Hey I got an assignment on which I need to calculate the transfer function. It is about the following circuit:


simulate this circuit – Schematic created using CircuitLab

So my approach (note all capicitors values are the same and so referred to as C):

  1. At node A: i1 = i2 + i3 + i5
  2. Ideal opamp so: i3 = i4 $$\frac{Ua}{Zc}=-\frac{Uo}{R4}$$ $$Ua=-\frac{UoZc}{R4}$$

  3. Filling in 1.: $$\frac{Ui-Ua}{Zc} = \frac{Ua}{R2}+\frac{Uo}{R4}+\frac{Ua-Uo}{Zc}$$

$$\frac{Ui+\frac{UoZc}{R4}}{Zc}=-\frac{\frac{UoZc}{R4}}{R2}-\frac{Uo}{R4}-\frac{\frac{UoZc}{R4}-Uo}{Zc}$$ $$UiZc+\frac{UoZc^2}{R4}=-\frac{UoZcR2}{R4}-\frac{Uo}{R4}-\frac{UoZc^2-UoZcR4}{R4}$$ $$UiZc=\frac{-UoZcR2-Uo-UoZc^2-UoZc^2-UoZcR4}{R4}$$ $$Ui=\frac{-UoZcR2-Uo-UoZc^2-UoZc^2-UoZcR4}{R4Zc}$$ $$\frac{Uo}{Ui}=\frac{R4Zc}{-2Zc^2-ZcR2-ZcR4-1}$$ $$Zc=\frac{1}{jwC}$$


So the question is is this correct? I hope it is however I tried testing my results by simulating the design in LTSpice, taking R4 = R2 = 1K and C = 1uF and then I entered the formula, with the same component values, in matlab however the graphs are not the same.. :(

LT Spice matlab

EDIT Okay now it is confirmed the answer is wrong how would I derive such circuit? I tried it again but I'm yet again not getting the right answer.. I won't post it here since it takes to much time to type it out.

  • \$\begingroup\$ iVision, looking at your final function only one can see that it cannot be right: In the denominator you are summing "R" and "C". This is impossible! \$\endgroup\$
    – LvW
    Apr 21 '16 at 13:32

The formula for a multiple feedback high-pass filter should look like this: -

enter image description here

Don't worry about "s" if you've not come across it - replace it with jw but I can see your formula must be wrong i.e....


It can't be right because the jwC parts cancel out (top and bottom) leaving a TF that is not frequency dependent. Also the component names in your picture do not match your formula names.

The picture in my answer came from this website and you can also use it to double-check any results because it uses a design tool where you plug in the values.

  • \$\begingroup\$ Okay it is wrong I see. But how would I derive it? And the jw doesn't cancel out, the two is outside the inner brackets, right? \$\endgroup\$
    – joell
    Apr 21 '16 at 13:46
  • \$\begingroup\$ Oh OK yes I see that - that makes a big difference. However I think the numerator shouldn't have jwC due to the definition of Zc if I follow your last two formulas correctly. \$\endgroup\$
    – Andy aka
    Apr 21 '16 at 13:56
  • \$\begingroup\$ The 3rd line (point 3) is false (at least as far as the units are concerned): Uo/R4. \$\endgroup\$
    – LvW
    Apr 21 '16 at 14:57
  • \$\begingroup\$ @LvW me or the OP? \$\endgroup\$
    – Andy aka
    Apr 21 '16 at 14:58
  • \$\begingroup\$ Every time I try it I find a new mistake.. Is there any reference on solving such circuit? I found this one: extras.springer.com/2006/978-0-387-28340-1/wbapdx72.pdf it is a LPF one but I still fail to get the right result.. \$\endgroup\$
    – joell
    Apr 21 '16 at 15:11

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