RL High-Pass Filter: Analysis and Applications

Why Inductors Make High-Pass Filters

Most introductory filter problems use resistors and capacitors, but the inductor offers a second, equally valid route to frequency shaping. The RL high-pass filter passes high-frequency signals and attenuates low-frequency ones, making it the natural complement to the RC low-pass filter. Learning the RL version is worth the effort for two reasons. First, it forces you to reason about inductive impedance, which behaves in exactly the opposite way to capacitive impedance. Second, the inductor is the ingredient that makes resonance possible, so mastering it here is the bridge to RLC circuits and high-Q tuned networks later on.

The intuition rests on the inductor's impedance, $Z_L = sL = j\omega L$. At DC the impedance is zero, so an inductor placed from the output node to ground behaves like a short circuit and pins the output to zero. As frequency rises the impedance grows without bound, so eventually the inductor dominates the divider and the full input voltage appears across it. That rising impedance with frequency is precisely what produces the high-pass shape.

Deriving the Transfer Function

Place the resistor $R$ in series with the signal path and the inductor $L$ from the output node to ground, taking the output across the inductor. The circuit is a voltage divider with $Z_R = R$ as the series element and $Z_L = sL$ as the shunt element:

Dividing numerator and denominator by $L$ puts the transfer function into its canonical first-order high-pass form, where the corner frequency is exposed directly in the denominator:

This has a zero at the origin ($s = 0$) and a pole at $s = -R/L$. The zero at the origin is what kills the DC response: at $s = 0$, $H = 0$. As $s \to \infty$ the ratio approaches 1, so the passband gain is unity. Notice that if you instead measured the output across the resistor with the inductor in series, you would get the low-pass response $H(s) = R/(R + sL)$; once again, where you tap the output decides the filter type.

Cutoff Frequency and a Worked Example

The pole at $s = -R/L$ sets the cutoff. In radians per second $\omega_c = R/L$, and in hertz:

Take $R = 1\;\text{k}\Omega$ and $L = 10\;\text{mH}$. Then:

Signals above 15.9 kHz pass with near-unity gain; signals below it are attenuated at 20 dB per decade. To check the slope, evaluate the magnitude one decade below cutoff, at 1.59 kHz, where $\omega L / R = 0.1$. The magnitude is $|H| = 0.1/\sqrt{1 + 0.1^2} \approx 0.0995$, which is about −20 dB — exactly the expected first-order behavior. At the cutoff itself the gain is $1/\sqrt{2} = 0.707$ (−3 dB) and the phase is +45°. Suppose instead you needed a 1 kHz cutoff with the same inductor; solving $R = 2\pi f_c L$ gives $R = 2\pi \cdot 1000 \cdot 0.01 \approx 62.8\;\Omega$.

Reading the Bode Plot

The RL high-pass Bode magnitude plot is the mirror image of the RC low-pass. Below cutoff it rises at +20 dB per decade as frequency increases toward the corner; at $f_c$ it sits at −3 dB; and above cutoff it flattens to a 0 dB passband. The phase starts at +90° far below cutoff — the output leads the input because of the inductor — passes through +45° at $f_c$, and falls toward 0° far above. The positive phase is the signature of a single zero at the origin combined with one pole.

It helps to connect the asymptotes to physical reasoning. Far below cutoff the inductor's impedance is tiny, so the output is roughly $\omega L / R$ times the input, and that ratio doubles every time the frequency doubles — which is the +20 dB/decade rising skirt. As the frequency climbs through the corner, the inductor's impedance grows to match the resistor, the divider becomes balanced, and the curve bends over toward the flat unity passband. The phase tells the same story from the energy-storage side: at low frequency the inductor's voltage leads its current by a full 90°, and as the resistor begins to dominate the loop the leading angle shrinks toward zero. Reading magnitude and phase together gives you a complete and intuitive picture of the filter without ever solving a single numerical point.

Where RL High-Pass Filters Show Up

Because inductors are bulky and lossy at low frequencies, you mostly meet RL high-pass behavior where inductors are already present for other reasons. In power electronics, the inductors used for energy storage and current smoothing naturally form high-pass and low-pass networks with the surrounding resistances, and understanding the corner frequency keeps switching ripple where you want it. In radio-frequency work, small air-core or ferrite inductors are compact enough to be practical, and RL sections appear in bias-tee networks, matching circuits, and chokes that block low-frequency content while passing the RF signal. Motor and relay drive circuits also exhibit RL high-pass edges, which is why flyback diodes and snubbers are needed to tame the fast voltage transients the inductance produces. In every case the single governing number is the same corner $f_c = R/(2\pi L)$, so the analysis you learned for a clean two-element filter transfers directly to the messy real circuit.

RL High-Pass vs RC High-Pass

An RC high-pass filter uses a series capacitor and a shunt resistor, with the output across $R$, giving:

Mathematically the RL and RC high-pass filters are identical in form — both are first-order with one pole, one zero at the origin, the same −20 dB/decade skirt, and a unity passband. The choice between them is practical, not theoretical. The table below contrasts the two, and the deeper trade-offs are summarized beneath it.

Capacitors are smaller, cheaper, and lighter than inductors, especially when low cutoffs demand large component values, which is why RC filters dominate signal-level design. Real inductors also carry parasitic series resistance and winding capacitance that distort the response at the extremes, whereas capacitors tend to behave more ideally. Inductors win in power electronics, where they tolerate large currents, and they are indispensable whenever you need resonance.

Common Mistakes

• Using the RC cutoff formula for an RL filter. The cutoff is $R/(2\pi L)$, not $1/(2\pi RC)$. Note that here larger $R$ raises the cutoff, the opposite of the RC case.
• Forgetting the inductor is a short at DC. Beginners sometimes expect the output to follow the input at low frequency; in fact the inductor clamps it to zero.
• Ignoring the inductor's DC resistance. A real inductor's winding resistance appears in series with $sL$ and prevents the response from reaching a true zero at DC.
• Reversing the phase sign. An RL high-pass produces a leading (positive) phase in the stopband; confusing it with the lagging phase of a low-pass leads to errors in feedback analysis.
• Expecting a sharp cutoff. Like any first-order filter, the roll-off is a gentle 20 dB/decade, not a steep wall.

Try It in CircuitMath

Drop a resistor into the series path and an inductor from the output node to ground, add a source and a ground, and analyze. CircuitMath returns $H(s)$ in LaTeX so you can confirm the cutoff equals $R/(2\pi L)$. Open the editor and try swapping the inductor and resistor positions to watch the response flip from high-pass to low-pass.

Related Tutorials

• RC Low-Pass Filter — the complementary first-order filter using a capacitor.
• RLC Resonant Circuit — combine R, L, and C for resonance and quality factor.
• Transfer Functions in Circuit Analysis — understand poles and zeros behind $H(s)$.