Take the 2-minute tour ×
Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. It's 100% free, no registration required.

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):

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

share|improve this question
add comment

2 Answers

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.)

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.