2
\$\begingroup\$

For the question enter image description here

I arrived at the following answer (I have assumed V1 btw)

enter image description here

Is my solution right? I just have a big doubt if whether I can use the voltage division rule at point A when there is an RC network after it.

\$\endgroup\$
  • 1
    \$\begingroup\$ No, you can't do that. There are loading effects. You could do (output impedance)/(input impedance). \$\endgroup\$ – Chu May 12 at 12:11
  • \$\begingroup\$ yep, that's the formula for the unloaded voltage divider. But: assuming you can just cascade RC filters like that is a mistake we've all been through at least once. I sympathize! I think in this situation: try with a star-triangle transform, simplify a bit and then you have a single, different voltage divider that you can analyze. \$\endgroup\$ – Marcus Müller May 12 at 12:17
  • \$\begingroup\$ @Marcus Miller, is there any special formula for a loaded voltage divider? \$\endgroup\$ – noorav May 12 at 13:00
  • \$\begingroup\$ not that I'd be aware of. Do the full network analysis! \$\endgroup\$ – Marcus Müller May 12 at 13:01
  • \$\begingroup\$ You would need a buffer (opamp G=1) in between sections A and B, to be able to multiply transfer functions, like in your example. \$\endgroup\$ – Marko Buršič May 12 at 13:34
3
\$\begingroup\$

From the following schematic:

schematic

simulate this circuit – Schematic created using CircuitLab

I get the following set of expressions using nodal analysis:

$$\begin{align*} \frac{V_\text{O}}{R_2} + s\: C_2\: V_\text{O} &= \frac{V_\text{X}}{R_2}\\\\ \frac{V_\text{X}}{R_1} + \frac{V_\text{X}}{R_2} + s\: C_1\:V_\text{X} &=\frac{V_\text{I}}{R_1} + \frac{V_\text{O}}{R_2} \end{align*}$$

Solving for \$V_\text{I}\$ and \$V_\text{O}\$ and dividing, I get:

$$\begin{align*} H\left(s\right)=\tfrac{V_\text{O}}{V_\text{I}}&=\frac{1}{R_1\,R_2\,C_1\,C_2\,s^2+\left(R_1\,C_1+R_1\,C_2+R_2\,C_2\right)s+1} \end{align*}$$

Since \$C=C_1=C_2\$ and \$R=R_1=R_2\$ this simplifies into:

$$\begin{align*} H\left(s\right)=\tfrac{V_\text{O}}{V_\text{I}}&=\frac{1}{R^2\,C^2\,s^2+3 R\,C\,s+1} \end{align*}$$

If you set \$\omega_{_0}=\frac{1}{R\,C}\$ then:

$$\begin{align*} H\left(s\right)=\tfrac{V_\text{O}}{V_\text{I}}&=\frac{\omega_{_0}^2}{s^2+3\,\omega_{_0}\,s+\omega_{_0}^2} \end{align*}$$

The standard low-pass 2-pole form (see equation (3) here for my reasoning of its development) is:

$$\begin{align*} H\left(s\right) &=\frac{\omega_{_0}^2}{s^2+2\,\zeta\,\omega_{_0}\,s+\omega_{_0}^2} \end{align*}$$

So it follows that \$\zeta=1.5\$ and that the system is over-damped (greater than 1.) Expected for a passive low-pass.

\$\endgroup\$
  • \$\begingroup\$ I'm confused, in the first equation, shouldn't it be Vo/sC2 instead of Vo*sC2. Because the current through the C2 capacitor due to voltage Vo would be 1/C2 * integral of Vo(dt). So taking the Laplace transform would leave us with Co/sC2 right? \$\endgroup\$ – noorav May 12 at 19:30
  • \$\begingroup\$ @noorav What does division by \$s\$ mean to you? What does multiplication by \$s\$ mean to you? To me, differentiation is multiplication by \$s\$ and integration is division by \$s\$. How do you see them? \$\endgroup\$ – jonk May 12 at 19:34
  • \$\begingroup\$ Yeah that's how I see them too. So integral would mean that s comes in the denominator right? \$\endgroup\$ – noorav May 12 at 19:39
  • \$\begingroup\$ @noorav The current through a capacitor is \$I_\text{C}=C\frac{\text{d} }{\text{d}t}\,V_\text{C}\$. So when applying nodal analysis, \$I_\text{C}=C\,s\,V_\text{C}\$ because \$s\,V_\text{C}\$ is differentiation of \$V_\text{C}\$. \$\endgroup\$ – jonk May 12 at 19:40
  • \$\begingroup\$ Ah Jesus, I confused the equation for current through a capacitor with current through an inductor. My bad, apologies. \$\endgroup\$ – noorav May 12 at 19:42
4
\$\begingroup\$

This exercise is a classic and you can see that applying brute-force algebra leads to a complex expression, difficult to factor. Furthermore, you can make mistake as you did. First, assign proper labels to each of the components, \$R_1\$, \$C_1\$ etc. It is important for the correct arrangements of results. If you want to solve this transfer function the hard way, transform the first stage with an equivalent Thévenin generator taken across the first capacitor. Its output impedance is \$Z_{th}(s)=\frac{1}{sC_1}||R_1\$ while the voltage is \$V_{th}(s)=\frac{\frac{1}{sC_1}}{\frac{1}{sC_1}+R_1}\$. Then, solve a resistive divider implying \$R_{th}\$, \$R_2\$ and \$C_2\$. Good luck with that!

The easiest and fastest way is to apply the fast analytical techniques or FACTs. When dealing with passive or active elements, they can't be beaten. First, consider the dc case, \$s=0\$. What is the gain when all caps are open-circuited? It is 1 in absence of a loading resistance: \$H_0=1\$.

Then, remove the capacitors while you set \$V_{in}\$ to 0 V: replace it by a short circuit and "look" into each capacitor's connecting terminals to determine the resistance you see. This resistance multiplied by the considered capacitor forms a time constant \$\tau=RC\$, the basis of the FACTs. You do that as illustrated in the below drawings and you have \$b_1=\tau_1+\tau_2\$ in a few seconds.

Then, select one of the caps and set it in its high-frequency state (a short circuit) then "look" into the terminals of the second cap. Combine this new time constant with one that you have already determined (\$\tau_1\$ for instance) and you have the second-order term: \$b_2=\tau_1\tau_{12}\$.

enter image description here

Once done, you can write the denominator \$D(s)=1+sb_1+s^2b_2\$ and express the transfer function as in below Mathcad sheet. The equation as it is is difficult to use, you have no insight in what is does. Apply the low-\$Q\$ approximation (look here) and factor it with two cascaded poles. You end-up with a low-entropy expression, exactly what the FACTs will always naturally lead you to.

enter image description here

\$\endgroup\$
  • \$\begingroup\$ I really appreciate you giving this answer. But this is the first time I've heard about FACTs and this hasn't been taught in my college either. So even if I did understand this and implement such a solution in my exam, it'll most likely be deemed wrong because it wouldn't be as per the answer key. Is there any other way this could be done? Using simple, plain old transfer function equations? \$\endgroup\$ – noorav May 12 at 13:03
  • \$\begingroup\$ I understand your comment. Yes, the other way is by expressing the Thévenin generator involving the two left-side elements. This is the classic approach and already a first step towards FACTs. This is proposed in the first part of the answer. If you have time, explore the FACTs and once you have the skill, you"ll never go back to the "old" way : ) You can also have a look at this seminar: cbasso.pagesperso-orange.fr/Downloads/PPTs/… \$\endgroup\$ – Verbal Kint May 12 at 13:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.