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I am an engineering student currently learning how to design low-pass active filters. Currently, I am having trouble matching my calculations with the simulation values. My design has the following parameters: fc = 250Hz, Q = 0.70 (Butterworth approx.) My current circuit is shown below:

enter image description here

I am using Multisim conducting AC Sweep analysis to see the frequency response of the circuit. The response is shown below:

enter image description here

Using the formula 1/2piRC (R1=R2, C1=C2), I calculated the resistor value choosing 47nF initially. Supposedly, it should have fc = 250Hz based on the equation. Based on the simulation, the -3dB point is already pass that (x2 = 359, y2 = -3dB). I used the gain 1.588dB, which is the approx. value for a butterworth filter, and calculated for R3 with 10k chosen Rf value.

Am I missing something here? Is there a design consideration I am missing?

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    \$\begingroup\$ The resistors are correct for gain = 1.588, not 1.588 dB. (So, about 4dB, do the math yourself). So, at what frequency is the gain about 1 dB? 250Hz? It's doing what you expect then. \$\endgroup\$ – user_1818839 Jan 27 at 19:53
  • \$\begingroup\$ Hi. The frequency is approx. 250 Hz. I also observed this awhile ago, but didn't really understand why or couldn't explain if its probably the gain's fault. \$\endgroup\$ – Guorishix Jan 27 at 19:55
  • \$\begingroup\$ Perhaps that is a non-ideal op-amp, which has finite slew rate and bandwidth? Is there an ideal op-amp in the simulator you can try? \$\endgroup\$ – Justme Jan 27 at 19:58
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    \$\begingroup\$ @Elmark Corpus I wondered if you've actually understood Brian's comment? The cut off frequency is defined as being 3dB down from the pass band. The gain in the pass band is +4dB. Therefore at the true cut-off frequency, the gain will be +4dB - 3dB = 1dB. You are actually assuming the cut-off frequency is where the gain is 7dB down from the pass band. \$\endgroup\$ – James Jan 27 at 20:38
  • \$\begingroup\$ @James, thank you for clarifying. I actually didn't understood it at first. \$\endgroup\$ – Guorishix Jan 28 at 0:06
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Well, first of all it is not very hard to find the transfer function of this circuit. But I will use the result from Okawa Electric Design:

$$\mathscr{H}\left(\text{s}\right):=\frac{\text{v}_\text{out}\left(\text{s}\right)}{\text{v}_\text{in}\left(\text{s}\right)}=\frac{\frac{\text{G}}{\text{C}_1\text{C}_2\text{R}_1\text{R}_2}}{\text{s}^2+\left(\frac{1}{\text{C}_1\text{R}_1}+\frac{1}{\text{C}_1\text{R}_2}+\frac{1-\text{G}}{\text{C}_2\text{R}_2}\right)\text{s}+\frac{1}{\text{C}_1\text{C}_2\text{R}_1\text{R}_2}}\tag1$$

Where \$\text{G}:=1+\frac{\text{R}_4}{\text{R}_3}\$.

And the circuit is given by:

enter image description here

Using your values, we can see that:

$$\mathscr{H}\left(\text{s}\right)=\frac{1080000000000000000}{381969\text{s}(721497\text{s}+1600000000)+680000000000000000}\tag2$$

Now, we can find the cut-off frequency \$\omega_0\$ by finding:

$$\left|\underline{\mathscr{H}}\left(\text{j}\omega_0\right)\right|=\frac{1}{\sqrt{2}}\cdot\left|\underline{\hat{\mathscr{H}}}\left(\text{j}\omega\right)\right|\tag3$$

This leads to:

$$\omega_0=\frac{200000000 \sqrt{24 \sqrt{145}+1}}{2164491}\approx1573.52\space\text{rad/sec}\tag4$$

In Hertz, this gives:

$$\text{f}_0=\frac{100000000 \sqrt{24 \sqrt{145}+1}}{2164491 \pi }\approx250.433\space\text{Hz}\tag5$$

I found that last result, by using the following Mathematica-code:

In[1]:=Clear["Global`*"];
G = 1 + (R4/R3);
R1 = 13545;
R2 = 13545;
R3 = 17000;
R4 = 10000;
C1 = 47*10^(-9);
C2 = 47*10^(-9);
s = I*2*Pi*f;
x = ((G)/(C1*C2*R1*
       R2))/(s^2 + ((1/(C1*R1)) + (1/(C1*R2)) + ((1 - G)/(C2*R2)))*
      s + (1/(C1*C2*R1*R2)));
y = FullSimplify[
   Sqrt[ComplexExpand[Re[x]]^2 + ComplexExpand[Im[x]]^2]];
solution = FullSimplify[Solve[{D[y, f] == 0, f > 0}, f]];
solutionForF = f /. solution[[1]];
n = Solve[{y == (1/Sqrt[2])*Limit[y, f -> solutionForF], f > 0}, f];
Flatten[n] && Flatten[N[n]]

Out[1]={f -> (100000000 Sqrt[1 + 24 Sqrt[145]])/(2164491 \[Pi])} && {f -> 
   250.433}
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  • \$\begingroup\$ As we can see from the transfer function (general form) the pole frequency is 1/SQRT(R1R2C1C2). On the other hand, for Butterworth responses we know that wp=wc. Hence, for equal componenets: wc=1/RC. \$\endgroup\$ – LvW Feb 9 at 14:16
  • \$\begingroup\$ The OP had already calculated the expected frequency response, that was not the question. The question was how to interpret the plot produced by the simulation, which Brian Drummond answered correctly over a month ago. \$\endgroup\$ – Elliot Alderson Mar 2 at 20:57

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