are effectively zero. $\endgroup$ – Arvo Oct 28 at 16:26
  • $\begingroup$ The superposition principle makes it easier for this linear system. $\endgroup$ – Peter Mortensen Oct 29 at 2:30
1
$\begingroup$

Picture a parallel plate capacitor made of perfect electrical conductor, separated by a perfect insulator. Clearly if you hook up a DC voltage source to either side of this no current will flow, because it's an open circuit. There will, however, be a voltage across the plates, which is exactly the voltage of the source.

Time for a suspension of disbelief: imagine you paused time and replaced that voltage source with the same voltage, but reversed the polarity. If you then un-paused time you would have the plates each being the negative of the voltage at the terminals of your source and a short circuit connecting these plates to that source.

Slowing down this polarity reversal and making it periodic would just be typical sinusoidal AC current (and if you can imagine this working with a sinusoid I can use Fourier series to stretch your understanding to arbitrary(ish) signals).

If you have finite resistance between the ideal cap and the voltage source you can see why lower frequencies will pass less current than our extreme example where we flipped the source over in an instant. The difference in voltage between the cap plate and the voltage source is less (after a uniform interval of time) if the voltage is changing slower, so less current will flow.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks. I like this answer, but I'm going to need to read it 100 times to fully get my head around it :) $\endgroup$ – gingerbreadboy Oct 31 at 9:39
1
$\begingroup$

A capacitor along with a resistor can act like a filter because its impedance is frequency dependent and by division of voltage between resistor and capacitor it works.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

FWIW, I think of it as the capacitor having lower impedance at higher frequencies. As the frequency component of a signal gets higher, the capacitor in the RC filter diagram above looks more and more like a piece of wire, thus allowing more of the signal amplitude to be developed across the resistor. At low frequencies, the cap impedance is high, compared to the resistance of the resistor, so more signal appears across the capacitor, and less across the resistor. (We're taking our output across the resistor.)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks. I feel like this fills in a few more gaps in my foundation. $\endgroup$ – gingerbreadboy Oct 31 at 9:50
1
$\begingroup$

A capacitor is an open circuit. Direct current can't flow through it because the plates of the capacitor don't contact.

However, when the current is alternating (or a signal), the electric field induced by one plate induces a current in the other plate.

That current is proportional to the capacitance and the time rate of change of the voltage.

I = C * dV/dT

Capacitors are sometimes used with DC power sources to remove high frequency noise - they don't conduct the low frequency DC because they are open.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

29 October 2020

Let us address the crux of the question - the physics.

Electrical energy is stored in electric and magnetic field flux. In a capacitor, this energy is stored in the electric field lines that bridge the dielectric between the plates.

The charge on each plate is, conceptually, equal and opposite for a given measured potential. In fact, on one plate there is an excess of electrons (over and above the charge balance in the protons and electrons comprising the atoms of the metal), and on the other plate there is a deficit of electrons (conceptually 'holes' is you like). The opposite charges are attracted to one another, but can not neutralize owing to the surface energy barrier at the junction between the metal plates and the dielectric. The field lines express the force each charge experiences. The distance between the plates times the force is the work (energy) required to separate the charge, scaled by the permittivity of the dielectric.

The density of electrons on the negative plate is larger proximal to the dielectric, and is likewise smaller on the positive plate.

If you connect a capacitor in series to an infinite impedance amplifier input, the input will not load the connected plate. The signal on the amplifier input will faithfully reproduce the input signal at all frequencies for an ideal capacitor, irrespective of the charge state of the capacitor. The voltage across the capacitor will be fixed and unchanging.

In practice, there is leakage with some characteristic high impedance, typically on the order of a mega-ohm or more. Therefore low frequencies will be attenuated, since the slowly changing waveform will permit time for the plate attached to the amplifier to charge and discharge. This is a high pass filter, and it is commonly found coupling different stages of a multistage amplifier.

If the capacitor loads a signal line by connecting one capacitor terminal to ground, or any fixed voltage, a low pass filter will result. For example the distributed capacitance of a transmission line reacts with the distributed resistance to attenuate high frequency signals. In a transmission line there is also distributed series inductance which acts as an additional low pass filter.

The resistance, or impedance, is a measure of the average collision frequency of a charge subject to an electric field traversing the material. In a low impedance material the drift velocity is higher owing to infrequent collisions, and in a high impedance material the drift velocity is lower so charging takes longer, i.e. is delayed. When a signal is changing rapidly (high frequency) the charge carriers are mostly just wobbling back and forth due to collisions.

In the first case, this prevents charging from attenuating a signal, and in the second case, this causes a signal passed by the line to be attenuated by preventing the distributed stored charge from changing appreciably.

Ultimately, these delays, inductive or capacitive, are a consequence of the necessity to change the stored energy which is done by changing the magnetic or electric field. The limiting factor is the speed of light which will be lower than the free space value in a high permeability or high permittivity material.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.