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voltage gain and cutoff frequency of the low pass filter

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anesziere

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Good evening,

I am having some troubles finding the expressions for voltage gain and the cutoff frequency of the -40 db/ decade low pass Butterworth filter shown in the figure below.

38_1157204789.GIF


I am a bit confused because they used the capacitor C2...

Thanks for your time.

Take care.
~Anesziere
 

The filter is a Sallen-key low pass filter. Since no current will flow into the opamp, the inverting terminal of the opamp will be at the output voltage, Vo.

If u label the node voltage between R1 and R2 as Vx, you can write

V0 = [1 /(R2C1s + 1)] * Vx

After that you should apply KCL at node Vx, and you can find Vo/Vi to find the transfer function, after which you put s = jω, and you will get the cutoff frequency and gain.

The gain is 1, and cutoff is ωo = 1/R√(C1C2)

The function of cap C2 is to provide feedback. This feedback must be effective only near the cutoff frequency. At low frequencies, the impedance is too large so there is little feedback. At the higher frequencies, there is feedback, but now C1 is effectively a short, so the output voltage is low (close to zero), and hence it doesn't affect the rest of the circuit.

[Discussions and dervations will be given in an filter book like Huelsman. I used Franco]
 

Thanks ajitg for your prompt reply. :D:D

I tried applying KCL at node Vx in order to find Vx in function of Ei (Vi) but the gain wasn't equal to 1. Could please show me how you found Vx?

As for the cutoff frequency, I do not understand by we used both of the capacitors...

I would really appreciate if you could clarify me on these points.

Thanks again!!
~Anesziere
 

Good evening!

Thanks vfone for the link :D However, we didn't study the transfer function. So I was wondering if it is possible to find the gain without it.

Thanks again ;)
 

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