Calculate RL High-pass Filter

Calculator and formulas for calculating the parameters of an RL high-pass filter

Calculate RL High-pass Filter

RL High-pass Filter

This function calculates the properties of a high-pass filter made from a resistor and an inductor. It calculates the output voltage, attenuation, and phase shift at the given frequency.

Results
Reactance XL:
Output voltage:
Attenuation dB:
Phase shift φ:

RL High-pass Filter

RL High-pass Circuit
Parameters
\(\displaystyle L\) = Inductance [H]
\(\displaystyle R\) = Resistance [Ω]
\(\displaystyle U_1\) = Input voltage [V]
\(\displaystyle U_2\) = Output voltage [V]
\(\displaystyle X_L\) = Inductive reactance [Ω]
\(\displaystyle f_c\) = Cutoff frequency [Hz]
\(\displaystyle φ\) = Phase angle [°]

Formulas for RL High-pass Filter

Calculate Voltage Ratio

The output voltage U2 of an RL high-pass filter is calculated using the following formula.

Output Voltage
\(\displaystyle U_2=U_1 \cdot\frac{2 \cdot \pi \cdot f \cdot L} {\sqrt{R^2 + (2 \cdot \pi \cdot f \cdot L)^2}}\)

or simpler, when XL is known:

\(\displaystyle U_2=U_1 \cdot\frac{X_L}{\sqrt{R^2 + X_L^2}}\)
Reactance
\(\displaystyle X_L=2 \pi \cdot f \cdot L\)

The inductive reactance increases proportionally with frequency.

Attenuation in Decibels

The attenuation is 3dB at the cutoff frequency. It can be calculated for different frequencies using the formulas below.

Attenuation (simple)
\(\displaystyle V_u=20 \cdot \lg \left(\frac{U_2}{U_1} \right)\)

When input and output voltages are known.

Attenuation (complex)
\(\displaystyle V_u=20\cdot\lg\left(\frac{\omega \cdot L} {\sqrt{R^2 + (\omega \cdot L)^2}}\right)\)

Direct calculation from R, L and ω.

Phase Shift

In an RL high-pass filter, the output voltage leads the input voltage by 0° - 90° depending on frequency. At the cutoff frequency, the phase shift is 45°.

Phase angle (simple)
\(\displaystyle \phi=\arccos \left(\frac{U_2}{U_1} \right)\)
Phase angle (complex)
\(\displaystyle \phi= \arctan \left(\frac{R}{\omega \cdot L}\right)\)
High-pass Phase Response

Cutoff Frequency

At cutoff frequency fc, the value of the amplitude-frequency response equals 0.707. This corresponds to –3dB.

Cutoff Frequency Formulas
\(\displaystyle \omega_c= \frac{R}{L} \Rightarrow f_c=\frac{R}{2\cdot\pi\cdot L}\)
\(\displaystyle R=2\cdot\pi\cdot f_c\cdot L\)
\(\displaystyle L=\frac{R}{2\cdot\pi\cdot f_c}\)

Calculation Examples

Example 1: Output Voltage

Given: R = 10Ω, L = 2mH, f = 500Hz, U₁ = 10V

1. \(X_L = 2\pi \cdot 500 \cdot 0.002 = 6.28Ω\)
2. \(U_2 = 10 \cdot \frac{6.28}{\sqrt{10^2 + 6.28^2}} = 5.33V\)

Result: Output voltage = 5.33V

Example 2: Cutoff Frequency

Given: R = 1kΩ, L = 100mH

\(f_c = \frac{R}{2\pi \cdot L} = \frac{1000}{2\pi \cdot 0.1} = 1592Hz\)

Result: Cutoff frequency ≈ 1.59kHz

Example 3: Attenuation

Given: U₁ = 10V, U₂ = 7.07V

\(V_u = 20 \cdot \lg\left(\frac{7.07}{10}\right) = 20 \cdot \lg(0.707) = -3dB\)

Result: Attenuation = -3dB (cutoff frequency)

Example 4: Phase Shift

Given: R = 100Ω, ω×L = 100Ω

\(\phi = \arctan\left(\frac{R}{\omega \cdot L}\right) = \arctan\left(\frac{100}{100}\right) = 45°\)

Result: Phase shift = 45° (at cutoff frequency)

Practical Applications

Audio Technology
  • Treble filters
  • Crossover networks
  • DC decoupling
  • Rumble filters
Signal Processing
  • Differentiator
  • Edge enhancement
  • AC coupling
  • Noise filters
Measurement Technology
  • Common mode rejection
  • Offset filters
  • High-frequency couplers
  • Interference suppression

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Inductance  •  Reactance of a coil  •  Cutoff frequency R/L  •  Differentiator R/L  •  Highpass filter R/L  •  Lowpass filter R/L  •  Series circuit R/L  •  Parallel circuit R/L  •  Transformer  •