--- title: "Transfer function of RC low-pass filter" subject: "Signals and Systems" chapter: 5 key: "semester_4/signals_and_systems:rc-lowpass-transfer-function" source: https://www.bctengineer.com/q/semester_4/signals_and_systems/rc-lowpass-transfer-function site: BCT Engineering --- # Transfer function of RC low-pass filter Signals and Systems · Chapter 5: Continuous-Time Systems - Asked in: 2082 Kartik, 5 marks; 2083 Baisakh (new course), 4 marks (asked 2 times) - Video: [Transfer Function of RC Circuits: High Pass Filter and Low Pass Filter](https://www.youtube.com/watch?v=37zMAyxrPy8) — Engineering Funda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/rc-lowpass-transfer-function - Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-5 ## Question Derive the transfer function of a RC low pass filter and plot the magnitude and phase. ## 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. ## Answer ```text R x(t) o--/\/\/--+--------o + | + === C y(t) - | - ---o-----------+--------o ``` #### Transfer function By KVL, with $i(t) = C\,\dfrac{dy}{dt}$: $$ x(t) = RC\frac{dy(t)}{dt} + y(t) $$ Taking the Fourier transform (with $\frac{d}{dt} \leftrightarrow j\omega$): $$ \begin{aligned} X(j\omega) &= j\omega RC\,Y(j\omega) + Y(j\omega) \\ H(j\omega) &= \frac{Y(j\omega)}{X(j\omega)} = \frac{1}{1 + j\omega RC} \end{aligned} $$ In Laplace form, $H(s) = \dfrac{1}{1 + sRC}$. (Same result from the voltage divider: $H = \dfrac{1/j\omega C}{R + 1/j\omega C}$.) #### Magnitude and phase $$ |H(j\omega)| = \frac{1}{\sqrt{1 + (\omega RC)^2}}, \qquad \angle H(j\omega) = -\tan^{-1}(\omega RC) $$ | $\omega$ | $\lvert H\rvert$ | $\lvert H\rvert$ (dB) | $\angle H$ | |---|---|---|---| | $0$ | $1$ | $0$ | $0^\circ$ | | $1/RC$ | $1/\sqrt2 = 0.707$ | $-3$ | $-45^\circ$ | | $10/RC$ | $0.0995$ | $-20$ | $-84.3^\circ$ | | $\to\infty$ | $\to 0$ | slope $-20$ dB/dec | $\to -90^\circ$ | ```text |H| angle H 1 |----. 0 |----. | `. | `. .707|- - - -* -45|- - - -* | `-. | `-.___ | `---. -90 |- - - - - - - - +------+-----------> w +------+--------> w 1/RC 1/RC ``` The cutoff (half-power, $-3$ dB) frequency is $\omega_c = 1/RC$; the circuit is a first-order low-pass filter. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. H(jω) of the RC low-pass is: A. jωRC/(1 + jωRC) B. 1/(1 + jωRC) C. 1 + jωRC D. RC/(1 + jω) Answer: B. Fourier transform of x = RC y' + y gives this. (Key point: Transfer function) 2. |H(jω)| equals: A. 1/sqrt(1 + (ωRC)^2) B. 1/(1 + ωRC) C. sqrt(1 + (ωRC)^2) D. ωRC/sqrt(1 + (ωRC)^2) Answer: A. Magnitude of 1/(1 + jωRC). (Key point: Magnitude) 3. At ω = 1/RC, the gain and phase are: A. 0.5 and -90° B. 1 and 0° C. 0.707 (-3 dB) and -45° D. 0.707 and +45° Answer: C. At ω = 1/RC, |H| = 1/√2 and ∠H = -tan^{-1}(1) = -45°. (Key point: Cutoff) 4. As ω -> ∞, the phase and slope approach: A. 0° and 0 dB/decade B. +90° and +20 dB/decade C. -45° and -40 dB/decade D. -90° and -20 dB/decade Answer: D. First-order low-pass rolls off at -20 dB/dec with phase -> -90°. 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