Worked solution · University · Circuit theory and network analysis

RC low-pass filter: cutoff frequency and Bode plot

Deriving the cutoff frequency of an RC low-pass filter from first principles, then sketching the Bode plot it produces.

A low-pass filter built from a single resistor and capacitor is the simplest frequency-selective circuit there is, and it turns up everywhere from anti-aliasing filters to audio tone controls. Given real component values, we derive the cutoff frequency from the voltage-divider transfer function, then read the Bode plot straight off the result.

Problem. A resistor R = 3.3 kΩ is placed in series with a capacitor C = 47 nF, with the output voltage taken across the capacitor. Find the cutoff frequency of this low-pass filter, and sketch its Bode magnitude plot.

R and C form a voltage divider. Vout is taken across the capacitor.

Setting up the transfer function

H(j\omega) = \dfrac{V_{out}}{V_{in}} = \dfrac{Z_C}{R + Z_C}

R and C form a voltage divider; the capacitor's impedance takes the place of a second resistor.

Z_C = \dfrac{1}{j\omega C}

A capacitor's impedance falls as frequency rises, which is what gives the filter its shape.

H(j\omega) = \dfrac{1}{1 + j\omega RC}

Substitute Zc, then multiply top and bottom by jωC to clear the fraction.

Finding the cutoff frequency

|H(j\omega)| = \dfrac{1}{\sqrt{1 + (\omega RC)^2}}

The magnitude of a complex ratio is the ratio of the magnitudes.

|H(j\omega_c)| = \dfrac{1}{\sqrt2} \ \Rightarrow\ \omega_c RC = 1

Cutoff is defined as the point where output power has halved: gain has fallen to 1/√2.

f_c = \dfrac{\omega_c}{2\pi} = \dfrac{1}{2\pi RC}

Converting angular frequency ω (rad/s) to frequency f (Hz).

f_c = \dfrac{1}{2\pi (3300)(47\times10^{-9})}

R = 3.3 kΩ, C = 47 nF.

f_c \approx 1026\ \text{Hz} = 1.03\ \text{kHz (3 s.f.)}

This is the −3 dB point: the frequency at which the gain has fallen to 1/√2 of its low-frequency value.

Phase response

The same transfer function also fixes the phase shift between output and input, which is the other half of a full Bode plot:

\phi(\omega) = -\arctan(\omega RC)

At fc the phase lag is exactly 45°. Well below fc it is close to 0°; well above it, close to 90°.

The Bode plot

Flat below fc, then falling 20 dB per decade above it — the actual curve passes 3 dB below this asymptote right at fc.

The same voltage-divider method extends directly to high-pass and band-pass filters — only which component the output is taken across changes. It's the building block behind most of circuit theory's frequency-response questions, and behind signals and systems' treatment of filtering more generally.

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