In the given RC circuit, there are 3 capacitors, out of which the voltages of 2 capacitors can be uniquely determined.

This implies that it is a second-order circuit. But when I found the expression, I got only one pole i.e first-order response.

Where am I wrong in determining the order of the circuit?

enter image description here

  • \$\begingroup\$ Looks right to me. \$\endgroup\$
    – jonk
    Sep 1 at 18:54
  • 3
    \$\begingroup\$ I've noticed that all questions previously raised have not attracted any formal answer acceptance. I've also noticed that at least one question hasn't had its answer followed by either a comment or an upvote. If other folk also notice this, it might be a reason for what seems to be a decline in help offered on more recent questions. People give help for free and it's not clear-cut how those people will react in the face of no prospect of upvotes and no prospect of answer acceptance. Just saying. \$\endgroup\$
    – Andy aka
    Sep 1 at 19:00
  • \$\begingroup\$ @Andyaka My mistake. Sorry for that. I didn't think much about that and I was new to this platform. So, I don't know that this would have an impact. Thanks for letting me know. \$\endgroup\$
    – prashanth
    Sep 1 at 19:34
  • \$\begingroup\$ @jonk Which is right? First order or second? \$\endgroup\$
    – prashanth
    Sep 1 at 19:35
  • 1
    \$\begingroup\$ @prashanth I thought you wondered if your simplification was valid, treating all of the R and C values as equal. If they are exactly equal, then I do find your simplification through cancellation. If, however, you want to see the full transfer function in all its detail without cancellation, then yes the result is 2nd order and has both high pass and low pass terms. \$\endgroup\$
    – jonk
    Sep 2 at 2:06

1 Answer 1


The schematic with symbols and nothing assumed about equality of any of them is:


simulate this circuit – Schematic created using CircuitLab

Using freely available SymPy find, through KVL, the following steps towards a solution:

var('r1 r2 c1 c2 r3 c3 vi vop vom s')
eq1 = Eq( vop/r1 + vop/r2 + vop/(1/s/c1) + vop/(1/s/c2), vi/r1 + vom/r2 + vom/(1/s/c2) )
eq2 = Eq( vom/r3 + vom/r2 + vom/(1/s/c3) + vom/(1/s/c2), vi/(1/s/c3) + vop/r2 + vop/(1/s/c2) )
a2 = solve( [ eq1, eq2 ], [ vom, vop ] )
tf2( simplify( (a2[vop]-a2[vom]) / vi ) )

The tf2 function yields the following 2nd order result:

$$\begin{align*} \omega_{_0} =& \frac{\sqrt{R_1 + R_2 + R_3}}{\sqrt{R_1\,R_2\,R_3}\sqrt{C_1\,C_2 + C_1\,C_3 + C_2\,C_3}} \\\\ \zeta =&\frac12\cdot\frac{R_1\,C_1\left(R_2 + R_3\right) + R_2\,C_2\left(R_1 + R_3\right) + R_3\,C_3\left(R_1 + R_2\right)}{\sqrt{R_1 + R_2 + R_3}\sqrt{R_1\,R_2\,R_3}\sqrt{C_1\,C_2 + C_1\,C_3 + C_2\,C_3}} \\\\ \frac{v_{_\text{OUT}}}{v_{_\text{IN}}}=G_s =& -\left[\frac{1}{1+C_2\left(\frac{1}{C_1}+\frac{1}{C_3}\right)}\right]\frac{\left(\frac{s}{\omega_{_0}}\right)^2}{\left(\frac{s}{\omega_{_0}}\right)^2+2\zeta\left(\frac{s}{\omega_{_0}}\right)+1}\\\\&+\left[\frac{R_2}{R_1 + R_2 + R_3}\right]\frac{1}{\left(\frac{s}{\omega_{_0}}\right)^2+2\zeta\left(\frac{s}{\omega_{_0}}\right)+1} \end{align*}$$

If you set \$R=R_1=R_2=R_3\$ and \$C=C_1=C_2=C_3\$ then, technically, you wind up with:

$$\begin{align*} \omega_{_0} =& \frac{1}{R\,C} \\\\ \zeta =&1 \\\\ \frac{v_{_\text{OUT}}}{v_{_\text{IN}}}=G_s =& -\left[\frac{1}{3}\right]\frac{\left(\frac{s}{\omega_{_0}}\right)^2}{\left(\frac{s}{\omega_{_0}}\right)^2+2\zeta\left(\frac{s}{\omega_{_0}}\right)+1}\\\\&+\left[\frac{1}{3}\right]\frac{1}{\left(\frac{s}{\omega_{_0}}\right)^2+2\zeta\left(\frac{s}{\omega_{_0}}\right)+1} \end{align*}$$

In this case, if you plot your reduced result against the fuller 2nd order result the plots should be the same. (They should overlap.) The reason is simple to see if you set \$R=1\$ and \$C=1\$. Then you have \$\left(1-s\right)\left(1+s\right)\$ divided by \$\left(1+s\right)^2\$ and the factor \$\left(1+s\right)\$ found in both numerator and denominator can then cancel, effectively leaving your result.

Hopefully, this helps you resolve questions about this case. (That is, assuming you haven't already done so.)


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