I've the following filter circuit:


simulate this circuit – Schematic created using CircuitLab

The transfer function can be written as follows (I think):


question: is my equation right?

  • \$\begingroup\$ This looks a lot like homework. Please show more work. It will help a lot to keep things straight if you use C1 and C2, and R1 and R2 and equate them in the end. \$\endgroup\$ – TimWescott Apr 29 '19 at 17:31
  • \$\begingroup\$ Looks ok, but when factored out, look for roots s²=1/(L*C2) and Q ~ X(s)/R ~1 / ζ \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 29 '19 at 17:46
  • 1
    \$\begingroup\$ Wow your equation has really \$tiny_{tiny_{tiny_{tiny_{tiny_{tiny_{tiny_{letters}}}}}}} \$ .... I contend that you could rewrite your fraction to contain less fractions in the fraction. \$\endgroup\$ – KingDuken Apr 29 '19 at 18:32
  • \$\begingroup\$ What is the transfer function you want? In other words, you need to show where the stimulus is (the input) and where do you observe a response (the output). If you stimulate on the left side (\$V_{in}\$) and observe the voltage response across the right-side capacitor (\$V_{out}\$), there is no zero in this transfer function and the denominator is in the form of \$D(s)=1+b_1s+b_2s^2+b_3s^3\$. It can be obtained in a few minutes using the fast analytical circuits techniques (see cbasso.pagesperso-orange.fr/Downloads/PPTs/…). \$\endgroup\$ – Verbal Kint Apr 30 '19 at 6:25

The formula you propose is actually that of a brute-force approach. If I apply this technique to your network, I will transform the \$RC\$ front-end filter and the input voltage into a Thévenin generator whose output impedance is \$R_1||C_1\$. The brute-force transfer function is thus:

enter image description here

However, should you try to develop this expression, you will end up in an ugly arm-long formula with many cross-products, with the obvious risk of making a lot of errors (I surely would!). The best and easiest way is to apply the fast analytical techniques or FACTs. The principle is straightforward: determine the time constants of your circuit when the excitation source is set to zero. Here, it is a voltage source \$V_{in}\$ and setting it to 0 V means replacing it by a short circuit. When done, temporarily disconnect each of the energy-storing elements and "look" through its connections to determine the resistance driving it. The below drawing shows the approach:

enter image description here

You now inspect these little individual sketches - no algebra, just look at the drawing - and collect all the time constants while you alternately place one element in its dc state (a short or an open respectfully for an inductor or a cap) or in its hi-frequency state (the opposite). You then capture all these numbers in a Mathcad sheet or equivelent:

enter image description here

So you can see that you naturally obtain a denominator whose shape obeys that of a formalized form: \$D(s):1+b_1s+b_2s^2+b_3s^3\$. The trick is then to factor this polynomial recognizing a dominant low-frequency pole followed by a second-order polynomial. This is the right way to write the transfer function in a low-entropy way, having a resonant frequency immediately visible and a quality factor. Finally, if you compare the brute-force expression and the one obtained with the FACTs, they are identical:

enter image description here

You see the approach here. The FACTs are not only the simplest and often the fastest tool to get the symbolic answer, they naturally lead to a low-entropy (read factored) form where gains, poles and zeroes (if any) are immediately observable. Also, if at the end you observe a difference between the brute-force approach and the FACTs, you just need to fix a simple drawing to make things right. With the brute-force approach, you re-start from scratch and yell : )

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