As far as I know only this are two lowpass filters (R-C1, L-C2) of order 1 and 2 respectively. Is it possible to denote this circuit as Low-pass filter of third order? Or is it possible to transform this circuit into a standard RLC-filter form?

# Electrical – RC-LC network equals lowpass filter of third order

circuit analysisfilterpassive-networks

#### Related Solutions

@Confused: Sorry - but I have to start with some general comments:

In principle, for an 8th order filter you have two basic alternatives:

(a) **Direct realization** (active topology derived from a passive and tabulated reference structure, and (b) **Cascade realization** as a series connection of 4 second-order stages.

I suppose, you are following the latter approach - and here you have again several alternatives (how the various 2nd order stages are realized). It seems that you have decided to use Sallen-Key realizations because you have mentioned finite gain values.

But also in this case, you again have alternatives: Unity gain approach, gain-of-two approach or equal-component approach (with gain values lower than "3"). Independent on these 3 alternatives, you must know that all 4 stages look different: They are individually designed for equal pole frequencies (applies only for Butterworth response) but for different pole-Q values to be found in filter tables. Hence, you will NOT have 4 identical 2nd-order stages but each of the 4 must bedesigned separately.

I am not sure if this answers all of your questions - perhaps it helps if you could give us some more detail of your envisaged design.

**EDIT 1:** The following link leads you to a document (from TI) which gives you the Q values for your 8th-order filter on page 8 (**correction**: page 9)

http://www.ti.com/lit/an/sloa049b/sloa049b.pdf

**EDIT 2:** For your convenience, here are the formulas for designing the 4 stages (equal pole frequencies wp, different Qp values) - to be applied for the gain-of-two version:

**C1=C2=C** and **wp=1/[C*Sqrt(R1R3)]** and **Qp=Sqrt(R1/R3)** with R1: Most left resistor(connected to input signal). For a gain of "2" you can use any two equal resistors in the negative feedback path.

Your question has a simple answer, but allow me explain a few things that will help you understand how to design Direct Coupled filters.

1) Cohn described the four inverter circuits I show on the top row. These circuits contain negative component values which, as you pointed out, are absorbed into the tank elements.

It is clear however that a negative capacitor generates inductive reactance and a negative inductor generates capacitive reactance. So all of Cohn's inverters have an alternate equivalent that I show in the 2nd row, where the negative components are replaced with positive valued components, which can also be absorbed into the tank circuit.

2) These inverter circuits can be thought of as impedance matching circuits. As a simple example, if you are designing an inverter to go between 2 tanks with characteristic impedances of 100 and 200 Ohms, then a quarter wave transmission line with a characteristic impedance of sqrt(100*200) can be used as an inverter.

With this in mind, it is possible to generate many possible inverter topologies.

3) If the tank circuits on the ends have characteristic impedances equal to the source and load resistances respectively, then the two inverters on the ends are not needed. But this isn't usually the case. You can force this to happen, but it doesn't usually generate a desirable circuit in terms of component values and component Q requirements.

So, one possibility for the end inverters is to use one of the inverters shown in the 2nd row.

It is more common however, to use a 2 element matching circuit to get from the end tank impedance to the source (or load) impedance. These are designed with standard matching techniques, using either formulas or a Smith Chart.

4) You should remember that impedance inverters are equivalent circuits at the design frequency only. At all other frequencies they only approximate the desired response. Consequently, this design method can only be used for narrow band filters, and then, the overall response might not be what you were expecting, as shown below.

Since we usually want a narrow band response however, this design method is widely used by RF engineers and has proven to be quite powerful because of the large number of ways (some examples) the various tank, inverter, and matching topologies can be combined to form a desirable response.

## Best Answer

Yes. It has three reactive components, and no two of those components are in pure series or parallel configuration.

At a single frequency, this circuit can be transformed. However, as a

filter, i.e. a circuit that reacts differently to different frequencies, it is not equivalent to a 2nd order RLC filter.A 2nd order filter has an asymptotic roll-off of -40 dB / decade. A 3rd order filter, such as this, has an asymptotic roll-off of -60 dB / decade.