Scaling a Low-Pass Prototype to 50 Ohm and Real Frequency
Scale each prototype element with L = R0·g/ωc and C = g/(R0·ωc). At 1 GHz and 50 Ω, a g = 1, 2, 1 Butterworth becomes a 3.18 pF shunt C, a 15.9 nH series L and a 3.18 pF shunt C.
For practitioners
A low-pass prototype table gives you numbers like 1, 2, 1. Those are not henries or farads. They describe a filter terminated in 1 Ω with a cut-off of 1 rad/s. To build something you can solder onto a 50 Ω board, you have to scale it twice: once for impedance and once for frequency. This page covers both steps and works through a complete 1 GHz, 3rd-order Butterworth design.
What the prototype actually represents
A normalized low-pass prototype is a ladder of alternating series inductors and shunt capacitors. Its element values g1 to gN are set for a source resistance of 1 Ω and a cut-off angular frequency of 1 rad/s. For an odd-order Butterworth, the load is also 1 Ω.
The prototype fixes the shape of the response: passband flatness, roll-off rate, and where the −3 dB point sits relative to cut-off. Scaling does not change any of that. It only moves the response to a new impedance level and a new frequency. So every property you picked when you chose the filter type and order carries over unchanged.
Each g-value is used in one of two ways. In the shunt-first form, g1 is a shunt capacitor, g2 a series inductor, g3 a shunt capacitor, and so on. In the series-first (dual) form, the roles swap. Both forms give the same response.
Impedance scaling: from 1 Ω to R0
To move from 1 Ω to a system impedance R0, multiply every impedance in the network by R0. An inductor's impedance is jωL, so inductances go up by R0. A capacitor's impedance is 1/(jωC), so capacitances go down by R0. Resistive terminations become R0.
This keeps every impedance ratio in the circuit the same. Since the response depends only on those ratios, the filter shape is preserved.
Frequency scaling: from 1 rad/s to ωc
To move the cut-off from 1 rad/s to ωc, every reactance that used to occur at 1 rad/s now has to occur at ωc. Reactance depends on the product ωL or ωC, so both L and C get divided by ωc.
Put the two steps together and you get the two scaling formulas you will use for every element:
The most common mistake is to put fc into these formulas where ωc belongs. That makes every value 2π times too large, and the filter cuts off at about one-sixth of the frequency you intended.
A useful sanity check: reactance at cut-off
The formulas lead to a simple rule. At ωc, a scaled inductor has reactance ωc·L = g·R0. A scaled capacitor has reactance 1/(ωc·C) = R0/g. So if g = 1, the element's reactance at cut-off equals R0.
This gives you a quick way to check a design without redoing the arithmetic. Compute each element's reactance at fc and compare it with g·R0 (inductors) or R0/g (capacitors). If a value is off by about 6.28, an fc slipped in where ωc belonged. If it is off by about 50, the impedance scaling was applied in the wrong direction.
Worked example: 3rd-order Butterworth at 1 GHz, 50 Ω
Specification: a 3rd-order Butterworth low-pass, R0 = 50 Ω, fc = 1 GHz. The prototype values are g1 = 1, g2 = 2, g3 = 1. Use the shunt-first form.
Step 1: cut-off angular frequency. ωc = 2π × 1×10^9 = 6.283×10^9 rad/s.
Step 2: scale g = 1. As an inductor: L = 50 / 6.283×10^9 = 7.96 nH. As a capacitor: C = 1 / (50 × 6.283×10^9) = 3.18 pF.
Step 3: assign elements.
- C1 (shunt, g1 = 1): 3.18 pF
- L2 (series, g2 = 2): 2 × 7.96 nH = 15.9 nH
- C3 (shunt, g3 = 1): 3.18 pF
Step 4: check reactances at 1 GHz. C1 and C3: 1/(ωc·C) = 1/(6.283×10^9 × 3.18×10^-12) ≈ 50 Ω, which equals R0/g with g = 1. L2: ωc·L = 6.283×10^9 × 15.9×10^-9 ≈ 100 Ω, which equals g·R0 with g = 2.
Both checks pass. Because the response shape comes straight from the Butterworth prototype, this network is −3 dB at 1 GHz with a maximally flat passband below it.
If you prefer fewer capacitors, use the series-first dual instead: series L1 = 7.96 nH, shunt C2 = 2/(50 × 6.283×10^9) = 6.37 pF, series L3 = 7.96 nH. The ideal response is identical. The choice usually comes down to which part values are easier to source and lay out.
Scaling to other frequencies and impedances
Since every value is inversely proportional to ωc, moving the cut-off is just a ratio. Doubling fc to 2 GHz halves every L and C: 1.59 pF, 7.96 nH, 1.59 pF. Halving fc doubles them.
Impedance works in opposite directions for the two element types. A higher R0 raises inductances and lowers capacitances by the same ratio. So keep one set of 1 GHz, 50 Ω values for g = 1 (7.96 nH and 3.18 pF) and multiply by g and by the frequency ratio. For any 50 Ω low-pass, that covers most of the arithmetic.
In plain words · for students and beginners
Filter scaling in plain words
An everyday analogy
Think of the prototype values as a recipe written for one person. It tells you the ratio of ingredients, not how much to buy. To cook for a crowd, you multiply everything. Scaling a filter works the same way: the g-values are the ratios, and the scaling formulas turn them into real amounts.
The analogy breaks down in one place. In a recipe, everything is multiplied by the same number. In a filter, inductors and capacitors move in opposite directions when you change impedance. A higher impedance means bigger inductors but smaller capacitors. A higher frequency makes both smaller.
The same example, in simple steps
The drawing shows the starting point: three parts in a ladder, a capacitor, then an inductor, then a capacitor, with values 1, 2 and 1. Here a ladder means parts that alternate between going "across" to ground (shunt) and "along" the signal path (series).
- Choose the system impedance: 50 Ω. This is the standard resistance most RF cables and equipment expect.
- Choose the cut-off frequency: 1 GHz. This is where the filter starts cutting the signal down.
- Turn frequency into angular frequency: ωc = 2π × 1 GHz = 6.283×10^9 rad/s. (Angular frequency is just regular frequency times 2π, measured in radians per second.)
- A part with value 1 becomes 7.96 nH if it is an inductor, or 3.18 pF if it is a capacitor. (nH means nanohenry, a unit of inductance. pF means picofarad, a unit of capacitance.)
- A part with value 2 is twice as large: the inductor becomes 15.9 nH.
- Final filter: 3.18 pF to ground, 15.9 nH in series, 3.18 pF to ground.
Check: at 1 GHz, the 3.18 pF capacitor "resists" the signal with 50 Ω of reactance, the same as the system impedance. (Reactance is the AC opposition a capacitor or inductor shows at a given frequency.)
Mistakes beginners make
- Using fc instead of ωc. Forgetting the 2π makes every part about 6.28 times too large. The filter then cuts off far too low.
- Mixing up the formulas. R0 goes on top for inductors and on the bottom for capacitors. If your capacitor comes out in microfarads at 1 GHz, something is upside down.
- Losing track of units. Write the powers of ten clearly. 7.96×10^-9 H is 7.96 nH, not 7.96 μH.
- Thinking the math is the finished filter. Real parts and board traces add extra inductance and capacitance. Always simulate and measure.
Remember
- Inductors: L = R0·g/ωc. Capacitors: C = g/(R0·ωc). Always use ωc = 2π·fc.
- At 1 GHz and 50 Ω, g = 1 means 7.96 nH or 3.18 pF. Multiply by g for other values.
- Check your answer: at fc, each part's reactance should be g·R0 for inductors and R0/g for capacitors.
Related pages on RF View
Frequently asked questions
How do I convert a g-value to an inductor value?
Use L = R0·g/ωc, where R0 is the system impedance and ωc = 2π·fc. At 50 Ω and 1 GHz, g = 1 gives 7.96 nH.
How do I convert a g-value to a capacitor value?
Use C = g/(R0·ωc). At 50 Ω and 1 GHz, g = 1 gives 3.18 pF, and that capacitor has a reactance of 50 Ω at the cut-off.
What are the component values for a 3rd-order Butterworth low-pass at 1 GHz in 50 Ω?
With g = 1, 2, 1 in the shunt-first form, you get a 3.18 pF shunt capacitor, a 15.9 nH series inductor and a second 3.18 pF shunt capacitor.
How can I quickly check scaled filter values?
At fc, each inductor should have a reactance of g·R0 and each capacitor a reactance of R0/g. If the numbers don't match, a scaling step went wrong.