A telephone filter keeps the 300 to 3,400 Hz voice band and rejects everything else, and a single resistor-capacitor section is too gentle for the job: at twice its cutoff frequency it has dropped only about 7 dB. A resonant LC circuit with a Q of 10 centred on 2,000 Hz holds the neighbouring 1,000 Hz and 3,000 Hz channels down by about 23.5 and 18.5 dB, enough for telegraph carriers but not for clean voice channels.
Part of Telecommunications
Low-pass, high-pass and band-pass circuits built from resistors, capacitors and inductors, sized for telephone voice and carrier channels.
A Filter Chooses Frequencies, And The Telephone Choice Is 300 To 3,400 Hz
Sound Wave Basics gives the reason: the 300 Hz lower cutoff rejects hum from power systems and the 3,400 Hz upper cutoff rejects hiss, while the band in between keeps the vowel resonances and enough consonant energy. Everything outside is something to remove, and a long line, a rectifier or a neighbouring channel all supply some. The basic circuits and their formulas are in RC and RL Circuits and Capacitance and Inductance; this article sizes them for the telephone job.
One RC Section Cuts Gently
For a low-pass RC filter the corner frequency is f_c = 1 / (2π × R × C). The response relative to the input is 1 / √(1 + (f / f_c)²).
Design a 3,400 Hz low-pass with C = 10 nF (0.01 µF).
- R = 1 / (2π × 3,400 × 10 × 10⁻⁹)
- 2π × 3,400 = 21,362; × 10⁻⁸ = 2.136 × 10⁻⁴
- R = 1 / 2.136 × 10⁻⁴ = 4,681 Ω, so use 4.7 kΩ
How much does it reject?
- At f_c: 1 / √2 = 0.707, which is 3 dB down
- At 2 × f_c = 6,800 Hz: 1 / √5 = 0.447, which is 7.0 dB down
- At 10 × f_c = 34,000 Hz: 1 / √101 = 0.0995, which is 20 dB down
Seven decibels one octave beyond the voice band is a gentle roll-off. It keeps hiss under control only slowly.
A matching high-pass at 300 Hz. For C = 0.1 µF, R = 1 / (2π × 300 × 10⁻⁷) = 1 / (1.885 × 10⁻⁴) = 5,305 Ω. Mains hum at 60 Hz is 60 / 300 = 0.2 of the corner. The response is 0.2 / √(1 + 0.2²) = 0.196, which is 14 dB down, so hum is reduced but still audible. A single section is not enough; hum control needs more sections in cascade.
A Resonant LC Circuit Gives A Narrow Band
A coil and capacitor resonate at f₀ = 1 / (2π√(LC)). Choose a 2,000 Hz carrier slot as in Relay Circuits and C = 0.1 µF:
- L = 1 / ((2π × 2,000)² × C)
- (2π × 2,000)² = (12,566)² = 1.579 × 10⁸
- L = 1 / (1.579 × 10⁸ × 10⁻⁷) = 1 / 15.79 = 0.0633 H, about 63 mH
The width of the band is f₀ / Q. A good LC circuit has a Q of 100 to 500 (Oscillator Circuits), which at 2,000 Hz would be a band only 4 to 20 Hz wide. That is far narrower than a voice, and good for a telegraph carrier. A loaded Q of 10 gives 2,000 / 10 = 200 Hz.
For a resonant circuit the response is 1 / √(1 + Q² × (f / f₀ − f₀ / f)²). With Q = 10 and f₀ = 2,000 Hz:
- At 1,000 Hz: f/f₀ − f₀/f = 0.5 − 2 = −1.5; Q × 1.5 = 15; response 1 / √226 = 0.0665, which is 23.5 dB down
- At 3,000 Hz: 1.5 − 0.667 = 0.833; × 10 = 8.33; response 1 / √70.4 = 0.119, which is 18.5 dB down
- At 1,500 Hz and 2,500 Hz the response is only 15.4 and 13.3 dB down, so channels closer than 1,000 Hz apart would leak
A voice channel is 3,100 Hz wide. At a centre frequency f₀ the Q to match that width is f₀ / 3,100, a low figure, and a single low-Q circuit has gently sloping sides. Real channel filters therefore use several sections.
The Loading Coil Is An Accidental Low-Pass Filter
Inductance in series and capacitance in shunt is a low-pass section (Capacitance and Inductance). A loaded line is a ladder of them, which is why it passes voice and blocks higher frequencies; see Wire Line Systems.
What You Need
- Capacitors with tolerance you have checked on a meter; film or ceramic, not electrolytic, for filters
- Inductors wound on a core, with the inductance measured against a known capacitor by finding the resonant frequency
- An audio oscillator and a voltmeter to sweep a response (Oscillator Circuits)
- A table of measured level at each test frequency, since calculation only gets the design near
Method: Prove A Band-Pass Filter
Step 1: Build. Wind the coil to the calculated inductance, using the measured inductance, not the turns count alone.
Step 2: Sweep. Feed a steady tone through the filter at the centre frequency and note the level. Move to each neighbouring slot and note how far down it is.
Step 3: Compare with the calculation. A difference of several dB usually means a low-Q coil or a loading effect from the next stage.
Step 4: Adjust. Trim the capacitor to centre the response on the slot.
Common Mistakes
- Expecting a steep cutoff from one section. The 7 dB at twice the corner is the answer.
- Ignoring Q. A sloppy coil makes the filter wide and shallow.
- Letting the next stage load the filter. Its input resistance changes Q and shifts the band.
- Judging by ear. Measure the neighbouring slot.
What’s Next
- Learn next: Carrier Systems, where filters separate channels
- Combine with: Modem Concepts, where a pair of filters tells two tones apart