1
\$\begingroup\$

I have an assignment for school and one of the questions is asking me to find the "bandwidth" of a "pi filter." This teacher miss words things sometimes and I know it's actually a T filter and a low pass filter so the bandwidth is really cutoff frequency. But I can't find a formula to find it any where. Does anyone know the formula to calculate cutoff frequency? enter image description here

edit: some of these answers really helped thank you. This particular assignment is about thermal noise in power supplies. But I was stuck on one question asking me to "calculate the filter's bandwidth." The filter had no resistors in it and I didn't think the rest of the circuit was very relevant. But here is a picture of the entire thing for anyone who is curious. enter image description here

\$\endgroup\$
3
  • \$\begingroup\$ To verify your calculations, consider simulating this Cauer filter in Qucs or QucsStudio. These simulators give you a choice between S-parameter simulation and plain AC simulation (bode plot output, customizable). For an S-parameter simulation, you need to attach a power source (with a nominal impedance) at either of the two ports. For a plain AC simulation, use a voltage source with a series terminator on the input, and also terminate the output. If you were not given a nominal impedance, just try with 50 Ohms as the most typical value in RF technology. \$\endgroup\$
    – frr
    Sep 30, 2020 at 8:16
  • 1
    \$\begingroup\$ @frr I'm probably pedantic, but it might be better to name it a "Cauer network", as opposed to "Cauer filter", since the latter is a synonym to "elliptic filter", or a type of filter, while the former refers to the LC ladder, or a network topology, as opposed to a type of filter. \$\endgroup\$ Sep 30, 2020 at 13:37
  • \$\begingroup\$ ...and yet, despite more than one advice about how your filter needs terminations, you still simulate it improperly. \$\endgroup\$ Oct 1, 2020 at 6:53

3 Answers 3

4
\$\begingroup\$

You should find a different teacher if he teaches you wrong words, which serve as a basis for future understanding. That said, the circuit is meant to be either doubly terminated -- one resistance as load and one as the source's resistance --, or singly terminated -- only one resistance as either load or source resistance, and the other end either shorted (or voltage source) or open (or current source). This is what I mean:

types

A is the doubly-terminated case: the source has a resistance, and the load is a resistance. It can also be a \$\Pi\$, but I left that out. B through E are singly-terminated. B and C have a finite resistance as load, but B has zero input resistance (voltage source) while C has infinite input resistance (current source). And D and E have finite input resistance but the output is either zero (short, D), or infinite (open, E).

Notice how the topologies change to a \$\Pi\$ when using singly-terminated cases when one side is opened. That is to ensure that either the first element is not in series with the current source (C), or the curent will flow through the last element (E).

This changes the transfer function for each case, since they are solved differently. I won't go through all the cases and derivations, but I'll show for the doubly-terminated case (A), which I presume is what your teacher meant to show. There are various methods that can be used, here a simple mesh analysis which gives you the following:

$$\begin{align} \left\{ \begin{array}{x} R_o&=sL_1I_1+\frac{I_1-I_2}{sC_1}+R_iI_1 \\ \frac{I_1-I_2}{sC_1}&=sL_2I_2+RoI_2 \end{array} \right. \end{align}$$

Solve for the currents and keep only the \$I_2\$ solution (since that is of interest), which gives you the transfer function:

$$\begin{align} H(s)&=\frac{R_o}{C_1L_1L_2s^3+(C_1L_1R_o+C_1L_2R_i)s^2+(C_1R_iR_o+L_2+L_1)s+R_o+R_i} \\ {}&=\frac{\frac{R_o}{L_1L_2C_1}}{{{s}^{3}}+\frac{\left( {L_1}\, \mathit{R_o}+{L_2}\, \mathit{R_i}\right) \, {{s}^{2}}}{{L_1}\, {L_2}}+\frac{\left( {C_1}\, \mathit{R_i}\, \mathit{R_o}+{L_2}+{L_1}\right) s}{{C_1}\, {L_1}\, {L_2}}+\frac{\mathit{R_o}+\mathit{R_i}}{{C_1}\, {L_1}\, {L_2}}} \end{align}$$

Since this is a passive filter, there will be an attenuation at the output in the form of \$\frac{R_o}{R_o+R_i}\$, which can be seen in the transfer function above as the numerators of the coefficients of the zeroth power in both the numerator and the denominator, leaving you with the term \$\omega^3=\frac{1}{L_1L_2C_1}\$, which is the corner (cutoff) frequency.


As a slight generalization, you'll find out that, at least for LC ladder topologies (Cauer networks), the \$\omega^n\$ (or corner frequency) is usually the product of the LC elements: for 2nd order it's \$\frac{1}{L_1C_1}\$, for 3rd it's above, for 4th \$\frac{1}{L_1L_2C_1C_2}\$, etc. This is also valid for pole-zero transfer functions; Cauer networks are for all-pole, only.

\$\endgroup\$
0
4
\$\begingroup\$

Building on the answer given by a concerned citizen, I have applied the fast analytical circuits techniques or FACTs as an alternative to the series of equations given in the text. By splitting circuit A in a series of small individual drawings, you can determine this 3rd-oder transfer function by inspection, without writing a line of algebra. The process is described in the book I wrote on the subject.

The principle is simple: determine the natural time constants of the circuit for a zeroed excitation (\$V_{in }=0\$, the source is replaced by a short circuit) in different configurations. All the sketches are gathered in the below drawing:

enter image description here

Nothing mysterious here, just break the circuit into small pieces where its energy-storing elements are either set in their dc state (a cap. is open circuited and an inductor is replaced by a wire) or in their high-frequency state (a cap. is short circuited while an inductor is replaced by an open circuit). The exercise consists of "looking" through the considered energy-storing element terminals to determine the resistance \$R\$ driving that element. The time constant is therefore \$\tau=RC\$ or \$\tau=\frac{L}{R}\$. Then, assemble all these elements to define the denominator expressed as:

\$D(s)=1+sb_1+s^2b_2+s^3b_3=1+s(\tau_1+\tau_2+\tau_3)+s^2(\tau_1\tau_{12}+\tau_1\tau_{13}+\tau_2\tau_{23}+s^3\tau_1\tau_{12}\tau_{123}\$.

To test this expression, I usually determine the brute-force formula using a Thévenin generator and I plot the differences in magnitude and phase. If the difference with Mathcad lies in the pico range, this is good. If not, I identify the guilty sketch and fix it. The rest remains untouched and that is the beauty of the FACTs: should you make a mistake, you do not restart from scratch and only modify the part guilty of deviation. The Mathcad sheet is below:

enter image description here

I have then tried to rework the 3rd-oder polynomial form into a cascaded 1st-order filter followed by a second-order polynomial considering that one root dominates at high frequency while two other are quite close (with the adopted components values of course). Remember, the goal is a design-oriented analysis or D-OA: the equation must reflect the designer needs to calculate components values like a cutoff frequency or a known attenuation at dc. It is thus important to format the result into a low-entropy form naturally unveiling these parameters. The result is shown below:

enter image description here

Finally, as always, I check that the time-constants-based response is exactly the same as what the brute-force Thévenin-based gives me:

enter image description here

\$\endgroup\$
1
\$\begingroup\$

Firstly, your problem is under-defined in that you need to present a load impedance on one side and a source impedance on the other. These impedances are normally regarded as resistors. Then, before jumping to the analysis of the full T-network you split the problem into two halves i.e. you split the problem into two L-Pads like this: -

enter image description here

And, you recognize that this is an impedance transformer where the source impedance on the left is matched to a higher impedance on the right. This website calculator gives you the formulas, the derivation and a calculator for the basic L-PAD: -

enter image description here

And the next part of the process is to recognize that two of these L-Pad filters placed back-to-back form a T-network like this: -

enter image description here

\$R_X\$ is a ghost impedance that represents the output impedance of the left L-Pad and the input impedance of the reversed right-hand L-Pad. It doesn't play a part in the final design but can reshape the response so it does need to be factored in. \$R_X\$ is higher than both \$R_{IN}\$ and \$R_L\$.

Once you have derived the formula for a single L-Pad, you combine back-to-back L-Pads to make a T-network (noting that the two capacitors in parallel become one component). In the final T-network (with input and output impedance the same), \$L\$ is the same value as calculated in the L-Pad and \$C\$'s value doubles (because it combines two capacitors into one).

The only variables are: -

  • \$\omega\$, the frequency where matching is truly resistive and also represents the frequency where the low-pass filtering kicks in. This is possibly the formula that your teacher is asking for.
  • \$R_L\$ - usually matched with \$R_{IN}\$
  • \$R_{IN}\$
  • \$R_X\$ - the ghost impedance representing the output impedance of the left L-Pad and the input impedance of the reversed right-hand L-Pad.

And normally \$R_{IN}\$ and \$R_L\$ are usually kept the same. This allows multiple T-networks to be cascaded to achieve phenomenal filter performance with very little extra math - the main math is in the basic L-Pad.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.