I am doing a filter analysis of a simple network filter, requiring drawing some equivalent circuit to simplify the expression and analysis of the transfer function.

The issue I've got is regarding a simplification step on 2 RC cell in parallel :

RC // RC equivalent circuit

In one hand, the equivalence seems right : I've got two identical RC cells in parallel with the same voltage across each of them. So in each branch I have the same impedance, thus the same current, thus each cap will charge at the same speed. This will be equal to a 2 time bigger cap we a 2 time weaker resistor.

However, in the other hand, when I try to calculate the equivalent impedance of the 2 parallel cells, I definitely see that the result is different than an \$ (R/2)(2C) \$ impedance :

$$ \frac{(R + 1/jCw)(R + 1/jCw)}{(2R + 2/jCw)} =! \frac{R}{2} + \frac{1}{j2Cw} $$

So, is this equivalence suitable in a filter analysis purpose, or did I miss something with my calculations/ understanding ?

Thank you very much,


Start with the individual series networks: -

$$Z = R + \dfrac{1}{sC} = \dfrac{sCR + 1}{sC}$$

In case you were not sure s = jω.

Now, if you add the two networks as admittances then take the reciprocal you get: -

$$Z_{PARALLEL} = \dfrac{1}{\dfrac{sC}{sCR+1} + \dfrac{sC}{sCR+1}}$$

$$ = \dfrac{sCR +1}{2\cdot sC}$$

did I miss something with my calculations/ understanding ?

Your understanding was fine but your calculation went wrong somewhere.

  • 1
    \$\begingroup\$ Thank you very much, bad calculation skill I guess. \$\endgroup\$ – Valentin BEYNARD Apr 2 '20 at 14:27

I think that something goes wrong with your calculation.

Since the impedance Z is the same( \$Z=\frac{1}{jwC}+R\$), the equivalent parallel of the two cells is \$Z_{//}=\frac{1}{2}Z\$.

\$Z_{//}=\frac{1}{2}Z=\frac{1}{2}[\frac{1}{jwC}+R]=\frac{1}{2}[\frac{1+jwRC}{jwC}]=\frac{j}{j}\frac{1}{2}[\frac{1+jwRC}{jwC}]=-\frac{j}{2}[\frac{1+jwRC}{wC}]=-\frac{j}{2}[\frac{1+jwRC}{wC}]=\frac{wRC}{2wC}-j\frac{1}{2wC}=\frac{R}{2}-j\frac{1}{2wC}= \frac{R}{2}+\frac{1}{jw2C}\$.

That's it!

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    \$\begingroup\$ I think, the last two lines are not necessary. Instead: Z/2=(R+1/sC)/2. qed. \$\endgroup\$ – LvW Apr 2 '20 at 14:50

or did I miss something with my calculations/ understanding ?


When you said,

This will be equal to a 2 time bigger cap we a 2 time weaker resistor.

you were using the formulas for capacitors in parallel and resistors in parallel.

But the two capacitors are not in parallel. They aren't connected to each other on both ends.

And the two resistors are not in parallel, because they aren't connected to each other on both ends either.

So you have to calculate the impedance of the RC combination, and then what that is in parallel with itself, rather than assume the result is the same as two parallel capacitors in series with two parallel resistors.


Your calculations, and those of the other answers, are correct, it's the interpretations that are wrong (and yours got wrong somewhere along the road, the start was fine). When you double the value of a capacitor, you double the value of C in the denominator, thus having \$\frac{1}{2sC}\$ means twice the value of the capacitor.

To verify this, here's a little test:


C1 and C2, with both their Rser set, are the network in your picture, left side. C3 is what you condidered theoretically, same picture, right side. C4 is what you think you're getting, and the Laplace source is the mathematical expression, for verification. Notice that V(a) and V(d) are slightly displaced for better seeing that V(a) (the parallel network), V(b) (your theoretical result), and V(d) (the Laplace mathematical expression) are identical, but V(c) (your interpretation), is not.

Which concludes that, indeed, as the math shows in your OP and all the other answers, the equivalent network will be comprised of R/2 and 2C, or R||R and C||C. The intepretation was wrong, that's all. Note that this is only valid for R=R and C=C, otherwise the resulting formula is a 2nd order transfer function.


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