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How do RC high-pass and low-pass filters select frequencies, and what sets the cut-off frequency?

Passive filters: the RC low-pass and high-pass filter, the cut-off (break) frequency, the frequency response and the half-power point, gain in decibels, and the Bode plot.

A focused answer to WJEC A-Level Electronics passive filters, covering the RC low-pass and high-pass filter, the cut-off (break) frequency, the frequency response and the minus 3 dB half-power point, gain expressed in decibels, and the Bode plot.

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  1. What this dot point is asking
  2. The answer
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What this dot point is asking

Filters select which frequencies pass and which are blocked, and they appear throughout communications, audio and signal conversion. WJEC expects you to describe the RC low-pass and high-pass filter, calculate the cut-off frequency, explain the frequency response and the minus 3 dB half-power point, express gain in decibels, and sketch a Bode plot. The cut-off calculation and the decibel/half-power argument are reliable, high-mark exam content.

The answer

Low-pass and high-pass RC filters

The cut-off frequency

The cut-off (break) frequency is where the capacitor's reactance equals the resistance. At this frequency the output power has fallen to half its passband value, hence the half-power point.

Decibels and the minus 3 dB point

At the cut-off, the output power is half the input, which is 10log⁡10(0.5)=−3 dB10\log_{10}(0.5) = -3\,\text{dB}. In voltage terms the output is 12=0.707\frac{1}{\sqrt{2}} = 0.707 of the input, also −3 dB-3\,\text{dB}. This is why the cut-off is the minus 3 dB point.

The Bode plot and roll-off

A Bode plot graphs gain in decibels against frequency on a logarithmic scale. Beyond the cut-off, a single-RC filter rolls off at 6 dB per octave (the gain drops by 6 dB each time the frequency doubles), a straight line on the log plot.

Examples in context

Example 1. Removing mains hum from audio
A high-pass filter with a cut-off around 20 Hz lets music through but attenuates 50 Hz mains hum picked up on the cables. Choosing the cut-off just below the lowest wanted note keeps the bass while removing the rumble.
Example 2. The reconstruction filter on a DAC
A low-pass filter after a DAC removes the high-frequency steps of the staircase output, smoothing it to a continuous waveform. Its cut-off is set above the highest signal frequency but below the sampling frequency, so the signal passes and the stepping is removed.
Example 3. Tone control in an amplifier
A simple treble control is a high-pass filter whose cut-off can be varied, boosting or cutting the high frequencies relative to the rest. The Bode plot makes the effect clear: moving the cut-off shifts where the gain begins to fall.

Try this

Q1. An RC low-pass filter has R=10 kΩR = 10\,\text{k}\Omega and C=10 nFC = 10\,\text{nF}. Find the cut-off frequency. [2 marks]

  • Cue. fc=12πRC=12π×104×10−8=1,590 Hzf_c = \frac{1}{2\pi RC} = \frac{1}{2\pi \times 10^4 \times 10^{-8}} = 1{,}590\,\text{Hz} (about 1.6 kHz1.6\,\text{kHz}).

Q2. State the output voltage as a fraction of the input at the cut-off frequency, and the corresponding gain in decibels. [2 marks]

  • Cue. 0.7070.707 of the input (that is 12\frac{1}{\sqrt{2}}), which is minus 3 dB.

Exam-style practice questions

Practice questions written in the style of WJEC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WJEC Eduqas 20215 marksAn RC low-pass filter uses a resistor of 1.6 kΩ1.6\,\text{k}\Omega and a capacitor of 100 nF100\,\text{nF}. Calculate the cut-off frequency, and explain what happens to signals well above and well below this frequency.
Show worked answer →

The cut-off frequency of an RC filter is where the resistance equals the reactance.

Cut-off: fc=12πRC=12π×1600×100×10−9=11.0×10−3=1,000 Hzf_c = \dfrac{1}{2\pi R C} = \dfrac{1}{2\pi \times 1600 \times 100 \times 10^{-9}} = \dfrac{1}{1.0 \times 10^{-3}} = 1{,}000\,\text{Hz}.

Well below 1 kHz1\,\text{kHz} the capacitor's reactance is high, so most of the signal appears across it (the output), and the filter passes the signal with little attenuation.

Well above 1 kHz1\,\text{kHz} the reactance is low, so little voltage appears across the capacitor and the output falls; the filter attenuates high frequencies, rolling off at 6 dB per octave.

Markers reward the cut-off of 1 kHz1\,\text{kHz}, low frequencies passed, and high frequencies attenuated above the cut-off.

WJEC Eduqas 20194 marksA filter has an output that is half the power of its input at the cut-off frequency. Express this as a gain in decibels, and explain why the cut-off is called the minus 3 dB point.
Show worked answer →

Gain in decibels for a power ratio is G=10log⁡10(PoutPin)G = 10 \log_{10}\left(\dfrac{P_{out}}{P_{in}}\right).

At the cut-off the output power is half the input: G=10log⁡10(0.5)=10×(−0.301)=−3.0 dBG = 10 \log_{10}(0.5) = 10 \times (-0.301) = -3.0\,\text{dB}.

It is called the minus 3 dB point because the power has dropped to half (a 3 dB fall) at that frequency. In voltage terms the output has fallen to 12=0.707\dfrac{1}{\sqrt{2}} = 0.707 of the input, which also corresponds to minus 3 dB.

Markers reward the decibel calculation giving minus 3 dB, the half-power meaning, and the 0.707 voltage ratio.

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