The corner frequency of a circuit is the point where the output signal drops to 70.71% of its maximum voltage, which corresponds to a 3 dB reduction in gain. For a simple RC circuit, you find it with the formula: corner frequency = 1/(RC), where R is resistance in ohms and C is capacitance in farads. For an RL circuit, it’s R/L. But the concept extends well beyond these basic cases, and knowing how to find it from a formula, a transfer function, or a Bode plot will serve you in almost any filter design situation.
The RC and RL Formulas
For a first-order RC filter (low-pass or high-pass), the corner frequency in radians per second is:
ω_c = 1 / (R × C)
If you need the answer in hertz instead of radians per second, divide by 2π:
f_c = 1 / (2π × R × C)
For a first-order RL filter, the formula swaps in inductance:
ω_c = R / L
In both cases, the corner frequency depends entirely on the passive component values. A 1 kΩ resistor paired with a 1 μF capacitor gives ω_c = 1,000,000 / 1,000 = 1,000 rad/s, or about 159 Hz. Increasing the resistance or capacitance lowers the corner frequency; decreasing either one raises it.
What Happens at the Corner Frequency
The corner frequency isn’t an arbitrary choice. It marks the specific point where the output power drops to exactly half the maximum value. In voltage terms, the output is 1/√2 (about 0.707) of the input. In decibels, that’s -3 dB, which is why you’ll also hear it called the “3 dB frequency” or “half-power point.”
At this same frequency, the phase shift between input and output is exactly 45° for a first-order filter. Below the corner frequency, the phase shift is small. Above it, the shift approaches 90°. The 45° midpoint is another reliable way to confirm you’ve identified the right frequency.
Finding It From a Transfer Function
When you’re working with a transfer function rather than a physical circuit, the corner frequency lives at the pole of the function. A first-order low-pass transfer function looks like this:
H(s) = 1 / (1 + s/ω_c)
The trick is to look at the denominator and find where the real part equals the imaginary part. For an RC filter, the denominator contains the term (1 + jωRC). Setting the real part (1) equal to the imaginary part (ωRC) gives you ω_c = 1/RC. The frequency where those two parts are equal is your corner frequency, and it corresponds to the pole of the transfer function.
This principle scales to more complex circuits. Each pole in a transfer function creates a corner frequency, and each one produces its own break in the frequency response. A second-order system has two poles and can have two distinct corner frequencies, or two poles at the same frequency depending on the design.
Reading It From a Bode Plot
On a Bode magnitude plot, the corner frequency is where two straight-line asymptotes meet. For a simple low-pass filter, the plot is flat (0 dB slope) at low frequencies and drops at 20 dB per decade at high frequencies. The corner frequency sits right at the intersection of those two lines.
Here’s how the asymptotic approximation works. At frequencies well below the corner frequency, the gain is essentially flat because the reactive component (capacitor or inductor) has little effect. At frequencies well above the corner frequency, the gain falls steadily at 20 dB for every tenfold increase in frequency. The piecewise approximation follows the flat line up to the break point, then switches to the falling line. The actual response curve is smooth and sits about 3 dB below the intersection point at the corner frequency itself.
If you’re measuring a real circuit with a frequency sweep, look for the point where the output has dropped 3 dB from its passband level. That’s your measured corner frequency, and it should match the calculated value if your components are close to their rated values.
Second-Order and Active Filters
For second-order filters like the Sallen-Key topology, the corner frequency depends on all four passive components. The general relationship is:
ω_c² = 1 / (R1 × R2 × C1 × C2)
A common design approach starts by normalizing the cutoff frequency to 1 radian per second, choosing component ratios that give the desired filter shape (Butterworth, Bessel, Chebyshev), and then scaling all the capacitor values down by a factor of 1/ω_0 to shift the corner frequency to your target. The relationship ω_0 = 2π × f_c converts between angular frequency and hertz.
Second-order filters roll off at 40 dB per decade beyond the corner frequency, twice as steep as a first-order filter. The Bode plot shows the same asymptote intersection principle, but the high-frequency line falls twice as fast.
Cascaded Filter Stages
When you connect multiple identical filter stages in series to get a steeper rolloff, the overall corner frequency shifts lower than the corner frequency of any individual stage. This catches many designers off guard. If you cascade several first-order filters that each have the same cutoff, the combined 3 dB point ends up at a lower frequency because the small losses from each stage compound.
As a rough guideline, the overall corner frequency of a cascaded system is reduced by a factor of about 2.3 compared to the individual stages. To hit a specific overall corner frequency, you need to set each individual stage’s cutoff proportionally higher, then let the cascading effect bring the combined response down to your target.
Quick Reference for Common Cases
- First-order RC filter: ω_c = 1/(RC), or f_c = 1/(2πRC)
- First-order RL filter: ω_c = R/L, or f_c = R/(2πL)
- Second-order Sallen-Key: ω_c = 1/√(R1 × R2 × C1 × C2)
- From a Bode plot: find the intersection of the flat and sloped asymptotes, or measure the point 3 dB below the passband
- From a transfer function: find where the real and imaginary parts of the denominator are equal, or identify the pole locations
Whichever method you use, the result should agree. The formula, the transfer function pole, the Bode plot asymptote intersection, and the measured 3 dB drop all point to the same frequency. If they don’t line up, check your component tolerances or verify that parasitic capacitance and inductance aren’t shifting the response at higher frequencies.