Introduction to Electronics - Filters: RL high-pass filter behaviour

An RL high-pass filter is designed to allow high-frequency signals to pass through while attenuating low-frequency signals. In this configuration, a resistor and an inductor are arranged so that the output is taken across the inductor. At low frequencies, the inductor exhibits a low impedance, causing most of the input voltage to drop across the resistor and resulting in a very small output across the inductor. As the frequency increases, the impedance of the inductor rises, allowing more of the input voltage to appear at the output. This behaviour creates a distinct separation between the passband (high frequencies) and the stopband (low frequencies), with the cutoff frequency marking the transition between the two regions.

Circuit Diagram

The following schematic represents a basic RL high-pass filter:

Mathematical Underpinning

If you are allergic to mathematics, you can go straight to the end of this segment and checkout the cutoff frequency function as you will need it later. If you are not allergic, then continue. The behaviour of the RL high-pass filter is described by its transfer function. The transfer function is given by:

[math]H(s)=\frac{sL}{R+sL}[/math]

Substituting the complex frequency [math]s[/math] with [math]j\omega[/math], where [math]\Omega[/math] is the angular frequency in radians per second, the transfer function becomes:

[math]H(j\omega)=\frac{j\omega L}{R+j\omega L}[/math]

The magnitude of the transfer function, which represents the filter’s frequency response, is:

[math]|H(j\omega)|=\frac{\omega L}{\sqrt{R^2+(\omega L)^2}}[/math]

At low frequencies (when [math]\omega L \ll R[/math]), the magnitude is near 0, indicating strong attenuation of low-frequency signals. As the frequency increases and [math]\omega L[/math] becomes much larger than [math]R[/math], the magnitude approaches 1, meaning that high-frequency signals pass with little attenuation. The cutoff frequency, defined as the frequency at which the output drops to [math]\frac{1}{\sqrt{2}}[/math] of the input (approximately -3 dB), is calculated as:

[math]f_c=\frac{R}{2\pi L}[/math]

This formula is derived by setting [math]|H(j\omega)|=\frac{1}{\sqrt{2}}[/math] and solving for [math]\omega[/math].

Frequency Response Plot

The following Python code snippet demonstrates how to plot the frequency response of an RL high-pass filter and mark the cutoff frequency on the plot:

import numpy as np
import matplotlib.pyplot as plt

# Define resistor and inductor values
R = 1e3 # Resistance in ohms
L = 1e-3 # Inductance in henrys

# Calculate the cutoff frequency
fc = R / (2 * np.pi * L)

# Define frequency range for the plot (10 Hz to 1 MHz)
frequencies = np.logspace(1, 6, num=500)
omega = 2 * np.pi * frequencies

# Calculate the magnitude of the transfer function for the high-pass filter
H_mag = (omega * L) / np.sqrt(R**2 + (omega * L)**2)
H_db = 20 * np.log10(H_mag)

# Create the plot
plt.figure(figsize=(8, 4))
plt.semilogx(frequencies, H_db)
plt.xlabel("Frequency (Hz)")
plt.ylabel("Magnitude (dB)")
plt.title("Frequency Response of RL High-Pass Filter")
plt.grid(True, which="both", linestyle="--")

# Mark the cutoff frequency with a red dashed line
plt.axvline(x=fc,
color="red",
linestyle="--",
label=f"Cutoff Frequency: {fc:.1f} Hz")
plt.legend()

plt.show()

Executing this script will produce a plot that displays how the RL high-pass filter attenuates low frequencies and passes high frequencies, with a red dashed line indicating the cutoff frequency at [math]f_c=\frac{R}{2\pi L}[/math].

Key Takeaways

• In an RL high-pass filter, the resistor and inductor form a voltage divider where the output is taken across the inductor. Low frequencies result in a low output due to the low impedance of the inductor, while high frequencies yield a higher output.
• The filter is characterised by the transfer function [math]H(j\omega)=\frac{j\omega L}{R+j\omega L}[/math], which shows the frequency-dependent behaviour of the circuit.
• The magnitude of the frequency response, [math]|H(j\omega)|=\frac{\omega L}{\sqrt{R^2+(\omega L)^2}}[/math], demonstrates that the filter effectively attenuates low frequencies and allows high frequencies to pass.
• The cutoff frequency, given by [math]f_c=\frac{R}{2\pi L}[/math], marks the boundary between the stopband and the passband, serving as a key design parameter for the filter.

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