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Transfer function of RC low-pass filter

Short answer

Transfer function of RC low-pass filter: H(jω) = 1/(1 + jωRC). From KVL, RC dy/dt + y = x; taking Fourier transform gives H(jω). Cutoff frequency ω_c = 1/RC. Magnitude |H| = 1/√(1+(ωRC)²), phase = -arctan(ωRC). Low frequencies pass, high attenuated.

Derive the transfer function of a RC low pass filter and plot the magnitude and phase.

        R
 x(t) o--/\/\/--+--------o
  +             |        +
               === C    y(t)
  -             |        -
 ---o-----------+--------o

Transfer function

By KVL, with i(t)=C dydti(t) = C\,\dfrac{dy}{dt}:

x(t)=RCdy(t)dt+y(t)x(t) = RC\frac{dy(t)}{dt} + y(t)

Taking the Fourier transform (with ddt↔jω\frac{d}{dt} \leftrightarrow j\omega):

X(jω)=jωRC Y(jω)+Y(jω)H(jω)=Y(jω)X(jω)=11+jωRC\begin{aligned} X(j\omega) &= j\omega RC\,Y(j\omega) + Y(j\omega) \\ H(j\omega) &= \frac{Y(j\omega)}{X(j\omega)} = \frac{1}{1 + j\omega RC} \end{aligned}

In Laplace form, H(s)=11+sRCH(s) = \dfrac{1}{1 + sRC}. (Same result from the voltage divider: H=1/jωCR+1/jωCH = \dfrac{1/j\omega C}{R + 1/j\omega C}.)

Magnitude and phase

∣H(jω)∣=11+(ωRC)2,∠H(jω)=−tan⁡−1(ωRC)|H(j\omega)| = \frac{1}{\sqrt{1 + (\omega RC)^2}}, \qquad \angle H(j\omega) = -\tan^{-1}(\omega RC)
ω\omega∣H∣\lvert H\rvert∣H∣\lvert H\rvert (dB)∠H\angle H
0011000∘0^\circ
1/RC1/RC1/2=0.7071/\sqrt2 = 0.707−3-3−45∘-45^\circ
10/RC10/RC0.09950.0995−20-20−84.3∘-84.3^\circ
→∞\to\infty→0\to 0slope −20-20 dB/dec→−90∘\to -90^\circ
|H|                         angle H
 1 |----.                  0 |----.
   |     `.                   |     `.
.707|- - - -*                -45|- - - -*
   |         `-.              |         `-.___
   |             `---.    -90 |- - - - - - - -
   +------+-----------> w     +------+--------> w
        1/RC                       1/RC

The cutoff (half-power, −3-3 dB) frequency is ωc=1/RC\omega_c = 1/RC; the circuit is a first-order low-pass filter.

✓ Answer checked by Claude Opus 5.5 (high effort)

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