Calculator and formulas for calculating the parameters of an RL high pass
This function calculates the properties of a highpass filter consisting of a resistor and a coil. The output voltage, attenuation and phase shift are calculated for the given frequency.

\(\displaystyle L\) = Inductance [H]
\(\displaystyle R\) = Resistance [Ω]
\(\displaystyle U_1\) = Input voltage [V]
\(\displaystyle U_2\) = Output voltage [V]
\(\displaystyle X_L\) = Reactance [Ω]
\(\displaystyle f_g\) = Cutoff frequency [Hz]
\(\displaystyle φ\) = Phase angle [°]
The output voltage U_{2} of an RL high pass is calculated using the following formula.
\(\displaystyle U_2=U_1 ·\frac{2 · π · f ·L} {\sqrt{R^2 + (2 · π · f · L)^2}}\)
or simply if X_{L} is known
\(\displaystyle U_2=U_1 ·\frac{X_L}{\sqrt{R^2 + X_L^2}}\)
\(\displaystyle X_L=2 π · f ·L\)
At the resonance frequency, the damping is 3 dB. It can be calculated for the various frequencies using the formulas below. If the input and output voltage are known, the attenuation can easily be calculated using the following formula.
\(\displaystyle V_u=20 · lg \left(\frac{U_2}{U_1} \right) \)
If the voltages are not known, the following formula is used.
\(\displaystyle V_u=20·lg\left(\frac{2 · π · f ·L} {\sqrt{R^2 + (2 · π · f · L)^2}}\right)\)
or simply shown
\(\displaystyle V_u=20·lg\left(\frac{ω · L} {\sqrt{R^2 + (ω · L)^2}}\right)\)
In an RL high pass, the output voltage leads the input voltage by 0 °  90 °, depending on the frequency. At the resonance frequency, the phase shift is 45 °. At frequencies that are higher than the cutoff frequency, it tends to 0. At lower frequencies, the phase shifts in the direction of 90 °. The phase shift can be calculated using the following formula.
\(\displaystyle φ=acos \left(\frac{U_2}{U_1} \right)\)
\(\displaystyle φ= arctan \left(\frac{R}{ω ·L}\right)\)
At the cutoff frequency f_{g} or ω_{g} the value of the amplitudefrequency response (ie the magnitude of the transfer function) is 0.707, which corresponds to 3dB.
\(\displaystyle 0.707= \frac{1}{\sqrt{2}}\)
\(\displaystyle ω_g= \frac{R}{L} ⇒ f_g=\frac{R}{2·π·L}\)
\(\displaystyle R=2·π·f_g·L\)
\(\displaystyle L=\frac{R}{2·π·f_g}\)
