Why in a low-pass filter, the higher the input frequency, the lower the frequency passed by the capacitor?

I know that there is a formula Z(c) = 1/omegajC.

The higher the omega, the higher the frequency

But I need physical meaning

  • \$\begingroup\$ on an abstract level, magnitude of response = |1/(1+jω/ω_0)| , where ω_0 is the corner frequency, and this shows it directly. In a circuit, think of it like a voltage divider , |Vout| = Vin * |Z1 / (Z1+Z2)| = Vin * |R / (R + 1/jωC)|, it follows from that, you should get ω_0 = 1/RC ...... at higher frequency, the impedance of the "bottom" of the voltage divider is bigger, so it divides more \$\endgroup\$
    – Pete W
    May 15, 2021 at 19:15
  • 1
    \$\begingroup\$ @PeteW Thank you very much \$\endgroup\$
    – aleksandr
    May 15, 2021 at 20:14
  • 1
    \$\begingroup\$ @Александр2222 What do you mean by needing a "physical meaning?" Do you mean to understand the low-pass and high-pass filters in some kind of "physical way?" Or do you want to understand a capacitor in a physical way? If the latter, we have physical models that operate at various levels. We do NOT know ultimate reality, yet. (Probably never will.) But at what model level do you want to understand a capacitor, if that's it? Or.. can you just write more about what you want to understand? \$\endgroup\$
    – jonk
    May 15, 2021 at 20:46
  • \$\begingroup\$ @jonk Thanks. Answer is below \$\endgroup\$
    – aleksandr
    May 16, 2021 at 3:20

1 Answer 1


High frequencies pass through capacitors easily. Low frequencies don't.

For an input with both low and high frequencies a low pass filter will pass the low frequencies through, while the high frequencies are shunted to ground through the capacitor.

enter image description here

A high pass filter swaps the R and C:

enter image description here

Let's look at a low pass example. You can think of a capacitor as a frequency-dependent resistor. Its resistance is, as you noted in your question, 1/omega * C.

Omega is 2 * pi * freq. So for example, a 1uF capacitor's resistance at 100kHz is: 1/(2 * pi * 100000 * 1e-6) = 1.6 ohms, which is very low. At 10Hz that same capacitor has 16K ohms.

If our R is say, 1K ohm, then our two frequencies see two different circuits. Most of the 100KHz is attenuated, and most of the 10Hz gets through.

enter image description here

Finally, a capacitor is different from a resistor in that it introduces a phase delay, but that's a new topic.

  • \$\begingroup\$ Thanks. If we look at high-pass filter? \$\endgroup\$
    – aleksandr
    May 15, 2021 at 20:13
  • \$\begingroup\$ Thank you very much \$\endgroup\$
    – aleksandr
    May 16, 2021 at 3:21

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