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How to Pick Filter Order From a Rejection Requirement

By John · Updated · 6 min read

A Butterworth needs N ≥ log10(10^(LAs/10) − 1) / (2·log10(ωs/ωc)), so 40 dB at twice the cut-off needs N = 7. A Chebyshev with 0.5 dB ripple meets the same 40 dB at ratio 2 with N = 5.

For practitionersNew to this? Plain-language version

For practitioners

A rejection requirement usually comes as a pair of numbers: an attenuation LAs in dB, and a frequency where you need it. To size a filter, turn that pair into one dimensionless quantity, the ratio ωs/ωc between the stopband frequency and the passband edge. Then use the closed-form order equation for the response type you want. This page goes through that arithmetic for Butterworth and Chebyshev low-pass prototypes and works one example from start to finish.

Step one: normalize the specification

The order formulas use only the ratio ωs/ωc, not absolute frequencies. A spec of "40 dB at 2 GHz with a 1 GHz cut-off" and a spec of "40 dB at 200 MHz with a 100 MHz cut-off" are the same problem: LAs = 40 dB at ωs/ωc = 2.

Before you compute anything, pin down what ωc means for the response type:

  • Butterworth: ωc is the 3 dB point.
  • Chebyshev: ωc is the edge of the equiripple passband. It is the highest frequency where the attenuation still equals the ripple LAr, not the 3 dB point.

If you mix up these two definitions, the ratio is wrong before you start. The order you get will then be wrong too, even though the arithmetic looks correct.

The Butterworth order equation

A Butterworth low-pass prototype with its 3 dB point at ωc has this attenuation:

A(ω)=10log⁡10(1+(ωωc)2N)A(\omega) = 10\log_{10}\left(1 + \left(\frac{\omega}{\omega_c}\right)^{2N}\right)

Set A(ωs) ≥ LAs and solve for N:

N≥log⁡10(10LAs/10−1)2log⁡10(ωs/ωc)N \geq \frac{\log_{10}\left(10^{L_{As}/10} - 1\right)}{2\log_{10}\left(\omega_s/\omega_c\right)}

The numerator depends only on how much rejection you need. The denominator depends only on how far from the edge you need it. Doubling the rejection in dB roughly doubles the numerator. Moving the stopband frequency farther out makes the denominator grow only logarithmically. That is why a tight transition band costs a lot of order.

Far from cut-off, the Butterworth response approaches a slope of 20·N dB per decade, which is about 6·N dB per octave. This gives a quick sanity check on any result. A seventh-order filter should give roughly 42 dB one octave above the 3 dB point, and it does.

Worked example: 40 dB at twice the cut-off

Take LAs = 40 dB at ωs/ωc = 2.

  1. Convert the rejection: 10^(40/10) − 1 = 10^4 − 1 = 9999.
  2. Numerator: log10(9999) = 4.0 (to the precision that matters here).
  3. Denominator: 2·log10(2) = 2·0.30103 = 0.602.
  4. Ratio: 4.0 / 0.602 = 6.64.
  5. Round up: N = 7.

Now check the result against the attenuation equation:

  • N = 7: 10·log10(1 + 2^14) = 42.1 dB. It passes with 2.1 dB to spare.
  • N = 6: 10·log10(1 + 2^12) = 36.1 dB. It misses by 3.9 dB.
  • N = 5: 10·log10(1 + 2^10) = 30.1 dB. It misses by almost 10 dB.
11.52340102030405060Normalized frequency ω/ωc (log scale)Attenuation (dB)N = 7N = 6N = 5All curves 3.01 dB at ω/ωc = 1Requirement: 40 dB at ω/ωc = 2At ω/ωc = 2:N = 7: 42.1 dB, passesN = 6: 36.1 dB, missesN = 5: 30.1 dB, misses
Butterworth attenuation for N = 5, 6 and 7 against the 40 dB requirement at ω/ωc = 2. Computed from the formula.

The plot shows all three curves passing through 3.01 dB at ω/ωc = 1 and then spreading apart. At the vertical line ω/ωc = 2, only the N = 7 curve is above the 40 dB requirement line. The N = 6 curve misses by a few dB. This is typical: the fractional result 6.64 tells you that the sixth order is close but not enough. No amount of tuning a sixth-order Butterworth will close that gap, because its shape is fixed once N and ωc are fixed.

The Chebyshev order equation

A Chebyshev response allows ripple of LAr dB in the passband. In exchange, it gets a steeper transition. Its attenuation is A(ω) = 10·log10(1 + k²·cosh²(N·arcosh(ω/ωc))) for ω > ωc, where k² = 10^(LAr/10) − 1. Solving for N gives:

N≥arcosh(10LAs/10−1k2)arcosh(ωs/ωc)N \geq \frac{\text{arcosh}\left(\sqrt{\frac{10^{L_{As}/10} - 1}{k^2}}\right)}{\text{arcosh}\left(\omega_s/\omega_c\right)}

The same spec with 0.5 dB ripple works out like this:

  1. k² = 10^(0.05) − 1 = 0.122.
  2. Inside the square root: 9999 / 0.122. The square root is 286.3.
  3. Numerator: arcosh(286.3) = 6.35.
  4. Denominator: arcosh(2) = 1.317.
  5. Ratio: 6.35 / 1.317 = 4.82, so N = 5.

That is two fewer reactive elements than the Butterworth. Remember that ωc means something different here. The Chebyshev passband is 0.5 dB down at ωc, while the Butterworth is 3 dB down at its ωc. If your system spec sets the passband edge as a maximum-loss point, the Chebyshev definition is often the more natural fit anyway.

Practical points before you commit to N

The order equations describe an ideal lossless prototype. Treat the computed N as a floor. Component tolerance, finite Q and layout parasitics all shift the real response, so leave margin rather than designing to exactly 40.0 dB.
  • Always round up. A result of 6.01 still means N = 7.
  • Check where the spec actually applies. If the rejection is required across a band, use the lowest stopband frequency. That is the worst case for a low-pass filter.
  • For band-pass designs, apply the formulas to the low-pass prototype after mapping the band-pass frequencies to an equivalent prototype ratio. The order you find is the prototype order.
  • Verify by measurement. Once the filter is built, the stopband shows up in S21. The page What S21 and S11 of a Low-Pass Filter Look Like explains how to read those traces.

In plain words · for students and beginners

Filter order in plain words

An everyday analogy

Think of a filter as a ramp that leads down from a flat road (the frequencies you keep) to a deep ditch (the frequencies you block). The "order" is how many sections the ramp has. More sections make the ramp steeper, so you reach the bottom of the ditch sooner. A rejection spec says: "by this distance from the road, you must be this deep."

The analogy breaks down in one place. A real ramp has a fixed slope, but a filter's slope keeps getting steeper the farther you go past the edge. Also, each "section" is a real part, such as a capacitor or an inductor. Each one adds a little loss and cost.

The same example, step by step

The spec is 40 dB of blocking at twice the cut-off frequency. "dB" is a log scale. 40 dB means the unwanted signal's power is cut to one ten-thousandth.

Step 1Step 2Step 3Step 4Step 5Spec:LAs = 40 dBat ωs/ωc = 2Compute10^(LAs/10) − 1= 10^4 − 1= 9999Divide log10(9999)= 4.0by 2·log10(2)= 0.602Result6.64Round up:N = 7side note: same specChebyshev,0.5 dB ripple (k² = 0.122):arcosh(286.3)/arcosh(2)= 6.35/1.317 = 4.82so N = 5
Step-by-step Butterworth order calculation for 40 dB of rejection at twice the cut-off frequency, with the Chebyshev 0.5 dB ripple result shown as a side note. Concept drawing.
  1. Write down the spec: 40 dB at a ratio of 2.
  2. Turn 40 dB into a plain number: 10^4 − 1 = 9999.
  3. Take log10 of that, which is about 4.0. Divide by 2·log10(2) = 0.602.
  4. The result is 6.64.
  5. You can't build part of a section, so round up: N = 7.

A Chebyshev filter allows a small wobble (ripple) in the passband. With 0.5 dB of wobble, the same job takes only N = 5. The side note in the diagram shows that calculation.

Mistakes beginners make

  • Rounding down. If you get 6.64, the answer is 7. Six sections give only 36.1 dB, which is short of 40.
  • Using the wrong edge for Chebyshev. For a Chebyshev filter, the "cut-off" in the formula is where the passband wobble ends, not the 3 dB point.
  • Mixing up ratio and frequency. The formula needs the ratio, for example 2, not the frequency in MHz.
  • Ignoring real parts. Real capacitors and inductors are not perfect, so give yourself a few dB of extra margin.

Remember

  • Butterworth: N ≥ log10(10^(LAs/10) − 1) / (2·log10(ωs/ωc)), and 40 dB at 2x gives N = 7.
  • Chebyshev with 0.5 dB ripple meets the same spec with N = 5.
  • Always round the order up, then check the result with the attenuation formula.

Frequently asked questions

What Butterworth order gives 40 dB rejection at twice the cut-off frequency?

The bound is log10(9999)/(2·log10(2)) = 6.64, so you round up to N = 7. That order gives 42.1 dB at ωs/ωc = 2, while N = 6 gives only 36.1 dB.

Why does a Chebyshev filter need a lower order for the same rejection?

A Chebyshev trades passband flatness for a steeper transition. With 0.5 dB ripple, 40 dB at ratio 2 needs N = 5 instead of the Butterworth's N = 7.

Is the Chebyshev cut-off frequency the 3 dB point?

No. In the Chebyshev order formula, ωc is the edge of the equiripple passband, where attenuation first equals the ripple LAr. Using the 3 dB frequency there gives the wrong order.

Should I round the filter order up or down?

Always up. The formula gives the minimum order that meets the spec, so any fractional result, even 6.1, means the next whole number.

References and further reading

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