Chebyshev Ripple and Passband Return Loss Relationship
A lossless equal-ripple filter has a minimum passband return loss of RL = 10·log10(1 + 1/k²), where k² = 10^(LAr/10) − 1. A 0.1 dB ripple gives 16.4 dB return loss and a VSWR of about 1.36. A 0.5 dB ripple drops the return loss to 9.6 dB.
Engineer’s note: That harmless-looking 0.5 dB ripple spec means return loss can sink to 9.6 dB at the peaks. Hand that number to whoever owns the amplifier or mixer port next to the filter, because they're probably sizing their match against a clean 50 Ω and won't get it.
For practitioners
A Chebyshev filter is usually specified by its passband ripple, for example 0.1 dB or 0.5 dB. That number reads like a transmission spec, and a small one looks harmless. For a lossless filter, though, the ripple and the input match are the same quantity. Every tenth of a dB missing from S21 has been reflected back toward the source. This page shows the exact link and gives the numbers.
Where the ripple comes from
An equal-ripple low-pass response has this transmission:
T_N is the Chebyshev polynomial of order N and ω is the frequency normalized to the ripple cutoff. Inside the passband, |ω| ≤ 1, T_N swings between −1 and +1. So |S21|² swings between 1 (where T_N = 0) and 1/(1 + k²) (where |T_N| = 1). The ripple factor k sets the depth of those swings. The ripple in dB is the attenuation at the bottom of each dip:
From ripple to worst-case return loss
A lossless two-port obeys |S11|² + |S21|² = 1. Wherever transmission dips, reflection peaks by exactly the same amount of power. The maximum passband reflection is:
Return loss is −10·log10 |S11|², so the minimum passband return loss is:
The curve is steep near zero ripple. Going from 0.5 dB to 0.1 dB of ripple improves the worst-case return loss from 9.6 dB to 16.4 dB. Getting past 20 dB takes ripple in the hundredths of a dB. For small ripple, k² ≪ 1, two approximations hold: LAr ≈ 4.343·k² dB and RL ≈ 10·log10(1/k²). Each 10× cut in ripple then buys about 10 dB of return loss.
The relation also runs the other way. Because 1 − |Γ|² = 1/(1 + k²), the ripple is exactly the mismatch loss at the worst-case reflection: LAr = −10·log10(1 − |Γ|²). The passband ripple of a lossless filter is a mismatch-loss curve plotted against frequency.
Worked example: 0.1 dB ripple
- k² = 10^(0.1/10) − 1 = 10^0.01 − 1 = 0.0233.
- Minimum |S21|² = 1/1.0233 = 0.977, which is −0.1 dB as expected.
- Maximum |S11|² = 0.0233/1.0233 = 0.0228. About 2.3 % of the incident power is reflected.
- RL = 10·log10(1 + 1/0.0233) = 10·log10(43.9) = 16.4 dB.
- |Γ| = √0.0228 = 0.151.
- VSWR = (1 + 0.151)/(1 − 0.151) = 1.151/0.849 ≈ 1.36.
The same steps for 0.5 dB give k² = 0.122, RL = 10·log10(9.20) = 9.6 dB, |Γ| ≈ 0.330 and VSWR ≈ 1.98.
| Ripple LAr | k² | Min RL | Max |Γ| | Max VSWR |
|---|---|---|---|---|
| 0.01 dB | 0.0023 | 26.4 dB | 0.048 | 1.10 |
| 0.1 dB | 0.0233 | 16.4 dB | 0.151 | 1.36 |
| 0.5 dB | 0.122 | 9.6 dB | 0.330 | 1.98 |
| 1 dB | 0.259 | 6.9 dB | 0.454 | 2.66 |
A filter spec of "0.1 dB ripple" therefore also means "VSWR up to 1.36 at the input". Any amplifier or mixer port next to the filter sees that load. It is a measurable mismatch, not a rounding error.
Where the reflection peaks sit
Take a 3rd-order filter. T3(ω) = 4ω³ − 3ω has zeros at ω = 0 and ω = √3/2 ≈ 0.866. It reaches |T3| = 1 at ω = 0.5 and ω = 1.0. At the zeros the filter is perfectly matched: S21 touches 0 dB and S11 falls toward −∞ dB (a deep null). At ω = 0.5 and ω = 1.0, S21 dips to −0.1 dB and S11 rises to −16.4 dB.
The S11 trace is not a flat line at −16.4 dB. It is a row of reflection zeros separated by peaks that all reach the same level. An order-N filter has N reflection zeros counted over ±ω, so on the positive axis you see about N/2 nulls. A filter measured on a VNA can therefore look excellent at one frequency and only 16 dB at another. Judge it by the envelope of the peaks, not by the depth of the nulls.
Choosing ripple from a return-loss target
Inverting the relation gives the ripple you can allow for a given return-loss requirement:
A 15 dB return-loss target allows about 0.14 dB of ripple. A 20 dB target allows about 0.044 dB. Lower ripple costs selectivity at a given order. So a tight match spec pushes toward a higher order, or toward accepting a softer skirt.
These numbers describe the lossless prototype. Real resonators dissipate power. Dissipation adds insertion loss and tilts S21, and it also tends to make the measured return loss look slightly better than the lossless prediction, because some of the power that would have been reflected is absorbed. Once dissipation matters, read the ripple-to-RL link as a design relation, not as a test you can check from a measured S21 alone.
In plain words · for students and beginners
Chebyshev ripple and return loss in plain words
An everyday analogy
Think of a shop window. Most light passes through the glass, but a little bounces back, which is why you see a faint reflection of yourself. A filter's passband works the same way. Most of the signal passes, a little is reflected back toward where it came from, and none is "used up" inside a perfect (lossless) filter. If 2 % of the light bounces off, only 98 % gets through.
The analogy breaks down in one place. A window reflects about the same amount at every color. A Chebyshev filter does not. At some frequencies it reflects almost nothing, and at others it reflects the maximum amount. That up-and-down pattern is the "ripple" (the small wave in the passband response).
The 0.1 dB example in simple steps
- Ripple: the most the signal drops inside the passband, here 0.1 dB.
- Step 1: Turn dB into a power ratio. A 0.1 dB drop means 97.7 % of the power gets through.
- Step 2: What did not get through was reflected: 100 % − 97.7 % = about 2.3 %.
- Step 3: Return loss (how far below the incoming signal the reflection is, in dB) is 10·log10(1/0.0228) = 16.4 dB.
- Step 4: Reflection coefficient |Γ| (the voltage ratio of reflected to incoming wave) is the square root of 0.0228, which is 0.151.
- Step 5: VSWR (voltage standing wave ratio, a different way to state the same mismatch) is (1 + 0.151)/(1 − 0.151) ≈ 1.36.
On this chart, 16.4 dB return loss sits at a VSWR of about 1.36. If the ripple grows to 0.5 dB, return loss falls to 9.6 dB and VSWR rises to almost 2.
Mistakes beginners make
- "0.1 dB is basically nothing." For transmission it is small. For the match it means 16.4 dB return loss, which many specs would count as only moderate.
- Mixing up ripple and insertion loss. Ripple is the wobble inside the passband. Insertion loss (the total drop through the filter) also includes power turned into heat by real parts.
- Reading the deepest S11 null. S11 has sharp dips. The number that matters is the highest peak, because the filter has to meet the spec at every passband frequency.
- Expecting a lossy filter to match the formula exactly. The formula is exact only for a lossless filter.
Remember
- Ripple and return loss are linked by RL = 10·log10(1 + 1/k²), with k² = 10^(ripple/10) − 1.
- 0.1 dB ripple means 16.4 dB return loss, |Γ| = 0.151, VSWR ≈ 1.36.
- S21 dips and S11 peaks happen at the same frequencies.
Related pages on RF View
Frequently asked questions
What return loss does a 0.1 dB ripple Chebyshev filter have?
A lossless 0.1 dB equal-ripple filter has a worst-case passband return loss of 16.4 dB. That is |Γ| = 0.151, or a VSWR of about 1.36.
How do I convert Chebyshev ripple in dB to return loss?
First compute k² = 10^(LAr/10) − 1. Then RL = 10·log10(1 + 1/k²). This holds for a lossless filter at the frequencies where the ripple reaches its maximum.
Why does S11 peak at the same frequencies where S21 dips?
In a lossless two-port |S11|² + |S21|² = 1. Any power that does not get through is reflected, so the minima of S21 and the maxima of S11 fall at the same frequencies.
What ripple should I design for if I need 20 dB return loss?
Invert the relation: LAr = −10·log10(1 − 10^(−RL/10)). For 20 dB return loss this gives about 0.044 dB of ripple.