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I have a question about the transfer function of the above RC-lowpass filter. If I switch places of the capacitor and the resistor to the right, and then re-make the circuit as a thevenin-equivalent with the capacitor as the load:


simulate this circuit

... Then I end up with the seemingly correct transfer function: $$\frac{\frac{1}{j\omega C}}{\frac{R}{2}+\frac{1}{j\omega C}}=\frac{1}{1+j\frac{R\omega C}{2}}$$ with cutoff frequency w=2/RC.

However, if I instead of turning part of the circuit into a Thevenin-equivalent just take C//R= $$\frac{\frac{R}{j\omega C}}{R+\frac{1}{j\omega C}}=\frac{R}{1+jR\omega C}$$

and then voltage division over C//R: $$\frac{\frac{R}{1+jR\omega C}}{R+\frac{R}{1+jR\omega \:C}}=\frac{1}{1+\frac{R}{\frac{R}{1+jR\omega \:C}}}=\frac{1}{2+j\omega RC}$$ ...I end up with the above. Where the cutoff frequency would have to be w=0 if I'm not mistaken. The second approach is the one I tried first but the result seemed too suspect, but I'm not sure why the second approach doesn't work. Is there anything logically wrong with finding the voltage over ab by combining the branches followed by voltage division this way or am I just messing up the math somewhere?

Edit: With the help of the comments below I've realized that the first approach is missing a factor of 1/2, making the results equal.

However, that brings up another issue: my current understanding of how we find the cutoff frequency is to check where the magnitude of the transfer function is equal to $$\frac{1}{\sqrt{2}}$$. With the magnitude of the correct transfer function being $$\frac{1}{\sqrt{4+\left(\omega RC\right)^2}}$$, thus having a denominator which can't become sqrt(2), I can't really make sense of it, which is why I assumed the approach must be wrong. Can anyone clear this up?

  • 1
    \$\begingroup\$ When \$s=0\$, open the capacitor and you have a dc gain of 0.5. Do you have this value in both of your equations for \$s=0\$? You may be interested by my seminar on FACTs which simplifies the determination of transfer functions. \$\endgroup\$ Oct 24, 2023 at 11:27

1 Answer 1


Your first approach produces the wrong result. Instead, following a proper process you would find that \$V_{_\text{OUT}}=\left[\frac12V_{_\text{IN}}\vphantom{\frac{\frac1{s\,C}}{\frac1{s\,C}+R}}\right]\cdot\left[\frac{\frac1{s\,C}}{\frac1{s\,C}+\frac{R}{2}}\right]\$ or else \$\frac{V_{_\text{OUT}}}{V_{_\text{IN}}}=\frac12\frac{\frac1{s\,C}}{\frac1{s\,C}+\frac{R}{2}}=\frac1{2+R\,C\,s}\$.

In Python/SymPy:

simplify( 1/2 * (1/s/c) / (1/s/c + r/2) )
1/(c*r*s + 2)

That is exactly the same result you got in the 2nd case, which is correct.

KCL also provides the same result:

solve( Eq( vout/r + vout*s*c + vout/r, vin/r ), vout)[0] / vin
1/(c*r*s + 2)

However, the standard formula here is \$K\cdot\frac{1}{\left(\frac{s}{\omega_{_0}}\right)+1}\$, with \$K\$ being the DC gain value.

Set \$\omega_{_0}=\frac2{R\,C}\$ (as you found already.) Then:

$$\begin{align*} &\frac1{2+R\,C\,s} \\\\ &\frac1{2+\frac{2\,s}{\omega_{_0}}} \\\\ &\frac{\frac12}{1+\frac{s}{\omega_{_0}}} \\\\ &\frac12\cdot\frac{1}{1+\frac{s}{\omega_{_0}}} \end{align*}$$

So \$K=\frac12\$ as expected when keeping \$\omega_{_0}=\frac1{\frac{R}2\,C}\$ as anticipated.

All correct approaches provide the same result.

(Forgot to mention, yes you can set \$\sigma=0\$ in this case so that \$s=j\omega\$.)

Added per question in comments below

if we take the magnitude of the function the denominator becomes sqrt(4+(wRC)^2), so the fact that this can't become sqrt(2) is what made me think it must be wrong. Any helpful input on this perhaps?

The problem is that \$K=\frac12\$. So the resulting magnitude isn't supposed to be \$\frac1{\sqrt{2}}\$. Instead it is supposed to be half that much, or \$\frac12\frac1{\sqrt{2}}\$.

That's your problem here. You just failed to take into account the DC gain. But that is likely also because you didn't know how to find it.

There is a way to solve these kinds of problems without knowing or caring about \$K\$ and by only looking at the denominator. If you take the 1st order denominator as \$b_1s+b_0\$ then \$\omega_{_0}=\frac{b_0}{b_1}\$. In your case where you have \$b_0=2\$ and \$b_1=R\,C\$, then \$\omega_{_0}=\frac{2}{R\,C}\$.

As I wrote earlier, the standard form is \$K\cdot\frac{1}{\left(\frac{s}{\omega_{_0}}\right)+1}\$, with \$K\$ being the DC gain value. Let's see if you can follow through here:

$$\begin{align*} &\frac1{R\,C\,s+2} \\\\ &\frac{a_0}{b_1s+b_0}, \text{where }a_0=1, b_1=R\,C\text{ and }b_0=2 \\\\ &\frac{\frac1{b_0}a_0}{\frac{b_1}{b_0}s+1}=\frac{\frac1{b_0}}{\frac1{b_0}}\cdot \frac{a_0}{b_1s+b_0} \\\\ &\frac{\frac1{b_0}a_0}{\frac{s}{\omega_{_0}=\frac{b_0}{b_1}}+1} \\\\ &K\cdot\frac1{\frac{s}{\omega_{_0}}+1}, \text{where }K=\frac{a_0}{b_0}\text{ and }\omega_{_0}=\frac{b_0}{b_1} \end{align*}$$

The above proves how it is that \$\omega_{_0}=\frac{2}{R\,C}\$ and, as a necessary consequence, how you also find that \$K=\frac12\$ falls out of the process.

In 2nd order denominators, when you come into them someday, for any \$b_2s^2+b_1s+b_0\$ find \$\omega_{_0}=\sqrt{\frac{b_0}{b_2}}\$ and \$Q=\frac{\sqrt{b_2 \,b_0}}{b_1}\$. There's similar simple algebraic reasons for arriving at this result for 2nd order. (It will also turn out that for a low-pass only 2nd order form, then \$K=\frac{a_0}{b_0}\$, too.)

Dig a little deeper and there are connections and relationships within and between calculus and linear algebra subspaces solving differential equations (particularly the nullspaces therein.)

Still more deeply, you'll find vast stretches of probability theory, graph theory, and additive combinatorics, combined with a side tour of Schur's Theorem and Schur's work trying to solve Fermat's Last Theorem using mod p (Schur not only failed, but he proved it must fail) and coloring schemes.

There's a profound iceberg of very powerful applied mathematics lurking below the tip you are just barely seeing.

  • \$\begingroup\$ Thanks for the detailed response. In reading this, I'm realizing that I forgot a factor of 1/2Vin in the first approach from the voltage division, so the result would in fact be the same. I'm pretty new to all this, so my only current approach to actually finding $$\omega_0$$ is to check at what value the magnitude of the transfer-function is equal to 1/sqrt(2). In the second (correct) case, if we take the magnitude of the function the denominator becomes sqrt(4+(wRC)^2), so the fact that this can't become sqrt(2) is what made me think it must be wrong. Any helpful input on this perhaps? \$\endgroup\$
    – iegbg
    Oct 24, 2023 at 12:40
  • 1
    \$\begingroup\$ @iegbg The problem is that \$K=\frac12\$. So the result isn't supposed to be \$\frac1{\sqrt{2}}\$. See my addition at the end of my answer above for more details. \$\endgroup\$ Oct 24, 2023 at 21:47
  • \$\begingroup\$ Thank you so much for taking the time with this. You're right, I had misunderstood how to look for the answer at all, but I realize now I was supposed to look for 1/sqrt(2) of the maximum value in this case, so everything makes a lot more sense now. I'm looking forward to digging further into this later on, pretty fascinating stuff \$\endgroup\$
    – iegbg
    Oct 25, 2023 at 10:38
  • \$\begingroup\$ @iegbg You should probably select the answer, then. Not for me. But to save time for others who may take the idea you still are seeking more answers. If you are still seeking such, then leave things as they are. \$\endgroup\$ Oct 25, 2023 at 11:12

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