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I have a circuit very similar to the inverting band pass filter circuit here: enter image description here

I was asked a question about the circuit's feedback transfer function. Is this the same as the transfer function for the whole circuit?

Which I calculated to be this (for the circuit above):

$$\frac{V_0}{V_i}=\frac{j \omega R_2 C_1}{1+j \omega R_2 C_2 + j \omega C_1 + \omega^2 C_1 C_2}$$

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2 Answers 2

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The feedback transfer function is not the same as that of the amplifier, because the overall circuit suppresses whatever is fed back. So roughly speaking, they are opposite. The transfer function in the negative feedback is a high pass filter, and therefore the amplifier is a low-pass filter. (The overall circuit is band-pass because of the high pass in front of it.)

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Something seems wrong with your transfer function calculation. There should be a minus sign in front of it and in front of \$\omega^2\$ as well (\$j^2=-1\$). A couple of R's also seem to be missing in the denominator.

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