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Turning a Low-Pass Prototype Into a Band-Pass Filter

By John · Updated · 6 min read

The band-pass mapping replaces each series inductor with a series LC and each shunt capacitor with a parallel LC, all resonant at ω0 = sqrt(ω1·ω2). For g = 1, 50 Ω, 1 GHz and Δ = 0.1, the series branch is 79.6 nH with 0.318 pF.

Engineer’s note: I treat values like 0.318 pF and 0.796 nH as the mapping telling you to change topology. At that size the pads and the part body are the component, and the resonator ends up tuned by layout. That's the point to move to coupled resonators.

For practitionersNew to this? Plain-language version

For practitioners

A low-pass prototype is a ladder of normalized g-values. It assumes a 1 Ω source and a cut-off of 1 rad/s. To get a band-pass filter you don't redesign anything. You change the frequency variable, and every reactive element turns into a resonator. The response shape you picked for the prototype (ripple, order, skirt slope) carries over unchanged. Only the frequency axis gets stretched and moved.

The frequency mapping

The low-pass to band-pass transformation replaces the prototype frequency variable with:

ω→1Δ(ωω0−ω0ω)\omega \rightarrow \frac{1}{\Delta}\left(\frac{\omega}{\omega_0} - \frac{\omega_0}{\omega}\right)

Two parameters set the mapping. The centre frequency is the geometric mean of the band edges:

ω0=ω1ω2\omega_0 = \sqrt{\omega_1 \omega_2}

The fractional bandwidth is the passband width relative to that centre:

Δ=ω2−ω1ω0\Delta = \frac{\omega_2 - \omega_1}{\omega_0}

Check the mapping at a few points. At ω = ω0 the bracket is zero, so the band-pass centre maps to prototype DC. At ω = ω2 the bracket equals (ω2 − ω1)/ω0, because ω0²/ω2 = ω1. Dividing by Δ gives exactly +1, which is the prototype cut-off. At ω = ω1 you get −1. The prototype's response from −1 to +1, mirrored around DC, becomes the passband from ω1 to ω2. As ω goes to 0 or to infinity, the mapped variable goes to minus or plus infinity. So the prototype stopband becomes two stopbands, one on each side.

Low-pass prototype (normalized ω)ω = −1cut-off ω = 1−3−2−10123ω0−10−20−30−40|S21| (dB)ω = +1 → f2ω = 0 → f0ω = −1 → f1f1f0f20.80.91.01.11.2f (GHz)0−10−20−30−40|S21| (dB)Band-passf1 = 0.951 GHzf0 = 1 GHzf2 = 1.051 GHzf2 − f1 = 0.100 GHzΔ = (f2 − f1)/f0 = 0.1f1·f2 = f0²(geometric symmetry)
The low-pass prototype response and the band-pass response it maps to with f0 = 1 GHz and Δ = 0.1: prototype DC maps to f0 and the edges at −1 and +1 map to f1 and f2. Computed from the formula.

The figure shows the consequence. With f0 = 1 GHz and Δ = 0.1, solving f2 − f1 = 0.1 GHz together with f1·f2 = f0² gives f1 = 0.951 GHz and f2 = 1.051 GHz. The passband is not centred arithmetically. The upper edge sits 51 MHz above f0 and the lower edge 49 MHz below. This is geometric symmetry. It holds for every point on the response, not just the edges. Two frequencies with equal attenuation always multiply to f0². At Δ = 0.1 the asymmetry is small. For wider bands it grows, and the skirts look visibly lopsided on a linear frequency axis.

What happens to a series inductor

Take a prototype series inductor with normalized value L (its g-value). Its normalized reactance is ωL. Substitute the mapping and scale the impedance by Z0:

jX=jLZ0Δ(ωω0−ω0ω)=jωLZ0Δω0+1jωΔω0Z0LjX = j\frac{L Z_0}{\Delta}\left(\frac{\omega}{\omega_0} - \frac{\omega_0}{\omega}\right) = j\omega\frac{L Z_0}{\Delta \omega_0} + \frac{1}{j\omega \frac{\Delta}{\omega_0 Z_0 L}}

The first term is an inductor. The second is a capacitor. Both are in series in the same branch:

L′=LZ0Δω0,C′=Δω0Z0LL' = \frac{L Z_0}{\Delta \omega_0}, \qquad C' = \frac{\Delta}{\omega_0 Z_0 L}

The product L'·C' = 1/ω0², so the series LC resonates at ω0. At resonance the branch is a short. That matches a prototype inductor at DC, which has zero reactance.

What happens to a shunt capacitor

The dual argument applies to a shunt capacitor of normalized value C. Its admittance ωC maps to a capacitive susceptance plus an inductive one. Both sit in parallel across the line:

C′=CΔω0Z0,L′=ΔZ0ω0CC' = \frac{C}{\Delta \omega_0 Z_0}, \qquad L' = \frac{\Delta Z_0}{\omega_0 C}

Again L'·C' = 1/ω0². At ω0 the parallel tank is an open circuit, just as a prototype shunt capacitor is open at DC. Away from ω0 the tank's susceptance grows, and it shunts the signal to ground. This is the band-pass equivalent of the capacitor's rising susceptance above cut-off.

Worked example: g = 1, 50 Ω, 1 GHz, Δ = 0.1

Use a prototype series inductor with g = 1, then a shunt capacitor with g = 1. Set Z0 = 50 Ω, f0 = 1 GHz, so ω0 = 2π × 10⁹ ≈ 6.283 × 10⁹ rad/s, and Δ = 0.1.

Series branch:

  • L' = 1 × 50 / (0.1 × 6.283 × 10⁹) = 79.6 nH
  • C' = 0.1 / (6.283 × 10⁹ × 50 × 1) = 0.318 pF

Shunt branch:

  • C' = 1 / (0.1 × 6.283 × 10⁹ × 50) = 31.8 pF
  • L' = 0.1 × 50 / (6.283 × 10⁹ × 1) = 0.796 nH

Check the series resonator: 1/(2π·sqrt(79.6 nH × 0.318 pF)) = 1.00 GHz. The shunt pair gives the same result, since 31.8 pF × 0.796 nH is the same product.

1. Low-pass prototypeZ0 = 50 ΩL: g = 1C: g = 1inout2. Frequency mappingω → (1/Δ)(ω/ω0 − ω0/ω)ω0 = sqrt(ω1·ω2)Δ = (ω2 − ω1)/ω0 = 0.1f0 = 1 GHz3. Band-pass resultZ0 = 50 ΩL' = 79.6 nHC' = 0.318 pFL' = 0.796 nHC' = 31.8 pFinoutseries LC resonatesat f0 = 1.00 GHzparallel LC resonatesat f0 = 1.00 GHz
Low-pass to band-pass transformation of a g = 1 series inductor and g = 1 shunt capacitor at Z0 = 50 Ω, f0 = 1 GHz and Δ = 0.1. Concept drawing.

Reading the numbers

The worked values show the main practical problem with this direct mapping. The series inductor and capacitor are divided by Δ or multiplied by Δ, and the shunt pair goes the opposite way. So the ratio between the two resonators grows as 1/Δ². The series branch has a large inductor and a tiny capacitor. The shunt branch has a large capacitor and a tiny inductor. At Δ = 0.1, 0.318 pF is close to the size of parasitic pad capacitance. 0.796 nH is similar to the inductance of a short trace or a component's own body. Narrower bandwidths make this spread worse.

The formulas give ideal values. Any finite resonator Q adds loss that rises as Δ shrinks, because the stored energy in each resonator scales as 1/Δ. A narrow band-pass built from low-Q lumped parts will show much more insertion loss than its low-pass parent, even with the same g-values.

This is why narrowband designs usually move to coupled-resonator forms with impedance or admittance inverters, which make all the resonators the same type. The direct mapping is still the starting point. It shows where the resonator values come from and why fractional bandwidth controls them.

In plain words · for students and beginners

Low-pass to band-pass in plain words

An everyday picture

A low-pass filter is like a door that lets slow walkers through and stops runners. A band-pass filter is like a turnstile that only turns for people walking at one particular pace, give or take a little. To build the turnstile you keep the same plan as the door. You just swap each part for a part that "likes" one speed.

In the filter, a coil (inductor, a part that resists fast changes in current) and a capacitor (a part that stores charge and resists slow changes) each get replaced by a tuned pair. A tuned pair is a coil and a capacitor working together that respond strongly at one frequency, called the resonant frequency. The analogy breaks down in one place: the "allowed pace" is not centred in a simple way. The band reaches a little further above the centre than below it.

inoutcoil Lcapacitor Clow-pass: lets lowfrequencies througheach part becomes a pairtuned to 1 GHzpair in seriespair inparallelband-pass: lets only aband around 1 GHz throughpass band0.9511centre1.051frequency (GHz)width = 10% ofthe centre frequency
How each part of a low-pass prototype becomes a resonator tuned to 1 GHz, which gives a band-pass filter with a 10% wide pass band. Concept drawing.

The top row is the low-pass filter. The bottom row is the band-pass version, where each part became a pair tuned to 1 GHz. The strip under it shows the pass band from 0.951 to 1.051 GHz. Its width is 10% of the centre frequency.

The same example, step by step

  1. Pick the centre frequency: 1 GHz.
  2. Pick the width as a fraction of the centre: 10%, so Δ = 0.1.
  3. Pick the system impedance: 50 Ω (the standard resistance level RF circuits are built around).
  4. Start with a prototype coil of value 1. It becomes a coil of 79.6 nH in series with a capacitor of 0.318 pF.
  5. A prototype capacitor of value 1 becomes a capacitor of 31.8 pF in parallel with a coil of 0.796 nH.
  6. Check: both pairs resonate at 1.00 GHz.

Mistakes beginners make

  • Taking the centre as the plain average of the band edges. It is the square root of their product. That is why the edges are 0.951 and 1.051 GHz, not 0.95 and 1.05.
  • Mixing up which pair goes where. A series coil becomes a series pair, and a shunt capacitor (one connected to ground) becomes a parallel pair.
  • Forgetting the 2π. The formulas use ω = 2π·f, not f.
  • Trusting tiny values. A 0.318 pF capacitor is about as small as the stray capacitance of the board itself.

Remember

  • Every coil and capacitor becomes a tuned pair, and every pair resonates at the centre frequency.
  • The width as a fraction of the centre, Δ, sets all the part values.
  • Narrower bands give more extreme part values and more loss.

Frequently asked questions

How do you convert a low-pass prototype to a band-pass filter?

Substitute ω → (1/Δ)·(ω/ω0 − ω0/ω) into the prototype. Each series inductor becomes a series LC, and each shunt capacitor becomes a parallel LC. Every resonator is tuned to ω0.

Why is the band-pass centre frequency the geometric mean of the band edges?

The mapping sends the prototype edges ω = −1 and ω = +1 to band edges ω1 and ω2 with ω1·ω2 = ω0². So ω0 = sqrt(ω1·ω2), not the arithmetic average.

What is fractional bandwidth in filter design?

Fractional bandwidth is Δ = (ω2 − ω1)/ω0, the passband width divided by the centre frequency. It appears directly in every element formula and sets how far apart the L and C values in each resonator end up.

What are the band-pass element values for g = 1 at 1 GHz with 10% bandwidth in 50 ohms?

The series branch is 79.6 nH in series with 0.318 pF. The shunt branch is 31.8 pF in parallel with 0.796 nH. Both resonate at 1.00 GHz.

References and further reading

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