# Calculating transfer function of second order high pass filter

I'm struggling with calculating the transfer function H(s) of a highpass filter, calculating a transfer function of a first order filter is no problem but I haven't mastered second order filters yet.

I assume I should use KCL to do it as efficient as possible?

Best regards

• This is a 3rd-order filter as you have 3 energy-storage elements having independent state variables. KCL and KVL will get you there (considering a Thévenin generator involving $R$ and $L_1$) but good luck to factor this out in a meaningful low-entropy expression. I am computing this transfer function using fast analytical circuits techniques, stay tuned. – Verbal Kint Oct 13 '17 at 18:04

Let V the voltage at the junction of $R$, $C$ and $L_1$. Use KCL at this junction and the output. This will give the following two equations.

$\frac{u_1-V}{R}=C s \left(V-u_2\right)+\frac{V}{L_1 s}$

$C s \left(V-u_2\right)=\frac{u_2}{L_2 s}$

These two equations have 2 unknowns $V$ and $u_2$. In this case you are interested only in $u_2$. You can solve for $V$ in terms of $u_2$ from the second equation and substitute that in the first, to get the solution for $u_2$.

I always use an algebra system to deal with these cumbersome and error-prone calculations. They will return the answer in a blink.

I will apply the fast analytical circuits techniques or FACTs described here to determine this transfer function. The principle is to find the circuit time constants when the excitation ($V_{in}$) is reduced to 0 V (replaced by a short circuit). The schematic is below:

The story behind the FACTs is to alternatively set the energy-storing elements in their dc or high-frequency state: respectively open- and short-circuited for a cap, respectively short- and open-circuited for an inductor. We first start by determining the resistance when all elements are observed in dc: you disconnect the selected energy-storing element and you find the resistance offered by its connecting terminals while the other elements remain in their dc state. When the resistance is infinite, we adopt a finite value $R_{inf}$ and when it is 0, we adopt also a finite value equal to $R_{small}$. We do that to avoid indeterminacies. When this done, we continue by setting one of the three elements in its high-frequency state while we determine the resistance from another energy-storing element. $\tau_{12}$ for example means that we set $L_1$ in its high-frequency state and we determine the resistance offered by $L_2$'s terminals while $C_3$ remains in its dc state (open circuited). Finally, we set two elements in their high-frequency state and determine the resistance offered by the 3rd one. Then, you can assemble the denominator as:

$D(s)=1+s(\tau_1+\tau_2+\tau_3)+s^2(\tau_1\tau_{12}+\tau_1\tau_{13}+\tau_2\tau_{23})+s^3\tau_2\tau_{23}\tau_{321}$

The determination of the numerator can be done either using a null double injection (NDI) or via the calculation of simple gains. In this approach, the energy-storing elements are now alternatively set in their high-frequency state and you can reuse the time constants already determined in the denominator to form $N(s)$. You end-up with slightly more complicated terms than with a NDI but it is often perceived as a much simpler exercise. Also, as in this case, a lot of these gains are 0 which truly simplifies the approach. The circuits are shown below:

For instance, the notation $H^1$ means that $L_1$ is in its high frequency state (open circuited) while the two other elements are in their dc state. What is the gain in this mode? 0 as the cap. is open-circuited. $H^{12}$ means elements 1 and 2 are in their high-frequency state while element 3 is in its dc state: what is the gain? 0. If you do the exercise ok, you can assemble the numerator as follows:

$N(s)=H_0+s(\tau_1H^1+\tau_2H^2+\tau_3H^3)+s^2(\tau_1\tau_{12}H^{12}+\tau_1\tau_{13}H^{13}+\tau_2\tau_{23}H^{23})+s^3\tau_2\tau_{23}\tau_{321}H^{321}=s^3b_3H^{321}$

Here, the dc gain $H_0$ is equal to 0 (two inductors are shorted, the cap is open). The final transfer function is obtained by combining the above numbers and rearranging the result under a nice polynomial form as shown in the below Mathcad sheet. You can see that I have determined the reference transfer function using Thévenin as suggested in my comment. If you develop this expression by hand, you may make mistakes and find it difficult to factor it properly. With the techniques that I shown, if you spot a mistake, go back to the small intermediate sketches and fix the guilty one in a snapshot. As you can see, I did not write a single line of algebra to determine this transfer function which is kind of cool. This is the power of the FACTs that I encourage students to learn and acquire.

• I think you accedently wrote denominator twice – Mike Oct 13 '17 at 21:58
• @Mike, hello, what do you mean? In the text or in the sheet? – Verbal Kint Oct 14 '17 at 7:07
• see here: "The determination of the denominator involves calculating simple gains..." did you mean the nominator? – Mike Oct 14 '17 at 7:35
• Oh I see, good catch. I meant the numerator of course and I slightly updated the text. Thanks for pointing this out to me. – Verbal Kint Oct 14 '17 at 7:45
• This can be derived by 'normal' methods in a few lines. What's the advantage of this method? – Chu Oct 14 '17 at 13:46

I think, the most simple method is to use - as a first step - general impedances Z and to apply the known rules of voltage division.

Therefore: Lets rename R>>Z3 , C>>Z4, L1>>Z1 and L2>>Z2.

Now - use the voltage divider rule for finding the transfer function at the node between Z3 and Z4.

Obviously, we find: V34/V1=[Z1||(Z2+Z4)]/[Z3+Z1||(Z2+Z4)].

To get V2/V1, this expression is to be multiplied with (voltage division Z2-Z4): Z4/(Z2+Z4).

As a last step, introduce the appropriate expressions for Z1...Z4. Thats all.