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I am trying to calculate the frequency response of the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab The two extra GND points are used to show that Vi and Vo are measured in relation to the ground.

Here are 2 methods for this:

Method 1
enter image description here

Method 2
enter image description here

The first method is the one I thought. As you can see, I used the formula of voltage divider two times. But the frequency response I found is different from the one in the 2nd method. At this point, let me clarify that the 2nd method supposes R1=R2=R and C1=C2=C but either way the frequency response is different. Basically, the 2nd method finds the same function as in here: http://sim.okawa-denshi.jp/en/CRCRkeisan.htm I suppose one of the methods is wrong, but why?

Sorry for not typing the equations but I thought this would be a waste of time. If something is not clear please ask me to clarify it.

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  • \$\begingroup\$ The TF function due to method 1 is not correct. The s-term in the middle of the denominator must - in addition - contain a "mixed" term R1C2. The error is that you have used Z1=1/jwC1. You have forgotten that the first common node is loaded also by R2-C2. \$\endgroup\$
    – LvW
    Commented Dec 27, 2014 at 16:48
  • \$\begingroup\$ Can you be more specific? What should Z1 be equal to? \$\endgroup\$
    – mgus
    Commented Dec 27, 2014 at 16:59
  • \$\begingroup\$ When my class covered this in Circuit Analysis, our professor also pointed out that you could put a buffer (an amplifier with gain = 1, Zin = infinity, Zout = 0) between the two RC stages. Then, the circuit would act like you'd expect from Method 1. \$\endgroup\$
    – Greg d'Eon
    Commented Dec 28, 2014 at 5:54
  • \$\begingroup\$ You can check the exact transfer function here electronics.stackexchange.com/questions/220050/… \$\endgroup\$ Commented May 28, 2017 at 17:16

1 Answer 1

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Your answer according to "Method 1" would be correct if you defined \$Z_1\$ as

$$Z_1=\frac{1}{s C_1}||(R_2+\frac{1}{s C_2})=\frac{\frac{1}{s C_1}(R_2+\frac{1}{s C_2})} {\frac{1}{s C_1}+R_2+\frac{1}{s C_2}}=\frac{1+sR_2C_2}{s^2R_2C_1C_2+s(C_1+C_2)}\tag{1}$$

Use Eq. (1) combined with the relation between \$V_0\$ and \$V_1\$ and you will obtain the correct transfer function.

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