# Homework Help: Band pass filter and op-amp problem

1. May 7, 2007

### FunkyDwarf

1. The problem statement, all variables and given/known data
Hey guys, quick EE question for ya.

This is the end of a signals and noise assignment and i think i've got it but im not sure so i thought id check. Attached is the question, doing the 2nd part of 13

The part that isnt shown that we need is that the band pass is between 1 and 10 kilohertz
2. Relevant equations
Cutoff frequency for an intergrator= 1/2piRC
Cutoff frequcny for a high pass filter = same

3. The attempt at a solution
Ok heres my twisted reasoning. Looking just at the right hand side for a second, C1 is obviously only going to let high freqs through and given that the whole circuit must run between 1 and 10kHz C1 must let through from 1kHz up (i know i havent talked about Ci but ill get to that in a minute). So viewing the 2nd amp bit as a voltage divider with a bandpass built in i said at high frequencies C2 is going to have, as far as were concerned, infinite impedance as its the low pass bit of the filter. So current flows through C1 then R1 then gets divided across R2. Now i said that we can regard the resistor needed for the equation as simply R2 (although theyre equal anyway) because, i think, you can just put the capacitor infront of R1 and you have a voltage divider just across R2, as R1 wont affect any frequency stuff because its..well a resistor (im not 100% on all of this so forgive any obvious errors i make).

Then using the integrator equation i say that you can, (in the frequency band) assume C1 to be a piece of wire and go abound finding C2 given its the low pass and has to let through up to 10kHz.

So doing all that i get C2 = 3.386E-10 farads and C1 3.386E-9 Farands.

What i DONT get is how Ci and Ri come into it? Surely i can set the capacitance to next to zero and the resistance to nearly infinity and be done with it right?

Hope this makes sense and you guys can help!

Thanks
-G

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Last edited: May 7, 2007
2. May 9, 2007

### Staff: Mentor

Yeah, I don't get the Ci part of the question either, at least not from the info you've posted. It sounds like it should be the input cable capacitance of 10x100pF, but so what. If they don't give you the output impedance of whatever is driving into this amplifier, then there's not much you can do with Ri and Ci.

I didn't check your math on the rest, but it looks reasonable. BTW, it's good practice to put your answers for component values into engineering notation (exponents are powers of 3), and use the standard prefixes for those powers. Like your answer of C2 = 3.4E-10F should be written as C2 = 0.34nF = 3400pF.

3. May 10, 2007

### FunkyDwarf

Oh ok. This unit is part of my physics major so forgive my engineering ignorance :P