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RC high pass filter operation
Short answer
An RC high-pass filter has series C and shunt R. It passes high frequencies (Xc small, Vout≈Vin) and attenuates low frequencies (Xc large, Vout≈0). Transfer function H(jω)=jωRC/(1+jωRC), cut-off fc=1/(2πRC), gain rises at +20 dB/decade below fc.
Draw the circuit diagram of RC high pass filter and explain its operation with the help of frequency dependent response at the output.
Circuit diagram
Vin o----| |----+----o Vout
C |
[ R ]
|
GND o-----------+----o GND
A capacitor C is in series with the signal path and the resistor R is connected across the output. A high pass filter (HPF) passes frequencies above a cut-off frequency and attenuates lower frequencies.
Operation
- The reactance of the capacitor is Xc = 1/(2πfC).
- At low frequency Xc is very large. Most of the input voltage drops across C and very little appears across R, so Vout ≈ 0. For dc (f = 0) the capacitor is an open circuit and Vout = 0.
- At high frequency Xc is very small. The capacitor behaves like a short circuit, the whole input appears across R and Vout ≈ Vin.
Transfer function
Using the voltage divider (the output is taken across R):
H(jω) = Vout/Vin = R / (R + 1/jωC) = jωRC / (1 + jωRC)
With cut-off frequency fc = 1/(2πRC) (ωc = 1/RC):
- |H| = (ω/ωc) / √(1 + (ω/ωc)²)
- Phase φ = 90° − tan⁻¹(ω/ωc)
Frequency response
| Frequency | Magnitude | Phase |
|---|---|---|
| f ≪ fc | ≈ f/fc, rises at +20 dB/decade | → +90° |
| f = fc | 0.707 (−3 dB) | +45° |
| f ≫ fc | 1 (0 dB) | → 0° |
|H|
1 | ____________
| .-'
0.707|- - - - - .'
| .-'
| __.-'
+-----------|---------------- f
fc
- Below fc the output rises at 20 dB per decade.
- At fc the gain is 0.707 of the maximum (−3 dB point).
- Above fc the gain is nearly flat at 1. The output leads the input in phase.
Uses: coupling and dc-blocking capacitors in amplifiers, removing low-frequency hum or drift, and as a differentiator circuit when RC is small.
✓ Answer checked by Claude Opus 5.5 (high effort)
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