Why Is an RC Filter's Cutoff Frequency -3 dB?
Alternative titles
- RC Filters at Cutoff: Why the Signal Is 0.707, Not 0.5
- What the -3 dB Point Actually Means in a First-Order RC Filter
- From RC to 159 Hz: Understanding the First-Order Filter Transition
Short description / excerpt
The familiar RC cutoff formula gives a frequency, but the -3 dB label explains what that frequency means. Using a 10 kΩ resistor and 100 nF capacitor, this article connects the calculation to amplitude, power, low-pass and high-pass behaviour, and real circuit limits.
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electronics, embedded, engineering, beginners
Article
An RC filter is often introduced with one compact equation:
fc = 1 / (2πRC)
Put 10 kΩ and 100 nF into that equation and the result is 159.15 Hz. That part is easy to calculate. The more useful question is what the number means in a circuit.
Why is the point called “-3 dB”? Why not call it the frequency where the signal is half? And why do low-pass and high-pass filters share the same cutoff expression even though they pass opposite parts of the spectrum?
The answer is in the word transition. A first-order RC filter does not behave like an on/off switch. Its cutoff frequency is the conventional marker in a gradual change from passband to attenuation.
Start with one concrete RC value
For a resistor of 10 kΩ and a capacitor of 100 nF:
R = 10,000 Ω
C = 100 × 10^-9 F
fc = 1 / (2π × 10,000 × 100 × 10^-9)
= 159.154943... Hz
So a sensible engineering display is 159.15 Hz. This is the same cutoff frequency whether the parts are wired as a basic RC low-pass or a basic RC high-pass. The response around that point is different, but the RC time scale is the same.
The key number at cutoff is 1/√2
For an ideal first-order RC low-pass, the output magnitude is:
|HLP(f)| = 1 / √(1 + (f / fc)^2)
At f = fc, the ratio f / fc is 1. Substituting it gives:
|HLP(fc)| = 1 / √(1 + 1^2)
= 1 / √2
≈ 0.70710678
The high-pass magnitude is:
|HHP(f)| = (f / fc) / √(1 + (f / fc)^2)
At the same frequency, it also becomes 1 / √2. In other words, a low-pass has fallen to about 70.7% of its passband voltage amplitude at cutoff, while a high-pass has risen to about 70.7% of its passband amplitude.
That is why the same fc is meaningful for both topologies.
Why 0.707 becomes -3 dB
Decibels for a voltage or amplitude ratio are calculated as:
20 log10(Vout / Vin)
At cutoff, the amplitude ratio is 1 / √2:
20 log10(1 / √2) = -3.0103 dB
Engineers usually shorten that to -3 dB. The approximation is close enough for the label, but the distinction matters when explaining it.
“-3 dB” does not mean the voltage amplitude is half. Half voltage would be approximately -6.02 dB. At the cutoff of this RC filter, voltage amplitude is about 0.707 of the passband value.
Power is proportional to voltage squared when impedance is unchanged. Squaring the cutoff amplitude ratio gives:
(1 / √2)^2 = 1 / 2 = 0.5
So at the -3 dB point, the power is half of the passband power for the same impedance. This is the origin of the common “half-power point” phrase.
Reading the low-pass response
In a low-pass RC circuit, low frequencies see little attenuation. At frequencies well below 159.15 Hz in this example, the output is close to the input. At 159.15 Hz, the magnitude is 0.707. Above that, the capacitor increasingly shunts the output toward ground and the output falls gradually.
Far above cutoff, an ideal first-order low-pass rolls off at roughly -20 dB per decade. That is a slope, not a vertical edge. A 1 kHz component is attenuated more than a 200 Hz component, but neither disappears simply because it is above 159 Hz.
Reading the high-pass response
A high-pass RC circuit tells the complementary story. Very low frequencies are attenuated because the series capacitor presents a large reactance. At 159.15 Hz, the output has reached 0.707 of its high-frequency passband amplitude. At much higher frequencies, the capacitor’s reactance becomes small and the output approaches the input.
Far below cutoff, the ideal high-pass magnitude changes at about +20 dB per decade as frequency rises. Again, cutoff is the center of a transition, not a brick wall.
A useful way to check the intuition
It helps to calculate a few frequencies rather than treating fc as a magic boundary. For the 159.15 Hz low-pass:
- At 15.9 Hz, one tenth of cutoff, the amplitude is close to 0.995.
- At 159.15 Hz, it is 0.707.
- At 1.59 kHz, ten times cutoff, it is close to 0.0995.
The high-pass has those endpoint magnitudes reversed: close to 0.0995 at one tenth of cutoff and close to 0.995 at ten times cutoff. These values make the gradual transition visible.
You can vary R, C, topology, and a frequency probe with the free CircuitClarity RC Filter Cutoff Calculator. Watching the response update is a useful complement to the equation.
The ideal formula is a starting point
The formula assumes an ideal source and an ideal load. In a physical circuit, the source resistance can become part of the effective R, and the next stage’s input impedance can load the output. Resistor and capacitor tolerances shift the actual cutoff. Capacitor ESR and parasitic capacitance or inductance matter more as frequency rises.
Those effects do not invalidate fc = 1 / (2πRC). They explain why it is best used as the nominal design value, followed by a check of the full circuit and the component data.
The -3 dB point is therefore not a failure threshold. It is a clear, reproducible reference: the frequency where a first-order RC filter’s amplitude is 1 / √2, its same-impedance power ratio is one half, and its transition is well underway.
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