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

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VinnyCee
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Homework Statement



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

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

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

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

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

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...

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

[tex]\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}}[/tex]So that means that...

[tex]a_1\,=\,\frac{1}{C_1\,C_2\,R}[/tex]

[tex]a_2\,=\,\frac{2}{C_1}[/tex]

[tex]a_3\,=\,\frac{1}{C_1\,C_2\,R}[/tex]

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

[tex]s^2\,+\,2000\,s\,+\,\frac{1}{C_2}\,=\,0[/tex]

EQUAL TO...

[tex]-1\,\pm\,j\,\sqrt{399}[/tex]

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: 589
  • Quad.gif
    Quad.gif
    1.1 KB · Views: 538
So the C2 value for part 2 is ... calculating ... brb!
 
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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.