For the following circuit i am having trouble finding the following respective cutoff frequencies as i am unsure on if i need to take in to account R3 for the high pass portion. Also for the low pass portion do i need to take into account R1 and R2 or not as i know what Vin1 will be?

R1 and R2 are used as a voltage divider as Vout1 has to be less than a certain voltage due to the use it entering an amp after.

Hence what are the formulas i need to use to calculate the cutoff frequencies for say 15Hz - 50kHz? Can i just do for the highpass fc=1/(2pi*(R1+R2)*C) and for the lowpass portion fc=1/(2*pi*R3*C)? To find Vout1 can i just do Vin1*(R/(R+1/jwc))?

Also the high pass is used to block the dc bias from the input.enter image description here

Also so are R1 and R2 in series? Then if they are what is R3 in terms of R1 and R2 (as in parallel OR series configuration)

UPDATED:enter image description here For this given example when i analyse it like the the first plot it gets the correct cutoff frequencies that i am expecting. The problem is that i dont get how to calculate the resistor values for when the voltage divider is used. As it can be seen when i tried, the cutoff frequencies are not what i want and are far off. Hence what equations do i use to calculate R1,R2 and R3 as i assume for the second plot all resistors matter for each filter. First plot gets 20Hz and 15kHz for the -3db cutoff however the second plot is around 20Hz and 5kHz?

Thus how i calculated the lowpass resistor value is wrong and was wondering what formula is needed for fc for the lowpass.

  • \$\begingroup\$ What is the load for this voltage divider ? The time constant for C2 is R3+R1||R2 = 10.6kOhm+20.5kOhm = 31.1k Ohm and $$F \approx \frac{0.16}{C_2 * (R_3+R_1||R_2)} \approx \frac{0.16}{1nF*31.1k\Omega} = 5.14kHz$$ and for the C1 the time constant is R1+R2 if we ignore the voltage divider load. \$\endgroup\$
    – G36
    Commented May 13, 2017 at 6:39
  • \$\begingroup\$ 1.73V is Vin. Also i dont get why R3 is in series with R1||R2? So if i wanted to get a cutoff of 20kHz how would i go about that for the resistor value. I found that when calculating the high pass cutoff frequency of 20Hz i could just use R1 +R2 for the Req but i dont understand why i dont need to use R3? Also where did you get 0.16 from? If the voltage divider where not the same resistances would they still be in parallel? \$\endgroup\$
    – Student
    Commented May 13, 2017 at 6:41

1 Answer 1


To be able to find the pole frequency we need to find the equivalent resistance seen from the \$C_2\$ capacitor point of view. In short, we need to find the Thevenin's resistance seen from \$C_2\$ capacitor terminals. So, to do this we "short" all the voltage sources \$V_1\$ and \$C_1\$.


simulate this circuit – Schematic created using CircuitLab

And from the inspection we see that \$R_{th} = R_3+R_1||R_2\$

And the pole frequency is:

$$F = \frac{1}{2 \pi * C_2 * R_{th}} \approx \frac{0.16}{C_2 R_{th}}$$

And you can do the same thing if you want to find \$C_1\$ pole.

But this time we have an open circuit at the end of the voltage divider.

Hence, if we ignore the voltage divider load we can see that \$R_{th}\$ for hign pass filter is \$R1+R2\$

We can use this approximation method because \$C_1\$ >>\$C_2\$

  • \$\begingroup\$ Thank you so much. So as you said we can do this same approximation to find c1 why does R3 not matter. As in when plotting it it never changed the cutoff frequency for the lowpass?Also so what do you mean when you state we can use this approximation because C1>>C2? \$\endgroup\$
    – Student
    Commented May 13, 2017 at 7:08
  • \$\begingroup\$ To be more precise in your circuit we have two filters connected in series, so they affect each other. But in this case, because the pole frequency are split (away from each other p1 =20Hzz and p2 = 5.1kHz)) we do not have to worry about the influence of one filter into the other. Also, I edit my answer. \$\endgroup\$
    – G36
    Commented May 13, 2017 at 7:22
  • \$\begingroup\$ Oh ok, yeah I forgot that it would do that. thanks \$\endgroup\$
    – Student
    Commented May 13, 2017 at 8:05
  • \$\begingroup\$ When calculating Vout for each filter, for the lowpass do i still use Rth and do i use Vin1? Or can i just ude Vin1*C2/(R3+C2)? \$\endgroup\$
    – Student
    Commented May 13, 2017 at 13:55
  • \$\begingroup\$ I don't quite understand can you elaborate? \$\endgroup\$
    – G36
    Commented May 13, 2017 at 17:33

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