First-Order RC and RL Filter Calculator: Cutoff, Phase and Response
Introduction
A first-order filter is the simplest building block there is: a
resistor and a reactive element, a single pole, with the entire behavior
summarized by one number—the cutoff frequency. From it follow the
20 dB-per-decade slope, −3 dB, and 45° phase shift at the
cutoff point.
These calculators quickly handle the calculations repeated every
time at the bench: sizing a filter, choosing the nearest standard
capacitor value, and knowing in advance what to expect from a
measurement at a given frequency.
| Circuit | Cutoff frequency | Time constant |
|---|---|---|
| RC | $$ f_c = \frac{1}{2\pi R C} $$ | $$ \tau = R C $$ |
| RL | $$ f_c = \frac{R}{2\pi L} $$ | $$ \tau = \frac{L}{R} $$ |
| General relationship | $$ f_c = \frac{1}{2\pi\tau} $$ | $$ \tau = \frac{1}{2\pi f_c} $$ |
Relationship between the two, valid for both:
$$ f_c = \frac{1}{2\pi\tau} $$Card 1 — RC: from R and C to cutoff frequency
RC low-pass and high-pass filters share the same cutoff frequency: what changes is which of the two components the output is taken across, not where the cutoff occurs. This card therefore applies to both.
$$ f_c = \frac{1}{2\pi R C} \qquad \tau = R C $$
Card 2 — Inverse RC: from fc and R to the capacitor
This is the design case: you know where you want the cutoff, you have chosen the resistor, and you need the capacitor. The theoretical result almost never matches a commercially available value, so the calculator also shows the nearest E12 value and—more importantly—the cutoff frequency you actually obtain when using it.
For a 1 kHz cutoff with R = 10 kΩ, the required value would be 15.9 nF: with an actual 15 nF capacitor the cutoff shifts to about 1.06 kHz, roughly 6% higher. If that 6% matters, the practical solution is to adjust the resistor rather than search for a capacitor value that does not exist.
$$ C = \frac{1}{2\pi f_c R} $$
Card 3 — RL: from R and L to cutoff frequency
The structure is the same as in the RC case, but the time constant is formed in the opposite way: it increases with inductance and decreases with resistance. An RL circuit with R = 100 Ω and L = 10 mH has the same 100 µs time constant as an RC circuit with 1 kΩ and 100 nF, and therefore the same cutoff frequency.
$$ f_c = \frac{R}{2\pi L} \qquad \tau = \frac{L}{R} $$
Card 4 — Response at a selected frequency
Given an RC pair and a test frequency, this card calculates attenuation and phase shift. This is the calculation needed to compare a Bode measurement with theory: if one decade above cutoff the low-pass response is not about −20 dB and −84°, something in the measurement chain is not behaving as expected.
At the cutoff frequency the familiar values apply: −3.01 dB and 45° of phase shift, lagging for the low-pass filter and leading for the high-pass filter.
Low-pass:
$$ \left|H(f)\right| = \frac{1}{\sqrt{1 + \left(f/f_c\right)^2}} \qquad \varphi = -\arctan\left(\frac{f}{f_c}\right) $$
High-pass:
$$ \left|H(f)\right| = \frac{f/f_c}{\sqrt{1 + \left(f/f_c\right)^2}} \qquad \varphi = \arctan\left(\frac{f_c}{f}\right) $$
Conversion to decibels:
$$ \left|H\right|_{\mathrm{dB}} = 20 \log_{10}\left|H(f)\right| $$
Reference values
| $$ f / f_c $$ | Low-pass | High-pass |
|---|---|---|
| 0.1 | −0.04 dB · −5.7° | −20.04 dB · +84.3° |
| 1 | −3.01 dB · −45.0° | −3.01 dB · +45.0° |
| 10 | −20.04 dB · −84.3° | −0.04 dB · +5.7° |
Transfer functions
| Filter | Magnitude | Phase |
|---|---|---|
| Low-pass | $$ \left|H(f)\right| = \frac{1}{\sqrt{1 + \left(f/f_c\right)^2}} $$ | $$ \varphi = -\arctan\left(\frac{f}{f_c}\right) $$ |
| High-pass | $$ \left|H(f)\right| = \frac{f/f_c}{\sqrt{1 + \left(f/f_c\right)^2}} $$ | $$ \varphi = \arctan\left(\frac{f_c}{f}\right) $$ |
| In decibels | $$ \left|H\right|_{\mathrm{dB}} = 20 \log_{10}\left|H(f)\right| $$ | — |