I've got a trouble with this filter: enter image description here

I tried calculating transfer function from four-pole matrix, but can't handle it :( Any help much appreciated. That shaggy drawed element near U1 source is coil. Sorry for my drawing.

What I've done so far (I think it is wrong): enter image description here

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    \$\begingroup\$ Welcome to the site. Please realise that this is not a free design house, homework-answering service or an on-line technical encyclopedia, copied out to you on demand. People will help you take the next step if your question shows that you've done as much as you possibly could on your own - which your post doesn't, I'm afraid. You have not even labeled the components! Please revise your question showing your work and findings so far, in considerable detail. Again, a warm welcome to the site. \$\endgroup\$ – Oldfart Jan 5 '18 at 17:31
  • \$\begingroup\$ Sure, I'm sorry. Just thought that my work is trash. Thanks for the tip! My notes are included from now. \$\endgroup\$ – needtobe Jan 5 '18 at 17:58

This is the problem with the classical analysis, you end up in algebraic paralysis and getting control back is difficult. The easiest and fastest way is to consider the FACTs. By considering the physical time constants of this two energy-storing elements (2nd-order circuit), I can derive the transfer function with minimum effort and get it factorized almost immediately. The denominator obeys \$D(s)=1+sb_1+s^2b_2\$. There is a gain \$H_0\$ for \$s=0\$ obtained when the capacitor is open-circuited and the inductor replaced by a short: \$H_0=1\$. Then, reduce the excitation to 0 V (replace the input source by a short circuit) and "look" at the resistance offered by the capacitor (while \$L_1\$ is short circuited) and the inductor (while \$C_2\$ is open circuited) connecting terminals. You have two time constants \$\tau_2=R_2C_2\$ and \$\tau_1=\frac{L_1}{R_1}\$ and you can form \$b_1=\tau_1+\tau_2\$. The second-order time constant is obtained by considering \$L_1\$ open circuited while "looking" into \$C_2\$'s terminals: \$\tau_{12}=C_2(R_1+R_2)\$ leading to \$b_2=\tau_1\tau_{12}\$. The below drawing illustrates how to conduct this analysis:

enter image description here

There is no zero in this network and the expression fitting the second-order polynomial is: \$H(s)=H_0\frac{1}{1+\frac{s}{Q\omega_0}+(\frac{s}{\omega_0})^2}\$. Now, if you compute the quality factor \$Q\$, it is very low and the whole network can be replaced by two cascaded poles: one dominates at low frequency (closer to the origin) while the second shapes the response at high frequency. This is called the low-\$Q\$ approximation. As such, the transfer function can be rewritten as \$H(s)=H_0\frac{1}{(1+\frac{s}{\omega_{p1}})(1+\frac{s}{\omega_{p2}})}\$ with \$\omega_{p1}=Q\omega_0\$ and \$\omega_{p2}=\frac{\omega_0}{Q}\$. The below Mathcad sheet shows you the whole process here and compares the factored expression with the raw, brute-force expression (upper right corner):

enter image description here

For the raw expression, I identified a Thévenin generator driving the left side of \$R_2\$ and affected by an output impedance made of \$L_1||R_1\$. You then apply an impedance divider expression and there you go. The problem with this method and yours is that you may make mistakes when unwinding the whole expression and trying to rearrange it in a friendly (read factored) format. By applying the FACTs, you see that you simply split the circuit in a succession of simple sketches for which time constants are determined by inspection, no algebra! Then, you assemble the pieces to form the denominator you want, already in a canonical form. Should you later spot a mistake, go back to the sketch, fix it and the rest remains untouched. I really encourage you to dig these FACTs and determine transfer functions in the fastest way ever!

  • \$\begingroup\$ I've got one more question. In this Href(s) equation, what is in parallel? s*L1 with R1 ? \$\endgroup\$ – needtobe Jan 5 '18 at 21:56
  • \$\begingroup\$ Hi, if you want to apply the brute-force algebra, you can determine a Thévenin generator at the left side of \$R_2\$. So you see an input source \$V_{in}\$ followed by \$L_1\$ and loaded by \$R_1\$. The equivalent generator has an amplitude equal to \$V_{in}\frac{R_1}{R_1+sL_1}\$ and an output impedance equal to \$R_1||sL_1\$, the parallel combination of \$R_1\$ and \$L_1\$. \$\endgroup\$ – Verbal Kint Jan 5 '18 at 22:31

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