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I'm repairing some instrumentation equipment and I've come across a filter I do not fully understand. It seems to be a band pass filter and the only documentation that came with this device was an old handwritten note on the filter break frequencies and gain. I have copied this as neatly as I can and attached it below.

enter image description here

As you can see, there is no transfer function derivation, and only a graph showing which components set gains and break frequencies. I have simulated this in Spice and it seems to really obey those equations.

As I have not come across this type of filter before I set out to understand it better by deriving the transfer function and those equations as shown on the graph above myself.

I (believe) I have managed to derive the TF (attached below) but now I'm stuck on how to proceed and derive the equations as shown on the graph above.

enter image description here

My question is how to proceed with deriving the filter gains and break frequencies? Any tool I can use to help me? Does anyone know what topology/kind of filter this would be classified as? I have not found anything on the Internet of this topology but maybe my search terms are wrong.

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  • \$\begingroup\$ What is your main problem? To understand why/if it is a kind of bandpass or to derive the corner frequencies - or to verify the transfer function?. Well, after the first look I can say it is a very poor "bandpass" (with finite gain for very low/high frequencies), which consists of a series connection of a (poor) highpass with a (poor) lowpass (more correct: phase-lead resp. phase-lag units). Because It is a very uncommon circuit it cannot be "classified" in filter terms. However, perhaps it fulfills the particular requirements for the application. \$\endgroup\$ – LvW Jun 19 '14 at 14:36
  • \$\begingroup\$ The designer appears to have used a couple assumptions in this filter approximations. The main one appears to be this: 1/(C2R3) >> 1/(R1C1) Without that assumption, there would be no way he/she could have assumed that R1C1 would be the lower frequency pole. \$\endgroup\$ – horta Jun 19 '14 at 17:37
  • \$\begingroup\$ @LvW I would like to derive the corner frequencies myself and hence verify what the original designer noted in the diagram is correct (it seems to be from simulations) and learn something along the way. The application is sensing voltage across an RTD biased by a precision current source (not shown in the diagrams). I'm not sure what kind of measurements they are interested in but I believe the circuit fulfills their needs adequately. I will attempt to re-derive the transfer function and see if I can obtain the same corner frequencies. \$\endgroup\$ – IgorEE Jun 19 '14 at 20:37
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In my first answer I have described how you can find the exact solution for the 2 zeros and the 2 pole frequencies (which are identical to the wanted break frequencies). However, here is a good approach which should be sufficient for the shown circuit. In principle, I follow the way as outlined already by Dave Tweed´s answer: Simplification of the circuit. In the present case, you can create three different (simplified) circuits of first order only which easily can be analysed.

1.) For the first rising region of the transfer function the high pass part with C1 is responsible (C2 causes the falling part and can be neglected). Furthermore, for very low frequencies (including DC) the gain Ao=1+R3/R2 is assumed to be not much larger than unity which is the possible minimum.

Hence, for acceptable filtering it is assumed that R2>>R3. As an equivalent diagram for the lower frequency range (without C2 and R2) we arrive at a circuit with only the three components R1, R3 and C1. It is a simple task to find the relevant time constants (invers to the corresponding break frequencies):
Using your indices, we thus find T2=(R1+R3)C1 and T1=R1C1.

2.) Above the frequency f1 the capacitor C1 is not effective any more (and the capacitor C2 is assumed to be not yet effective). Hence, we have a simple non-inverting amplifier with the gain (maximum of the transfer function) Amax= 1+R3/Rp with Rp=R1||R2.

3.) For rising frequencies, the low pass part with capacitor C2 becomes effective (C1 is considered as a short). Hence, the feedback path consists of R3||C2 and Rp only.

The time constant T3 (pole frequency) can be derived as T3=R3C2 and the last break frequency (zero) is determined by T4=R3C2/(1+R3/Rp).


Finally, it is to be noted, that all results are in agreement with the values given in the scetched BODE diagram. This can be verified using the well-known relations for a 20dB gain slope (as used in the graph with G1/G2=f1/f2).

Final remark: Thus, it can be concluded that the information contained in the scetched BODE diagram (break frequencies) also are only approximations.

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The trick is to know which terms in your derived equation are significant and which are not. The original designer's equations are simplifications that make the assumption that the various cutoff frequencies are "far enough" from each other that they don't significantly affect each other. Your "full" derivation includes terms for those interactions. There's really no rigorous mathematical way to get from the latter to the former; you mainly need to put real values in for the components and then decide whether each term is large enough to matter.

Bottom line: Use the designer's equations to help you pick component values, then use your equations to see what the actual response is going to be. Make small adjustments to the values as needed.

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  • \$\begingroup\$ So do you think the original designer just used intuition and experience to quickly sketch up that graph with equations? My guess would have been that it is a standard filter configuration (such as Sallen Key) with well known equations, and consulting any book one could quickly sketch up that graph. However since I can not find this configuration anywhere I am left thinking it must have been just intuition and experience that enabled the original designer to sketch up that graph. \$\endgroup\$ – IgorEE Jun 19 '14 at 17:19
  • \$\begingroup\$ Yes, it's exactly how I would approach designing such a circuit. I would use the simplified equations to do the design, then verify the results either on a breadboard, in simulation, or (last choice) working out the full equations. \$\endgroup\$ – Dave Tweed Jun 19 '14 at 17:27
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"My question is how to proceed with deriving the filter gains and break frequencies? Any tool I can use to help me?"

At first the good news: The transfer function looks OK.

Regarding your question: If you really want to find the correct expressions for gain (maximum) and the corner frequencies (poles and zeros of the transfer function) you have to proceed as follows:

1.) Recalculate the transfer function (search for main denominator) with the aim to have only one single fraction (the "1" must disappear).

2.) As a result, you will have a transfer function consisting of a 2nd-order numerator and a 2nd-order denominator.

3.) Find the zeros of the numerator (zeros of the transfer function) as well as for the denominator (poles of the transfer function). Then you have the expressions for the 4 corner frequencies.

Regarding a corresponding tool: I know only about a tool which can find the pole/zero distribution for a given function with parts values (that means: not as a general expression/formula).

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