Low-Pass Prototype Element Values: What g1 to gN Mean
Low-pass prototype g-values are element values normalized to a 1 Ω source and ωc = 1 rad/s, with g0 and g(N+1) as the source and load terms. For Butterworth, g_k = 2·sin((2k − 1)π/(2N)), so N = 3 gives 1, 2, 1.
For practitioners
A filter design table gives you a row of plain numbers: g0, g1, g2 and so on up to g(N+1). There are no henries, no farads and no ohms in it. That row is a low-pass prototype, and once you know what each number stands for, it works for any impedance level and any cut-off frequency.
What a prototype row actually is
A low-pass prototype is a lossless LC ladder between a source resistor and a load resistor. It is normalized in two ways:
- The source impedance is 1 Ω.
- The cut-off frequency is ωc = 1 rad/s.
Within that normalized world, g1 to gN are the reactive element values. In a ladder, each one is either a series inductance or a shunt capacitance, and the two types alternate along the ladder. If g1 is a shunt capacitor, g2 is a series inductor, g3 is a shunt capacitor again, and so on. If g1 is a series inductor, the pattern flips.
So a g-value is a number of henries or farads, but only in a circuit that runs from a 1 Ω source with a cut-off at 1 rad/s. It is not tied to any real frequency or impedance until you scale it. That is why one row serves every design: the response shape lives in the ratios between the g-values, and scaling only moves that shape to a new impedance level and frequency.
g0 and g(N+1): the terminations
The two end values are not reactive elements. g0 is the source and g(N+1) is the load. Each one is either a resistance or a conductance, and which one it is depends on its neighbour:
- g0 is a resistance if g1 is a shunt capacitor, and a conductance if g1 is a series inductor.
- g(N+1) is a resistance if gN is a shunt capacitor, and a conductance if gN is a series inductor.
The rule follows from duality. A ladder that starts with a shunt C is naturally described with a source resistance. Its dual, which starts with a series L, swaps every impedance for an admittance, so the termination becomes a conductance. When g0 = 1 the difference disappears, because 1 Ω and 1 S are the same termination. When a termination value is not 1, getting this wrong inverts the load.
A table always lists g(N+1) in its own column. Read it from the table; do not assume it is 1 for every response family.
Computing Butterworth values directly
For the maximally flat (Butterworth) response you do not need a table. The element values have a closed form:
with g0 = g(N+1) = 1. At ω = 1 rad/s the normalized prototype is 3 dB down, which is the standard Butterworth cut-off.
Worked example, N = 3:
- g1 = 2·sin(π/6) = 2·sin 30° = 1
- g2 = 2·sin(3π/6) = 2·sin 90° = 2
- g3 = 2·sin(5π/6) = 2·sin 150° = 1
- g0 = g4 = 1
Worked example, N = 5 (angles in steps of 36°, starting at 18°):
- g1 = 2·sin 18° = 0.618
- g2 = 2·sin 54° = 1.618
- g3 = 2·sin 90° = 2
- g4 = 2·sin 126° = 1.618
- g5 = 2·sin 162° = 0.618
- g0 = g6 = 1
The values are symmetric about the middle element. That comes straight from the formula: replacing k with N + 1 − k changes the angle to π − (2k − 1)π/(2N), and sin(π − x) = sin x. For odd N the middle element sits at k = (N + 1)/2, where the angle is π/2, so it is always exactly 2. For even N there is no middle element and no value reaches 2. The outer elements are the smallest because the angle is closest to 0 or π there.
From one row to any impedance and cut-off
To move the prototype to a system impedance R0 and a cut-off ωc = 2π·fc, scale each element by its type:
Terminations scale as R = g·R0 when g is a resistance, and R = R0/g when g is a conductance.
Inductors scale up with impedance and capacitors scale down. Both scale down with frequency. That is the full content of impedance and frequency scaling for a low-pass filter.
Take the N = 3 Butterworth row in a 50 Ω system, starting with a shunt capacitor:
- C1 = C3 = 1/(50·ωc)
- L2 = 2·50/ωc = 100/ωc
- Source and load: 50 Ω each
The dual form, starting with a series inductor, uses the same row:
- L1 = L3 = 50/ωc
- C2 = 2/(50·ωc) = 1/(25·ωc)
Both ladders give the same response. Pick a value for ωc and you have real component values. The response itself is easier to check from S21 and S11 than from the element list; see Low-Pass Filter S-Parameters Explained.
Common reading errors
Other mistakes come up often:
- Treating g-values as picofarads or nanohenries. They have no unit prefix. They are henries and farads in a 1 Ω, 1 rad/s circuit.
- Using f instead of ω. The prototype is normalized to 1 rad/s. Divide by 2π·fc, not by fc.
- Assuming g(N+1) = 1. That holds for Butterworth. For other families, read the table column.
- Mixing up the terminations. With a series-L start, g0 is a conductance. If it is not 1, then R = R0/g0, not R0·g0.
In plain words · for students and beginners
Low-pass prototype values in plain words
A recipe written for one person
A prototype table is like a recipe written for exactly one serving. Every number in it assumes one standard situation. Here that situation is a 1 Ω source and a cut-off of 1 rad/s. A cut-off is the frequency where the filter starts blocking signals. rad/s (radians per second) is a way of measuring frequency, and 1 rad/s is about 0.16 Hz. To cook for a crowd, you scale the recipe. To build a real filter, you scale the g-values to your real impedance (often 50 Ω) and your real cut-off frequency.
Where the analogy breaks down: in a recipe every ingredient grows the same way. In a filter, inductors (coils) get bigger when you raise the impedance, but capacitors get smaller. Both get smaller when you raise the frequency. Each type has its own scaling rule.
The three-element example, step by step
A ladder is a chain of parts: one part goes in series (in line with the signal path), the next goes in shunt (from the signal path down to ground), and so on, taking turns.
- Pick the order: N = 3, meaning three reactive parts (parts that store energy, here coils and capacitors).
- Use the Butterworth formula, 2 times the sine of an angle. The angles for N = 3 are 30°, 90° and 150°.
- 2·sin 30° = 1, so g1 = 1. 2·sin 90° = 2, so g2 = 2. 2·sin 150° = 1, so g3 = 1.
- The ends are g0 = 1 (the source) and g4 = 1 (the load). In the drawing they are resistors, because the parts next to them are shunt capacitors.
- The result: shunt capacitor 1, series coil 2, shunt capacitor 1, between two 1 Ω resistors.
To make it real in a 50 Ω system: each capacitor becomes 1/(50·ωc), the coil becomes 100/ωc, and both resistors become 50 Ω. Here ωc is 2π times your cut-off frequency in hertz.
Mistakes beginners make
- Reading g-values as real component values. A "2" is not 2 nH or 2 pF. It means nothing in hardware until you scale it.
- Forgetting the 2π. The formulas use ωc = 2π·fc. If you leave out the 2π, every part comes out about six times too large.
- Forgetting that the row works both ways. The same numbers can start with a capacitor or with a coil. Decide which form you are building first, then scale each part by its own type.
- Treating g0 and g(N+1) like coils or capacitors. They are the source and load ends of the filter, not filter parts.
Remember
- g-values are normalized: 1 Ω source, cut-off 1 rad/s, no units until you scale.
- g0 and g(N+1) are the terminations. g1 to gN take turns between series coil and shunt capacitor.
- Butterworth values follow 2·sin((2k − 1)π/(2N)): 1, 2, 1 for N = 3, and 0.618, 1.618, 2, 1.618, 0.618 for N = 5.
Related pages on RF View
Frequently asked questions
What do g0 and g(N+1) mean in a low-pass prototype table?
g0 is the source termination and g(N+1) is the load termination. Each one is a resistance when the element next to it is a shunt capacitor and a conductance when the element next to it is a series inductor.
How do I calculate Butterworth g-values without a table?
Use g_k = 2·sin((2k − 1)π/(2N)) for k = 1 to N, with g0 = g(N+1) = 1. For N = 3 this gives 1, 2, 1.
What units do low-pass prototype g-values have?
None until you scale them. They are normalized to a 1 Ω source and a cut-off of 1 rad/s, so a g-value turns into henries, farads or ohms only after you apply impedance and frequency scaling.
Can the same g-values start with either a series inductor or a shunt capacitor?
Yes. The two ladder forms are duals and give the same response. Only the meaning of each g changes: inductance or capacitance, resistance or conductance.