# Filter transfer function

I've the following filter circuit:

simulate this circuit – Schematic created using CircuitLab

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

$$\frac{\frac{1}{\text{sC}}}{\frac{1}{\text{sC}}+\frac{1}{\frac{1}{\frac{1}{\text{sC}}+\text{sL}+\text{R}}+\frac{1}{\frac{1}{\text{sC}}}}+\text{R}}$$

question: is my equation right?

• 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. Apr 29, 2019 at 17:31
• Looks ok, but when factored out, look for roots s²=1/(L*C2) and Q ~ X(s)/R ~1 / ζ Apr 29, 2019 at 17:46
• 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.
– user103380
Apr 29, 2019 at 18:32
• 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/…). Apr 30, 2019 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:
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 $$\V_{in}\$$ and setting it to 0 V means replacing it by a short circuit. When done, temporarily disconnect each of the energy-storing elements and "look" through its connections to determine the resistance driving it. The below drawing shows the approach:
So you can see that you naturally obtain a denominator whose shape obeys that of a formalized form: $$\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: