Engineering Solving a Circuit Transfer Function: Find C2 for R,C1

AI Thread Summary
The discussion focuses on finding the transfer function of a given circuit and determining the value of capacitor C2. The transfer function is expressed as V_o/V_i = a1/(s^2 + a2s + a3), with a1, a2, and a3 derived in terms of resistances and capacitances. Participants confirm the values of a1, a2, and a3 as a1 = 1/(C1C2R), a2 = 2/C1, and a3 = 1/(C1C2R). To achieve complex conjugate poles at -1 ± j√399, the correct coefficients must be substituted into the quadratic formula, leading to the calculation of C2. The conversation also touches on technical aspects of posting mathematical expressions in the forum.
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Homework Statement



http://img252.imageshack.us/img252/410/prelab4problem1tz5.jpg

Find transfer function of the circuit above (i.e. - \frac{V_o(s)}{V_i(s)})

\frac{V_o(s)}{V_i(s)}\,=\,\frac{a_1}{s^2\,+\,a_2\,s\,+\,a_3}

1) Find a1, a2, a3 in terms of R, C1 and C2

2) Given that C_1\,=\,100\,\mu\,F and R\,=\,10\,K\Omega, find C_2 such that the system has a pair of complex conjugate poles located at -1\,\pm\,j\,\sqrt{399}.

Homework Equations



KCL, OP Amp rules, complex numbers.

The Attempt at a Solution



Ok, I went through a nodal analysis, I'm not going to post the steps here, but here are the results...

\frac{V_o}{V_i}\,=\,\frac{1}{C_1\,C_2\,R\,s^2\,+\,2\,C_2\,R\,s\,-\,1}

\frac{V_o}{V_i}\,=\,\frac{\frac{1}{C_1\,C_2\,R}}{s^2\,+\,\frac{2}{C_1}\,s\,-\,\frac{1}{C_1\,C_2\,R}}So that means that...

a_1\,=\,\frac{1}{C_1\,C_2\,R}

a_2\,=\,\frac{2}{C_1}

a_3\,=\,\frac{1}{C_1\,C_2\,R}

That's for part one, does that seem right?For part two, we want to MAKE the roots of the following equation (denominator):

s^2\,+\,2000\,s\,+\,\frac{1}{C_2}\,=\,0

EQUAL TO...

-1\,\pm\,j\,\sqrt{399}

How do I make that happen?
 
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The first problem is that you've got the transfer function wrong. See the first attachment for the correct transfer function.

And as to your last question, don't you remember the quadratic formula? See the second attachment.

Substitute the coefficients a, b and c, and then select C2 to get the roots you want.
 

Attachments

  • Expr.gif
    Expr.gif
    750 bytes · Views: 547
  • Quad.gif
    Quad.gif
    1.1 KB · Views: 502
So the C2 value for part 2 is ... calculating ... brb!
 
Last edited:
Well, what did you get?

And, now, you can answer a question for me. How do you paste those mathematical expressions into your post? When I right click on one and select properties, it appears that it is a Latex image.
 

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