Low-Pass to High-Pass Transformation: Swapping L and C
The mapping ω → −ωc/ω turns a series-inductor element g into a series capacitor C' = 1/(Z0·ωc·g) and a shunt-capacitor element g into a shunt inductor L' = Z0/(ωc·g). For g = 1, 50 Ω and 1 GHz that gives 3.18 pF and 7.96 nH.
Engineer’s note: Watch the shunt inductors in a high-pass ladder. They short the line to ground at DC, which kills any bias you meant to feed through it. At 1.59 pF the series cap is small enough that pad capacitance moves the cut-off, so leave yourself room to retune.
For practitioners
The frequency mapping
A low-pass prototype is designed at a normalized frequency. Call it Ω. Its cut-off is at Ω = 1 and its terminations are 1 Ω. Its element values g1 to gN are the inductances and capacitances that give the wanted response, such as Butterworth or Chebyshev.
To get a high-pass filter, replace the prototype frequency with a function of the real frequency ω:
The left side is the normalized prototype frequency. The right side is a ratio of two real frequencies in rad/s. Both sides are dimensionless, which is what the prototype variable needs to be. The figure further down writes this on its arrow as ω → −ωc/ω. There, the left-hand ω means the prototype's normalized frequency.
Look at three points to see what the mapping does:
- Prototype DC (Ω = 0) maps to ω → ∞. The low-pass passband edge at DC becomes the top of the high-pass passband.
- Prototype cut-off (|Ω| = 1) maps to ω = ωc. The cut-off does not move.
- Prototype stopband (Ω → ∞) maps to ω → 0. Strong attenuation near DC is what a high-pass filter needs.
The minus sign has no effect on the magnitude. For a ladder of real L and C, |S21| at −Ω equals |S21| at +Ω. The sign is there so that the reactances come out as physical parts, as the next section shows.
Why each L becomes a C
Take a series inductor with normalized value L in the prototype. Its impedance is jΩL. Substitute the mapping:
A series capacitor C' has impedance 1/(jωC') = −j/(ωC'). Match the two expressions and you get:
So the inductor's reactance now falls as frequency goes up, which is how a capacitor behaves. Without the minus sign the result would be a negative inductance, which you cannot buy.
Do the same with a shunt capacitor C in the prototype. Its admittance is jΩC, which maps to −jωc·C/ω. That is the admittance of a shunt inductor, 1/(jωL'):
The topology does not change. Series branches stay series and shunt branches stay shunt. Only the element type flips, and the new value is the reciprocal of the old one divided by ωc.
Adding impedance scaling
Real filters work at Z0, usually 50 Ω, not at 1 Ω. Impedance scaling multiplies every impedance by Z0. Inductances are multiplied by Z0 and capacitances are divided by Z0. If you combine that with the high-pass mapping, the design formulas for a prototype element g are:
A large g gives a small high-pass element. A big prototype inductor has a large reactance. A small capacitor also has a large reactance. The swap keeps the reactance level at cut-off the same: at ω = ωc, the new element has the same magnitude of reactance as the old one, with the opposite sign.
The response: same cut-off, mirrored passband
Since |Ω| = 1 maps to ω = ωc, the attenuation at cut-off carries over unchanged. For a Butterworth prototype that is −3 dB. A Butterworth low-pass of order n has |S21|² = 1/(1 + Ω^2n). For the high-pass version this becomes:
The skirt mirrors on a logarithmic frequency axis. The low-pass attenuation at 2ωc equals the high-pass attenuation at ωc/2. For n = 3 that is 10·log10(1 + 2⁶) = 10·log10(65), about 18.1 dB. The filter order and the steepness per octave do not change. The skirt just faces the other way.
In the plots, the low-pass response is flat at 0 dB from 0 to 1 GHz and falls above. The high-pass response falls below 1 GHz and is flat above. Both pass the −3 dB point at fc = 1 GHz. The passband and the stopband have changed sides around the same dashed cut-off line.
Worked example: 3rd-order Butterworth at 50 Ω, 1 GHz
The 3rd-order Butterworth prototype has g1 = 1, g2 = 2, g3 = 1. Take the shunt-first form of the low-pass ladder: shunt C1 = 1, series L2 = 2, shunt C3 = 1, between 1 Ω terminations. Apply the swap at ωc = 1 rad/s and 1 Ω:
- shunt C1 = 1 becomes shunt L1 = 1/g1 = 1.000
- series L2 = 2 becomes series C2 = 1/g2 = 0.500
- shunt C3 = 1 becomes shunt L3 = 1/g3 = 1.000
The drawing shows this normalized high-pass ladder: a shunt inductor L1, a series capacitor C2 and a shunt inductor L3. The source and load marked "1" are 1 Ω terminations, because the prototype is normalized to 1 Ω.
Now scale to Z0 = 50 Ω and fc = 1 GHz, so ωc = 2π × 10⁹ rad/s:
- L1 = Z0/(ωc·g1) = 50/(2π × 10⁹ × 1) = 7.96 nH (shunt)
- C2 = 1/(Z0·ωc·g2) = 1/(50 × 2π × 10⁹ × 2) = 1.59 pF (series)
- L3 = 7.96 nH (shunt)
For a single g = 1 element at the same Z0 and fc, the formulas give C' = 3.18 pF from a series-inductor element and L' = 7.96 nH from a shunt-capacitor element. The middle element above is half of 3.18 pF because its g is 2.
In plain words · for students and beginners
Low-pass to high-pass transformation in plain words
An everyday analogy
Think of a road gate that lets slow cars through and stops fast ones. Swap the parts inside and the gate now lets fast cars through and stops slow ones. The speed where it switches stays the same. An inductor blocks high frequencies and a capacitor blocks low frequencies. So swapping one for the other flips which frequencies get through.
Where the analogy breaks down: you do not just move the same parts around. Each new part needs a new value from a formula. The swap also keeps the filter's shape, meaning which parts are in series (in the signal path) and which are shunt (from the signal path to ground).
The same example in simple steps
Start with the simplest low-pass filter. It has one inductor in the signal path and one capacitor to ground. Both have the prototype value g = 1. (A prototype is a ready-made table design for 1 Ω and a cut-off of 1 rad/s.) The goal is a filter for 50 Ω cables with a cut-off at 1 GHz.
- Find ωc, the cut-off in radians per second: ωc = 2π × 1 GHz, about 6.28 × 10⁹.
- The series inductor becomes a series capacitor: C' = 1/(Z0·ωc·g) = 1/(50 × 6.28 × 10⁹ × 1) = 3.18 pF.
- The shunt capacitor becomes a shunt inductor: L' = Z0/(ωc·g) = 50/(6.28 × 10⁹ × 1) = 7.96 nH.
- Check the cut-off. It is still 1 GHz. Now signals above 1 GHz pass and signals below are blocked.
The top row of the drawing is the low-pass filter: a 7.96 nH inductor in series and a 3.18 pF capacitor to ground. The bottom row is the high-pass filter: a 3.18 pF capacitor in series and a 7.96 nH inductor to ground. The arrow label is the frequency swap that connects the two rows. Its left side means the prototype's normalized frequency, which is just a number with no units.
Mistakes beginners make
- Moving the cut-off. The swap keeps fc where it was. Only the side that passes signals changes.
- Swapping the numbers but not using the formula. For g = 1 the values only seem to trade places. For g = 2 the new capacitor is half as large, 1.59 pF instead of 3.18 pF.
- Changing the layout. A part in series stays in series and a part to ground stays to ground. Only the type of part changes.
- Forgetting the 2π. The formulas need ωc in rad/s, not fc in Hz.
Remember
- Every inductor becomes a capacitor and every capacitor becomes an inductor, in the same position.
- Series capacitor C' = 1/(Z0·ωc·g); shunt inductor L' = Z0/(ωc·g).
- The cut-off stays at ωc. The passband moves from below it to above it.
Related pages on RF View
Frequently asked questions
How do I convert a low-pass filter prototype to a high-pass filter?
Replace the normalized prototype frequency with −ωc/ω. Every series inductor becomes a series capacitor C' = 1/(Z0·ωc·g), and every shunt capacitor becomes a shunt inductor L' = Z0/(ωc·g).
Does the cut-off frequency change in a low-pass to high-pass transformation?
No. The cut-off stays at ωc. The passband moves from below ωc to above it, and the attenuation at ωc is the same as in the low-pass design.
What are the high-pass element values for g = 1 at 50 Ω and 1 GHz?
A series-inductor element with g = 1 becomes a 3.18 pF series capacitor. A shunt-capacitor element with g = 1 becomes a 7.96 nH shunt inductor.
Why is there a minus sign in the mapping ω → −ωc/ω?
The minus sign makes a positive inductive reactance turn into the negative reactance of a capacitor, so the result can be built from real capacitors and inductors. It does not change the magnitude response, because |S21| is the same at +ω and −ω.