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I'm working on a project and I need to find the transfer function fhe following filter (as a function of the value of the capacitors and resistors). How can I find that?

enter image description here

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    \$\begingroup\$ Use spice to analyze. \$\endgroup\$ – tilz0R Nov 6 '17 at 13:35
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    \$\begingroup\$ What did your web research show? You need to show the results of your effort in your question or it will probably be closed as homework with no attempt at a solution. Note that you can save the CircuitLab schematic directly into the question rather than a screenshot. This has the advantage that it remains editable. \$\endgroup\$ – Transistor Nov 6 '17 at 13:40
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    \$\begingroup\$ So - less of a 'project' and more of a 'homework assignment' by the looks of it. \$\endgroup\$ – brhans Nov 6 '17 at 13:41
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    \$\begingroup\$ What do you think spice does if it's not doing it mathematically? It doesn't build an actual circuit and measure it ... \$\endgroup\$ – brhans Nov 6 '17 at 13:55
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    \$\begingroup\$ I'm voting to close this question as off-topic because it is homework, without effort shown. \$\endgroup\$ – Brian Carlton Nov 6 '17 at 22:26
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R1 and C1 just form a normal low pass filter, then the opamp R2, R3, C2, and C3 form a low pass Sallen-Key filter. Multiply the results, and you're done :)

A good reference for Sallen-Key designs is this application note, and the Wikipedia page for an RC filter is pretty good. As noted in the comments, the SK will load the RC filter so you will need to calculate the impedance of the SK as well.

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  • \$\begingroup\$ I do not now how to write it mathematically :( \$\endgroup\$ – Klopq Nov 6 '17 at 13:38
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    \$\begingroup\$ Hmm, if it were that simple. The SK-filter part does add a load to the RC filter (R1, C1) so it is a bit more complex than simply muliplying the two unless you take the load of the SK-part into account with the RC filter. \$\endgroup\$ – Bimpelrekkie Nov 6 '17 at 13:39
  • \$\begingroup\$ @Bimpelrekkie I want to take everything into account \$\endgroup\$ – Klopq Nov 6 '17 at 13:40
  • \$\begingroup\$ @Bimpelrekkie That is correct, yes. You'd need the impedance of the SK as well to be accurate, but for simplicity one could use a low impedance RC followed by a high impedance SK. Edited accordingly :) \$\endgroup\$ – awjlogan Nov 6 '17 at 13:42
  • \$\begingroup\$ ....for simplicity...? Of course, you can simplify everything - however, the result will be wrong! \$\endgroup\$ – LvW Nov 6 '17 at 14:54
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As Sammy Hagar says, "there's only one way to rock" and to determine the transfer function of this circuit, the way to go is to apply the FACTs. First, we redraw the original circuit in a simpler form to make it look more friendly. The op amp is just buffering the output voltage to bias the right side of \$C_3\$ and \$R_4\$ is off the picture considering a 0-\$\Omega\$ output impedance:

enter image description here

There are three energy-storing elements with independent state variables: it is a 3rd-order system. I can see there is no zero in this circuit because if I short any of these caps in the original schematic featuring the op amp, the response is always 0 despite the presence of a stimulus. Therefore, the transfer function obeys \$H(s)=H_0\frac{1}{1+sb_1+s^2b_2+s^3b_3}\$

First, we start with the dc transfer function, when \$s=0\$: open the caps and determine the gain linking \$V_{out}\$ to \$V_{in}\$. This is 1 and \$H_0=1\$. Now, reduce the excitation to 0 V (replace the stimulus \$V_{in}\$ by a short circuit) and determine the resistance "seen" from each capacitor. The below circuit shows you how to do it:

enter image description here

This works by inspection most of the time, no need to write a single line of algebra! Look, to determine the resistance driving \$C_1\$, imagine in your head that you probe with an ohm-meter \$C_1\$'s terminals. The resistance you see in this case is \$R_1\$ so the first time constant \$\tau_1=C_1R_1\$. Repeat the operation for \$C_2\$ and find \$\tau_2=C_2(R_1+R_2+R_3)\$. For \$\tau_3\$, you should find \$\tau_3=0\$. You sum these time constants to form \$b_1=\tau_1+\tau_2+\tau_3\$.

For \$b_2\$, we will determine \$\tau_{12}\$, \$\tau_{13}\$ and \$\tau_{23}\$. This time, for \$\tau_{12}\$ for instance, I will short \$C_1\$ and "look" at the resistance offered by \$C_2\$'s terminals. A shown in the picture, this is \$\tau_{12}=C_2(R_1+R_2)\$. Continue with the two other time constants and form \$b_2=\tau_1\tau_{12}+\tau_1\tau_{13}+\tau_2\tau_{23}\$.

For the final lap, determine \$\tau_{123}\$ for which \$C_1\$ and \$C_2\$ are replaced by a short circuit while looking through \$C_3\$'s terminals. You should find \$\tau_{123}=C_3(R_2||R_3)\$ and form \$b_3=\tau_1\tau_{12}\tau_{123}\$.

The transfer function appears in the below Matchcad sheet and confirms the 3rd-order behavior. I have rewritten the raw expression using well-separated cascaded poles for the sake of the example:

enter image description here

enter image description here

Edit: If you now use the on-line tool and program a 1-kHz characteristic frequency, the transfer function in which the component values clearly appear in \$H_2(s)\$ below confirms the 3rd-order Bessel dynamic response:

enter image description here

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  • \$\begingroup\$ I was also trying to answer this question. But in your work you've a value \$\text{R}_\text{s}=10^{-12}\space\Omega\$, but where did you find that? And I got a cut off frequency of \$\approx 153\$ Hz. \$\endgroup\$ – Jan Nov 7 '17 at 7:45
  • \$\begingroup\$ The \$R_s\$ value is there to avoid a potential indeterminacy but here, it can be replaced by 0 without a problem. Regarding cutoff, I have selected arbitrary values for the components so the poles are spread, giving multiple locations. I am sure you can group some of them and obtain a different response. However, in this case, the cascaded pole factorization no longer works (the raw expression \$H_1\$ is still ok) and you have to resort to a different expression. \$\endgroup\$ – Verbal Kint Nov 7 '17 at 9:45
  • \$\begingroup\$ Ok, but this system has one cut off frequency (because if you plot the absolute value , as you did) at the -3 dB point and that is about \$153\$ \$\endgroup\$ – Jan Nov 7 '17 at 9:48
  • \$\begingroup\$ You mean the -3-dB cutoff frequency? You may be right, I did not look at it, just derived the transfer function. \$\endgroup\$ – Verbal Kint Nov 7 '17 at 9:53
  • \$\begingroup\$ My answer was downvoted and I would be glad to know why? \$\endgroup\$ – Verbal Kint Nov 7 '17 at 11:53
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The most simple way for computing (by hand) the transfer function is the following:

1.) Find the voltage at the non-inv. opamp input up (taking into account Vin and Vout)

2.) Find the voltage at the inv. opamp input un (taking into account Vin and Vout)

3.) Equalize up=un (for idealized opamp). Now you have two equations for the two unknowns up(=un) and Vout/Vin. Solve this set of two equations.

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  • \$\begingroup\$ I think, it would be just fair to explain why you have downvoted my answer. It seems you never have tried this way for finding the transfer function? What do YOU propose instead? It is not a good behaviour to downvote an answer without presenting the reason or even a better alternative. I do not care about the "-1" - however, it may confuse the questioner. That`s the point!! \$\endgroup\$ – LvW Nov 7 '17 at 9:23
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enter image description here

The answer is here, you just need to seperate Vi and Vo, then the transfer function is Vo/Vi. Just remember, it will become much more easier if you solve this kind of problems in laplace domain.

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  • \$\begingroup\$ You can use latex in an answer... \$\endgroup\$ – Voltage Spike Nov 6 '17 at 22:19

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