Low Pass Filters
A low pass filter is one of the most useful circuits in electronics. As the name suggests, it passes low-frequency signals while attenuating (reducing) high-frequency signals. Below a certain frequency — the cutoff frequency — signals pass through with little loss. Above the cutoff frequency, signals are increasingly attenuated the higher their frequency goes.
A low pass filter passes frequencies below a chosen cutoff and attenuates those above it. It is one of the most widely used circuits in electronics: it removes hiss from audio, smooths the output of a PWM signal into a steady voltage, cleans the switching ripple from a power supply, strips harmonics from an oscillator or an inverter, and stops an analog-to-digital converter from seeing frequencies it cannot handle. Understanding how these filters work — and how to calculate them — is a core practical skill.
Low pass filter frequency response on a logarithmic scale. The passband is flat up to the cutoff frequency fc. At fc, the output is −3 dB (0.707× the input voltage). Above fc, the single-pole filter rolls off at 20 dB per decade — each factor of 10 increase in frequency reduces the output by another 20 dB.
View LargerHow a Low Pass Filter Works
To understand a low pass filter, start with the concept of a voltage divider. A basic voltage divider uses two resistors in series — one connected to the input, one to ground — and takes the output from the junction between them. The output voltage is a fixed fraction of the input, regardless of frequency.
Now replace the bottom resistor with a capacitor. A capacitor's impedance (XC) is high at low frequencies and low at high frequencies. At low frequencies, XC is large compared to the series resistor R — most of the voltage appears across the capacitor (the output), and little is dropped across R. At high frequencies, XC is small compared to R — most of the voltage is dropped across R, and little appears at the output.
This is the essence of an RC low pass filter: the capacitor divides the signal voltage in a way that is frequency-dependent, passing low frequencies and attenuating high frequencies. The same principle applies to an RL low pass filter, where the inductor acts as the top element (high impedance at high frequency, low impedance at low frequency) with a resistor to ground at the output.
The RC Low Pass Filter
The RC low pass filter consists of a resistor in series with the input and a capacitor from the junction to ground. The output is taken across the capacitor.
The transfer function (ratio of output to input voltage) as a function of frequency is:
Vout / Vin = XC / √(R² + XC²) = 1 / √(1 + (f/fc)²)
where the cutoff frequency fc is:
fc = 1 / (2πRC)
At f = fc: Vout/Vin = 1/√2 = 0.707 = −3 dB
The cutoff frequency fc is the half-power frequency — the frequency at which output power has fallen to half its maximum value, or equivalently, voltage has fallen to 0.707 of maximum (same −3 dB definition as bandwidth). Below fc, the filter is in the passband. Above fc, the filter is in the stopband (rolloff region).
At the cutoff frequency, R = XC, meaning the resistor's impedance equals the capacitor's impedance. This gives a phase shift of exactly 45° between input and output at fc.
An audio circuit picks up high-frequency hiss and switching noise. You want a low pass filter with a cutoff frequency of 3,400 Hz (top of the telephone voice band) using a 10 kΩ resistor. Find the required capacitance.
Rearranging fc = 1/(2πRC) for C:
C = 1 / (2π × fc × R) = 1 / (2π × 3,400 × 10,000)
= 1 / (213,628,300) = 4.68 nF ≈ 4.7 nF (standard value)
Verification: fc = 1/(2π × 10,000 × 4.7×10⁻⁹) = 1/295,310 = 3,387 Hz ✓
Attenuation at 60 Hz: Vout/Vin = 1/√(1 + (60/3387)²) = 1/√(1.000314) = 0.9998 — essentially zero attenuation at 60 Hz!
Correction: A filter with fc = 3,400 Hz passes 60 Hz freely. To attenuate 60 Hz, you would need fc far below 60 Hz — like 10 Hz — using a much larger capacitor: C = 1/(2π × 10 × 10,000) = 1.59 µF. At 3,400 Hz, attenuation would then be: Vout/Vin = 1/√(1 + (3400/10)²) = 1/√115,601 = 0.00294 = −50.6 dB. This effectively removes audio above 10 Hz, which is not useful for voice.
The lesson: filter cutoff frequency selection requires understanding what you want to pass and what you want to reject — the cutoff lies between them.
RC Low Pass Cutoff Frequency Calculator
RC Low Pass Filter Cutoff Frequency Calculator
Enter R and C to find the cutoff frequency, or enter any two known values to solve for the third. Attenuation at any frequency is also calculated.
The RL Low Pass Filter
The RL low pass filter replaces the capacitor with an inductor. The inductor is placed in series with the signal path (not the resistor as in RC filters), and the resistor is connected from the output to ground. The inductor has low impedance at low frequencies (allowing signal to pass) and high impedance at high frequencies (blocking the signal), while the resistor provides the output load.
Wait — that sounds backwards from the RC filter. Let us clarify both configurations:
| Filter Type | Series Element | Shunt Element | Why It Low-Passes |
|---|---|---|---|
| RC Low Pass | Resistor (R) | Capacitor (C) to ground | At high f, C shorts the output to ground; at low f, C is open and output ≈ input |
| RL Low Pass | Inductor (L) | Resistor (R) to ground | At high f, L blocks the signal; at low f, L passes signal freely to the R output |
The cutoff frequency formula for an RL low pass filter is:
fc = R / (2πL)
where R is in ohms and L is in henrys
At f = fc: XL = R (inductor and resistor impedances are equal), output = −3 dB
The RL filter is less common than the RC filter in general electronics because inductors are bulkier and more expensive than capacitors for the same cutoff frequency. However, RL filters are used in power supply output stages (where the inductor also stores energy) and in RF circuits where inductors are more natural elements than large capacitors.
A regulated power supply includes an output RL filter. The load resistance is 10 Ω. You want the cutoff frequency to be 100 Hz so that the 120 Hz ripple component from the rectifier is attenuated by at least 20 dB. Find the required inductance and verify the attenuation at 120 Hz.
Step 1 — Find L for fc = 100 Hz:
L = R / (2π × fc) = 10 / (2π × 100) = 10 / 628.3 = 15.9 mH
Step 2 — Attenuation at 120 Hz:
Vout/Vin = 1/√(1 + (f/fc)²) = 1/√(1 + (120/100)²) = 1/√(1 + 1.44) = 1/√2.44 = 1/1.562 = 0.640
Attenuation = 20 log10(0.640) = −3.87 dB
Problem: Only 3.87 dB of ripple attenuation — far less than the required 20 dB. A single-pole RL filter with fc = 100 Hz gives only 3.87 dB at 120 Hz because 120 Hz is barely above cutoff.
Solution: To get 20 dB at 120 Hz from a single pole, you need fc much lower, OR use a two-pole LC filter (which rolls off at 40 dB/decade). With a two-pole filter, 20 dB at 120 Hz requires fc ≈ 38 Hz. This illustrates why power supply filters often use multiple L-C stages.
RL Low Pass Cutoff Frequency Calculator
RL Low Pass Filter Cutoff Frequency Calculator
Enter R and L to find the cutoff frequency. Optionally enter a test frequency to find attenuation.
Rolloff Rate and Filter Order
A single RC or RL section is called a first-order or single-pole filter. Above the cutoff frequency, its attenuation increases at a rate of 20 dB per decade — for every ten-fold increase in frequency above fc, attenuation increases by 20 dB. This slope corresponds to −6 dB per octave (per doubling of frequency).
−20 dB/decade = −6 dB/octave
At 10× fc: −20 dB attenuation
At 100× fc: −40 dB attenuation
At 1000× fc: −60 dB attenuation
For many applications, −20 dB/decade is not steep enough. Adding more filter sections — each adding one pole — increases the rolloff rate:
| Filter Order (Poles) | Rolloff Rate | Attenuation at 10× fc | Components Needed |
|---|---|---|---|
| 1st order (1 pole) | −20 dB/decade | −20 dB | 1 R + 1 C, or 1 L + 1 R |
| 2nd order (2 poles) | −40 dB/decade | −40 dB | 1 L + 1 C (LC filter) |
| 3rd order (3 poles) | −60 dB/decade | −60 dB | 2 L + 1 C, or 1 L + 2 C |
| 5th order (5 poles) | −100 dB/decade | −100 dB | 3 L + 2 C (typical LC low pass) |
| 7th order (7 poles) | −140 dB/decade | −140 dB | 4 L + 3 C (high-performance LC) |
Filters that must remove harmonics or switching noise typically use 5 or 7 poles to achieve 40 dB or more of suppression with a cutoff frequency close to the wanted signal. Placing the cutoff close to the wanted frequency provides more attenuation at the second harmonic while still passing the fundamental with minimal loss.
LC Low Pass Filters
The RC and RL low pass filters described above work well for small signals at audio and low frequencies, but filters that carry power — the output filter of a class D amplifier or an inverter, or a filter at the output of a high-frequency oscillator — use LC (inductor-capacitor) circuits. The reason is efficiency: resistors dissipate power as heat, and even a 1 Ω resistor in the output of a 100 W amplifier carrying 1.4 A would waste I²R = 1.4² × 1 = 2 W unnecessarily. Inductors and capacitors (ideal) dissipate no power — they only store and return it.
The standard LC low pass filter is the Chebyshev or Butterworth LC ladder network. These networks alternate shunt capacitors (to ground) and series inductors, creating a cascade of poles. The component values are calculated from filter design tables based on the cutoff frequency and desired impedance.
For a 7-element (7-pole) Chebyshev low pass filter designed for 50 Ω systems, the cutoff frequency formula for each element follows:
For a low pass filter with cutoff fc and system impedance Z0:
Cn = cn / (2π × fc × Z0)
Ln = ln × Z0 / (2π × fc)
where cn and ln are normalized values from filter design tables
For common applications, published filter designs and online filter calculators provide component values directly. These have been optimized and verified — there is no need to derive them from tables for typical applications. The Filters and Impedance Matching module covers Butterworth and Chebyshev design in detail.
A 5-element Chebyshev low pass filter has fc = 10 MHz in a 50 Ω system. It follows an oscillator running at 7.1 MHz. Estimate attenuation at the second harmonic (14.2 MHz) and third harmonic (21.3 MHz).
At the second harmonic (14.2 MHz):
Frequency ratio: 14.2/10 = 1.42 — only 42% above cutoff
For a 5-pole Chebyshev filter, attenuation ≈ 5 × 20 × log10(1.42) ≈ 5 × 20 × 0.152 = 15.2 dB
(Actual Chebyshev attenuation is higher due to equiripple design, typically 25–35 dB at 1.42× cutoff for 5 poles with 0.5 dB ripple.)
At the third harmonic (21.3 MHz):
Frequency ratio: 21.3/10 = 2.13
Approximate attenuation: 5 × 20 × log10(2.13) ≈ 5 × 20 × 0.328 = 32.8 dB
(Actual Chebyshev: typically 45–55 dB at 2.13× cutoff.)
Conclusion: A single 5-pole LC low pass filter gives excellent harmonic suppression for a 7.1 MHz signal. Combined with the oscillator's own tank circuit, total harmonic suppression typically exceeds 50 dB — a very clean sine wave.
Low Pass Filters in Practice
Anti-Aliasing Filters
An analog-to-digital converter sampling at 44.1 kHz can only represent frequencies below 22.05 kHz. Anything higher folds back as false lower frequencies (aliasing). A low pass filter in front of the ADC removes those frequencies before they are sampled — every sound card, data logger and digital oscilloscope has one.
PWM to Analog
A microcontroller can make an adjustable DC voltage by switching a pin rapidly on and off (PWM) and passing the result through an RC low pass filter. With the cutoff well below the PWM frequency, only the average voltage remains, set by the duty cycle. The same idea, with an LC filter, turns the switched output of a class D amplifier back into smooth audio.
Power Supply and Inverter Filtering
The output of a switching power supply contains ripple at its switching frequency, and a pure-sine inverter builds its output from fast switching. LC low pass filters remove those switching frequencies, leaving smooth DC or a clean 50/60 Hz sine wave. At these power levels the filter inductors and capacitors must be rated for the full current and voltage.
Audio Hiss and Subwoofer Filters
Audio equipment uses RC or active low pass filters to roll off hiss above the audible range and to feed only the bass to a subwoofer (a typical subwoofer crossover is 80–120 Hz). Combined with a high pass filter (covered in the next lesson), a low pass filter forms a band-pass filter that passes only a chosen range of frequencies.
Experiment: Build and Test an RC Audio Low Pass Filter
You need: 10 kΩ resistor, 4.7 nF capacitor, breadboard, audio signal generator or PC sound card with signal generator software, oscilloscope or audio voltmeter.
Theory: fc = 1/(2π × 10,000 × 4.7×10⁻⁹) = 3,387 Hz.
Build: Connect 10 kΩ from input to junction. Connect 4.7 nF from junction to ground. Output is taken at the junction.
Test:
- Apply 1 V at 100 Hz. Measure output — should be nearly 1 V (well below cutoff).
- Apply 1 V at 3,387 Hz. Measure output — should be 0.707 V (−3 dB at cutoff).
- Apply 1 V at 10 kHz. Measure output — should be 0.323 V (−9.8 dB, close to single-pole prediction of about −9.5 dB).
- Apply 1 V at 33.87 kHz (10× fc). Measure output — should be 0.0995 V (about −20 dB).
Expected result: Measured values should be within 5–10% of theoretical at each frequency. Any large discrepancy suggests a wiring error or component out of specification. This experiment demonstrates the −20 dB/decade rolloff of a single-pole filter directly.
Frequently Asked Questions
Why is the cutoff frequency defined at −3 dB rather than where the signal is completely blocked?
A simple RC or RL filter never completely blocks any frequency — attenuation increases gradually and approaches but never reaches infinite attenuation. The −3 dB point (0.707 × voltage, half power) was chosen as the standard cutoff because it represents the half-power frequency and sits at the mathematically significant point where the reactive impedance equals the resistance. It is a precise, measurable, and universally agreed-upon definition. In practice, "useful" signal passes below this frequency and is "significantly attenuated" above it, making it a practical engineering boundary.
Can a low pass filter also match impedances?
A simple low pass filter does not change impedance. However, LC low pass filter topologies (like the L-network) can simultaneously filter harmonics and transform impedances when designed for that purpose. The L, Pi, and T networks covered in a later lesson are effectively impedance-transforming filters that serve both functions at once.
Does a low pass filter affect the signal I want to keep?
A well-designed low pass filter with its cutoff comfortably above the highest wanted frequency passes the signal with negligible loss and does not affect its quality. A filter with a cutoff too close to the wanted frequencies attenuates them and shifts their phase. For a 7.1 MHz signal, a filter with fc = 10 MHz provides good harmonic suppression with minimal effect on the fundamental; for audio, a cutoff of about 20–30 kHz removes ultrasonic noise without affecting what you hear.
Test Your Knowledge
Answer the questions below to check your understanding of this lesson. Every answer can be found in the lesson above.