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Cutoff Frequency of an RC Low-Pass Filter

A first-order RC low-pass filter has $R = 1.0\ \text{k}\Omega$ and $C = 1.0\ \mu\text{F}$ .

Find the $-3\ \text{dB}$ cutoff frequency in hertz.

Given

R	1 kohm	Resistance
C	1 uF	Capacitance

Hint 1

The cutoff is where the reactance of the capacitor equals the resistance.

Hint 2

f_c = 1/(2πRC).

Hint 3

RC = 1 kΩ × 1 μF = 1 ms, so the answer is near 159 Hz, not 1000.

Worked solution — try the problem first

$f_c = \frac{1}{2\pi RC}$

Convert to base units:

$RC = (1.0\times10^{3})(1.0\times10^{-6}) = 1.0\times10^{-3}\ \text{s}$

$f_c = \frac{1}{2\pi(1.0\times10^{-3})} = \frac{1}{6.283\times10^{-3}} = 159.2\ \text{Hz}$

$\boxed{f_c = 159\ \text{Hz}}$

What happens at $f_c$ . The output is $1/\sqrt2 = 0.707$ of the input in amplitude — that is the definition of the $-3$ dB point, since $20\log_{10}(0.707) = -3.01$ dB. The phase shift there is exactly $-45°$ . Above cutoff, a first-order filter rolls off at $-20$ dB/decade.

Angular vs ordinary frequency. $\omega_c = 1/RC = 1000$ rad/s. Reporting that as hertz overstates the cutoff by $2\pi$ .

Concepts:RC filtersCutoff frequencyFrequency responseDecibels

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