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I have designed an active inverting bandpass filter, and when comparing its pole zero plot to a passive bandpass fitler with the same cutoff frequencys, the pole zero plot is exactly the same, even though the bode plots for both are different. I will run through my calculations and results below. enter image description here

With a gain of 2.5, Lower cutoff of 75Hz, and uppper of 31kHz, I found component values to be: \$R_1 = 2122\Omega\$, \$R_2 = 5305\Omega\$, \$C_1 = 1\mu\text{F}\$, \$C_2 = 0.967\text{nF}\$. From which I derived the transfer function and came find the following:

$$H(s) = -\frac{R_2C_1s}{(1+C_1R_1s)(1+C_2R_2s)}$$

$$=> H(s) = -\frac{5.305\times 10^{-3}s}{1.088\times 10^{-8}s^2 + 2.127\times 10^{-3}s + 1}$$

Using Matlab, I got the following result (Bode and Pole Zero Map) enter image description here

enter image description here

To compare how the system responded compared to a passive bandpass filter seen below, I used derived the transfer function and placed the same values used for the other filter, thus giving the same cut-off frequencies.

enter image description here

$$H(s) = \frac{R_2C_2s}{(1+R_2C_2s)(1+R_1C_1s+ \frac{R_2}{R_1}R_1C_1s)}$$

$$=> H(s) = \frac{5.13\times 10^{-6}s}{3.81\times 10^{-8}s^2 + 7.432\times 10^{-3}s + 1}$$

When processing this in MATLAB, I get the following plots:

enter image description here enter image description here

What is didn't expect or understand is the pole zero response. What do the poles and zeros represent in this system, and also shouldn't the opamp introduce its own poles and zeros and its is adding gain to the system. I tried reading on the pole zeros responses of low pass and high pass filters on google, and they look nothing like my results! Added to that, I was originally trying to understand at what point of the system the gain is applied too, from looking at this response, is it correct to assume the system is LowPass>Gain>HighPass?

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  • \$\begingroup\$ I think there was a missing \$s\$ in the middle term of the denominator of the first equation. I added it, but please edit your post to remove it if it should not be there. \$\endgroup\$ – Null Jan 30 '18 at 17:55
  • \$\begingroup\$ Well spotted, thanks. I have learned how to add the equations now; I added a couple more. \$\endgroup\$ – user160063 Jan 30 '18 at 18:34
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To compare how the system responded compared to a passive bandpass filter seen below, I used derived the transfer function and placed the same values used for the other filter, thus giving the same cut-off frequencies.

Wrong! You can't do this to compare your filters because now, there is an interaction between components that you would not get with a virtual-ground inverting op-amp configuration. It's even reflected in your bode plots - look at the mid-point frequencies - the op-amp circuit is about 10 kHz whilst the passive one is about 1 kHz. Look at the gains - they are massively different.

No, you can't do this and obtain anything useful to say.

What do the poles and zeros represent in this system, and also shouldn't the opamp introduce its own poles and zeros and its is adding gain to the system.

A good designer will normally choose an op-amp so that its unwanted characteristics are avoided so no, most op-amp circuits do not rely significantly on the op-amps non-idealities.

Pole zero tutorial for a 2nd order system: -

enter image description here

The top three pictures are example bode plots of a 2nd order low pass filter. The bottom right is the traditional pole-zero diagram. The bottom left shows how the bode plot and pole-zero diagram are interlinked in a 3D way.

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For the first circuit the transfer function is \$\frac {Z_2}{Z_1}\$; for the second circuit it’s \$\frac{Z_2}{Z_1+Z_2}\$, as there’s no buffer between the lpf and hpf.

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  • \$\begingroup\$ I pulled the equation from: engineering.purdue.edu/ME365/Textbook/chapter7.pdf Page 7-7, equation (15) \$\endgroup\$ – user160063 Jan 30 '18 at 19:15
  • \$\begingroup\$ The poles cannot be the same. \$\endgroup\$ – Chu Jan 30 '18 at 22:08
  • \$\begingroup\$ Care to explain why? \$\endgroup\$ – user160063 Jan 31 '18 at 0:32
  • \$\begingroup\$ The transfer function denominators are not the same: \$Z_1\$ is not equal to \$Z_1+Z_2\$, as demonstrated by your two transfer functions \$\endgroup\$ – Chu Jan 31 '18 at 8:22
  • \$\begingroup\$ oh, for the first circuit, sorry misread. It is correct, you can see how its derived here: electronics.stackexchange.com/questions/351811/… \$\endgroup\$ – user160063 Jan 31 '18 at 18:06

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