Butterworth LPF prototyping by insertion loss

In the insertion loss method, the filter response is defined by the power loss ratio:

PLR = P_available / P_delivered = 1 / (1 − |Γ(ω)|²)
IL (dB) = 10 log PLR

A physically realisable PLR must be a ratio of even polynomials in ω, so the designer chooses a polynomial that gives the wanted response.

Procedure for a Butterworth LPF prototype

1. Specify the filter: cut-off frequency fc, source/load impedance R0, and the minimum attenuation A (dB) at some stopband frequency ω.
2. Select the maximally flat response:

PLR = 1 + k²(ω/ωc)^(2N)

k² = 10^(Lr/10) − 1 sets the loss at ωc; for the usual 3 dB cut-off k = 1. The response is flat at ω = 0 and rolls off at 20N dB/decade.

3. Find the order N:

N ≥ log10(10^(A/10) − 1) / (2·log10(ω/ωc))
Example: A = 20 dB at ω = 2ωc
N ≥ log10(99)/(2×0.301) = 3.31  → N = 4

4. Get the normalised element values (R0 = 1 Ω, ωc = 1 rad/s) from the table, or

g0 = 1,  gk = 2 sin[(2k − 1)π/(2N)],  g(N+1) = 1

5. Draw the ladder: gk alternate between shunt C and series L. Two equivalent forms exist: starting with a shunt C (π type) or a series L (T type).

   Rs   L2       L4        (π-type, N=4)
 o-/\/-+--UUU--+--UUU--+--o
       |       |       |
      C1      C3       RL
       |       |       |
 o-----+-------+-------+--o

6. Impedance and frequency scaling to R0 and ωc:

L = gk·R0/ωc,   C = gk/(R0·ωc),   RL = g(N+1)·R0

Example (N = 3, R0 = 50 Ω, fc = 2 GHz): C1 = C3 = 1.59 pF, L2 = 7.96 nH.

7. Transform (if needed) to high-pass, band-pass or band-stop using the standard frequency transformations.
8. Implement at microwave frequency: Richards' transformation (L → short-circuited stub, C → open-circuited stub, λ/8 at fc) with Kuroda identities, or stepped-impedance lines. Simulate and tune.

Butterworth is chosen when a flat passband is needed; Chebyshev gives a sharper cut-off for the same N at the cost of ripple.