High Pass Filters
A high pass filter does exactly the opposite of a low pass filter: it passes high frequencies while attenuating low frequencies. Below the cutoff frequency, signals are increasingly blocked. Above the cutoff frequency, signals pass with minimal loss.
In amateur radio, the high pass filter has several critical applications. The most important is broadcast interference rejection — strong AM broadcast stations in the 550–1700 kHz range can overload a receiver's front end, causing intermodulation products that appear as spurious signals across the HF bands. A high pass filter with a cutoff around 2 MHz between the antenna and receiver eliminates broadcast-band signals before they reach the sensitive receiver input. This is one of the most effective and affordable improvements you can make to an HF station in an urban environment.
High pass filter frequency response on a logarithmic scale. Below the cutoff frequency fc, attenuation increases at 20 dB per decade. At fc, output is −3 dB. Above fc, the filter is in the passband and output is essentially equal to input.
View LargerHow a High Pass Filter Works
Recall from the low pass filter lesson that an RC low pass filter places the resistor in series (from input to junction) and the capacitor as a shunt element (from junction to ground). Output is taken across the capacitor.
A high pass filter swaps the positions: the capacitor goes in series (from input to junction) and the resistor goes to ground. Output is taken across the resistor.
At low frequencies, the capacitor has high impedance (XC = 1/(2πfC) is large). Most of the input voltage is dropped across the capacitor's high impedance, leaving little for the resistor (the output). At high frequencies, XC is small — little voltage is dropped across the capacitor, and most of the input appears at the output resistor. This is high-pass behavior: low frequencies are attenuated, high frequencies pass through.
The physical reason is the same as for all capacitive circuits: capacitors block DC and strongly impede low-frequency AC. They provide a high-impedance path for these signals. At high frequencies, the capacitor's impedance falls and it effectively becomes a short circuit — signals pass through the capacitor to the output resistance with minimal loss.
The RC High Pass Filter
The transfer function for an RC high pass filter is:
Vout / Vin = R / √(R² + XC²) = (f/fc) / √(1 + (f/fc)²)
where the cutoff frequency fc is:
fc = 1 / (2πRC)
This is identical to the low pass cutoff formula — the same R and C values produce the same cutoff frequency whether used as a high pass or low pass filter.
At f = fc: Vout/Vin = 0.707 = −3 dB (same half-power point)
The fact that the cutoff frequency formula is the same for both RC filter types is a fundamental and elegant result. The difference between a high pass and low pass filter is purely in which component you take the output from — the capacitor (low pass) or the resistor (high pass). The frequency at which the transition occurs is always fc = 1/(2πRC).
Below fc, the high pass filter rolls off at −20 dB/decade (toward lower frequencies). Above fc, the output is flat. Multi-pole high pass filters using multiple capacitor-resistor sections increase the rolloff rate to 40, 60, or more dB/decade below cutoff.
You operate on the 160-meter band (1.8–2.0 MHz) and live near an AM broadcast station at 1,040 kHz that overloads your receiver. Design an RC high pass filter to attenuate the broadcast station by at least 20 dB while passing 1.8 MHz. Available resistor: 50 Ω (to match your coax system impedance — noting that a proper RF filter would use LC, but this demonstrates the calculation).
Step 1 — Choose fc: You want to pass 1.8 MHz and reject 1,040 kHz. A cutoff of 1,500 kHz (1.5 MHz) would be between the two, but let us check attenuation. At fc = 1.5 MHz, attenuation at 1.04 MHz:
Vout/Vin = (f/fc) / √(1 + (f/fc)²) = (1.04/1.5) / √(1 + (1.04/1.5)²) = 0.693 / √(1.480) = 0.693 / 1.217 = 0.570 = −4.9 dB
Only 4.9 dB — insufficient. A single-pole RC filter cannot give 20 dB at 1.04 MHz while passing 1.8 MHz — the frequencies are too close together.
Step 2 — Reality check: With a single RC pole and fc = 1.04 MHz (at the interference frequency):
At 1.04 MHz: −3 dB (by definition)
At 1.8 MHz: Vout/Vin = (1.8/1.04) / √(1 + (1.8/1.04)²) = 1.731 / √(3.996) = 1.731 / 1.999 = 0.866 = −1.25 dB (good passband)
Conclusion: A single-pole RC filter cannot achieve 20 dB rejection of a nearby frequency. You need multiple poles (LC filter) for steep rolloff between close frequencies. The lesson: RC filters are useful when the frequencies to be rejected are far from the passband (10× or more). For close-in filtering, use LC or active filter designs.
Practical solution: A 3-pole LC high pass filter designed for 2 MHz cutoff in 50 Ω gives ~35 dB at 1.04 MHz (almost an octave below cutoff).
RC High Pass Cutoff Frequency Calculator
RC High Pass Filter Cutoff Frequency Calculator
Enter R and C to find the cutoff frequency and attenuation at any test frequency. Remember: the same RC values give the same fc as the equivalent low pass filter.
Phase Response of the RC High Pass Filter
Like the low pass filter, the high pass filter introduces a phase shift between input and output. However, the direction is opposite: where the low pass filter introduces a phase lag (output lags input), the high pass filter introduces a phase lead (output leads input).
θ = +arctan(XC/R) = +arctan(fc/f)
At f << fc: θ → +90° (output leads input by 90°)
At f = fc: θ = +45° (output leads input by 45°)
At f >> fc: θ → 0° (no phase shift — output equals input)
In the passband (high frequencies), the high pass filter introduces negligible phase shift — output and input are essentially in phase. In the transition region near fc, the +45° phase lead is present. Well below cutoff, the phase approaches +90° even though the magnitude is greatly attenuated.
This phase response is important in audio and feedback amplifier design. An RC high pass filter in a feedback path can introduce a phase lead that affects stability. In RF applications, the phase shift is generally not critical as long as the signal of interest is well above the cutoff frequency.
Low Pass vs. High Pass: Component Swap
The relationship between low pass and high pass RC filters is remarkably simple — swapping the positions of R and C changes the filter type while keeping the cutoff frequency the same:
| Property | RC Low Pass | RC High Pass |
|---|---|---|
| Series element (input to junction) | Resistor (R) | Capacitor (C) |
| Shunt element (junction to ground) | Capacitor (C) | Resistor (R) |
| Output taken across | Capacitor (low impedance at high f blocks output) | Resistor (high impedance at low f drops input) |
| Cutoff frequency formula | fc = 1/(2πRC) | fc = 1/(2πRC) — identical |
| Phase at fc | −45° (output lags input) | +45° (output leads input) |
| Rolloff direction | −20 dB/decade above fc | −20 dB/decade below fc |
| DC behavior | Passes DC (capacitor open, R bypassed) | Blocks DC (capacitor in series) |
The DC blocking property of the high pass filter — the fact that the series capacitor blocks DC entirely — is extremely useful in electronic circuits independent of any filtering function. A single capacitor in series with a signal path blocks DC bias while allowing AC signals to pass. This is called a coupling capacitor or DC blocking capacitor. The capacitor value is chosen so that fc (determined by C and the load resistance R) is well below the lowest signal frequency of interest.
An audio preamplifier output has a DC bias of 4.5 V. The next stage has an input resistance of 47 kΩ. You want the coupling capacitor to have fc below 20 Hz (the lowest audio frequency). Find the minimum capacitance needed.
From fc = 1/(2πRC), solving for C:
C = 1 / (2π × fc × R) = 1 / (2π × 20 × 47,000)
= 1 / (5,906,194) = 0.169 µF
Use the next standard value up: 0.22 µF
Verification: fc = 1/(2π × 0.22×10⁻⁶ × 47,000) = 1/64,977 = 15.4 Hz
This passes audio from 15 Hz upward, blocks the 4.5 V DC bias, and introduces only −3 dB at 15 Hz (negligible effect on voice frequencies starting at ~300 Hz).
Attenuation at 300 Hz: Vout/Vin = (300/15.4) / √(1 + (300/15.4)²) = 19.48 / √(379.4) = 19.48 / 19.48 = 1.000 = 0 dB. Essentially perfect at 300 Hz.
High Pass Filters in Your Station
Broadcast Interference Rejection
AM broadcast stations (550–1700 kHz) operate at high power — 50 kW in some cases — and can be only miles from an amateur station. Even if you cannot hear the broadcast station on your receiver, its signal may be overloading your front-end amplifier, causing intermodulation distortion that appears as ghost signals, mixing products, and elevated noise floor across your HF bands. A high pass filter with fc around 2–3 MHz inserted between the antenna feedline and the receiver eliminates this problem.
Commercial high pass filter modules are available from suppliers like DX Engineering and ICE (International Crystal Electronics) for exactly this purpose. They use multi-pole LC designs for steep rolloff. Understanding this lesson helps you evaluate which filter provides adequate suppression for your situation.
TV Interference — TVI Filter
A transmit-side TVI (television interference) filter placed between your transmitter and antenna suppresses your transmitter's signal in the TV frequency bands. Older analog TV used channels 2–6 (54–88 MHz) and 7–13 (174–216 MHz). Digital TV now occupies 470–806 MHz. High pass filters with cutoff at 30 MHz block HF transmitter fundamentals from reaching the TV frequency range.
Antenna DC Blocking
When phantom power or DC is fed through a coax cable to a remote preamplifier or bias-T, a coupling (DC blocking) capacitor is needed at each end of the RF path to keep the DC from reaching the receiver or transmitter. This is a direct application of the single-series-capacitor high pass filter — the capacitor's value is chosen to pass the lowest RF frequency of interest while blocking the DC supply voltage.
CW Audio Filter
Combined with a low pass filter, a high pass filter forms a bandpass filter around the CW sidetone frequency. A high pass with fc at 400 Hz combined with a low pass at 800 Hz creates a 400 Hz bandpass centered near 600 Hz — the typical CW pitch. This dramatically improves weak-signal CW copy by rejecting audio outside the Morse note frequency.
Experiment: Compare RC Low Pass and High Pass Filters Side by Side
You need: Two 10 kΩ resistors, two 10 nF capacitors, breadboard, audio signal generator, oscilloscope or AC voltmeter.
Theory: fc = 1/(2π × 10,000 × 10×10⁻⁹) = 1,592 Hz for both filters.
Build Filter 1 (Low Pass): 10 kΩ from input to junction, 10 nF from junction to ground. Output across capacitor.
Build Filter 2 (High Pass): 10 nF from input to junction, 10 kΩ from junction to ground. Output across resistor.
Test both at the same frequencies and record:
- 100 Hz: Low pass ≈ 1 V (passes); High pass ≈ 0.063 V (blocks)
- 1,592 Hz (fc): Both should read ≈ 0.707 V (−3 dB)
- 15,920 Hz (10× fc): Low pass ≈ 0.100 V (−20 dB); High pass ≈ 0.995 V (near flat)
Expected result: The two filters have identical cutoff frequencies but opposite response shapes. At fc, both measure 0.707 × Vin. Below fc, the low pass passes and the high pass attenuates. Above fc, the high pass passes and the low pass attenuates. This experiment demonstrates the component-swap relationship between the two filter types.
Frequently Asked Questions
Why do RC high pass and low pass filters have the same cutoff frequency formula?
The cutoff frequency occurs where the capacitive reactance XC equals the resistance R — that is, where 1/(2πfC) = R. Solving for f gives fc = 1/(2πRC), which contains only R and C. Whether the R is the series element and C is the shunt (low pass) or C is the series element and R is the shunt (high pass), the balance condition XC = R still occurs at the same frequency. The component positions change which output is "high" at which frequency, but not the frequency of transition.
Can I use a high pass filter to block DC without affecting my RF signal?
Yes — this is the coupling capacitor application. A single capacitor in series with the signal path blocks DC completely (a capacitor passes no DC in steady state) while passing RF signals with negligible attenuation if the capacitor's reactance XC is much smaller than the circuit impedance at the signal frequency. For a 50 Ω RF circuit at 7 MHz, a 100 pF capacitor has XC = 227 Ω — too high. Use 10 nF: XC = 2.27 Ω — nearly transparent to RF while blocking DC completely.
Should I use a high pass or low pass filter to suppress broadcast interference?
Use a high pass filter. AM broadcast stations operate at 550–1700 kHz — below the HF amateur bands. A high pass filter with a cutoff of 2–3 MHz blocks the broadcast band while passing all amateur HF frequencies (3.5 MHz and above). A low pass filter would do the opposite: block HF and pass the broadcast frequencies, which is exactly the wrong result for this application.
Test Your Knowledge
Answer the questions below to check your understanding. Every answer can be found in the lesson above.