I have a summer midterm project due this Thursday. The assignment is to get this circuit: http://i.stack.imgur.com/D05wY.jpg

enter image description here

to create this graph of dB vs frequency at the voltage probe point of N1; I am giving it 1 V AC, but 0 V DC as shown http://i.stack.imgur.com/ir7f8.jpg

enter image description here

it is a Bandpass filter for 0.5 Hz and the program used is B2Spice

I need help choosing the correct values for the resistors and capacitors; I already have the transfer function and component relationships -- see my comment below for picture

enter image description here

I am not sure how to tackle this problem and my professor didn't help me very much or answer my questions very well. I need to find the corner frequency and RC time constant? There are 3 poles in the target graph? Any help is appreciated, thank you.

  • \$\begingroup\$ i.stack.imgur.com/qhj4c.jpg \$\endgroup\$ – user26419 Jul 15 '13 at 23:55
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    \$\begingroup\$ Turn your time constants to cutoff freqs; T=1/(2*pi*f) \$\endgroup\$ – Scott Seidman Jul 15 '13 at 23:59
  • \$\begingroup\$ OK, that is 1/pi --> 0.31831 seconds \$\endgroup\$ – user26419 Jul 16 '13 at 0:33
  • \$\begingroup\$ how do I use that number to find the RC values? from the second picture, I found the slope of the straight line part to 0.2564; something my professor told me to find \$\endgroup\$ – user26419 Jul 16 '13 at 0:44
  • \$\begingroup\$ Without working too hard (its not my homework) I'd try to see if I could use the -40dB gain at f= infinity to constrain the problem \$\endgroup\$ – Scott Seidman Jul 16 '13 at 1:32

Since this is not an answer, it should be a comment. But, this "comment" is too long and detailed for a comment.

(1) The equations for the filter in the 3rd image do not include the effect of load resistor in the 1st image schematic.

(2) This is not so much a bandpass filter but, rather, the sum of a 2nd order low-pass and a 1st order bandpass (as can be seen by writing the transfer function as two separate fractions).

(3) I'm not sure what "corner frequency" is in this context. For a high-pass or low-pass filter, the context is clear. For this, I'm not so sure.

Have you left out some details of the problem?

(Also, did you put all this off to nearly the last minute?)

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  • \$\begingroup\$ Well, here is the exact assignment: dl.dropboxusercontent.com/u/47822189/… ; there are five of us to a group too, our professor said that this should be easy. Our group will be pouring over your comment tomorrow, lol. You are right though, essentially we are tacking on another resistor to the complicated RC filter. I made some MATLAB code to help me solve the problem. As far as I know, if I put in the right R and C values, I'll get the right graph out: dropbox.com/sh/c7yyubu6w1h06h8/3kB7xMbH_j/Midterm_Project.m \$\endgroup\$ – user26419 Jul 16 '13 at 1:53
  • \$\begingroup\$ @user26419, I don't think the circuit given can give you the response in the graph. \$\endgroup\$ – Alfred Centauri Jul 16 '13 at 2:20
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    \$\begingroup\$ My group has noticed that no matter what values we put for the resistors and capacitors, we cannot get the Bode plot of dB vs freq to look like the assignment; I just emailed my professor and may send him this discussion link. Thanks for your help Alfred. I would give you rep, but I don't have enough myself. This seems like a cool website (better than yahoo answers --> engineering) \$\endgroup\$ – user26419 Jul 16 '13 at 2:50
  • \$\begingroup\$ @user26419, my initial thinking about this may have been off (it's late and I'm bushed). When the frequency is high enough, the impedance of C3 is much smaller than the stuff to the right so that stuff can be ignored. Assuming that the impedance of C1 is also much smaller than R1, I think the high frequency gain becomes: \$\dfrac{C_1}{C_1 + C_3}\$ \$\endgroup\$ – Alfred Centauri Jul 16 '13 at 3:15
  • \$\begingroup\$ @user26419, I've now verified via simulation that the high frequency gain shelves to the value given in my previous comment so that's a constraint as pointed out by another commenter. Good night and good luck. \$\endgroup\$ – Alfred Centauri Jul 16 '13 at 3:34

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