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What have I done wrong in determining the transfer function for this RCL circuit?

enter image description here

(1) $$\dfrac{Eo(s)}{Ei(s)}=\dfrac{s^2R_2CL+sL}{s^2CL(R_1+R_2)+s(R_1R_2C+L)+R_1} $$

Regardless, of how many times that I try, I get the same result. However, my TF response doesn't match the circuit response, as can be seen.

Using complex impedance method

Defining $$z_1 = R1 $$ $$z_s=\dfrac{1}{sC} + R_2 = \dfrac{1+sR_2C}{sC} $$

$$z_p = \dfrac{z_sz_L}{z_s+z_L}=\dfrac{\dfrac{1+sR_2C}{sC}sL}{\dfrac{1+sR_2C}{sC}+sL} =\dfrac{\dfrac{s^2R_2CL+sL}{sC}}{\dfrac{1+sR_2C}{sC}+sL}$$

Multiplying numerator and deniminator through by sC

$$z_p=z_2=\dfrac{s^2R_2CL+sL}{s^2CL+sR_2C+1} $$

Therefore, we have that

$$\dfrac{Eo(s)}{Ei(s)}=\dfrac{z_2}{z_1+z_2}= \dfrac{\dfrac{s^2R_2CL+sL}{s^2CL+sR2C+1}}{{R1}+\dfrac{s^2R_2CL+sL}{s^2CL+sR2C+1}}$$

Simplifying

$$\dfrac{Eo(s)}{Ei(s)}=\dfrac{s^2R_2CL+sL}{{R_1(s^2CL+sR_2C+1)}+{s^2R_2CL+sL}}$$

$$\dfrac{Eo(s)}{Ei(s)}=\dfrac{s^2R_2CL+sL}{s^2CL(R_1+R_2)+s(R_1R_2C+L)+R1}$$

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  • \$\begingroup\$ Using nodal analysis, I don't get the \$s^2R_2CL\$ term in the numerator. \$\endgroup\$
    – Chu
    Mar 30, 2021 at 18:27
  • \$\begingroup\$ When determining the output impedance, L1 is in parallel with R1. \$\endgroup\$
    – Chu
    Mar 30, 2021 at 18:37

3 Answers 3

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Well, we are trying to analyze the circuit. When we use and apply KCL, we can write the following set of equations:

$$\text{I}_{\text{R}_1}=\text{I}_\text{L}+\text{I}_{\text{R}_2}\tag1$$

When we use and apply Ohm's law, we can write the following set of equations:

$$ \begin{cases} \text{I}_{\text{R}_1}=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_\text{L}=\frac{\text{V}_1}{\text{sL}}\\ \\ \text{I}_{\text{R}_2}=\frac{\text{V}_1-\text{V}_2}{\text{R}_2}\\ \\ \text{I}_{\text{R}_2}=\text{sC}\text{V}_2 \end{cases}\tag2 $$

Substitute \$(2)\$ into \$(1)\$, in order to get:

$$ \begin{cases} \frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_1}{\text{sL}}+\frac{\text{V}_1-\text{V}_2}{\text{R}_2}\\ \\ \frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_1}{\text{sL}}+\text{sC}\text{V}_2 \end{cases}\tag3 $$

Now, the transfer function is given by:

$$\frac{\text{V}_2}{\text{V}_\text{i}}=\frac{\text{sL}}{\text{CL}\left(\text{R}_1+\text{R}_2\right)\text{s}^2+\left(\text{L}+\text{CR}_1\text{R}_2\right)\text{s}+\text{R}_1}\tag4$$

Where I used the following Mathematica-codes:

In[1]:=Clear["Global`*"];
FullSimplify[
 Solve[{IR1 == IL + IR2, IR1 == (Vi - V1)/R1, IL == V1/(s*L), 
   IR2 == (V1 - V2)/R2, IR2 == s*c*V2}, {IR1, IR2, IL, V1, V2}]]

Out[1]={{IR1 -> (Vi + c s (R2 + L s) Vi)/(
   R1 + (L + c R1 R2) s + c L (R1 + R2) s^2), 
  IR2 -> (c L s^2 Vi)/(R1 + (L + c R1 R2) s + c L (R1 + R2) s^2), 
  IL -> (Vi + c R2 s Vi)/(R1 + (L + c R1 R2) s + c L (R1 + R2) s^2), 
  V1 -> (L s (1 + c R2 s) Vi)/(
   R1 + (L + c R1 R2) s + c L (R1 + R2) s^2), 
  V2 -> (L s Vi)/(R1 + (L + c R1 R2) s + c L (R1 + R2) s^2)}}
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  • \$\begingroup\$ Hi Jan, many thanks for your assistance. Please note that I try to stay away from using KVL and KCL because the algebra can get a little messy. And to combine it with complex impedances is a step too far for me at the moment. However, I'll try to keep this method in mind going forward! \$\endgroup\$
    – aLoHa
    Mar 30, 2021 at 20:05
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When you determine a transfer function (TF), it can be for serving different purposes: solving a pure mathematical exercise for the fun of it or obtaining a transfer function for a practical application. If you look for the second option, you need a swift approach which, in the end, delivers a meaningful result that you can use for design purposes. This is the design-oriented analysis or D-OA in which Dr. Middlebrook always underlined the hyphen inserted between the two words.

In my opinion, nothing can beat the fast analytical circuits techniques or FACTs as illustrated in my book on the subject. The FACTs lend themselves very well to analyzing all sorts of linear circuits and, in particular, passive filters of any order. In the present case, I can derive the transfer function without writing a line of algebra by splitting the sketch in several small drawings that I individually inspect. Inspection meaning that I look at the circuit and I can see, in my head, what is the resistance \$R\$ "seen" from the connecting terminals of one of the energy-storing elements, \$L\$ or \$C\$. Once you have that resistance, you can form the time constants of the circuit (there are two here for a second-order filter) as \$\tau=RC\$ and \$\tau=\frac{L}{R}\$.

Let's see how it works with the below drawing:

enter image description here

You first start with \$s=0\$ and this is a dc bias: the inductor is replaced by a short circuit and the capacitor is open-circuited. The gain is obviously zero. Then, you turn the excitation off (\$V_{in}=0\$) and you "look" through the connecting terminals of each energy-storing element to determine the resistance \$R\$. When doing that, the second element is left in its dc state (a short for the inductor and an open circuit for the cap.). No need for equations in this mode and the time constants come easily.

You proceed with high frequency gains obtained when the inductor and the cap. are alternatively set in their respective high-frequency state (an inductor is open circuited while a cap. is replaced by a short). You have the \$H\$ gains.

Finally, assemble everything in a Mathcad sheet, and voilà, you have the TF you want in a few minutes. And if you spot an error because of the deviation in magnitude and phase between the TF you've found with the FACTs and the brute-force expression, then you can quickly review the small sketches and fix the guilty one: no need to restart from scratch as with the other methods:

enter image description here

If this expression is correct, there is still not much you can do with it and you need to rearrange it. What matters to determine the components values are the resonant frequency and the transfer function magnitude at resonance. Without these elements, there is nothing you can do with the TF we have to design your filter. We first rewrite the expression using a normalized second-order polynomial form in the denominator. When done, we can quickly identify a quality factor \$Q\$ and a resonant frequency \$\omega_0\$:

enter image description here

This form is still not acceptable and further factorization is necessary. When you factor the zero in the numerator, you finally obtain the low-entropy version of the transfer function. And this is the ultimate goal of the exercise:

enter image description here

You see that writing a transfer function without correctly factoring it into a meaningful form that you can exploit for design purposes is useless in my opinion. You certainly can plot the response with a mathematical solver but you won't infer anything from the TF you have determined. On the other hand, the FACTs lead you straight to a result that you can quickly rearrange in a low-entropy format and use it for designing your circuit. And if you can do that without writing a single equation, what more can I say? : )

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  • \$\begingroup\$ Thanks a lot for your input. Please note that the 1st half is a little bit too much for me to grasp at the moment, i.e., having to take on another form of analysing circuits in Mathcad, but the 2nd half looks like information that will prove valuable to me in the near future. So, your contribution is very much appreciated. \$\endgroup\$
    – aLoHa
    Apr 1, 2021 at 1:13
  • \$\begingroup\$ With pleasure. I can see in one of your comments that you were looking for a new method for exploring electrical analysis. This is a perfect timing to learn the FACTs then! : ) Have a look at my APEC 2016 seminar and you will see how powerful this technique can be. With a certain habit, you can "see" where the poles and zeroes hide in a circuit and it's truly useful - and enjoyable - in many cases. \$\endgroup\$ Apr 1, 2021 at 8:40
  • \$\begingroup\$ Thank you for sharing the presentation. I'm not too ingrained in any one particular method at this particular point in time to have a preference for any, so I will be certainly looking into it! :) \$\endgroup\$
    – aLoHa
    Apr 1, 2021 at 17:19
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Jan's answer gives you the result, but the reason you didn't get the correct answer is because you didn't consider the intermediary stage.

schematic

simulate this circuit – Schematic created using CircuitLab

Given your approach, the first thing to do is determine the voltage at point a by considering the whole L || (R2 + C) as an impedance, and only then use that voltage with the divider formed by R2 and C. I won't expand the equations, it looks like you're good with them:

$$\begin{align} Z_{RC}&=R_2+\dfrac{1}{sC}\tag{1} \\ Z_a&=\dfrac{1}{\dfrac{1}{sL}+\dfrac{1}{Z_{RC}}}\tag{2} \\ V_a&=V_1\cdot\dfrac{Z_a}{Z_a+R_1}\tag{3} \\ V_x&=V_a\cdot\dfrac{\dfrac{1}{sC}}{\dfrac{1}{sC}+R_2}\tag{4} \end{align}$$

And you get the correct transfer function. When you'll be expanding the above, you'll realize that it can get messy, that's why for these cases, it's better to use KCL/KVL/FACTS/etc, because the resulting system of equations is easier to analyze. In the end, it looks like you could have done this yourself if you payed a bit more attention.


Given your comment, I'll try to expand a bit. You have two things: the way you want to solve this is by using the voltage dividers, and the circuit is a Cauer network (also called ladder). Since you're interested in finding out the voltage at the output (V(x)), you first need to find out the previous node voltage, V(a). For that, you have to look at the circuit from a different perspective:

schematic

simulate this circuit

V(a) divides the input voltage and it does so at point a, so that's how the circuit looks like from the input's perspective (blue arrow). Which means to calculate V(a) you first need to determine the equivalent impedance formed by L, R2, and C, which is R2 in series with C (1), all in parallel with L (2). Expanded and continued from the equations above:

$$\begin{align} Z_a&=\dfrac{1}{\dfrac{1}{sL}+\dfrac{sC}{sR_2C+1}}=\dfrac{s^2R_2LC+sL}{s^2LC+sR_2C+1}\tag{5} \\ V_a&=V_1\cdot\dfrac{s^2R_2LC+sL}{s^2(R_1+R_2)LC+s(R_1R_2C+L)+R_1}\tag{6} \end{align}$$

And it's this (6) that needs to be used in (4) in order to find V(x), so you can see, it gets pretty messy. If you compare these with KCL/KVL/FACTS (in Verbal Kint's answer), you can't really say that they are less messy. BTW, Verbal Kint used a different approach by using the Thevenin equivalent source and then the output (also hinted at by Chu's comment), which means he went from output towards the input; different way, same results. Anyway, if you plot this transfer function agains V(a), you will get the same response:

test

I have kept the default Rser=1m for the inductances because it makes the comparison of the results better (otherwise they would have completely overlapped).

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  • \$\begingroup\$ Thanks you for your input, which is always welcome. I have to admit that at this particular moment in time that I'm not quite following your explanation. Think it's a case of memory overload. So, I'll come back to it later when i've had a chance to clear my head! \$\endgroup\$
    – aLoHa
    Mar 30, 2021 at 20:24
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    \$\begingroup\$ Hello a concerned citizen, one big difference with the FACTs in this example is that I can obtain the final expression in one shot (\$H_{20}\$) in my answer by associating the numerator and denominator determined painlessly. Without using KVL or KCL which, in my opinion, make things complicated and prone to errors. Finally, and this is my point, the important thing is truly to write the TF in the low-entropy form from which you can design your circuit. \$\endgroup\$ Mar 31, 2021 at 15:45
  • \$\begingroup\$ Bonjour a vous, aussi, @VerbalKint. I don't dare-, and won't disagree. The low-entropy form is always a bliss to read and understand. My comment (about the Thevenin source) was related to your brute-force approach, Href(s), as opposed to my approach, which was an attempt to follow in OP's steps. \$\endgroup\$ Mar 31, 2021 at 16:51
  • \$\begingroup\$ Bonjour @aconcernedcitizen, très bon français :-) I understand your comment related to the brute-force approach with Thévenin leading to a mess as the other ways using KVL and KCL. What is cool with Mathcad is that you can write this brute-force expression quite compactly with the || sign and the solver does the job for you. A bientôt ! \$\endgroup\$ Mar 31, 2021 at 17:50
  • \$\begingroup\$ @aconcernedcitizen your input as usual is always appreciated. But, my main issue at the moment is finding a method that I can use to analyze most (if not all) electrical circuits. I had hoped that the Complex impedance method would be the one. However, this particular circuit has proved to be lot more of a handful than I'd imagined. I have now tried to use the KVL method (and will be editing my question to show this), but the algebra gets a bit involved after a while, which is not bringing me any closer to the final TF. \$\endgroup\$
    – aLoHa
    Apr 1, 2021 at 1:22

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